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https://www.redblobgames.com/grids/hexagons-v1/
|
Hexagonal Grids
from Red Blob Games
Mar 2013, updated in Mar 2015
Hexagonal grids are used in some games but aren’t quite as straightforward or common as square grids. I’ve been collecting hex grid resources for nearly 20 years, and wrote this guide to the most elegant approaches that lead to the simplest code, largely based on the guides by Charles Fu and Clark Verbrugge. I’ll describe the various ways to make hex grids, the relationships between them, as well as some common algorithms. Many parts of this page are interactive; choosing a type of grid will update diagrams, code, and text to match.
Angles, size, spacing
Coordinate systems
Conversions
Neighbors
Distances
Line drawing
Range
Rotation
Rings
Field of view
Hex to pixel
Pixel to hex
Rounding
Map storage
Wraparound maps
Pathfinding
More reading
The code samples on this page are written in pseudo-code; they’re meant to be easy to read and understand. The implementation guide has code in C++, Javascript, C#, Python, Java, Typescript, and more.
Geometry#
Hexagons are 6-sided polygons. Regular hexagons have all the sides (edges) the same length. I’ll assume all the hexagons we’re working with here are regular. The typical orientations for hex grids are horizontal () and vertical ().
Hexagons have 6 edges. Each edge is shared by 2 hexagons. Hexagons have 6 corners. Each corner is shared by 3 hexagons. For more about centers, edges, and corners, see my article on grid parts (squares, hexagons, and triangles).
Angles#
In a regular hexagon the interior angles are 120°. There are six “wedges”, each an equilateral triangle with 60° angles inside. Corner i is at + 30°, size units away from the center. In code:
function hex_corner(center, size, i):
var angle_deg = 60 i + 30
var angle_rad = PI / 180 angle_deg
return Point(center.x + size cos(angle_rad),
center.y + size sin(angle_rad))
To fill a hexagon, gather the polygon vertices at hex_corner(…, 0) through hex_corner(…, 5). To draw a hexagon outline, use those vertices, and then draw a line back to hex_corner(…, 0).
The difference between the two orientations is that x and y are swapped, and that causes the angles to change: angles are 0°, 60°, 120°, 180°, 240°, 300° and angles are 30°, 90°, 150°, 210°, 270°, 330°.
Size and Spacing#
Next we want to put several hexagons together. In the horizontal orientation, the height of a hexagon is height = size
2. The vertical distance between adjacent hexes is vert height.
The width of a hexagon is width = sqrt(3)/2
height. The horizontal distance between adjacent hexes is horiz width.
Some games use pixel art for hexagons that does not match an exactly regular polygon. The angles and spacing formulas I describe in this section won’t match the sizes of your hexagons. The rest of the article, describing algorithms on hex grids, will work even if your hexagons are stretched or shrunk a bit.
Coordinate Systems#
Now let’s assemble hexagons into a grid. With square grids, there’s one obvious way to do it. With hexagons, there are multiple approaches. I recommend using cube coordinates as the primary representation. Use either axial or offset coordinates for map storage, and displaying coordinates to the user.
Offset coordinates#
The most common approach is to offset every other column or row. Columns are named col or q. Rows are named row or r. You can either offset the odd or the even column/rows, so the horizontal and vertical hexagons each have two variants.
Cube coordinates#
Another way to look at hexagonal grids is to see that there are three primary axes, unlike the two we have for square grids. There’s an elegant symmetry with these.
Let’s take a cube grid and slice out a diagonal plane at x y z. This is a weird idea but it helps us make hex grid algorithms simpler. In particular, we can reuse standard operations from cartesian coordinates: adding coordinates, subtracting coordinates, multiplying or dividing by a scalar, and distances.
Notice the three primary axes on the cube grid, and how they correspond to six hex grid diagonal directions; the diagonal grid axes corresponds to a primary hex grid direction.
Because we already have algorithms for square and cube grids, using cube coordinates allows us to adapt those algorithms to hex grids. I will be using this system for most of the algorithms on the page. To use the algorithms with another coordinate system, I’ll convert to cube coordinates, run the algorithm, and convert back.
Try: and: .
Study how the cube coordinates work on the hex grid. Selecting the hexes will highlight the cube coordinates corresponding to the three axes.
Switch to or back to hexagons.
Each direction on the cube grid corresponds to a line on the hex grid. Try highlighting a hex with z at 0, 1, 2, 3 to see how these are related. The row is marked in blue. Try the same for x (green) and y (purple).
Each direction on the hex grid is a combination of two directions on the cube grid. For example, northwest on the hex grid lies between the +y and -z, so every step northwest involves adding 1 to y and subtracting 1 from z.
The cube coordinates are a reasonable choice for a hex grid coordinate system. The constraint is that x + y + z =
0 so the algorithms must preserve that. The constraint also ensures that there’s a canonical coordinate for each hex.
There are many different valid cube hex coordinate systems. Some of them have constraints other than x + y + z = 0. I’ve shown only one of the many systems. You can also construct cube coordinates with x-y, y-z, z-x, and that has its own set of interesting properties, which I don’t explore here.
“But Amit!” you say, “I don’t want to store 3 numbers for coordinates. I don’t know how to store a map that way.”
Axial coordinates#
The axial coordinate system, sometimes called “trapezoidal”, is built by taking two of the three coordinates from a cube coordinate system. Since we have a constraint such as x + y + z = 0, the third coordinate is redundant. Axial coordinates are useful for map storage and for displaying coordinates to the user. Like cube coordinates, you can use the standard add, subtract, multiply, divide operations from cartesian coordinates.
Switch to or back to hexagons.
There are many cube coordinate systems, and many axial coordinate systems. I’m not going to show all of the combinations in this guide. I’m going to pick two variables, q for “column” and r as “row”. In the diagrams on this page, q is aligned with x and r is aligned with z but this alignment is arbitary, as you can rotate and flip the diagrams to make many different alignments.
The advantage of this system over offset grids is that the algorithms are cleaner. The disadvantage of this system is that storing a rectangular map is a little weird; see the map storage section for ways to handle that. There are some algorithms that are even cleaner with cube coordinates, but for those, since we have the constraint x + y + z = 0, we can calculate the third implicit coordinate and use that for those algorithms. In my projects, I name the axes q, r, s so that I have the constraint q + r + s = 0, and then I can calculate s = -q - r when needed.
Axes#
Offset coordinates are the first thing people think of, because they match the standard cartesian coordinates used with square grids. Unfortunately one of the two axes has to go “against the grain”, and ends up complicating things. The cube and axial systems go “with the grain” and have simpler algorithms, but slightly more complicated map storage. There’s also another system, called interlaced or doubled, that I haven’t explored here; some people say it’s easier to work with than cube/axial.
The axis is the direction in which that coordinate increases. Perpendicular to the axis is a line where that coordinate stays constant. The grid diagrams above show the perpendicular line.
Coordinate conversion#
It is likely that you will use axial or offset coordinates in your project, but many algorithms are simpler to express in cube coordinates. Therefore you need to be able to convert back and forth.
Axial coordinates#
Axial coordinates are closely connected to cube coordinates. Axial discards the third coordinate and cube calculates the third coordinate from the other two:
```
function cube_to_axial(cube):
var q = cube.x
var r = cube.z
return Hex(q, r)
function axial_to_cube(hex):
var x = hex.q
var z = hex.r
var y = -x-z
return Cube(x, y, z)
```
Offset coordinates#
Determine which type of offset system you use; -r are pointy top; -q are flat top:
shoves odd rows by +½ column
shoves even rows by +½ column
shoves odd columns by +½ row
shoves even columns by +½ row
```
function cube_to_oddr(cube):
col = cube.x + (cube.z - (cube.z&1)) / 2
row = cube.z
return Hex(col, row)
function oddr_to_cube(hex):
x = hex.col - (hex.row - (hex.row&1)) / 2
z = hex.row
y = -x-z
return Cube(x, y, z)function cube_to_evenr(cube):
col = cube.x + (cube.z + (cube.z&1)) / 2
row = cube.z
return Hex(col, row)
function evenr_to_cube(hex):
x = hex.col - (hex.row + (hex.row&1)) / 2
z = hex.row
y = -x-z
return Cube(x, y, z)function cube_to_oddq(cube):
col = cube.x
row = cube.z + (cube.x - (cube.x&1)) / 2
return Hex(col, row)
function oddq_to_cube(hex):
x = hex.col
z = hex.row - (hex.col - (hex.col&1)) / 2
y = -x-z
return Cube(x, y, z)function cube_to_evenq(cube):
col = cube.x
row = cube.z + (cube.x + (cube.x&1)) / 2
return Hex(col, row)
function evenq_to_cube(hex):
x = hex.col
z = hex.row - (hex.col + (hex.col&1)) / 2
y = -x-z
return Cube(x, y, z)
```
Implementation note: I use a&1 (bitwise and) instead of a%2 (modulo) to detect whether something is even (0) or odd (1), because it works with negative numbers too. See a longer explanation on my implementation notes page.
Neighbors#
Given a hex, which 6 hexes are neighboring it? As you might expect, the answer is simplest with cube coordinates, still pretty simple with axial coordinates, and slightly trickier with offset coordinates. We might also want to calculate the 6 “diagonal” hexes.
Cube coordinates#
Moving one space in hex coordinates involves changing one of the 3 cube coordinates by +1 and changing another one by -1 (the sum must remain 0). There are 3 possible coordinates to change by +1, and 2 remaining that could be changed by -1. This results in 6 possible changes. Each corresponds to one of the hexagonal directions. The simplest and fastest approach is to precompute the permutations and put them into a table of Cube(dx, dy, dz) at compile time:
```
var cube_directions = [
Cube(+1, -1, 0), Cube(+1, 0, -1), Cube( 0, +1, -1),
Cube(-1, +1, 0), Cube(-1, 0, +1), Cube( 0, -1, +1)
]
function cube_direction(direction):
return cube_directions[direction]
function cube_neighbor(cube, direction):
return cube_add(cube, cube_direction(direction))
```
Axial coordinates#
As before, we’ll use the cube system as a starting point. Take the table of Cube(dx, dy, dz) and convert into a table of Hex(dq, dr):
```
var axial_directions = [
, , ,
, ,
]
function hex_direction(direction):
return axial_directions[direction]
function hex_neighbor(hex, direction):
var dir = hex_direction(direction)
return Hex(hex.q + dir.q, hex.r + dir.r)
```
Offset coordinates#
With offset coordinates, the change depends on where in the grid we are. If we’re on an offset column/row then the rule is different than if we’re on a non-offset column/row.
As above, we’ll build a table of the numbers we need to add to col and row. However this time there will be two arrays, one for odd columns/rows and one for even columns/rows. Look at (1,1) on the grid map above and see how col and row change as you move in each of the six directions. Then do this again for (2,2). The tables and code are different for each of the four offset grid types, so pick a grid type to see the corresponding code.
```
var oddr_directions = [
[ , , ,
, , ],
[ , , ,
, , ]
]
function oddr_offset_neighbor(hex, direction):
var parity = hex. & 1
var dir = oddr_directions[parity][direction]
return Hex(hex.col + dir.col, hex.row + dir.row)
```
Pick a grid type:
Using the above lookup tables is the easiest way to to calculate neighbors. It’s also possible to derive these numbers, for those of you who are curious.
Diagonals#
Moving to a “diagonal” space in hex coordinates changes one of the 3 cube coordinates by ±2 and the other two by ∓1 (the sum must remain 0).
```
var cube_diagonals = [
Cube(+2, -1, -1), Cube(+1, +1, -2), Cube(-1, +2, -1),
Cube(-2, +1, +1), Cube(-1, -1, +2), Cube(+1, -2, +1)
]
function cube_diagonal_neighbor(cube, direction):
return cube_add(cube, cube_diagonals[direction])
```
As before, you can convert these into axial by dropping one of the three coordinates, or convert to offset by precalculating the results.
Distances#
Cube coordinates#
In the cube coordinate system, each hexagon is a cube in 3d space. Adjacent hexagons are distance 1 apart in the hex grid but distance 2 apart in the cube grid. This makes distances simple. In a square grid, Manhattan distances are abs(dx) + abs(dy). In a cube grid, Manhattan distances are abs(dx) + abs(dy) + abs(dz). The distance on a hex grid is half that:
function cube_distance(a, b):
return (abs(a.x - b.x) + abs(a.y - b.y) + abs(a.z - b.z)) / 2
An equivalent way to write this is by noting that one of the three coordinates must be the sum of the other two, then picking that one as the distance. You may prefer the “divide by two” form above, or the “max” form here, but they give the same result:
function cube_distance(a, b):
return max(abs(a.x - b.x), abs(a.y - b.y), abs(a.z - b.z))
In the diagram, the max is highlighted. Also notice that each color indicates one of the 6 “diagonal” directions.
Axial coordinates#
In the axial system, the third coordinate is implicit. Let’s convert axial to cube to calculate distance:
function hex_distance(a, b):
var ac = axial_to_cube(a)
var bc = axial_to_cube(b)
return cube_distance(ac, bc)
If your compiler inlines axial_to_cube and cube_distance, it will generate this code:
function hex_distance(a, b):
return (abs(a.q - b.q)
+ abs(a.q + a.r - b.q - b.r)
+ abs(a.r - b.r)) / 2
There are lots of different ways to write hex distance in axial coordinates, but no matter which way you write it, axial hex distance is derived from the Mahattan distance on cubes. For example, the “difference of differences” described here results from writing a.q + a.r - b.q - b.r as a.q - b.q + a.r - b.r, and using “max” form instead of the “divide by two” form of cube_distance. They’re all equivalent once you see the connection to cube coordinates.
Offset coordinates#
As with axial coordinates, we’ll convert offset coordinates to cube coordinates, then use cube distance.
function offset_distance(a, b):
var ac = offset_to_cube(a)
var bc = offset_to_cube(b)
return cube_distance(ac, bc)
We’ll use the same pattern for many of the algorithms: convert hex to cube, run the cube version of the algorithm, and convert any cube results back to hex coordinates (whether axial or offset).
Line drawing#
How do we draw a line from one hex to another? I use linear interpolation for line drawing. Evenly sample the line at N+1 points, and figure out which hexes those samples are in.
First we calculate N to be the hex distance between the endpoints.
Then evenly sample N+1 points between point A and point B. Using linear interpolation, each point will be A + (B - A) 1.0/N i, for values of i from 0 to N, inclusive. In the diagram these sample points are the dark blue dots. This results in floating point coordinates.
Convert each sample point (float) back into a hex (int). The algorithm is called cube_round.
Putting these together to draw a line from A to B:
```
function lerp(a, b, t): // for floats
return a + (b - a) t
function cube_lerp(a, b, t): // for hexes
return Cube(lerp(a.x, b.x, t),
lerp(a.y, b.y, t),
lerp(a.z, b.z, t))
function cube_linedraw(a, b):
var N = cube_distance(a, b)
var results = []
for each 0 ≤ i ≤ N:
results.append(cube_round(cube_lerp(a, b, 1.0/N i)))
return results
```
More notes:
There are times when cube_lerp will return a point that’s just on the edge between two hexes. Then cube_round will push it one way or the other. The lines will look better if it’s always pushed in the same direction. You can do this by adding an “epsilon” hex Cube(1e-6, 1e-6, -2e-6) to one or both of the endpoints before starting the loop. This will “nudge” the line in one direction to avoid landing on edge boundaries.
The DDA Algorithm on square grids sets N to the max of the distance along each axis. We do the same in cube space, which happens to be the same as the hex grid distance.
The cube_lerp function needs to return a cube with float coordinates. If you’re working in a statically typed language, you won’t be able to use the Cube type but instead could define FloatCube, or inline the function into the line drawing code if you want to avoid defining another type.
You can optimize the code by inlining cube_lerp, and then calculating B.x-A.x, B.x-A.y, and 1.0/N outside the loop. Multiplication can be turned into repeated addition. You’ll end up with something like the DDA algorithm.
I use axial or cube coordinates for line drawing, but if you want something for offset coordinates, take a look at this article.
There are many variants of line drawing. Sometimes you’ll want “super cover”. Someone sent me hex super cover line drawing code but I haven’t studied it yet.
Movement Range#
Coordinate range#
Given a hex center and a range N, which hexes are within N steps from it?
We can work backwards from the hex distance formula, distance = max(abs(dx), abs(dy), abs(dz)). To find all hexes within N steps, we need max(abs(dx), abs(dy), abs(dz)) ≤ N. This means we need all three of: abs(dx) ≤ N and abs(dy) ≤ N and abs(dz) ≤ N. Removing absolute value, we get -N ≤ dx ≤ N and -N ≤ dy ≤ N and -N ≤ dz ≤ N. In code it’s a nested loop:
var results = []
for each -N ≤ dx ≤ N:
for each -N ≤ dy ≤ N:
for each -N ≤ dz ≤ N:
if dx + dy + dz = 0:
results.append(cube_add(center, Cube(dx, dy, dz)))
This loop will work but it’s somewhat inefficient. Of all the values of dz we loop over, only one of them actually satisfies the dx + dy + dz = 0 constraint on cubes. Instead, let’s directly calculate the value of dz that satisfies the constraint:
var results = []
for each -N ≤ dx ≤ N:
for each max(-N, -dx-N) ≤ dy ≤ min(N, -dx+N):
var dz = -dx-dy
results.append(cube_add(center, Cube(dx, dy, dz)))
This loop iterates over exactly the needed coordinates. In the diagram, each range is a pair of lines. Each line is an inequality. We pick all the hexes that satisfy all six inequalities.
Intersecting ranges#
If you need to find hexes that are in more than one range, you can intersect the ranges before generating a list of hexes.
You can either think of this problem algebraically or geometrically. Algebraically, each region is expressed as inequality constraints of the form -N ≤ dx ≤ N, and we’re going to solve for the intersection of those constraints. Geometrically, each region is a cube in 3D space, and we’re going to intersect two cubes in 3D space to form a cuboid in 3D space, then project back to the x + y + z = 0 plane to get hexes. I’m going to solve it algebraically:
First, we rewrite constraint -N ≤ dx ≤ N into a more general form, xmin ≤ x ≤
xmax, and set xmin = center.x
- N and xmax = center.x + N. We’ll do the same for y and z, and end up with this generalization of the code from the previous section:
var results = []
for each xmin ≤ x ≤ xmax:
for each max(ymin, -x-zmax) ≤ y ≤ min(ymax, -x-zmin):
var z = -x-y
results.append(Cube(x, y, z))
The intersection of two ranges a ≤ x ≤ b and c ≤ x ≤ d is max(a, c) ≤ x ≤ min(b,
d). Since a hex region is expressed as ranges over x, y, z, we can separately intersect each of the x, y, z ranges then use the nested loop to generate a list of hexes in the intersection. For one hex region we set xmin = H.x - N and xmax = H.x + N and likewise for y and z. For intersecting two hex regions we set xmin = max(H1.x -
N, H2.x - N) and xmax =
min(H1.x + N, H2.x + N), and likewise
for y and z. The same pattern
works for intersecting three or more regions.
Obstacles#
If there are obstacles, the simplest thing to do is a distance-limited flood fill (breadth first search). In this diagram, the limit is set to moves. In the code, fringes[k] is an array of all hexes that can be reached in k steps. Each time through the main loop, we expand level k-1 into level k.
```
function cube_reachable(start, movement):
var visited = set()
add start to visited
var fringes = []
fringes.append([start])
for each 1 < k ≤ movement:
fringes.append([])
for each cube in fringes[k-1]:
for each 0 ≤ dir < 6:
var neighbor = cube_neighbor(cube, dir)
if neighbor not in visited, not blocked:
add neighbor to visited
fringes[k].append(neighbor)
return visited
```
Limit :
Rotation#
Given a hex vector (difference between one hex and another), we might want to rotate it to point to a different hex. This is simple with cube coordinates if we stick with rotations of 1/6th of a circle.
A rotation 60° right shoves each coordinate one slot to the right:
[ x, y, z]
to [-z, -x, -y]
A rotation 60° left shoves each coordinate one slot to the left:
[ x, y, z]
to [-y, -z, -x]
As you play with diagram, notice that each 60° rotation flips the signs and also physically “rotates” the coordinates. After a 120° rotation the signs are flipped back to where they were. A 180° rotation flips the signs but the coordinates have rotated back to where they originally were.
Here’s the full recipe for rotating a position P around a center position C to result in a new position R:
Convert positions P and C to cube coordinates.
Calculate a vector by subtracting the center: P_from_C = P - C = Cube(P.x - C.x, P.y - C.y, P.z - C.z).
Rotate the vector P_from_C as described above, and call the resulting vector R_from_C.
Convert the vector back to a position by adding the center: R = R_from_C + C = Cube(R_from_C.x + C.x, R_from_C.y + C.y, R_from_C.z + C.z).
Convert the cube position R back to to your preferred coordinate system.
It’s several conversion steps but each step is simple. You can shortcut some of these steps by defining rotation directly on axial coordinates, but hex vectors don’t work for offset coordinates and I don’t know a shortcut for offset coordinates. Also see this stackexchange discussion for other ways to calculate rotation.
Rings#
Single ring#
To find out whether a given hex is on a ring of a given radius, calculate the distance from that hex to the center and see if it’s radius. To get a list of all such hexes, take radius steps away from the center, then follow the rotated vectors in a path around the ring.
function cube_ring(center, radius):
var results = []
# this code doesn't work for radius == 0; can you see why?
var cube = cube_add(center,
cube_scale(cube_direction(4), radius))
for each 0 ≤ i < 6:
for each 0 ≤ j < radius:
results.append(cube)
cube = cube_neighbor(cube, i)
return results
In this code, cube starts out on the ring, shown by the large arrow from the center to the corner in the diagram. I chose corner 4 to start with because it lines up the way my direction numbers work but you may need a different starting corner. At each step of the inner loop, cube moves one hex along the ring. After 6 radius steps it ends up back where it started.
Spiral rings#
Traversing the rings one by one in a spiral pattern, we can fill in the interior:
function cube_spiral(center, radius):
var results = [center]
for each 1 ≤ k ≤ radius:
results = results + cube_ring(center, k)
return results
The area of the larger hexagon will be the sum of the circumferences, plus 1 for the center; use this formula to help you calculate the area.
Visiting the hexes this way can also be used to calculate movement range.
Field of view#
Given a location and a distance, what is visible from that location, not blocked by obstacles? The simplest way to do this is to draw a line to every hex that’s in range. If the line doesn’t hit any walls, then you can see the hex. Mouse over a hex to see the line being drawn to that hex, and which walls it hits.
This algorithm can be slow for large areas but it’s so easy to implement that it’s what I recommend starting with.
There are many different ways to define what’s “visible”. Do you want to be able to see the center of the other hex from the center of the starting hex? Do you want to see any part of the other hex from the center of the starting point? Maybe any part of the other hex from any part of the starting point? Are there obstacles that occupy less than a complete hex? Field of view turns out to be trickier and more varied than it might seem at first. Start with the simplest algorithm, but expect that it may not compute exactly the answer you want for your project. There are even situations where the simple algorithm produces results that are illogical.
Clark Verbrugge’s guide describes a “start at center and move outwards” algorithm to calculate field of view. Also see the Duelo project, which has an an online demo of directional field of view and code on Github. Also see my article on 2d visibility calculation for an algorithm that works on polygons, including hexagons. For grids, the roguelike community has a nice set of algorithms for square grids (see this and this and this); some of them might be adapted for hex grids.
Hex to pixel#
For hex to pixel, it’s useful to review the size and spacing diagram at the top of the page.
Axial coordinates#
For axial coordinates, the way to think about hex to pixel conversion is to look at the basis vectors. In the diagram, the arrow A→Q is the q basis vector and A→R is the r basis vector. The pixel coordinate is q_basis q + r_basis r. For example, B at (1, 1) is the sum of the q and r basis vectors.
If you have a matrix library, this is a simple matrix multiplication; however I’ll write the code out without matrices here. For the x = q, z = r axial grid I use in this guide, the conversion is:
function hex_to_pixel(hex):
x = size sqrt(3) (hex.q + hex.r/2)
y = size 3/2 hex.rx = size 3/2 hex.q
y = size sqrt(3) (hex.r + hex.q/2)
return Point(x, y)
Axial coordinates: or
The matrix approach will come in handy later when we want to convert pixel coordinates back to hex coordinates. All we will have to do is invert the matrix. For cube coordinates, you can either use cube basis vectors (x, y, z), or you can convert to axial first and then use axial basis vectors (q, r).
Offset coordinates#
For offset coordinates, we need to offset either the column or row number (it will no longer be an integer).
function oddr_offset_to_pixel(hex):
x = size sqrt(3) (hex.col + 0.5 (hex.row&1))
y = size 3/2 hex.rowx = size sqrt(3) (hex.col - 0.5 (hex.row&1))
y = size 3/2 hex.rowx = size 3/2 hex.col
y = size sqrt(3) (hex.row + 0.5 (hex.col&1))x = size 3/2 hex.col
y = size sqrt(3) (hex.row - 0.5 (hex.col&1))
return Point(x, y)
Offset coordinates:
Unfortunately offset coordinates don’t have basis vectors that we can use with a matrix. This is one reason pixel-to-hex conversions are harder with offset coordinates.
Another approach is to convert the offset coordinates into cube/axial coordinates, then use the cube/axial to pixel conversion. By inlining the conversion code then optimizing, it will end up being the same as above.
Pixel to Hex#
One of the most common questions is, how do I take a pixel location (such as a mouse click) and convert it into a hex grid coordinate? I’ll show how to do this for axial or cube coordinates. For offset coordinates, the simplest thing to do is to convert the cube to offset at the end.
First we invert the hex to pixel conversion. This will give us a fractional hex coordinate, shown as a small blue circle in the diagram.
Then we find the hex containing the fractional hex coordinate, shown as the highlighted hex in the diagram.
To convert from hex coordinates to pixel coordinates, we multiplied q, r by basis vectors to get x, y. You can think of this as a matrix multiply. Here’s the matrix for “pointy top” hexes:
⎡x⎤ ⎡ sqrt(3) sqrt(3)/2 ⎤ ⎡q⎤
⎢ ⎥ = size × ⎢ ⎥ × ⎢ ⎥
⎣y⎦ ⎣ 0 3/2 ⎦ ⎣r⎦
Converting from pixel coordinates back to hex coordinates is straightforward. We can invert the matrix:
⎡q⎤ ⎡ sqrt(3)/3 -1/3 ⎤ ⎡x⎤
⎢ ⎥ = ⎢ ⎥ × ⎢ ⎥ ÷ size
⎣r⎦ ⎣ 0 2/3 ⎦ ⎣y⎦
This calculation will give us fractional axial coordinates (floats) for q and r. The hex_round() function will convert the fractional axial coordinates into integer axial hex coordinates.
Here’s the code for “pointy top” axial hexes:
function pixel_to_hex(x, y):
q = (x sqrt(3)/3 - y / 3) / size
r = y 2/3 / size
return hex_round(Hex(q, r))
And here’s the code for “flat top” axial hexes:
function pixel_to_hex(x, y):
q = x 2/3 / size
r = (-x / 3 + sqrt(3)/3 y) / size
return hex_round(Hex(q, r))
That’s three lines of code to convert a pixel location into an axial hex coordinate. If you use offset coordinates, use return cube_to_{odd,even}{r,q}(cube_round(Cube(q, -q-r, r))).
There are many other ways to convert pixel to hex; see this page for the ones I know of.
Rounding to nearest hex#
Sometimes we'll end up with a floating-point cube coordinate (x, y, z), and we’ll want to know which hex it should be in. This comes up in line drawing and pixel to hex. Converting a floating point value to an integer value is called rounding so I call this algorithm cube_round.
With cube coordinates, x + y + z = 0, even with floating point cube coordinates. So let’s round each component to the nearest integer, (rx, ry, rz). However, although x + y + z = 0, after rounding we do not have a guarantee that rx + ry + rz = 0. So we reset the component with the largest change back to what the constraint rx + ry + rz = 0 requires. For example, if the y-change abs(ry-y) is larger than abs(rx-x) and abs(rz-z), then we reset ry = -rx-rz. This guarantees that rx + ry + rz = 0. Here’s the algorithm:
```
function cube_round(cube):
var rx = round(cube.x)
var ry = round(cube.y)
var rz = round(cube.z)
var x_diff = abs(rx - cube.x)
var y_diff = abs(ry - cube.y)
var z_diff = abs(rz - cube.z)
if x_diff > y_diff and x_diff > z_diff:
rx = -ry-rz
else if y_diff > z_diff:
ry = -rx-rz
else:
rz = -rx-ry
return Cube(rx, ry, rz)
```
For non-cube coordinates, the simplest thing to do is to convert to cube coordinates, use the rounding algorithm, then convert back:
function hex_round(hex):
return cube_to_axial(cube_round(axial_to_cube(hex)))
The same would work if you have oddr, evenr, oddq, or evenq instead of axial.
Implementation note: cube_round and hex_round take float coordinates instead of int coordinates. If you’ve written a Cube and Hex class, they’ll work fine in dynamically languages where you can pass in floats instead of ints, and they’ll also work fine in statically typed languages with a unified number type. However, in most statically typed languages, you’ll need a separate class/struct type for float coordinates, and cube_round will have type FloatCube → Cube. If you also need hex_round, it will be FloatHex → Hex, using helper function floatcube_to_floathex instead of cube_to_hex. In languages with parameterized types (C++, Haskell, etc.) you might define Cube<T> where T is either int or float. Alternatively, you could write cube_round to take three floats as inputs instead of defining a new type just for this function.
Map storage in axial coordinates#
One of the common complaints about the axial coordinate system is that it leads to wasted space when using a rectangular map; that's one reason to favor an offset coordinate system. However all the hex coordinate systems lead to wasted space when using a triangular or hexagonal map. We can use the same strategies for storing all of them.
Shape:
→ array[r][q + r/2]
Notice in the diagram that the wasted space is on the left and right sides of each row (except for rhombus maps) This gives us three strategies for storing the map:
Ignore the problem. Use a dense array with nulls or some other sentinel at the unused spaces. At most there’s a factor of two for these common shapes; it may not be worth using a more complicated solution.
Use a hash table instead of dense array. This allows arbitrarily shaped maps, including ones with holes. When you want to access the hex at q,r you access hash_table(hash(q,r)) instead. Encapsulate this into the getter/setter in the grid class so that the rest of the game doesn’t need to know about it.
Slide the rows to the left, and use variable sized rows. In many languages a 2D array is an array of arrays; there’s no reason the arrays have to be the same length. This removes the waste on the right. In addition, if you start the arrays at the leftmost column instead of at 0, you remove the waste on the left.
To do this for arbitrary convex shaped maps, you’d keep an additional array of “first columns”. When you want to access the hex at q,r you access array[r][q -
first_column[r]] instead. Encapsulate this into the getter/setter in the grid class so that the rest of the game doesn’t need to know about it. In the diagram first_column is shown on the left side of each row.
If your maps are fixed shapes, the “first columns” can be calculated on the fly instead of being stored in an array.
For rectangle shaped maps, first_column[r] ==
-floor(r/2), and you’d end up accessing array[r][q + r/2], which turns out to be equivalent to converting the coordinates into offset grid coordinates.
For triangle shaped maps as shown here, first_column[r] == 0, so you’d access array[r][q] — how convenient! For triangle shaped maps that are oriented the other way (not shown in the diagram), it’s array[r][q+r].
For hexagon shaped maps of radius N, where N = max(abs(x), abs(y), abs(z), we have first_column[r] == -N - min(0, r). You’d access array[r][q + N + min(0, r)]. However, since we’re starting with some values of r < 0, we also have to offset the row, and use array[r + N][q + N +
min(0, r)].
For rhombus shaped maps, everything matches nicely, so you can use array[r][q].
Wraparound maps#
In some games you want the map to “wrap” around the edges. In a square map, you can either wrap around the x-axis only (roughly corresponding to a sphere) or both x- and y-axes (roughly corresponding to a torus). Wraparound depends on the map shape, not the tile shape. To wrap around a rectangular map is easy with offset coordinates. I’ll show how to wrap around a hexagon-shaped map with cube coordinates.
Corresponding to the center of the map, there are six “mirror” centers. When you go off the map, you subtract the mirror center closest to you until you are back on the main map. In the diagram, try exiting the center map, and watch one of the mirrors enter the map on the opposite side.
The simplest implementation is to precompute the answers. Make a lookup table storing, for each hex just off the map, the corresponding cube on the other side. For each of the six mirror centers M, and each of the locations on the map L, store mirror_table[cube_add(M, L)] = L. Then any time you calculate a hex that’s in the mirror table, replace it by the unmirrored version. See stackoverflow for another approach.
For a hexagonal shaped map with radius N, the mirror centers will be Cube(2N+1, -N, -N-1) and its six rotations.
Pathfinding#
If you’re using graph-based pathfinding such as A or Dijkstra’s algorithm or Floyd-Warshall, pathfinding on hex grids isn’t different from pathfinding on square grids. The explanations and code from my pathfinding tutorial will work equally well on hexagonal grids.
Mouse over a hex in the diagram to see the path to it. Click or drag to toggle walls.
Neighbors. The sample code I provide in the pathfinding tutorial calls graph.neighbors to get the neighbors of a location. Use the function in the neighbors section for this. Filter out the neighbors that are impassable.
Heuristic. The sample code for A uses a heuristic function that gives a distance between two locations. Use the distance formula, scaled to match the movement costs. For example if your movement cost is 5 per hex, then multiply the distance by 5.
More#
I have an guide to implementing your own hex grid library, including sample code in C++, Java, C#, Javascript, Haxe, and Python.
In my Guide to Grids, I cover axial coordinate systems to address square, triangle, and hexagon edges and corners, and algorithms for the relationships among tiles, edges, and corners. I also show how square and hex grids are related.
The best early guide I saw to the axial coordinate system was Clark Verbrugge’s guide, written in 1996.
The first time I saw the cube coordinate system was from Charles Fu’s posting to rec.games.programmer in 1994.
DevMag has a nice visual overview of hex math including how to represent areas such as half-planes, triangles, and quadrangles. There’s a PDF version here that goes into more detail. Highly reocmmended! The GameLogic Grids library implements these and many other grid types in Unity.
James McNeill has a nice visual explanation of grid transformations.
Overview of hex coordinate types: staggered (offset), interlaced, 3d (cube), and trapezoidal (axial).
The Rot.js library has a list of hex coordinate systems: non-orthogonal (axial), odd shift (offset), double width (interlaced), cube.
Range for cube coordinates: given a distance, which hexagons are that distance from the given one?
Distances on hex grids using cube coordinates, and reasons to use cube coordinates instead of offset.
This guide explains the basics of measuring and drawing hexagons, using an offset grid.
Convert cube hex coordinates to pixel coordinates.
This thread explains how to generate rings.
The HexPart system uses both hexes and rectangles to make some of the algorithms easier to work with.
Are there pros and cons of “pointy topped” and “flat topped” hexagons?
Line of sight in a hex grid with offset coordinates, splitting hexes into triangles
Hexnet explains how the correspondence between hexagons and cubes goes much deeper than what I described on this page, generalizing to higher dimensions.
I used the PDF hex grids from this page while working out some of the algorithms.
Generalized Balanced Ternary for hex coordinates seems interesting; I haven’t studied it.
Hex-Grid Utilities is a C# library for hex grid math, with neighbors, grids, range finding, path finding, field of view. Open source, MIT license.
The Reddit discussion and Hacker News discussion and MetaFilter discussion have more comments and links.
The code that powers this page is written in a mixture of Haxe and Javascript: , , , , , and (for the cube/hex animation). However, if you are looking to write your own hex grid library, I suggest instead looking at my guide to implementation.
There are more things I want to do for this guide. I’m keeping a list on Trello. Do you have suggestions for things to change or add? Comment below.
Email me redblobgames@gmail.com, or tweet @redblobgames, or comment:
Endnotes
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Quotient Rings and the First Isomorphism Theorem for Rings (Algebra 1: Lecture 28 Video 1)
nckaplan math
928 subscribers
3 likes
Description
435 views
Posted: 11 Jan 2021
Lecture 28: We began this lecture by stating the First Isomorphism Theorem for Rings and also stating a theorem saying that every ideal of a ring is the kernel of a ring homomorphism. In order to prove such a statement, we need to understand the construction of quotient rings. We then reviewed how Dummit and Foote introduce the idea of a quotient group and emphasized that the presentation for quotient rings is completely parallel. We saw that for an additive subgroup I of R, the multiplication we wanted to put on R/I is well-defined if and only if I is an ideal of R. When I is an ideal of R we saw that R/I is a ring. We then proved the First Isomorphism Theorem for Rings and the statement about the correspondence between ideals and kernels of ring homomorphisms. In this way we saw that ideals and playing a role in the theory of rings somewhat analogous to the role that normal subgroups play in the theory of groups. We made some comments about how these statements and proofs change when you take the perspective that rings should have identities, subrings should contain the multiplicative identity of the larger ring, and ring homomorphisms should send the identity of the first ring to the identity of the second ring. We then gave several examples involving quotient rings and the First Isomorphism Theorem for Rings. Finally, we talked about the other isomorphism theorems for rings. We noted that they are very similar to the corresponding statements for groups.
Reading: This lecture closely followed pages 240-247 of Dummit and Foote with some additional inputs. We reviewed much of Section 3.1 of Dummit and Foote about quotient groups. The exact statements and page numbers are mentioned in the lecture videos. We also referenced some statements from Conrad's notes "Notes on Ideals". More specifically, we briefly discussed Theorems 3.4 and 4.1.
Dummit and Foote do not include details of the proofs of the 2nd, 3rd, and 4th/Lattice Isomorphism Theorems for Rings. I linked to notes of Zachary Scherr that provide these proofs in detail and also to notes of Bianca Viray that provide full proofs of the first three isomorphism theorems for rings if you use Conrad's definitions for rings, subrings, and ring homomorphisms.
Transcript:
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https://courses.lumenlearning.com/wm-prealgebra/chapter/translating-algebraic-expressions-from-words/
|
Module 2:The Language of Algebra
Translating Algebraic Expressions From Words
Learning Outcomes
Translate word phrases into algebraic expressions
Write an algebraic expression that represents the relationship between two measurements such as length and width or the amount of different types of coins
Translate Words to Algebraic Expressions
In the previous section, we listed many operation symbols that are used in algebra, and then we translated expressions and equations into word phrases and sentences. Now we’ll reverse the process and translate word phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. They are summarized below.
| Operation | Phrase | Expression |
---
| Addition | [latex]a[/latex] plus [latex]b[/latex] the sum of [latex]a[/latex] and [latex]b[/latex] [latex]a[/latex] increased by [latex]b[/latex] [latex]b[/latex] more than [latex]a[/latex] the total of [latex]a[/latex] and [latex]b[/latex] [latex]b[/latex] added to [latex]a[/latex] | [latex]a+b[/latex] |
| Subtraction | [latex]a[/latex] minus [latex]b[/latex] the difference of [latex]a[/latex] and [latex]b[/latex] [latex]b[/latex] subtracted from [latex]a[/latex] [latex]a[/latex] decreased by [latex]b[/latex] [latex]b[/latex] less than [latex]a[/latex] | [latex]a-b[/latex] |
| Multiplication | [latex]a[/latex] times [latex]b[/latex] the product of [latex]a[/latex] and [latex]b[/latex] | [latex]a\cdot b[/latex] , [latex]ab[/latex] , [latex]a\left(b\right)[/latex] , [latex]\left(a\right)\left(b\right)[/latex] |
| Division | [latex]a[/latex] divided by [latex]b[/latex] the quotient of [latex]a[/latex] and [latex]b[/latex] the ratio of [latex]a[/latex] and [latex]b[/latex] [latex]b[/latex] divided into [latex]a[/latex] | [latex]a\div b[/latex] , [latex]a/b[/latex] , [latex]\frac{a}{b}[/latex] , [latex]b\overline{)a}[/latex] |
Look closely at these phrases using the four operations:
the sum of [latex]a[/latex] and [latex]b[/latex]
the difference of [latex]a[/latex] and [latex]b[/latex]
the product of [latex]a[/latex] and [latex]b[/latex]
the quotient of [latex]a[/latex] and [latex]b[/latex]
Each phrase tells you to operate on two numbers. Look for the words of and and to find the numbers.
example
Translate each word phrase into an algebraic expression:
The difference of [latex]20[/latex] and [latex]4[/latex]2. The quotient of [latex]10x[/latex] and [latex]3[/latex]
Solution1. The key word is difference, which tells us the operation is subtraction. Look for the words of and and to find the numbers to subtract. [latex]\begin{array}{}\ \text{the difference of }20\text{ and }4\hfill \ 20\text{ minus }4\hfill \ 20 - 4\hfill \end{array}[/latex]
The key word is quotient, which tells us the operation is division. [latex]\begin{array}{}\ \text{the quotient of }10x\text{ and }3\hfill \ \text{divide }10x\text{ by }3\hfill \ 10x\div 3\hfill \end{array}[/latex] This can also be written as [latex]\begin{array}{l}10x/3\text{ or}\frac{10x}{3}\hfill \end{array}[/latex]
try it
example
Translate each word phrase into an algebraic expression:
How old will you be in eight years? What age is eight more years than your age now? Did you add [latex]8[/latex] to your present age? Eight more than means eight added to your present age.
How old were you seven years ago? This is seven years less than your age now. You subtract [latex]7[/latex] from your present age,a. Seven less than means seven subtracted from your present age,a.
Show Solution
Solution:
Eight more than [latex]y[/latex]2. Seven less than [latex]a[/latex]
The key words are more than. They tell us the operation is addition. More than means “added to”. [latex]\begin{array}{l}\text{Eight more than }y\ \text{Eight added to }y\ y+8\end{array}[/latex]
The key words are less than. They tell us the operation is subtraction. Less than means “subtracted from”. [latex]\begin{array}{l}\text{Seven less than }a\ \text{Seven subtracted from }a\ a - 7\end{array}[/latex]
try it
example
Translate each word phrase into an algebraic expression:
five times the sum of [latex]m[/latex] and [latex]n[/latex]2. the sum of five times [latex]m[/latex] and [latex]n[/latex]
Show Solution
Solution1. There are two operation words: times tells us to multiply and sum tells us to add. Because we are multiplying [latex]5[/latex] times the sum, we need parentheses around the sum of [latex]m[/latex] and [latex]n[/latex].
five times the sum of [latex]m[/latex] and [latex]n[/latex]
[latex]5(m+n)[/latex]
To take a sum, we look for the words of and and to see what is being added. Here we are taking the sum of five times [latex]m[/latex] and [latex]n[/latex].
the sum of five times [latex]m[/latex] and [latex]n[/latex]
[latex]5m+n[/latex]
Notice how the use of parentheses changes the result. In part 1, we add first and in part 2, we multiply first.
try it
Watch the video below to better understand how to write algebraic expressions from statements.
Later in this course, we’ll apply our skills in algebra to solving equations. We’ll usually start by translating a word phrase to an algebraic expression. We’ll need to be clear about what the expression will represent. We’ll see how to do this in the next two examples.
example
The height of a rectangular window is [latex]6[/latex] inches less than the width. Let [latex]w[/latex] represent the width of the window. Write an expression for the height of the window.
Show Solution
Solution
| | |
--- |
| Write a phrase about the height. | [latex]6[/latex] less than the width |
| Substitute [latex]w[/latex] for the width. | [latex]6[/latex] less than [latex]w[/latex] |
| Rewrite ‘less than’ as ‘subtracted from’. | [latex]6[/latex] subtracted from [latex]w[/latex] |
| Translate the phrase into algebra. | [latex]w - 6[/latex] |
try it
example
Blanca has dimes and quarters in her purse. The number of dimes is [latex]2[/latex] less than [latex]5[/latex] times the number of quarters. Let [latex]q[/latex] represent the number of quarters. Write an expression for the number of dimes.
Show Solution
Solution
| | |
--- |
| Write a phrase about the number of dimes. | two less than five times the number of quarters |
| Substitute [latex]q[/latex] for the number of quarters. | [latex]2[/latex] less than five times [latex]q[/latex] |
| Translate [latex]5[/latex] times [latex]q[/latex] . | [latex]2[/latex] less than [latex]5q[/latex] |
| Translate the phrase into algebra. | [latex]5q - 2[/latex] |
try it
In the following video we show more examples of how to write basic algebraic expressions from words, and simplify.
Candela Citations
CC licensed content, Original
Write Algebraic Expressions from Statements: Form ax+b and a(x+b). Authored by: Sousa, James (mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Write Basic Expressions from Words Modeling Situations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
CC licensed content, Shared previously
Question ID: 144907, 144916, 144917, 144918,146542,146541 . Authored by: Lumen Learning. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
CC licensed content, Specific attribution
Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at
Licenses and Attributions
CC licensed content, Original
Write Algebraic Expressions from Statements: Form ax+b and a(x+b). Authored by: Sousa, James (mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Write Basic Expressions from Words Modeling Situations. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
CC licensed content, Shared previously
Question ID: 144907, 144916, 144917, 144918,146542,146541 . Authored by: Lumen Learning. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
CC licensed content, Specific attribution
Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at
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https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/power.prop.test
|
power.prop.test function - RDocumentation
##### One week. Exclusive content. Lasting impact. Oct 6–10: How leaders are closing AI skills gaps. Unlock Access
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stats (version 3.6.2)
power.prop.test: Power Calculations for Two-Sample Test for Proportions
Description
Compute the power of the two-sample test for proportions, or determine parameters to obtain a target power.
Usage
power.prop.test(n = NULL, p1 = NULL, p2 = NULL, sig.level = 0.05, power = NULL, alternative = c("two.sided", "one.sided"), strict = FALSE, tol = .Machine$double.eps^0.25)
Arguments
n
number of observations (per group)
p1
probability in one group
p2
probability in other group
sig.level
significance level (Type I error probability)
power
power of test (1 minus Type II error probability)
alternative
one- or two-sided test. Can be abbreviated.
strict
use strict interpretation in two-sided case
tol
numerical tolerance used in root finding, the default providing (at least) four significant digits.
Value
Object of class "power.htest", a list of the arguments (including the computed one) augmented with method and note elements.
Details
Exactly one of the parameters n, p1, p2, power, and sig.level must be passed as NULL, and that parameter is determined from the others. Notice that sig.level has a non-NULL default so NULL must be explicitly passed if you want it computed.
If strict = TRUE is used, the power will include the probability of rejection in the opposite direction of the true effect, in the two-sided case. Without this the power will be half the significance level if the true difference is zero.
Note that not all conditions can be satisfied, e.g., for
power.prop.test(n=30, p1=0.90, p2=NULL, power=0.8, strict=TRUE)
there is no proportion p2 between p1 = 0.9 and 1, as you'd need a sample size of at least n=74 to yield the desired power for (p 1,p 2)=(0.9,1).
For these impossible conditions, currently a warning (warning) is signalled which may become an error (stop) in the future.
See Also
prop.test, uniroot
Examples
Run this code
```r
NOT RUN {
power.prop.test(n = 50, p1 = .50, p2 = .75) ## => power = 0.740
power.prop.test(p1 = .50, p2 = .75, power = .90) ## => n = 76.7
power.prop.test(n = 50, p1 = .5, power = .90) ## => p2 = 0.8026
power.prop.test(n = 50, p1 = .5, p2 = 0.9, power = .90, sig.level=NULL)
## => sig.l = 0.00131
power.prop.test(p1 = .5, p2 = 0.501, sig.level=.001, power=0.90)
## => n = 10451937
try(
power.prop.test(n=30, p1=0.90, p2=NULL, power=0.8)
) # a warning (which may become an error)
Reason:
power.prop.test( p1=0.90, p2= 1.0, power=0.8) ##-> n = 73.37
}
```
Run the code above in your browser usingDataLab
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404
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https://wiki.sei.cmu.edu/confluence/display/java/MSC06-J.+Do+not+modify+the+underlying+collection+when+an+iteration+is+in+progress
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Rule 49. Miscellaneous (MSC)
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MSC06-J. Do not modify the underlying collection when an iteration is in progress
Created by Dhruv Mohindra, last modified by David Svoboda on Aug 06, 2025
According to the Java API documentation [API 2014] for the Iterator.remove() method:
The behavior of an iterator is unspecified if the underlying collection is modified while the iteration is in progress in any way other than by calling this method.
Concurrent modification in single-threaded programs is usually a result of inserting or removing an element during iteration. Multithreaded programs add the possibility that a collection may be modified by one thread while another thread iterates over the collection. Undefined behavior results in either case. Many implementations throw a ConcurrentModificationException when they detect concurrent modification.
According to the Java API documentation [API 2014] for ConcurrentModificationException:
It is not generally permissible for one thread to modify a Collection while another thread is iterating over it. In general, the results of the iteration are undefined under these circumstances. Some Iterator implementations (including those of all the general purpose collection implementations provided by the JRE) may choose to throw this exception if this behavior is detected. Iterators that do this are known as fail-fast iterators, as they fail quickly and cleanly, rather that risking arbitrary, non-deterministic behavior at an undetermined time in the future.
...
Note that fail-fast behavior cannot be guaranteed because it is, generally speaking, impossible to make any hard guarantees in the presence of unsynchronized concurrent modification. Fail-fast operations throw ConcurrentModificationException on a best-effort basis. Therefore, it would be wrong to write a program that depended on this exception for its correctness: ConcurrentModificationException should be used only to detect bugs.
Reliance on ConcurrentModificationException is inadequate to prevent undefined behavior resulting from modifying an underlying collection while simultaneously iterating over the collection. The fail-fast behavior may occur only after processing an arbitrary number of elements. In Java Concurrency in Practice [Goetz 2006a], Goetz and colleagues note:
[Fail-fast iterators] are implemented by associating a modification count with the collection: if the modification count changes during iteration, hasNext or next throws ConcurrentModificationException. However, this check is done without synchronization, so there is a risk of seeing a stale value of the modification count and therefore...that the iterator does not realize a modification has been made. This was a deliberate design tradeoff to reduce the performance impact of the concurrent modification detection code.
Note that the enhanced for loop (for-each idiom) uses an Iterator internally. Consequently, enhanced for loops can also participate in concurrent modification issues, even though they lack an obvious iterator.
Noncompliant Code Example (Single-Threaded)
This noncompliant code example (based on Sun Developer Network SDN 2011 bug report 6687277) uses the ArrayList's remove() method to remove an element from an ArrayList while iterating over the ArrayList. The resulting behavior is unspecified.
class BadIterate { public static void main(String[] args) { List list = new ArrayList(); list.add("one"); list.add("two"); Iterator iter = list.iterator(); while (iter.hasNext()) { String s = (String)iter.next(); if (s.equals("one")) { list.remove(s); } } } }
Compliant Solution (iterator.remove())
The Iterator.remove() method removes the last element returned by the iterator from the underlying Collection. Its behavior is fully specified, so it may be safely invoked while iterating over a collection.
// ... if (s.equals("one")) { iter.remove(); } // ...
Noncompliant Code Example (Multithreaded)
Although acceptable in a single-threaded environment, this noncompliant code example is insecure in a multithreaded environment because it is possible for another thread to modify the widgetList while the current thread iterates over the widgetList. Additionally, the doSomething() method could modify the collection during iteration.
List widgetList = new ArrayList(); public void widgetOperation() { // May throw ConcurrentModificationException for (Widget w : widgetList) { doSomething(w); } }
Compliant Solution (Thread-Safe Collection)
This compliant solution wraps the ArrayList in a synchronized collection so that all modifications are subject to the locking mechanism.
List widgetList = Collections.synchronizedList(new ArrayList()); public void widgetOperation() { synchronized (widgetList) { // Client-side locking for (Widget w : widgetList) { doSomething(w); } } }
This approach must be implemented correctly to avoid starvation, deadlock, and scalability issues [Goetz 2006a].
Compliant Solution (Deep Copying)
This compliant solution creates a deep copy of the mutable widgetList before iterating over it:
List widgetList = new ArrayList(); public void widgetOperation() { List deepCopy = new ArrayList(); synchronized (widgetList) { // Client-side locking for (Object obj : widgetList) { deepCopy.add(obj.clone()); } } for (Widget w : deepCopy) { doSomething(w); } }
Creating deep copies of the list prevents underlying changes in the original list from affecting the iteration in progress. "Since the clone is thread-confined, no other thread can modify it during iteration, eliminating the possibility of ConcurrentModificationException. (The collection still must be locked during the clone operation itself)" [Goetz 2006a]. However, this approach is often more expensive than other techniques. There is also a risk of operating on stale data, which may affect the correctness of the code.
Compliant Solution (CopyOnWriteArrayList)
The CopyOnWriteArrayList data structure implements all mutating operations by making a fresh copy of the underlying array. It is fully thread-safe and is optimized for cases in which traversal operations vastly outnumber mutations. Note that traversals of such lists always see the list in the state it had at the creation of the iterator (or enhanced for loop); subsequent modifications of the list are invisible to an ongoing traversal. Consequently, this solution is inappropriate when mutations of the list are frequent or when new values should be reflected in ongoing traversals.
List widgetList = new CopyOnWriteArrayList(); public void widgetOperation() { for (Widget w : widgetList) { doSomething(w); } }
Risk Assessment
Modifying a Collection while iterating over it results in undefined behavior.
| Rule | Severity | Likelihood | Detectable | Repairable | Priority | Level |
--- --- ---
| MSC06-J | Low | Probable | No | No | P2 | L3 |
Automated Detection
Some static analysis tools can detect cases where an iterator is being used after the source container of the iterator is modified.
| Tool | Version | Checker | Description |
--- --- |
| Parasoft Jtest | 2024.2 | CERT.MSC06.ITMOD | Do not modify collection while iterating over it |
| PVS-Studio | 7.38 | V6053 |
Related Vulnerabilities
The Apache Harmony bug HARMONY-6236 documents an ArrayList breaking when given concurrent collections as input.
Bibliography
| | |
--- |
| [API 2014] | Class ConcurrentModificationException Interface Iterator |
| [Goetz 2006a] | Section 5.1.2, "Iterators and ConcurrentModificationException" |
| [SDN 2008] | Sun Bug database, Bug ID 6687277 |
draft
android-applicable
rule
msc
analyzable
6 Comments
Philip Miller
It seems that this recommendation should be moved to 11. Concurrency (CON).
Permalink
Nov 20, 2009
Dhruv Mohindra
Good suggestion because it almost exclusively talks about concurrency in the introduction. However, the NCE does not require multiple threads. I think I should make the introduction broader and keep this guideline here.
Permalink
Nov 20, 2009
Dhruv Mohindra
CopyOnWriteArrayList and CopyOnWriteArraySet could possibly be discussed here. The problem with them is that the iterator will work on "stale" copies of the collection if a thread modifies the collection when an iteration is in progress. However, this cannot be strictly an NCE or CS. Perhaps an NCE with an exception that these can be used if stale values are not a concern.
Permalink
Dec 26, 2009
Robert Seacord
This might belong here:
Although expensive in terms of performance, the CopyOnWriteArrayList and CopyOnWriteArraySet classes are sometimes used to create copies of the core Collection so that iterators do not fail with a runtime exception when some data in the Collection is modified. However, any updates to the Collection are not immediately visible to other threads. Consequently, the use of these classes is limited to boosting performance in code where the writes are fewer (or non-existent) as compared to the reads [JavaThreads 04]. In most other cases they must be avoided (see MSC13-J. Do not modify the underlying collection when an iteration is in progress for details on using these classes).
Permalink
May 13, 2011
James Ahlborn
I fixed the example "CS (Thread-Safe Collection)" to lock the collection around the iteration, as that is still necessary to safely iterate a synchronized collection. however, this example still does not avoid the possibility of a ConcurrentModificationException from the same thread caused by code in the doSomething() method modifying the widgetList. this should probably be called out in the text somewhere.
Permalink
Sep 19, 2012
David Svoboda
Findbugs has a checker PZ_DONT_REUSE_ENTRY_OBJECTS_IN_ITERATORS which warns if a Map.Entry object is preserved while iterating over a Map. The Map.Entry javadoc clearly forbids this. Can we generalize this rule to warn against stale objects like Map.Entry during iteration?
ED: Also the Findbugs checker DMI_ENTRY_SETS_MAY_REUSE_ENTRY_OBJECTS
Permalink
Aug 07, 2014
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5
What are the integer solutions to
x
2
−
1
=
2
y
2
?
9 Answers
Jakab Schrettner
·
9y
To find the positive integer solutions, i will rewrite the equation as:
x2−2y2=1
(x−√2y)(x+√2y)=1
We know that x=3 and y=2 is a solution, so substituting in these values gives us:
(3−2√2)(3+2√2)=1
Multiplying this equation with the one before yields:
[(x−√2y)(3−2√2)][(x+√2y)(3+2√2)]=1
[(3x+4y)−√2(2x+3y)][(3x+4y)+√2(2x+3y)]=1
(3x+4y)2−2(2x+3y)2=1
So if (x;y) is a solution, and we let x′=3x+4y and y′=2x+3y, then (x′;y′)also satisfies the equation.
We can use this to generate solutions:
(3;2)→(17;12)→(99;70)\rightarro
To find the positive integer solutions, i will rewrite the equation as:
x2−2y2=1
(x−√2y)(x+√2y)=1
We know that x=3 and y=2 is a solution, so substituting in these values gives us:
(3−2√2)(3+2√2)=1
Multiplying this equation with the one before yields:
[(x−√2y)(3−2√2)][(x+√2y)(3+2√2)]=1
[(3x+4y)−√2(2x+3y)][(3x+4y)+√2(2x+3y)]=1
(3x+4y)2−2(2x+3y)2=1
So if (x;y) is a solution, and we let x′=3x+4y and y′=2x+3y, then (x′;y′)also satisfies the equation.
We can use this to generate solutions:
(3;2)→(17;12)→(99;70)→(577;408)→(3363;2378)→…
I don’t know if this generates all positive integer solutions, but it generates an infinite number of them.
Also note that (1;0) is also a solution, and that x and y don't have to be positive.
Edit: changed (0;1) to (1;0)
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Senia Sheydvasser
number theorist
·
Author has 2.5K answers and 39.9M answer views
·
9y
As John Calligy and Jakab Schrettner correctly note, there are infinitely many solutions, and they give correct methods to produce more solutions starting from some initial solutions.
They essentially describe all of the solutions that exist. To prove this, it suffices to note that x2−2y2=1 is nothing more than a Pell Equation, which is more generally given by x2−Dy2=1, where D is any positive, squarefree integer.
The Pell equation is very well-studied—it is known that you can always find x0+y0√D such that
(x0+y0√D)(x0−y0√D)=x20−Dy20=
As John Calligy and Jakab Schrettner correctly note, there are infinitely many solutions, and they give correct methods to produce more solutions starting from some initial solutions.
They essentially describe all of the solutions that exist. To prove this, it suffices to note that x2−2y2=1 is nothing more than a Pell Equation, which is more generally given by x2−Dy2=1, where D is any positive, squarefree integer.
The Pell equation is very well-studied—it is known that you can always find x0+y0√D such that
(x0+y0√D)(x0−y0√D)=x20−Dy20=1, and
All other solutions (x,y) arise as x+√Dy=±(x0+y0√D)k, for some integer k.
Such a solution is known as a fundamental solution. (The fact that this such a fundamental solution must exist is really nothing more than a special case of Dirichlet's unit theorem.)
We can freely assume that x0>1,y0≥0, and it follows from that (x0+y0√D)k=x20+Ty20+2x0y0√D will have larger x and y components than the fundamental solution (for any positive k).
Therefore, there is a fundamental solution with x>1, y≥0, and it will be the one of such a form for which x is minimal.
It is easy to see that there are no solutions if x=2. Therefore, the solution x=3, y=2 is fundamental.
Ergo, every other solution x,y will be given by
x+y√2=±(3+2√2)k.
Assistant
Bot
·
1y
To find the integer solutions to the equation
x2−1=2y2,
we can rearrange it as follows:
x2=2y2+1.
This resembles a Pell-type equation. We can rewrite it as:
x2−2y2=1.
This is a standard Pell's equation of the form a2−Db2=1 where D=2. The fundamental solution to the equation x2−2y2=1 is (x1,y1)=(3,2).
The general solutions to this Pell's equation can be generated using the recurrence relations derived from the fundamental solution. The solutions can be expressed as:
xn+yn√2=(3+2√2)n.
Calculating a few iterations provides us with:
For n=0:
-
To find the integer solutions to the equation
x2−1=2y2,
we can rearrange it as follows:
x2=2y2+1.
This resembles a Pell-type equation. We can rewrite it as:
x2−2y2=1.
This is a standard Pell's equation of the form a2−Db2=1 where D=2. The fundamental solution to the equation x2−2y2=1 is (x1,y1)=(3,2).
The general solutions to this Pell's equation can be generated using the recurrence relations derived from the fundamental solution. The solutions can be expressed as:
xn+yn√2=(3+2√2)n.
Calculating a few iterations provides us with:
For n=0:
x0=1,y0=0
For n=1:
x1=3,y1=2
For n=2:
x2=17,y2=12
For n=3:
x3=99,y3=70
The pattern continues, and we can generate more solutions if needed.
The integer solutions to the equation x2−1=2y2 are thus given by the pairs (xn,yn):
(1,0)
(3,2)
(17,12)
(99,70)
etc.
These solutions can also be expressed as:
(x,y)=(3+2√2)n+(3−2√2)n
for n∈Z+, where both terms are approximations of the integer solutions. The negative solutions can be derived because both x and y can take negative values as well.
Thus, the complete set of integer solutions is:
(x,y)=(1,0),(3,2),(17,12),(99,70),(−1,0),(−3,−2),(−17,−12),(−99,−70),…
Related questions
What is the integer solution to the equation other than (1,0) {x^2 - 2y^2 =1}?
How does one find all integer solutions to
x
3
−
11
y
3
=
1
?
What are the integer solutions to
(
a
2
−
1
)
(
b
2
−
1
)
=
c
2
?
How can we find all integer solutions to
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+
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=
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?
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y
2
=
x
(
x
+
6
)
(
x
+
12
)
?
Ayush Manas
Maths freak and physics' lover
·
Author has 382 answers and 562.9K answer views
·
5y
The given equation is,
x2−2y2=1
Now, using Brahamgupta's identity,
(a+Nb)(c+Nd)= (a+N2bd)+N(ad+bc)
Put c=a=a^2, d=b=b^2,
(a2+Nb2)2=(a2−Nb2)2+N(2ab)2
Reshuffling,
(a2+Nb2)−N(2ab)2=(a2−Nb2)2
Thus you can see that eqn of the form
x2−Ny2=1 i.e. Pell's Equation are nothing but the special case of Brahamgupta's identity.
Thus,
We can solve Pell's Equation by Bhaskaracharya's Chakravala method which makes use of Brahamgupta's identity. This method is used to take only one pair of soln (x,y) let (p,q) then other pairs are obtained as,
(p2−Nq2)=1
=>(p+q√N)(p−q√N)=1
Now raising power n both sid
The given equation is,
x2−2y2=1
Now, using Brahamgupta's identity,
(a+Nb)(c+Nd)= (a+N2bd)+N(ad+bc)
Put c=a=a^2, d=b=b^2,
(a2+Nb2)2=(a2−Nb2)2+N(2ab)2
Reshuffling,
(a2+Nb2)−N(2ab)2=(a2−Nb2)2
Thus you can see that eqn of the form
x2−Ny2=1 i.e. Pell's Equation are nothing but the special case of Brahamgupta's identity.
Thus,
We can solve Pell's Equation by Bhaskaracharya's Chakravala method which makes use of Brahamgupta's identity. This method is used to take only one pair of soln (x,y) let (p,q) then other pairs are obtained as,
(p2−Nq2)=1
=>(p+q√N)(p−q√N)=1
Now raising power n both sides,
=>(p+q√N)n(p−q√N)n=1
And in general pairs (x_n,y_n) (n is in subscript) are given by,
((p+q√N)n+(p−q√N)n2,(p+q√N)n−(p−q√N)n2)
For your problem, N=2,
So, first pair can be hit and trial as (3,2)=(x,y).
Then other pairs can be found using above relation.
Brahmagupta's identity - Wikipedia
Chakravala method - Wikipedia
Thank You
Alvin Chen
Author has 124 answers and 499.9K answer views
·
9y
(1,0),(-1,0)
(3,2),(-3,2),(3,-2),(-3,-2)
(17,12),(-17,12),(17,-12),(-17,-12)
I think there’s infinite solutions, but the form seems complicated
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Awnon Bhowmik
I know a little about elementary Number Theory
·
Author has 3.7K answers and 11.2M answer views
·
9y
I will rewrite the equation as
x2=2y2+1
Trial and error method gives
(x,y)=(−1,0),(1,0),(−3,2),(3,2),(−3,−2),(3,−2),(−17,−12),(17,−12),(−17,12),(17,12)
x2 is odd, so x is odd
Let x=(2k−1)2
[math]\implies 2y^2=4k^2–4k[/math]
[math]\implies y^2=2k(k-1)[/math]
[math]\implies y^2=2k(k-1)[/math]
[math]\implies 4\mid y^2[/math]
[math]\implies 4l=2k(k-1)[/math]
[math]\implies 2l=k(k-1)[/math]
[math]\implies l=\dfrac{k(k-1)}{2}[/math]
[math]\implies l=\displaystyle \sum_{k=1}^{n} k[/math] ,[math]\forall k\in\N[/math]
Since [math]\N[/math] is an infinite set, we can choose any value [math]n\in\N[/math] to find [math]\sum k[/math] and this means that there can be infinite number of integer values for [math]y[/math], and subsequently for [math]x[/math]
Related questions
What are the integer solutions to
[math]12 \geq 2^{x+y} - 2^x[/math]
?
Can you find the integer solutions for
[math]X^3+Y^3+Z^3=10^3[/math]
?
What are the integer solutions to
[math]x^2+xy+2y^2=29[/math]
?
What are the integer solutions to
[math] m!=x^2-1,.. {x, m ..integers}[/math]
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What is the solution of (1-x-x^2-…) × (2-x-x^2-…) =0?
John Calligy
Author has 2.3K answers and 1.6M answer views
·
9y
There are infinitely many solutions. If
[math]x^2-2y^2=1[/math]
, there is another solution with
math^2-2(2x+3y)^2[/math]
Patrick Solé
I visit number theory for business and pleasure
·
Author has 1K answers and 579.2K answer views
·
9y
This is a well-known diophantine equation of a type called Pell,
Pell's equation
, that occurs in the search of units of number fields of the type Q(\sqrt(d)) with d>0. There is an infinity of solutions. Yours is the case d=2.
Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968)
·
Author has 16.6K answers and 29.3M answer views
·
9y
For any positive solutions, clearly their negatives are also solutions, so we only need to consider positive values. Each positive solution leads to three more solutions.
By inspection we can see that y = 0 gives a solution. Clearly we must have odd x and x = 3 works. Also 2y^2 + 1 must be an odd perfect square.
I suspect there are only the eight solutions above, but it needs a bit more work to prove.
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Jul-Dec 2018 - Volume 30 - Issue 2
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INTRODUCTION
CASE REPORTS
Case 1
Case 2
DISCUSSION
CONCLUSION
Declaration of patient consent
Financial support and sponsorship
Conflicts of interest
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Case Report
Biradicular mandibular canine
A review and report of two cases
Malik, Akanksha; Bansal, Parul 1,; Nikhil, Vineeta 1; Singh, Digvijay 2
Author Information
Department of Conservative Dentistry and Endodontics, AMC Dental College and Hospital, Ahmedabad, Gujarat, India
1 Department of Conservative Dentistry and Endodontics, Subharti Dental College, Meerut, Uttar Pradesh, India
2 MDS, Consultant Endodontist, Noida, Uttar Pradesh, India
Address for correspondence: Dr. Parul Bansal, 58/6, Jagriti Vihar, Meerut, Uttar Pradesh, India. E-mail: docparulbansal@gmail.com
This is an open access journal, and articles are distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as appropriate credit is given and the new creations are licensed under the identical terms.
Endodontology 30(2):p 159-162, Jul–Dec 2018. | DOI: 10.4103/endo.endo_92_17
Open
Abstract
Mandibular canine is generally considered to have one root and one canal, but it may possess two canals and even less frequently two roots and two or three canals. This article highlights the identification and endodontic management of two cases of a rare anatomical variation in mandibular canine with two roots and two canals.
INTRODUCTION
Missed canal/root is one of the main reasons for the failure of the root canal treatment. Variation in root and root canal morphology can be found in any tooth in the dental arch with varying degree of incidence. A thorough knowledge of the external and internal anatomy of teeth with its possible aberration is essential to ensure the success of endodontic treatment. Morphologically, mandibular canine is usually considered a monoradicular tooth with single root canal. Like other teeth in the dental arch, mandibular canine also does not always display the basic anatomy with one root and one canal. Several researchers have reported anatomical variations with varying degrees associated with mandibular canines [Table 1]. Occurrence of mandibular canines with one root and two root canals has been reported to be about 0%–24% in various ethnic groups [Table 2]. However, the occurrence of two roots with two or more canals is rare ranging from 0% to 5% [Table 2]. Despite the low occurrence of anatomical variations in mandibular canine, endodontists should be aware of possible variations with respect to the number of roots and root canals.
Table 1:
Case reports
Table 2:
Incidence reports
Conventional radiographs may not be able to give details about the root canal configuration in teeth with complex root canal morphology such as multiple root canals and extra roots, especially when present in buccolingual direction. Inadequate details may lead to poor prognosis. Advanced radiographic techniques such as the cone-beam computed tomography (CBCT) may result in three-dimensional evaluation of root canal configuration, thus predicting success.
This article reports endodontic management of mandibular canines with two roots and two canals. CBCT was done to confirm the internal anatomy of the tooth.
CASE REPORTS
Case 1
A 28-year-old female patient was referred to the Department of Conservative Dentistry and Endodontics with a complaint of pain in her lower right front tooth. The patient gave a history of root canal treatment done in relation to 43 by some general dentist, but as symptoms were not relieved, retreatment was attempted and gutta-percha was removed by the same dentist. On clinical examination, the tooth was tender on percussion. Radiographic examination created a suspicion of the presence of two roots with respect to 43 [Figure 1a]. CBCT examination confirmed the presence of two roots (one buccal and one lingual) and root bifurcation was present at the cervical-third of the roots [Figure 1c and d]. Under rubber dam isolation, access cavity was modified to access both the canal orifices (buccal and lingual) [Figure 1b]. Working length was measured with Canal Pro Apex Locator (Coltene, Whaledent GmbH, Germany) and confirmed radiographically [Figure 1e]. Canal preparation was done with ProTaper rotary system (Dentsply-Maillefer, Ballaigues, Switzerland) till size F3 for each canal. Irrigation with 3% NaOCl was done in between change of instruments. Calcium hydroxide (Prime dental PVT, India) was placed as an intracanal medicament for 1 week. Obturation was done with the corresponding ProTaper gutta-percha (Dentsply-Maillefer, Ballaigues, Switzerland) using AH Plus sealer (Dentsply) [Figure 1f and g]. Later on, postendodontic restoration was done with composite resin. The patient was asymptomatic during follow-up period.
Figure 1:
Case 1 –(a) Preoperative radiograph revealing the presence of two roots. (b) Cone-beam computed tomography image axial section confirming two roots. (c) Cone-beam computed tomography sagittal view showing root bifurcation.(d) Photograph showing two orifices. (e) Working length radiograph.(f) Master cone radiograph. (g) Postoperative radiograph
Case 2
A 50-year-old male patient presented with the history of severe pain in the lower left front tooth region for the past 2–3 days. On clinical examination, there was generalized attrition in the anterior teeth. The first premolar was endodontically treated and both the premolars were restored with crowns. The left lower canine (33) was tender on percussion. The tooth showed delayed response to pulp vitality tests as compared to healthy contralateral tooth. The diagnostic radiograph revealed widening of the periodontal ligament space in relation to 33 [Figure 2a]. Based on the clinical and radiographic examination, a diagnosis of irreversible pulpitis with acute apical periodontitis was established and root canal treatment was planned. The treatment was explained to the patient and consent was obtained.
Figure 2:
Case 2 – (a) Preoperative radiograph revealing the presence of two roots. (b) Photograph showing two orifices. (c) Working length radiograph.(d) Master cone radiograph. (e) Postoperative radiograph
Preoperative radiograph gave the suspicion of two roots in relation to left mandibular canine which was confirmed with multiple angled radiographs. Under local anesthesia, after rubber dam placement, access was made. Access was modified to expose two orifices, one buccal and one lingual [Figure 2b]. Working length was established using Canal Pro Apex Locator (Coltene, Whaledent GmbH, Germany) and confirmed radiographically for both the canals [Figure 2c]. Chemomechanical preparation was performed using ProTaper file system (Dentsply-Maillefer, Ballaigues, Switzerland) in a crown-down manner. About 3% solution of sodium hypochlorite and 17% ethylenediaminetetraacetic acid were used alternatively as irrigants at every change of instrument. The apical preparation was done to F2 file size in both the canals, and the canals were obturated with corresponding ProTaper gutta-percha cones (Dentsply-Maillefer, Ballaigues, Switzerland) [Figure 2d and e]. The patient was asymptomatic during the follow-up period of 10 months.
DISCUSSION
Proper diagnosis and identification of the possible permutations of the canal morphology are essential for the success of endodontic treatment. Failure to locate and treat an extra root/canal is one of the most common causes of root canal treatment failure. Detailed clinical and radiographic examinations are critical factor in detecting morphological variations. In mandibular canine with two roots, roots are generally positioned buccally and lingually which can be easily overlooked on preoperative radiograph due to superimposition of roots and canals. Radiographs taken with different horizontal angulations may help in identification or at least suspicion of root and root canal variations, but sometimes establishment of definite diagnosis may require CBCT.
Second root/canal in the mandibular canine can branch from the apical-, middle-, or coronal-thirds. Two roots may be observed separately radiographically when bifurcation is present either in the cervical or in middle thirds and in not in the line of the X-ray beam, whereas bifurcation which is present more apically is difficult to diagnose and treat. In the present case, the presence of the two roots was suspected on intraoral periapical as buccal root was present slightly mesially as compared to the lingual root and there was no exact superimposition of both the roots in angled radiograph.
In mandibular canine, usually one canal is present which exits as a single foramen at the apex. At times, two root canals or rarely three root canals with two roots, can be present. In the present case, we found two root canals, one buccal and one lingual in each buccal and lingual root.
Various researchers have carried out studies to demonstrate variation in the root/canal anatomy in different races using different methods. Variations in the internal anatomy among populations could be due to the differences in genetics and racial variations in the population, sample size, techniques, classification systems, and the researchers' judgment and diagnosis.
It is important to detect such anatomical variations before the initiation of endodontic treatment to prevent iatrogenic mishaps and to gain high success. The dental anatomical knowledge is an essential condition in the practice of endodontics. Hence, before the beginning of the treatment, a thorough knowledge of the root canal anatomy is important to achieve good results.
CONCLUSION
Although the incidence of mandibular canine with two roots and two or three canals is low, it can exist. A thorough knowledge of the tooth and root canal morphology, clinical exploration, and detailed radiographic interpretation as well as use of advanced radiographic technique may be helpful in detecting root canal aberrations and to achieve success.
Declaration of patient consent
The authors certify that they have obtained all appropriate patient consent forms. In the form the patient(s) has/have given his/her/their consent for his/her/their images and other clinical information to be reported in the journal. The patients understand that their names and initials will not be published and due efforts will be made to conceal their identity, but anonymity cannot be guaranteed.
Financial support and sponsorship
Nil.
Conflicts of interest
There are no conflicts of interest.
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Keywords:
Cone-beam computed tomography; mandibular canine; two canals; two roots
© 2018 Endodontology | Published by Wolters Kluwer – Medknow
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Biradicular mandibular canine: A review and report of two cases
Endodontology30(2):159-162, Jul-Dec 2018.
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Effect of Temperature Change on Reaction Rate
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Overview
The Arrhenius equation,
relates the activation energy and the rate constant, k, for many chemical reactions.
In this equation, R is the ideal gas constant, which has a value 8.314 J/mol·K, T is the temperature in kelvin, Ea is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.
The frequency factor, A, reflects how well the reaction conditions favor correctly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates.
The exponential term, e−Eₐ/RT, describes the effect of activation energy on the reaction rate. According to kinetic molecular theory, the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules—a lower activation energy results in a more significant fraction of adequately energized molecules and a faster reaction.
The exponential term also describes the effect of temperature on the reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy (RT) to overcome the activation barrier (Ea). This yields a higher value for the rate constant and a correspondingly faster reaction rate.
The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). The difference in activation energy required and the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of molecules, the reaction will occur slowly, since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the molecules' average kinetic energy, a large fraction of molecules will be adequately energetic, and the reaction will proceed rapidly.
Reaction diagrams are widely used in chemical kinetics to illustrate various properties of a reaction of interest. It shows how a chemical system's energy changes as it undergoes a reaction, converting reactants to products.
This text is adapted fromOpenstax, Chemistry 2e, Section 12.5: Collision Theory.
Transcript
A reaction's rate law defines the relationship between a reactant concentration and the reaction rate in terms of a rate constant, ‘k’.
The rate constant describes the relationship between temperature and kinetic parameters relating to the collision, orientation, and activation energy of reacting molecules via the Arrhenius equation. ‘A’ is a constant called the Arrhenius factor or frequency factor. ‘e’ is an exponential factor integrating activation energy measured in joules-per-mole, the gas constant, and the temperature in kelvin.
The parameters’ temperature dependence can be explained with the collision model, which states that reacting molecules should collide with sufficient energy in the correct orientation to initiate a chemical reaction.
The frequency factor constitutes two components—the collision frequency and the orientation factor. The collision frequency is the number of molecular collisions per unit time, whereas the orientation factor describes the probability of collisions with a favorable orientation.
Still, only a small fraction of collisions leads to a reaction. This is because the reacting molecules have to overcome an energy barrier, called the activation energy, to transform into products.
Only those molecules colliding with sufficient kinetic energy will have enough potential energy to bend, stretch, or break bonds to transform into a high-energy intermediate called the transition state, or the activated complex. The short-lived, unstable activated complex loses energy to form stable products, whose total energy is lower than that of the reactants.
The exponential factor in the Arrhenius equation represents the fraction of successful collisions resulting in products. An increase in temperature influences both the frequency factor and the exponential factor.
At elevated temperatures, molecules move faster, more forcefully, and with higher thermal energies, leading to more favorable collisions.
For most reactions, a temperature increase results in higher frequency and exponential factors, leading to a rise in the rate constant, consequently translating to an accelerated reaction rate.
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Trigonometry: y = a sin (bx) + c
Thomas Davis
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Posted: 18 Dec 2017
IGCSE 0606 syllabus, theme 10, Trigonometry: draw and use the graph of y = a sin (bx) + c.
For more math content and preparation for IGCSE exams 0580 and 0606, subscribe to my YouTube channel!
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hi this is Thomas welcome to trigonometry today we're looking at the trigonometric functions y equals sine X y equals cosine X and y equals tangent X and their graphic representations in the coordinate plane so let's begin with y equals sine X here we have a function our input is an x value and our output is the Y value now sine cosine tangent all trigonometric functions are periodic so there's a pattern of repetition we're going to look at one cycle of each of our functions so for sine one cycle is one complete revolution around the circle which goes from 0 to 2pi in radians and for sine at 0 the sine value is 0 and when we complete the revolution we end up at 2pi which also has a signed value of 0 halfway through we are at PI and halfway before there PI over 2 and then between PI and 2 pi is 3 PI over 2 and at these five critical points the curve for y equals sine X reaches one at PI over two goes back to zero at PI negative one at 3 PI over 2 and back to zero at two pi so our range on the y-axis is from negative one to positive one and this is one cycle or one period on the x axis from 0 to 2 pi for the function y equals sine X I can use my function either the algebra or the graph to identify my outputs for various inputs so for y equals sine of X if I have an input of 3 PI over 2 then I can see on the graph where 3 PI over 2 is on the x axis the y value is negative 1 or the input value of PI over 6 and PI over 6 is close to 0 far to the left of the curve at PI over 6 I can identify my Y value approximately with the graph or exactly with algebra as 1/2 at PI over 6 the sine value as we see on the graph as well as the unit circle gives us sine is 1/2 so these are two outputs for the input values into the function y equals sine of X there are three possible adjustments to the function which relate to amplitude period and vertical translation and we can see where these concepts lie in the more complete general sine function which is y equals a times sine BX plus C well identify our three characteristics and then look at an example amplitude is the value a and period is 2 pi divided by B and vertical displacement is C so these are three characteristics of the general sign function and we'll see how a given function results in variation in output compared to our first example so let's look at y equals two times sine of X and we'll compare that function to y equals sine of two x so beginning with y equals two times sine of X what I see is my amplitude is 2 so my amplitude is the upper limit as well as the lower limit of my function so my range is now negative 2 to 2 and I don't have any other adjustments on my to my function so I'm going from 0 to 2pi in one period and the adjustment is instead of going up to 1 and down to negative 1 I'm going up to 2 and down to negative 2 in one period so the amplitude is not 1 rather in y equals 2 sine X the amplitude is 2 now let's look at our next example y equals sine of 2x and in this case the change relates to the period so we can calculate the period as 2 pi divided by notice our B value is 2 so period 2 pi divided by 2 the period is pi thus instead of one period going from 0 to 2 pi one period will go from 0 to pi so from 0 to 2 pi we would actually have two periods for the function and the amplitude is 1 so our range will be negative 1 to 1 and we'll see what happens in this case is that in the range 0 to 2pi we have 2 cycles so I've completed one cycle from 0 to PI and then to continue I would complete a second cycle or a second period from PI to 2 pi so for the function y equals sine 2x the adjustment is that the period is decreased from 2 pi to pi and if we show the range of 0 to 2 pi we would have to complete periods in that range for y equals sine of 2x and finally let's look at the function y equals 5 sine of 1/3 X minus 2 so we have adjustments in this case to our amplitude our period and our vertical displacement let's calculate the period period is 2 PI over B and in this case our B value is 1 over 3 so 2 pi divided by 1 over 3 is 6 pi and we'll look at one period going to 6 pi on the x axis our halfway point will be at 3 pi now notice that we have vertical displacement of negative 2 so I'm going to mark my center point at negative 2 and the amplitude is 5 so I'm going to go up from negative 2 by 5 which would get me to a value of 3 and down from negative two by five which will get me to a value of negative seven and then we can also plot the zero or the x-axis up here as well so we can see that what we've done is we've shifted our graph since the vertical displacement centers us at negative two we're a little bit below the x-axis centering at x equals negative two so where will this graph and well we start at x equals zero and y equals negative two and I'll go up to three which is a movement or an amplitude of five come back down to three pi which is at x equal y equals negative two going down to seven so I reflect my amplitude of five from negative 2 to negative seven and then reaching negative two y equals negative two at the value for x of six pi so notice have completed one period I've extended the period because of the B value of one third the period is extended by a factor of three from 2 pi to 6 pi my amplitude is 5 so the range with the center line of y equals negative 2 my range is negative 7 on the y axis 2 3 on the y axis this is the function y equals 5 sine one-third X minus 2 now for each of these functions let's evaluate the input value of 11 PI over 6 function of 11 PI over 6 in all three cases so in our first function we'll have 2 times sine of 11 PI over 6 and sine of 11 PI over 6 is negative 1/2 so this is 2 times negative 1/2 with the final output value of negative 1 so for the function y equals 2 times sine of X an input value of 11 over 6 results in an output value of negative 1 in the second function y equals sine of 2 X we will have sine of 2 times 11 PI over 6 which is sine of 20 2 PI over 6 and the value of sine 22 PI over 6 is negative square root of 3 over 2 so the output value for the input 11 PI over 6 into the function y equals sine 2 X is negative square root of 3 over 2 and then finally the same input value into our third function y equals 5 sine one-third X minus 2 so here we'll have five times sine of one third times 11 PI over 6 minus 2 so we have five times sine of 11 PI over 18 minus 2 which is five times zero point nine three nine seven minus 2 which is two point six nine eight our final output value for the input of 11 PI over six into y equals five times sine one-third X minus two is two point six nine eight and to connect the algebra to or graphs I'm going to mark off on each of our three graphs 11 PI over six as well as the point on the curve which we've calculated as our output value in each of our three examples so we have 11 PI over six in both of our first cases we have a negative value on the curve which is in line with our calculations and then in the third case 11 PI over six gave us a positive output which we can see on the curve we're above the x-axis a y-value of 0 we're greater than that graphically as well as algebraically two point six nine eight is above the x-axis and in our first two examples we have a negative output values and we can see in both graphs that our point on the curve at 11 PI over six is below the x-axis so in this lesson we've learned about the sine function and in our next lesson we're going to learn about the cosine function calculating outputs algebraically as well as graphing the cosine function
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Formulary and Prescribing Guidelines SECTION 6: TREATMENT OF ADHD Section 6. Treatment of ADHD 2 Approved Medicines Management Group February 2020 6.1 Introduction Attention Deficit Hyperactivity Disorder (ADHD) is a neurobiological condition. The core symptoms are hyperactivity, impulsivity and inattention which may lead to educational and behavioural difficulties. The condition is often associated with learning difficulties, communication and motor co-ordination problems. Stimulant and other medications are recognised as part of the multi-modal management approach and help to improve concentration, reduce impulsivity and reduce hyperactivity. This guidance is not intended to be prescriptive and should be amended to the individual client and their circumstances. This includes advice and support for family and teachers in addition to specific psychological treatments for patients (such as behavioural therapy). It is worth noting that whilst these wider services are desirable, any shortfall in provision should not be used as a reason for delaying the appropriate use of medication. Treatment and care should take into account the individual’s needs and preferences. Good communication across all parties involved is essential for a treatment plan to be successful. 6.2 Treatment choice Children under 5 years Medication for pre-school children is not recommended. Parents should be offered group-based parent-training/education programmes as first line treatment1. Children aged 5 years and over and young people Offer medication for children aged 5 years and over and young people only if: their ADHD symptoms are still causing a persistent significant impairment in at least one domain after environmental modifications have been implemented and reviewed they and their parents and carers have discussed information about ADHD a baseline assessment has been carried out See the recommendations on medication below. ADHD frequently presents with other psychiatric conditions such as conduct disorder, oppositional defiant disorder, depression, anxiety disorders, tics and tourettes syndrome. These need to be identified during the initial assessment and appropriate management strategies offered which may include other medication. Many patients with ADHD (particularly during early adolescence) do not like taking regular medication and sometimes feel this is a punishment for their perceived disruptive behaviour. During this time professionals need to give a clear and consistent message to the patient (and family) about the progress the client is making. Using simple drug regimens, for example, once daily modified release doses may be useful to support adherence in such circumstances. Adults Section 6. Treatment of ADHD 3 Approved Medicines Management Group February 2020 Offer medication to adults with ADHD if their ADHD symptoms are still causing a significant impairment in at least one domain after environmental modifications have been implemented and reviewed. See the recommendations on medication choice below. Consider non-pharmacological treatment for adults with ADHD who have: made an informed choice not to have medication difficulty adhering to medication found medication to be ineffective or cannot tolerate it. 6.3 NICE Clinical Guidelines – MEDICATION All medication for ADHD should only be initiated by a healthcare professional with training and expertise in diagnosing and managing ADHD. Healthcare professionals initiating medication for ADHD should: be familiar with the pharmacokinetic profiles of all the short- and long-acting preparations available for ADHD ensure that treatment is tailored effectively to the individual needs of the child, young person or adult take account of variations in bioavailability or pharmacokinetic profiles of different preparations to avoid reduced effect or excessive adverse effects Baseline assessment Before starting medication for ADHD, people with ADHD should have a full assessment, which should include: a review to confirm they continue to meet the criteria for ADHD and need treatment a review of mental health and social circumstances, including: presence of coexisting mental health and neurodevelopmental conditions current educational or employment circumstances risk assessment for substance misuse and drug diversion care needs a review of physical health, including: a medical history, taking into account conditions that may be contraindications for specific medicines current medication Section 6. Treatment of ADHD 4 Approved Medicines Management Group February 2020 height and weight (measured and recorded against the normal range for age, height and sex) baseline pulse and blood pressure (measured with an appropriately sized cuff and compared with the normal range for age) a cardiovascular assessment An electrocardiogram (ECG) is not needed before starting stimulants, atomoxetine or guanfacine, unless the person has any of the features listed below or a co-existing condition that is being treated with a medicine that may pose an increased cardiac risk. 1 Refer for a cardiology opinion before starting medication for ADHD if any of the following apply: history of congenital heart disease or previous cardiac surgery history of sudden death in a first-degree relative under 40 years suggesting a cardiac disease shortness of breath on exertion compared with peers fainting on exertion or in response to fright or noise palpitations that are rapid, regular and start and stop suddenly (fleeting occasional bumps are usually ectopic and do not need investigation) chest pain suggesting cardiac origin signs of heart failure a murmur heard on cardiac examination blood pressure that is classified as hypertensive for adults (see NICE's guideline on hypertension in adults). Refer to a paediatric hypertension specialist before starting medication for ADHD if blood pressure is consistently above the 95th centile for age and height for children and young people. 6.4 NICE Clinical Guidelines –Children aged 5 years and over and young people NG87, published March 2018. Attention Deficit Hyperactivity Disorder: diagnosis and management1 Section 6. Treatment of ADHD 5 Approved Medicines Management Group February 2020 Before starting any treatment for ADHD, discuss the following with the person, and their family or carers as appropriate, encouraging children and young people to give their own account of how they feel: the benefits and harms of non-pharmacological and pharmacological treatments (for example, the efficacy of medication compared with no treatment or non-pharmacological treatments, potential adverse effects and non-response rates) the importance of adherence to treatment and any factors that may affect this (for example, it may be difficult to take medication at school or work, or to remember appointments). Record the person's preferences and concerns in their treatment plan. Reassure people with ADHD, and their families or carers as appropriate, that they can revisit decisions about treatments. Dietary advice Do not advise elimination of artificial colouring and additives from the diet as a generally applicable treatment for children and young people with ADHD. Do not advise or offer dietary fatty acid supplementation for treating ADHD in children and young people. Advise the family members or carers of children with ADHD that there is no evidence about the long‑term effectiveness or potential harms of a 'few food' diet for children with ADHD, and only limited evidence of short‑term benefits. Drug treatment Note on prescribing unlicensed medicines At the time of publication of NICE CG87 (March 2018), medicines used for the treatment of ADHD did not have a UK marketing authorisation for use in all indications. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. The recommendations below update NICE's technology appraisal guidance on methylphenidate, atomoxetine and dexamfetamine for ADHD in children and adolescents (TA98). At the time of publication (March 2018), medicines used for the treatment of ADHD did not have a UK marketing authorisation for use in children aged 5 years and under for this indication. Offer methylphenidate (either short or long acting) as the first line pharmacological treatment for children aged 5 years and over and young people with ADHD. Section 6. Treatment of ADHD 6 Approved Medicines Management Group February 2020 At the time of publication (March 2018), methylphenidate did not have a UK marketing authorisation for this indication in children aged 5 years or under. Consider switching to lisdexamfetamine for children aged 5 years and over and young people who have had a 6 week trial of methylphenidate at an adequate dose and not derived enough benefit in terms of reduced ADHD symptoms and associated impairment. At the time of publication (March 2018), lisdexamfetamine did not have a UK marketing authorisation for this indication in children aged 5 years. Consider dexamfetamine for children aged 5 years and over and young people whose ADHD symptoms are responding to lisdexamfetamine but who cannot tolerate the longer effect profile. At the time of publication (March 2018), dexamfetamine was only licensed for the treatment of ADHD in children and adolescents aged 6 to 17 years when response to previous methylphenidate treatment is considered clinically inadequate. Dexamfetamine is not licensed for the treatment of ADHD in children and adolescents aged 5 to 17 years who have responded to, but are intolerant to lisdexamfetamine. Offer atomoxetine or guanfacine to children aged 5 years and over and young people if: they cannot tolerate methylphenidate or lisdexamfetamine or their symptoms have not responded to separate 6 week trials of lisdexamfetamine and methylphenidate, having considered alternative preparations and adequate doses. At the time of publication (March 2018), atomoxetine or guanfacine did not have a UK marketing authorisation for this indication in children aged 5 years. 6.5 NICE Clinical Guidelines – ADULTS NG87, published March 2018. Attention Deficit Hyperactivity Disorder: diagnosis and management1 Drug treatment Note on prescribing unlicensed medicines At the time of publication of NICE CG87 (March 2018), medicines used for the treatment of ADHD did not have a UK marketing authorisation for use in all indications. The prescriber should follow relevant professional guidance, taking full responsibility for the decision. Informed consent should be obtained and Section 6. Treatment of ADHD 7 Approved Medicines Management Group February 2020 documented. See the General Medical Council's Prescribing guidance: prescribing unlicensed medicines for further information. Offer lisdexamfetamine or methylphenidate as first-line pharmacological treatment for adults with ADHD. At the time of publication (March 2018), lisdexamfetamine was licensed for use in adults with symptoms of ADHD that pre-existed in childhood. At the time of publication (March 2018), not all preparations of methylphenidate had a UK marketing authorisation for treating symptoms of ADHD in adults. Consider switching to lisdexamfetamine for adults who have had a 6 week trial of methylphenidate at an adequate dose but have not derived enough benefit in terms of reduced ADHD symptoms and associated impairment. Consider switching to methylphenidate for adults who have had a 6 week trial of lisdexamfetamine at an adequate dose but have not derived enough benefit in terms of reduced ADHD symptoms and associated impairment. Consider dexamfetamine for adults whose ADHD symptoms are responding to lisdexamfetamine but who cannot tolerate the longer effect profile. At the time of publication (March 2018), dexamfetamine did not have a UK marketing authorisation for this indication in adults. Offer atomoxetine to adults if: they cannot tolerate lisdexamfetamine or methylphenidate or their symptoms have not responded to separate 6 week trials of lisdexamfetamine and methylphenidate, having considered alternative preparations and adequate doses. At the time of publication (March 2018), atomoxetine was licensed for use in adults with symptoms of ADHD that pre-existed in childhood. 6.6 NICE Clinical Guidelines – FURTHER MEDICATION CHOICES Obtain a second opinion or refer to a tertiary service if ADHD symptoms in a child aged 5 years or over, a young person or adult are unresponsive to one or more stimulants and one non-stimulant. Section 6. Treatment of ADHD 8 Approved Medicines Management Group February 2020 Do not offer any of the following medication for ADHD without advice from a tertiary ADHD service: guanfacine for adults clonidine for children with ADHD and sleep disturbance, rages or tics atypical antipsychotics in addition to stimulants for people with ADHD and coexisting pervasive aggression, rages or irritability medication not included in recommendations in sections 6.4 (children aged 5 years and over and young people) and 6.5 (adults) At the time of publication (March 2018), guanfacine did not have a UK marketing authorisation for this indication. At the time of publication (March 2018), clonidine did not have a UK marketing authorisation for this indication. 6.7 NICE Clinical Guidelines – MEDICATION CHOICE – People with coexisting conditions Offer the same medication choices to people with ADHD and anxiety disorder, tic disorder or autism spectrum disorder as other people with ADHD. For children aged 5 years and over, young people and adults with ADHD experiencing an acute psychotic or manic episode: stop any medication for ADHD consider restarting or starting new ADHD medication after the episode has resolved, taking into account the individual circumstances, risks and benefits of the ADHD medication. 6.8 NICE Clinical Guidelines – Consideration when prescribing ADHD medication When prescribing stimulants for ADHD, consider modified-release once-daily preparations for the following reasons: convenience improving adherence reducing stigma (because there is no need to take medication at school or in the workplace) reducing problems of storing and administering controlled drugs at school the risk of stimulant misuse and diversion with immediate-release preparations their pharmacokinetic profiles. Immediate-release preparations may be suitable if more flexible dosing regimens are needed, or during initial titration to determine correct dosing levels. Section 6. Treatment of ADHD 9 Approved Medicines Management Group February 2020 When prescribing stimulants for ADHD, be aware that effect size, duration of effect and adverse effects vary from person to person. Consider using immediate- and modified-release preparations of stimulants to optimise effect (for example, a modified-release preparation of methylphenidate in the morning and an immediate-release preparation of methylphenidate at another time of the day to extend the duration of effect). Be cautious about prescribing stimulants for ADHD if there is a risk of diversion for cognitive enhancement or appetite suppression. Do not offer immediate-release stimulants or modified-release stimulants that can be easily injected or insufflated if there is a risk of stimulant misuse or diversion. Prescribers should be familiar with the requirements of controlled drug legislation governing the prescription and supply of stimulants. See NICE's guideline on controlled drugs. Dose titration During the titration phase, ADHD symptoms, impairment and adverse effects should be recorded at baseline and at each dose change on standard scales by parents and teachers, and progress reviewed regularly (for example, by weekly telephone contact) with a specialist. Titrate the dose against symptoms and adverse effects in line with the BNF or BNF for Children until dose optimisation is achieved, that is, reduced symptoms, positive behaviour change, improvements in education, employment and relationships, with tolerable adverse effects. Ensure that dose titration is slower and monitoring more frequent if any of the following are present in people with ADHD: neurodevelopmental disorders (for example, autism spectrum disorder, tic disorders, learning disability [intellectual disability]) mental health conditions (for example, anxiety disorders [including obsessive– compulsive disorder], schizophrenia or bipolar disorder, depression, personality disorder, eating disorder, post-traumatic stress disorder, substance misuse) physical health conditions (for example, cardiac disease, epilepsy or acquired brain injury). Shared care for medication After titration and dose stabilisation, prescribing and monitoring of ADHD medication should be carried out under Shared Care Protocol arrangements with primary care. Section 6. Treatment of ADHD 10 Approved Medicines Management Group February 2020 6.9 NICE Clinical Guidelines – Consideration when prescribing ADHD medication Maintenance and monitoring Monitor effectiveness of medication for ADHD and adverse effects, and document in the person's notes. Encourage people taking medication for ADHD to monitor and record their adverse effects, for example, by using an adverse effect checklist. Consider using standard symptom and adverse effect rating scales for clinical assessment and throughout the course of treatment for people with ADHD. Ensure that children, young people and adults receiving treatment for ADHD have review and follow up according to the severity of their condition, regardless of whether or not they are taking medication. Height and weight For people taking medication for ADHD: measure height every 6 months in children and young people measure weight every 3 months in children 10 years and under measure weight at 3 and 6 months after starting treatment in children over 10 years and young people, and every 6 months thereafter, or more often if concerns arise measure weight every 6 months in adults plot height and weight of children and young people on a growth chart and ensure review by the healthcare professional responsible for treatment. If weight loss is a clinical concern, consider the following strategies: taking medication either with or after food, rather than before meals taking additional meals or snacks early in the morning or late in the evening when stimulant effects have worn off obtaining dietary advice consuming high-calorie foods of good nutritional value taking a planned break from treatment changing medication. If a child or young person's height over time is significantly affected by medication (that is, they have not met the height expected for their age), consider a planned break in treatment over school holidays to allow 'catch up' growth. Consider monitoring BMI of adults with ADHD if there has been weight change as a result of their treatment, and changing the medication if weight change persists. Section 6. Treatment of ADHD 11 Approved Medicines Management Group February 2020 Cardiovascular Monitor heart rate and blood pressure and compare with the normal range for age before and after each dose change and every 6 months. Do not offer routine blood tests (including liver function tests) or ECGs to people taking medication for ADHD unless there is a clinical indication. If a person taking ADHD medication has sustained resting tachycardia (more than 120 beats per minute), arrhythmia or systolic blood pressure greater than the 95th percentile (or a clinically significant increase) measured on 2 occasions, reduce their dose and refer them to a paediatric hypertension specialist or adult physician. If a person taking guanfacine has sustained orthostatic hypotension or fainting episodes, reduce their dose or switch to another ADHD medication. Tics If a person taking stimulants develops tics, think about whether: the tics are related to the stimulant (tics naturally wax and wane) and the impairment associated with the tics outweighs the benefits of ADHD treatment. If tics are stimulant related, reduce the stimulant dose, or consider changing to guanfacine (in children aged 5 years and over and young people only), atomoxetine, clonidine, or stopping medication. At the time of publication (March 2018), atomoxetine was licensed for use in adults with symptoms of ADHD that pre-existed in childhood. At the time of publication (March 2018), clonidine did not have a UK marketing authorisation for this indication. Clonidine should only be considered for people under 18 years after advice from a tertiary ADHD service. Sexual dysfunction Monitor young people and adults with ADHD for sexual dysfunction (that is, erectile and ejaculatory dysfunction) as potential adverse effects of atomoxetine. Seizures If a person with ADHD develops new seizures or a worsening of existing seizures, review their ADHD medication and stop any medication that might be contributing to the seizures. After investigation, cautiously reintroduce ADHD medication if it is unlikely to be the cause of the seizures. Section 6. Treatment of ADHD 12 Approved Medicines Management Group February 2020 Sleep Monitor changes in sleep pattern (for example, with a sleep diary) and adjust medication accordingly. Worsening behaviour Monitor the behavioural response to medication, and if behaviour worsens adjust medication and review the diagnosis. Stimulant diversion Healthcare professionals and parents or carers should monitor changes in the potential for stimulant misuse and diversion, which may come with changes in circumstances and age. 6.10 NICE Clinical Guidelines – Adherence to treatment Use this guideline with NICE's guideline on medicines adherence to improve the care for adults with ADHD. The principles also apply to children and young people. Be aware that the symptoms of ADHD may lead to people having difficulty adhering to treatment plans (for example, remembering to order and collect medication). Ensure that people are fully informed of the balance of risks and benefits of any treatment for ADHD and check that problems with adherence are not due to misconceptions (for example, tell people that medication does not change personality). Encourage the person with ADHD to use the following strategies to support adherence to treatment: being responsible for their own health, including taking their medication as needed following clear instructions about how to take the medication in picture or written format, which may include information on dose, duration, adverse effects, dosage schedule (the instructions should stay with the medication, for example, a sticker on the side of the packet) using visual reminders to take medication regularly (for example, apps, alarms, clocks, pill dispensers, or notes on calendars or fridges) taking medication as part of their daily routine (for example, before meals or after brushing teeth) attending peer support groups (for both the person with ADHD and for the families and carers). Encourage parents and carers to oversee ADHD medication for children and young people. 6.11 NICE Clinical Guidelines – Review of medication and discontinuation Section 6. Treatment of ADHD 13 Approved Medicines Management Group February 2020 A healthcare professional with training and expertise in managing ADHD should review ADHD medication at least once a year and discuss with the person with ADHD (and their families and carers as appropriate) whether medication should be continued. The review should include a comprehensive assessment of the: preference of the child, young person or adult with ADHD (and their family or carers as appropriate) benefits, including how well the current treatment is working throughout the day adverse effects clinical need and whether medication has been optimised impact on education and employment effects of missed doses, planned dose reductions and periods of no treatment effect of medication on existing or new mental health, physical health or neurodevelopmental conditions need for support and type of support (for example, psychological, educational, social) if medication has been optimised but ADHD symptoms continue to cause a significant impairment. Encourage people with ADHD to discuss any preferences to stop or change medication and to be involved in any decisions about stopping treatments. Consider trial periods of stopping medication or reducing the dose when assessment of the overall balance of benefits and harms suggests this may be appropriate. If the decision is made to continue medication, the reasons for this should be documented. References 1. NICE NG87, published March 2018. Last updated: September 2019. Attention Deficit Hyperactivity Disorder: Diagnosis and management Accessed 29/1/2020.
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https://www.quora.com/When-should-I-use-the-ideal-gas-law-equation-vs-the-combined-gas-law-equation
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Ideal Gas
Gas Laws (physics)
Thermodynamics Physics
Science Physics
Combined Gas Law
Physical Chemistry
Real and Ideal Gases
5
When should I use the ideal gas law equation vs the combined gas law equation?
Fran Nolen
Taught high school honors physics and chemistry
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Author has 232 answers and 1.4M answer views
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8y
The Ideal Gas Law is used when you have one gas (or gas mixture) and a set temperature and pressure. PV=nRT
The combined gas law is actually the Ideal Gas Law written for one gas (or gas mixture) and two sets of temperature and pressure:
P2V2=nRT2 (the number of moles in each equation is the same)
Dividing the first equation by the second equation, the number of moles and the Universal Gas Constant, R, cancel out and you get
P1V1/(P2V2) = T1/T2 Combined Gas Law
This equation can be solved for any of the variables, given values for the others. Remember, temperature is always in Kelvin in ga
The Ideal Gas Law is used when you have one gas (or gas mixture) and a set temperature and pressure. PV=nRT
The combined gas law is actually the Ideal Gas Law written for one gas (or gas mixture) and two sets of temperature and pressure:
P1V1=nRT1
P2V2=nRT2 (the number of moles in each equation is the same)
Dividing the first equation by the second equation, the number of moles and the Universal Gas Constant, R, cancel out and you get
P1V1/(P2V2) = T1/T2 Combined Gas Law
This equation can be solved for any of the variables, given values for the others. Remember, temperature is always in Kelvin in gas problems.
Related questions
What is the difference between the combined gas law and the ideal gas law? Are they the same thing?
How is the ideal gas equation utilized to obtain combined gas law?
What is the combined and ideal gas law?
What are the ideal and combined gas law equations? What purpose do they serve?
Which equation agrees with the ideal gas law?
MNIN
Author has 1.5K answers and 2.5M answer views
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3y
the combined gas law is a derivative of the ideal gas law.
PV = nRT
PV/(nT) = R
since R is a constant.. ALL PV/(nT) are equal
P1V1 / (n1T1) = P2V2 / (n2T2)
at constant moles, n1 = n2 and we can cancel n’s so that
P1V1 / T1 = P2V2 / T2….. combined gas law
when do you use it? I don’t. I do this instead
start with P1V1 / (n1T1) = P2V2 / (n2T2)
rearrange for my desired unknown. V2 for example. V2 = V1 (P1/P2) (T2/T1) (n2/n1)
cancel anything held constant… moles for example… V2 = V1 (P1/P2) (T2/T1)
plug and chug
that way I’m never sitting around thinking… “ok… pressure and volume are constant… which
the combined gas law is a derivative of the ideal gas law.
PV = nRT
PV/(nT) = R
since R is a constant.. ALL PV/(nT) are equal
P1V1 / (n1T1) = P2V2 / (n2T2)
at constant moles, n1 = n2 and we can cancel n’s so that
P1V1 / T1 = P2V2 / T2….. combined gas law
when do you use it? I don’t. I do this instead
start with P1V1 / (n1T1) = P2V2 / (n2T2)
rearrange for my desired unknown. V2 for example. V2 = V1 (P1/P2) (T2/T1) (n2/n1)
cancel anything held constant… moles for example… V2 = V1 (P1/P2) (T2/T1)
plug and chug
that way I’m never sitting around thinking… “ok… pressure and volume are constant… which law is that?” then try to remember the exact law and write it down and solve the problem. Waste of time.
Jake Dickerman
HS Chemistry and Physics Teacher
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Author has 1.4K answers and 2.3M answer views
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7y
Every gas law - Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, Avogadro’s law, and the Combined Gas Law - can be derived using Ideal Gas Law. The assumption in the Combined Gas Law is that the moles of gas are remaining constant (as, of course, is the gas constant - that’s why we call it a constant) and two of the variables (temperature, volume, and pressure) are changing in ways that you know. In those circumstances, you will want to use the combined gas law just to make your life a little easier, since - again - the assumption is that nR is remaining constant.
Guy Clentsmith
Chemistry tutor... at Self-Employment (2018–present)
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Author has 26.5K answers and 19.7M answer views
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6y
When the circumstances warrant. The combined gas equation deals with EQUIMOLAR quantities of gas, under DIFFERENT conditions of temperature, pressure, and volume…
P1V1T1=P2V2T2
The Ideal Gas equation introduces the number of GASEOUS molecules, i.e. the moles of gas, under DIFFERENT conditions of temperature, pressure, and volume…
PV=nRT…
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Related questions
Is the ideal gas equation applicable for all fluids (gas or liquid)?
What is the ideal gas law?
Why do we use the ideal gas equation when instead van der Waals equation can be used as ideal gas equation is only for ideal gas?
Why is the combined gas law useful? How is it applied?
What is the ideal gas law and what is the equation that represents it?
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The choice between using the Ideal Gas Law and the Combined Gas Law depends on the specific conditions and information available in your problem. Here’s a breakdown of when to use each:
Ideal Gas Law
Equation: PV=nRT
When to Use:
- You have information about the number of moles of gas (n).
- You need to calculate one of the variables (pressure P, volume V, temperature T, or the number of moles n).
- The gas behaves ideally, meaning it follows the assumptions of the ideal gas law (high temperature, low pressure).
- You want to relate all four variables simultaneously.
Combined Gas Law
Equation: \f
The choice between using the Ideal Gas Law and the Combined Gas Law depends on the specific conditions and information available in your problem. Here’s a breakdown of when to use each:
Ideal Gas Law
Equation: PV=nRT
When to Use:
- You have information about the number of moles of gas (n).
- You need to calculate one of the variables (pressure P, volume V, temperature T, or the number of moles n).
- The gas behaves ideally, meaning it follows the assumptions of the ideal gas law (high temperature, low pressure).
- You want to relate all four variables simultaneously.
Combined Gas Law
Equation: P1V1T1=P2V2T2
When to Use:
- You are comparing two different states of the same gas (initial and final conditions).
- You do not know the number of moles of gas, or it's constant and thus cancels out.
- You need to analyze how changes in pressure, volume, and temperature affect each other.
- The gas behaves ideally in both states.
Summary
Use the Ideal Gas Law when you have the number of moles and want to find a specific variable.
Use the Combined Gas Law when comparing two states of a gas without needing to know the number of moles.
If you have a specific scenario in mind, feel free to share it for more tailored advice!
Steve Ruis
I have been thinking/reading about this for over 60 years
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If you look carefully, the ideal gas law has a term in it which the combined gas law does not. So, use it when you need to deal with that term, otherwise the CGL is fine.
Oh, and only when the conditions that allow those laws to make accurate predictions. Don’t do anything silly like trying at apply them to solids or condensing liquids.
Amy Chai
MD, Internal Medicine, MS, Epidemiology, author
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How does the ideal gas law work?
Does PV still equal nRT?
Ah such loveliness. We have the ideal gas law because we make a number of assumptions, hence the term, “ideal.” The answer, of course, is an approximation, assuming the gas molecules don’t interact with one another inside the adiabatic container.
So we do not actually have an ideal gas. We assume an ideal gas. That is how the ideal gas law works. It is good for most work, at least I never had cause to mess with other terms or anything like that. But I’m sure smarter, more advanced people use more advanced equations.
I am not sure if that was what you were asking, exactly,
Does PV still equal nRT?
Ah such loveliness. We have the ideal gas law because we make a number of assumptions, hence the term, “ideal.” The answer, of course, is an approximation, assuming the gas molecules don’t interact with one another inside the adiabatic container.
So we do not actually have an ideal gas. We assume an ideal gas. That is how the ideal gas law works. It is good for most work, at least I never had cause to mess with other terms or anything like that. But I’m sure smarter, more advanced people use more advanced equations.
I am not sure if that was what you were asking, exactly, but you did not ask how to calculate it, or what the terms mean, so…hope that helped answer your question.
If it didn’t, go to the Khan Academy and look for a tutorial about the Ideal Gas Law. This is not the place for a chemistry lecture.
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Alex Dowgialo
Former Head of Laboratory at Tuvinian Institute for Exploration of Natural Resources SD RAS
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What is the difference between the combined gas law and the ideal gas law? Are they the same thing?
The combined law and the ideal gas law are, strictly speaking, different things.
Combined gas law.
When Sir Robero Boyle conducted his volumetric tests in Oxford in 1662, it never occurred to him that he was dealing with “ideal gas”. Jacques Charles, who first discovered the law of volumes, also considered gas to be completely ordinary.
What today is called “combined gas law” in textbooks was formulated by Gay-Lusak (1822) and Sadi Carnot (1824), as a formula combining Boyle’s law and Charles’s law. However, the combination of the laws of state of gas was widely known thanks to the work of Clapey
The combined law and the ideal gas law are, strictly speaking, different things.
Combined gas law.
When Sir Robero Boyle conducted his volumetric tests in Oxford in 1662, it never occurred to him that he was dealing with “ideal gas”. Jacques Charles, who first discovered the law of volumes, also considered gas to be completely ordinary.
What today is called “combined gas law” in textbooks was formulated by Gay-Lusak (1822) and Sadi Carnot (1824), as a formula combining Boyle’s law and Charles’s law. However, the combination of the laws of state of gas was widely known thanks to the work of Clapeyron in 1834.
The improvement of the gas equation of state was required in connection with the advent of heat engines and their thermodynamic cycles. That is why Boyle's law was developed only after one and a half hundred years.
It is very important that Clapeyron’s law, which is the combined law, directly follows from the first law of thermodynamics in the Carnot formulation
(1)δQ=cv∂T∂pdp+cp∂T∂vdv
Here δQ is the quantity of heat, T is the temperature; cp, cv - heat capacity, respectively, at constant pressure and volume; p is the pressure; v is the specific volume.
A direct consequence of the Carnot law is the statement that the quantity of heat QC obtained in cycle C
exactly equal to the work of this cycle. For this to happen, the expression in parentheses from (2) should be [math]1[/math]
math \dfrac{\partial}{\partial p} c_p\dfrac{\partial T}{\partial v}-\dfrac{\partial}{\partial v} c_v\dfrac{\partial T}{\partial p}=1.[/math]
The simplest solution to the partial differential equation (3), at constant values of specific heat, is the function
math T=\dfrac{pv}{R},[/math]
where [math]R = c_p-c_v[/math] is the famous law of Robert Mayer (1845).
The ideal gas low.
The law of ideal gas is a terimine from the second half of the 19th century and is related to the studies of Maxwell and Boltzmann of their famous distribution function [math]f (u)[/math], and the formula directly follows from it for the average velocity of gas molecules [math]u_p[/math]
math u_p=\sqrt{\dfrac{2kT}{m}}=\sqrt{\dfrac{2RT}{M}}[/math]
Here [math]M = mN_a[/math], [math]m[/math] is the mass of the gas molecule, [math]N_a[/math] is the Avagadro number, and [math]k[/math] is the Boltzmann constant.
Total.
If the molecules do not interact with each other other than through collisions, the kinetic energy of their motion is directly related to the gas pressure. Under these conditions, equation (5) is converted to equation (4).
Throughout this story, equation (4) is confirmed by the experiments of Boyle and Gay-Lusak, and the distribution function of Maxwell-Boltzmann is confirmed by the experiments of Zartman. The idea of an “ideal gas” connects one with another.
Aakash Yadav
Studied at National College of Engineering, Talchhikhel-Lalitpur,Kathmandu (Graduated 2022)
·
Upvoted by
Surajit Karmakar
, M.Sc. Chemistry, The University of Burdwan (2019)
·
9y
Related
How is the Combined Gas Law derived?
According to Boyle’s Law “Temperature remaining constant, volume of the given mass of gas is inversely proportional to the pressure”.
Let ‘V’ be the volume and ‘P’ be the pressure, then at constant temperature,
V ∝ 1/P—————-(1)
According to Charles Law “Pressure remaining constant, volume of the given mass of gas is directly proportional to the temperature in kelvin scale or absolute scale.
Let ‘V’ be the volume of the given mass of gas, ‘T’ be the temperature, then at constant temperature’P’,
V ∝ T——————(2)
If pressure and temperature are simultaneously changed , tehn,
From (1) and (2)
V ∝ 1/T
or, V ∝
According to Boyle’s Law “Temperature remaining constant, volume of the given mass of gas is inversely proportional to the pressure”.
Let ‘V’ be the volume and ‘P’ be the pressure, then at constant temperature,
V ∝ 1/P—————-(1)
According to Charles Law “Pressure remaining constant, volume of the given mass of gas is directly proportional to the temperature in kelvin scale or absolute scale.
Let ‘V’ be the volume of the given mass of gas, ‘T’ be the temperature, then at constant temperature’P’,
V ∝ T——————(2)
If pressure and temperature are simultaneously changed , tehn,
From (1) and (2)
V ∝ 1/T
or, V ∝ T/P
or, V ∝ kT/P
∴PV/T = K ————-(3)(where k is the proportionality sign whose value depends upon mass of a gas and unit of pressure and volume.)
Let P1, V1, and T1 be the initial pressure, volume, and temperature (in kelvin scale) respectively. Similarly, P2, V2 and T2 be the final pressure, volume and temperature(in kelvin scale) of a given mass of gas. Then according to eqn(3).
We have,
P1V1/T1 = K ——-(4)
P2V2/T2=K ——-(5)
From (4) and (5) we get,
P1V1/T1 = P2V2/T2.
THIS IS THE FINAL FORM OF COMBINED GAS LAW EQUATION.
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Pratishtha Singh
Knows Hindi
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8y
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What is gas law?
Gas laws are the laws that explains the nature or behavior of the gas when certain conditions are applied to it like Change in pressure temperature or volume
.
sobasically they tell that how the gas is going to behave in those conditions whether it will show expansion or compression . Scientist gave 4 gas laws which provided information about gaseous states . These are
Boyle's law- this law states that at a constant temperature the pressure is inversely proportional to the volume for a given amount of gas .its a relationship between pressure and volume.
Suppose there is a piston gas with ain
Gas laws are the laws that explains the nature or behavior of the gas when certain conditions are applied to it like Change in pressure temperature or volume . sobasically they tell that how the gas is going to behave in those conditions whether it will show expansion or compression . Scientist gave 4 gas laws which provided information about gaseous states . These are
Boyle's law- this law states that at a constant temperature the pressure is inversely proportional to the volume for a given amount of gas .its a relationship between pressure and volume. Suppose there is a piston gas with ain it so initially there is only atmospheric pressure acting on it and it has some volume now if we apply some additional pressure there then volume decreases . Hence proved
Numerically. P=k×1/V. ( Where k is the constant of proportionality )
Or P×V= constant
Or P1× V1=P2×V2
2 Charles law -states that constant pressure, volume and temperature are directly proportional to each other for a given amount of gas .or for any one degree rise or fall in temperature the volume of gas increases or decreases by 1/273.15 of it's original volume. It's a relationship between volume and temperature. Suppose there is inflated balloon that is kept at a high temp so after some time an increase in volume can be observed whereas once kept at low temp a decrease in volume is observed.hece high temp high volume low temp low volume so it's a direct relationship.
Numerically ,. V=kT. ( Where k is constant of proportionality)
Or V/T=k
Or V1/T1 = V2/T2
Gay lussac's law states that at constant volume pressure is directly proportional to the temperaturefor a given amount of gas.
Numerically. P=kT (k is constant of proportionality)
Or P/T= constant
Or P1/V1 = P2/V2
4 . Avogadro's law states that under similar conditions of temperature and pressure equal volumeof gas have equal number of moles.
V=kn ( k is Avogadro's constant)
Now a gas following these laws strictly are called ideal gases which is a hypothetical concept.there is no such gas and are supposed to have no intermolecular forces between them and the gases which exist are real gases that show deviation from ideal behavior and follow these law only when the Temperature is high and pressure is low otherwise they follow an ideal gas equation that is a combination of 1,2,&4 law which gives the simultaneous effect of pressure temp and volume on a given amount of gas.
Ideal gas equation is PV= nRT or PV/T= CONSTANT
Aakash Yadav
Studied at National College of Engineering, Talchhikhel-Lalitpur,Kathmandu (Graduated 2022)
·
9y
Related
Gases:
Is there a modern equation that describes gas laws better than Ideal Gas laws?
Yes, Vander Waal’s Equation is your answer.
The ideal gas equation ( PV=nRT) is not obeyed by the real gases at high pressure or low temperature. However, Van der Waal deduced a modified gas equation ( Vand der Waal’s equation) which is obeyed by the real gases over a wide range of temperature and pressure, by making two important corrections to the usual gas equation, (PV=nRT).
Volume correction :
The equation is
{P + (ax^2)/v^2}(V-nb) = nRT
where, n = number of moles of a gas where ‘a’ and ‘b’ are constants called Van der Waal’s constant and are characteristics of the ga.
Constant ‘a’ is rel
Yes, Vander Waal’s Equation is your answer.
The ideal gas equation ( PV=nRT) is not obeyed by the real gases at high pressure or low temperature. However, Van der Waal deduced a modified gas equation ( Vand der Waal’s equation) which is obeyed by the real gases over a wide range of temperature and pressure, by making two important corrections to the usual gas equation, (PV=nRT).
Volume correction :
The equation is
{P + (ax^2)/v^2}(V-nb) = nRT
where, n = number of moles of a gas where ‘a’ and ‘b’ are constants called Van der Waal’s constant and are characteristics of the ga.
Constant ‘a’ is related to volume of the molecules and the constant ‘b’ to the inter-molecular forces.
For 1 mole of a gas
(P + a/v^2)(v-b) = RT
where,
(P + a/v^2) = pressure correction
(v-b) = volume correction for 1 mole of a gas.
Henry Gerard
PhD from Stanford University (Graduated 1970)
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Author has 2.2K answers and 805.1K answer views
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Why do real gases obey combined gas law but don't obey the ideal gas law?
At very high pressures and cryogenic temperatures the molecules of real gases have a weak molecule to molecule attractive force that can result in liquefaction.
Such gases are the same ones that act as ideal gases, under normal conditions, where the inter-molecular attractive force becomes negligible.
Learning App for Class 5-12 (Official Account)
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What are gas laws?
During high pressure or high-temperature conditions, a tyre inflated with air is at the risk of bursting. Or while climbing a mountain you start feeling problems to inhale? Why is it so? With changing physical conditions the behaviour of gaseous particles also deviates from their normal behaviour. The behaviour of a Gas can be studied by various laws known as the Gas laws. Let us see more!
The Gas Laws
All gases generally show similar behaviour when the conditions are normal. But with a slight change in physical conditions like pressure, temperature or volume these show a deviation. Gas laws are
During high pressure or high-temperature conditions, a tyre inflated with air is at the risk of bursting. Or while climbing a mountain you start feeling problems to inhale? Why is it so? With changing physical conditions the behaviour of gaseous particles also deviates from their normal behaviour. The behaviour of a Gas can be studied by various laws known as the Gas laws. Let us see more!
The Gas Laws
All gases generally show similar behaviour when the conditions are normal. But with a slight change in physical conditions like pressure, temperature or volume these show a deviation. Gas laws are an analysis of this behaviour of gases. The variables of state like the Pressure, Volume and Temperature of a gas depict its true nature. hence gas laws are relations between these variables. Let us study more about the important gas laws!
Boyle’s Law
Boyle’s law states the relation between volume and pressure at constant temperature and mass. Robert Boyle conducted an experiment on gases to study the deviation of its behaviour in changed physical conditions.
It states that under a constant temperature when the pressure on a gas increases its volume decreases. In other words according to Boyle’s law volume is inversely proportional to pressure when the temperature and the number of molecules are constant.
p ∝ 1/V
p = k1 1/V
k1 here is a proportionality constant, V is the Volume and p is the pressure. On rearranging, we get: k1= pV. Now, if a fixed mass of gas undergoes an expansion at constant temperature then the final volume and pressure shall be p2 and V2. The initial volume and initial pressure here is p1 and V1 then according to Boyle’s law: p1×V1 = p2×V2 = constant (k1)
p1/p2 = V2/V1
So according to Boyle’s law, if the pressure is doubled then at constant temperature the volume of that gas is reduced to half. The reason being the intermolecular force between the molecules of the gaseous substance. In a free state, a gaseous substance occupies a larger volume of the container due to the scattered molecules.
When a pressure is applied to the gaseous substance, these molecules come closer and occupy a lesser volume. In other words, the pressure applied is directly proportional to the density of the gas. Boyle’s law can be graphically represented as follows:
Charle’s Law
Jacques Charles in 1787 analyzed the effect of temperature on the volume of a gaseous substance at a constant pressure. He did this analysis to understand the technology behind the hot air balloon flight. According to his findings, at constant pressure and for constant mass, the volume of a gas is directly proportional to the temperature.
This means that with the increase in temperature the volume shall increase while with decreasing temperature the volume decreases. In his experiment, he calculated that the increase in volume with every degree equals 1/273.15 times of the original volume. Therefore, if the volume is V0 at 0° C and Vt is the volume at t° C then,
Vt = V0 +t/273.15 V0 ⇒ Vt = V0 (1+ t/273.15 )
⇒ Vt = V0 (273.15+ t/273.15 )
For the purpose of measuring the observations of gaseous substance at temperature 273.15 K, we use a special scale called the Kelvin Temperature Scale. The observations of temperature (T) on this scale is 273.15 greater than the temperature (t) of the normal scale.
T= 273.15+t
while, when T = 0° c then the reading on the Celsius scale is 273.15. The Kelvin Scale is also called Absolute Temperature Scale or Thermodynamic Scale. This scale is used in all scientific experiments and works. In the equation [ Vt = V0 (273.15+ t/273.15 ) ] if we take the values Tt = 273.15+t and T0 = 273.15 then:
Vt = V0 ( Tt / T0 )
which implies Vt/V0= ( Tt / T0 ), which can also be written as:
V2/V1= T2/ T1
or V1 /T1 = V2 / T2
V/T = constant = k2
Therefore, V= k2 T
The graphical representation of Charles law is shown in the figure above. Its an isobar graph as the pressure is constant with volume and temperature changes under observation.
Gay-Lussac’s law
Also referred to as Pressure-Temperature Law, Gay Lussac’s Law was discovered in 1802 by a French scientist Joseph Louis Gay Lussac. While building an air thermometer, Gay-Lussac accidentally discovered that at fixed volume and mass of a gas, the pressure of that gas is directly proportional to the temperature. This mathematically can be written as: p ∝ T
⇒ p/T = constant= k3
The temperature here is measured on the Kelvin scale. The graph for the Gay- Lussac’s Law is called as an isochore because the volume here is constant.
Avogadro’s Law
Amedeo Avogadro in 1811 combined the conclusions of Dalton’s Atomic Theory and Gay Lussac’s Law to give another important Gas law called the Avogadro’s Law. According to Avogadro’s law, at constant temperature and pressure, the volume of all gases constitutes an equal number of molecules. In other words, this implies that in unchanged conditions of temperature and pressure the volume of any gas is directly proportional to the number of molecules of that gas.
Mathematically, V ∝ n
Here, n is the number of moles of the gas. Hence, V= k4n
The number of molecules in a mole of any gas is known as the Avogadro’s constant and is calculated to be 6.022 1023. The values for temperature and pressure here are the standard values. For temperature, we take it to be 273.15 K while for the pressure it equals 1 bar or 105 pascals. At these Standard Temperature Pressure (STP) values, one mole of a gas is supposed to have the same volume. Now, n = m/M
According to Avogadro’s equation: V= k4 (m/M)
m/V= d (density); Therefore M=k4D
This means that at an unchanged temperature and pressure conditions, the molar mass of every gas is directly proportional to its density.
The above gas laws provide us with an indication of the various properties of gases at changed conditions of temperature, pressure volume and mass. These laws seem trivial but these find great importance in our day to day lives. From breathing to hot air balloons and vehicle tyres the deviation in gaseous behaviour in changed conditions may affect all. So the next time you are travelling just remember the effect change in physical conditions can have!
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[2308.01912] Triangles with integer sides and given perimeter
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arXiv:2308.01912 (math)
[Submitted on 26 Jun 2023]
Title:Triangles with integer sides and given perimeter
Authors:Tasos Patronis, Ioannis Rizos
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Abstract:In this paper we first study the isoperimetric problem in the case of integer triangles, as well as Alcuin's sequence and how it relates to the number of different integer triangles with a given perimeter. We then present and compare two different approaches to the above problem. The first approach is due to a university professor (a working mathematician), who uses a well-known technique of calculating the number of elements of finite sets, while the second one is due to a high school student with a special interest in mathematics and is purely number-theoretic. Furthermore, some related instructor perspectives are discussed.
Comments:in Greek language
Subjects:General Mathematics (math.GM)
Cite as:arXiv:2308.01912 [math.GM]
(or arXiv:2308.01912v1 [math.GM] for this version)
Focus to learn more
arXiv-issued DOI via DataCite
Journal reference:Mathematical Review, 2022, vol. 97-98, pp. 78-89
Submission history
From: Ioannis Rizos [view email]
[v1] Mon, 26 Jun 2023 19:11:21 UTC (312 KB)
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https://www3.cs.stonybrook.edu/~pfodor/LPCP2018/latin/latin.html
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(file name: latin.pl)
Tony schedules the conference talks for the conference. The aim is to fill in a 9 days of 9 slots grid with the talks of lengths from 1 through 9 minutes so that each digit 1 through 9 occurs exactly once in each row, exactly once in each column and exactly once in each of the 9 3-by-3 sub-squares subject to the following constraints. The constraints are on the sum of adjacent squares within each 3-by-3 sub-square. In the illustration below, the symbols <, = and > indicate that the sum of the values on either side (or above and below), the symbol must have sum less than 10, equal to 10 or greater than 10, respectively.
However, Tony finds out that the solutions for his scheduling is not unique. Your problem is to find how many different solutions satisfy all the constraints of the problem: row, column, 3-by-3 box and inequality constraints.
Input format
An input file for LP systems contains the following facts:
Facts of the form row(Number,S1,S2,S3,S4,S5,S6) containing the row number (1, 3, 5, 6, 8, 10, 11, 13 and 15) and 6 digits corresponding to constraints on the sum of values to the left and right of the symbol. In order to standardize the way we represent the three relations (<, = and >), we will use the integers -1, 0, 1 to represent <, = and >.
Facts of the form vertical(Number,S1,S2,S3,S4,S5,S6,S7,S8,S9) containing the row number (2, 4, 7, 9, 12 and 14) and 9 digits corresponding to constraints on the sum of values above and below the symbol.
Example input for the figure above:
row(1,-1,0,0,-1,0,1).
vertical(2,1,-1,1,1,-1,-1,1,-1,0).
row(3,1,0,-1,0,0,-1).
vertical(4,1,-1,-1,1,-1,1,1,-1,-1).
row(5,-1,-1,1,0,0,-1).
row(6,0,-1,0,1,0,1).
vertical(7,1,1,-1,-1,1,0,-1,1,-1).
row(8,1,-1,1,1,1,-1).
vertical(9,0,0,0,0,0,-1,-1,1,1).
row(10,-1,1,-1,-1,1,1).
row(11,-1,1,-1,1,-1,-1).
vertical(12,-1,1,1,-1,1,1,-1,1,0).
row(13,0,1,1,-1,0,1).
vertical(14,-1,1,-1,-1,1,-1,1,-1,1).
row(15,-1,1,0,1,0,-1).
Output format
The output should contain exactly one fact of the form solutions(S), where S is the number of different solutions for the input problem.
Acknowledgment: thanks to Annie Liu for suggesting related problems.
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Is a CT Within 6 Hours of Headache Onset Enough to Rule Out SAH? | AHA Blogs
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ORIGINALLY PUBLISHED March 22, 2016
Is a CT Within 6 Hours of Headache Onset Enough to Rule Out SAH?
+0 authors
Alexander E. Merkler, MD
Dubosh NM, Bellolio MF, Rabinstein AA, Edlow, JA. Sensitivity of Early Brain Computed Tomography to Exclude Aneurysmal Subarachnoid Hemorrhage: A Systematic Review and Meta-Analysis. Stroke. 2016
Subarachnoid hemorrhage (SAH) is the most devastating type of stroke – 50% of survivors are dead within six months and among those patients who survive, only 50% return to their previous level of functioning. For decades, classic neurology dogma has stated that in order to rule out a SAH, any patient who presents with a thunderclap headache (HA) must receive a lumbar puncture (LP) if the head CT is negative. However, recent data suggests that in neurologically intact patients, a CT is 100% sensitive to rule out SAH when performed within six hours using a modern generation CT scanner (16-slice or greater). Hence, is there is no longer a need to perform an LP after a negative head CT that is performed within six hours of HA onset?
In this manuscript, Dr. Dubosh et al perform a meta-analysis to determine the sensitivity of modern generation CT scanners to rule out SAH in patients presenting to an emergency department within six hours of thunderclap HA. The authors identified five articles that met their inclusion criteria; four were retrospective and one was prospective. In total, 8,907 patients with thunderclap HA underwent a CT within six hours.
Overall, thirteen out of the 8,907 patients had a missed SAH. The overall sensitivity of CT was 0.987 (95% CI 0.971-0.994), specificity was 0.999 (95% CI 0.993-1.0) and the likelihood ratio of a negative CT was 0.010 (95% CI 0.003-0.034). This equated to a miss rate of 1.5 per 1000 patients who present with thunderclap HA and receive a modern CT scan within six hours.
It is important to note that each of the five studies had certain limitations. For example, perimesencephalic hemorrhage 1 and SAH caused by a cervical arteriovenous malformation 2 were considered missed causes of SAH. In addition, in the one prospective study by Perry et al 3 (in which there were no documented missed cases of SAH), an LP was not performed in every patient who presented with thunderclap HA and had a negative CT. Although there was close follow-up using telephone interviews and monitoring coroner’s records, there may have been missed cases of SAH.
Modern CT performed within 6 hours of patients presenting with thunderclap HA is an extremely sensitive tool to rule-out SAH. As with most tests, it is impossible to say that it is 100% sensitive, but it certainly approaches it. Although perhaps very few cases of SAH may be missed, clinicians must weigh this against the potential consequences of performing an LP including time, anxiety, post-LP complications, unnecessary vascular imaging (CTA, MRA, angiography) and probably most importantly subsequent ramifications such as inappropriate procedures for incidentally found vascular lesions. Of course, missing a SAH may be life threatening and can lead to significant consequences including death.
References:
Blok KM, Rinkel GJ, Majoie CB, Hendrikse J, Braaksma M, Tijssen CC et al. CT within 6 hours of headache onset to rule out subarachnoid hemorrhage in nonacademic hospitals. Neurology. 2015;12:1927-193.
Backes D, Rinkel GJ, Kemperman H, Linn FH, Vergouwen MD. Time-dependent test characteristics of head computed tomography in patients suspected of nontraumatic subarachnoid hemorrhage. Stroke. 2012;43:2115-2119.
Perry JJ, Stiell IG, Sivilotti ML, Bullard MJ, Emond M, Symington C et al. Sensitivity of computed tomography performed within six hours of onset of headache for diagnosis of subarachnoid haemorrhage: prospective cohort study. BMJ. 2011;343:d4277.
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Functional equation
A functional equation, roughly speaking, is an equation in which some of the unknowns to be solved for are functions. For example, the following are functional equations:
Contents
1 Introductory Topics
1.1 The Inverse of a Function
2 Intermediate Topics
2.1 Cyclic Functions
2.2 Problem Examples
3 Advanced Topics
3.1 Functions and Relations
3.2 Injectivity and Surjectivity
4 See Also
Introductory Topics
The Inverse of a Function
The inverse of a function is a function that "undoes" a function. For an example, consider the function: . The function has the property that . In this case, is called the (right) inverse function. (Similarly, a function so that is called the left inverse function. Typically the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function.) Often the inverse of a function is denoted by .
Intermediate Topics
Cyclic Functions
A cyclic function is a function that has the property that:
A classic example of such a function is because . Cyclic functions can significantly help in solving functional identities. Consider this problem:
Find such that . Let and in this functional equation. This yields two new equations:
Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:
So, clearly,
Problem Examples
2006 AMC 12A Problem 18
2007 AIME II Problem 14
Advanced Topics
Functions and Relations
Given a set and , the Cartesian Product of these sets (denoted ) gives all ordered pairs with and . Symbolically,
A relation is a subset of . A function is a special time of relation where for every in the ordered pair , there exists a unique .
Injectivity and Surjectivity
Consider a function be a function from the set to the set , i.e., is the domain of and is the codomain of .
The function is injective (or one-to-one) if for all in the domain , if and only if . Symbolically, f(x)is injective⟺(∀a,b∈X,f(a)=f(b)⟹a=b).
The function is surjective (or onto) if for all in the codomain there exists a in the domain such that . Symbolically, f(x)is surjective⟺∀a∈Y,∃b∈X:f(b)=a.
The function is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically, f(x)is bijective⟺∀a∈Y,∃!b∈X:f(b)=a.
The function has an inverse function , where , if and only if it is a bijective function.
See Also
Functions
Cauchy Functional Equation
Retrieved from "
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INTRODUCTION E nteral nutrition is a commonly used modality in hospitalized patients and is preferred over par-enteral nutritional support. Although enteral nutri-tion is considered safe and cost effective, it is not with-out complications. Aspiration is considered the most serious tube feeding complication. It may be clinically unimportant or develop into respiratory failure. Hospi-talized patients who are intubated and receive tube feed-ing are at an especially high risk. The prevalence of aspi-ration pneumonia varies in the literature from 2% to 95%. Unfortunately, rigorous investigation of the inci-dence and risk factors for pulmonary aspiration are lack-ing. Discrepencies exist due to varying definitions of pulmonary aspiration, biased patient selection, differ-ences in illness severity, a variance in type of feeding tube and method, and short follow-up intervals to name a few. Astute assessment and use of preventive measures during delivery of enteral nutrition can help decrease the incidence of aspiration. PRACTICAL GASTROENTEROLOGY • APRIL 2003 89 Aspiration Risk and Enteral Feeding: A Clinical Approach NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Series Editor: Carol Rees Parrish, R.D., MS, CNSD Marianne Opilla, RN, Nutrition Support Clinician, Vir-ginia Commonwealth University Health System, MCV Hospitals and Physicians, Richmond, VA.
Enteral nutrition support is frequently delivered to the hospitalized patient who is unable to tolerate oral nutrition. It is considered safer and less expensive than par-enteral nutrition. The most commonly identified complication of enteral nutrition is aspiration. It is difficult to correlate and diagnose aspiration, pneumonia, and pneu-monitis as a direct result of enteral nutrition delivery. Studies defining aspiration com-plications related to tube feeding have lacked consistency in population and design.
Despite the fear of aspiration, the healthcare professional recognizes the importance of early nutritional intervention in the hospitalized patient. This article will review aspi-ration related to tube feedings, identification of risk factors, and prevention strategies that will enable clinicians to deliver safe enteral nutrition to the hospitalized patient.
Marianne Opilla PRACTICAL GASTROENTEROLOGY • APRIL 2003 90 NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Aspiration Risk and Enteral Feeding: A Clinical Approach ASPIRATION Aspiration occurs when material such as gastric con-tents, saliva, food, or nasopharyngeal secretions are inhaled into the airway or respiratory tract (1). An aspi-ration occurrence does not necessarily cause pneumonia.
In the healthy population, microaspiration is common and pulmonary complications seldom occur. The hospi-talized patient is at greater risk for developing respiratory compromise and pneumonia following an aspiration event because of impaired consciousness, altered airway defenses, and depressed immune function. Aspiration may present as silent, or with symptoms including coughing, choking, and acute respiratory distress (2).
PNEUMONIA Pneumonia accounts for increased incidence of morbid-ity and mortality in the hospitalized patient (3). Pneumo-nia occurs when the bacteria that normally exist in the oral, nasopharyngeal, and gastrointestinal tract are aspi-rated into the lung. The hospitalized patient is extremely vulnerable to developing pneumonia because the phar-ynx may be colonized with hospital flora (2). Pneumoni-tis is caused by aspiration of gastric contents. The acidic nature of the gastric regurgitation causes inflammation of the lung tissue (1). Diagnosis of pneumonia and pne-unonitis is confirmed by infiltrates on chest X-ray (3).
The actual event of aspiration leading to pneumonia or pneumonitis is often missed and correlated with symp-toms later.
REGURGITATION AND DYSPHAGIA Regurgitation and dysphagia may play a role in aspira-tion, but should be addressed separately from aspiration.
Regurgitation occurs when gastric contents reflux into the esophagus, pharynx, or oral cavity but do not enter the lungs. Many patients have gastric esophageal regur-gitation that does not result in pulmonary aspiration.
Dysphagia is difficult or dysfunctional swallowing (2). It may be present in the hospitalized patient with neurolog-ical compromise, sedation, or intubation (4). Dysphagia assessment by a speech pathologist can provide key information about the extent of dysphagia and whether silent aspiration is occurring. If a modified barium swal-low is performed, care must be taken with dysphagia and aspiration interpretation. Delivery of enteral nutrition, infused gastrically, or into the small bowel, does not affect aspiration potential if the source is oropharyngeal secretions or food by mouth.
COUGH AND GAG REFLEXES Assessment of the cough and gag reflex may be per-formed during dysphagia evaluation. The absence or presence of a gag reflex has not been shown to influence the risk of aspiration. Data show that patients with absent gag reflex do not necessarily aspirate and those with strong gag reflex may aspirate (5). Likewise, the cough reflex may or may not be elicited to prevent aspiration or signal an aspiration episode (silent aspiration). There-fore, diminished cough or gag reflexes are not reliable indicators of aspiration risk.
DETECTION METHODS Glucose Testing Enteral formulas contain significant carbohydrate (not necessarily glucose) and it is theorized that the presence of glucose in suctioned secretions at greater than 5mg/dL may indicate aspiration of formula (2). Reagent strips are used to test for the presence of glucose in suctioned tra-cheobronchial secretions. Studies of the validity of this method have shown that there is no reliable correlation between glucose levels in secretions and aspiration.
Patients were found to have elevated glucose levels even when not tube fed (6,7). It is, therefore, not recom-mended that glucose testing be considered a reliable method for detection of aspiration.
Blue Dye Test The blue dye test for aspiration is a traditional bedside practice that is based on the assumption that addition of blue food coloring or dye to the feeding formula helps the bedside clinician visualize aspiration events. In the 1980’s and 1990’s methylene blue was stored at the nurse’s sta-tion and syringed from a multi-dose vial. Spillage, stain-ing, and sterility became a concern and several products became available with FD&C Blue No. 1 blue food color-ing in 5 mL single dose sterile squeeze vials. (continued on page 92) PRACTICAL GASTROENTEROLOGY • APRIL 2003 92 NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Aspiration Risk and Enteral Feeding: A Clinical Approach The FD&C Blue No. 1 product was considered safe in healthy humans and animal studies. Although it was never approved for use as an additive to enteral feeding, it became popular in the hospital setting for coloring enteral formulas (8). Few institutions have protocols for use of blue coloring. In a survey study, Methany, et al found that the amount of blue coloring used is up to the discretion of the nurse (9). It varied from a few drops for pale blue to several milliters for a royal blue shade. In 2000, case reports of discoloration of skin and urine were reported after addition of FD&C Blue No. 1 to enteral feedings of hospitalized patients (10). Deaths were reported in several patients and autopsy revealed blue, discolored organs. The deaths occurred in patients with conditions that increase gut permeability, such as sepsis, severe burns, trauma, shock, vascular surgery, and renal failure (8,11). These findings certainly suggest that this method of detection is unreliable, extremely unsafe, and should be abandoned.
Many institutions are re-evaluating their protocols for use of blue coloring in enteral formula. Although there are no studies to date supporting the accuracy of this detection method, many clinicians are still using this method in their practice. They must do so with careful consideration of their patient population and recognize the potential for harm with blue dye use. Of note, many institutions use methylene blue in place of blue food dye. The cost difference is significant: Meth-ylene blue, 10 mL vial = $17.00; Steri-blue (recently taken off the market), 10 mL $3.50. Recommendations are to use FD&C Blue No. 1 available in single dose sterile squeeze vials and limit tinting of enteral for-mula to a few days only (12). See Table 1 for one insti-tution’s protocol.
ASPIRATION RISK FACTORS Ongoing patient assessment is paramount in the safe delivery of nutrients via feeding tubes. A proactive approach is to assess the patient for risk factors that may contribute to aspiration. This can be challenging for the clinician, since there is no defining single risk factor for anticipating aspiration. See Table 2 for a list of factors associated with aspiration. Supine Positioning Supine positioning is associated with increased aspira-tion events. Studies concur that there is less aspiration and respiratory compromise with elevation of the head of the bed 30 to 45 degrees during enteral nutrition delivery (13–15).
Impaired Level of Consciousness Level of consciousness (LOC) should be evaluated closely. Decreased LOC from sedation or illness can increase the risk of aspiration. The altered coordination Table 1 University of Virginia Health System Blue Dye Protocol After a review of the literature, the hospital Pharmacy and Therapeutics Committee approved the use of blue dye for certain patients.
Indications: • Suspected reflux of secretions from beyond the pylorus back into the stomach (due to j-arm feeding tip location in proximal duodenum or through gastrojejunostomy opening) • To help determine if j-arm of PEG/J has migrated back into stomach • Suspected aspiration of tube feeding • Suspected enterocutaneous fistula Duration: No more than 24 hours Amount: 2–4 drops (NOT mL) per 250 mL (1 can) of tube feeding formula Table 2 Aspiration Risk Factors • Supine Positioning • Impaired level of consciousness • Gastroesophageal reflux • Neurological deficits • Age >60 years • Enteral intubation • Malpositioned feeding tube • Bolus vs. continuous • Mechanical ventilation • Poor oral health • Inadequate nurse/patient ratio (continued from page 90) between breathing and swallowing interferes with the patient’s ability to protect the airway (16). Gastroesophageal Reflux Gastroesophageal reflux (GER) has also been identified as a risk factor for aspiration. Medical diagnosis and severity of illness may cause a decrease in esophageal sphincter pressure resulting in increased reflux and potential for aspiration (17). Drugs such as dopamine and acid suppressive agents, hyperglycemia, renal fail-ure, and sepsis may contribute to delayed gastric empty-ing in the hospitalized patient (4). The resulting gastric distention may cause more frequent episodes of GER and possible aspiration. GER in combination with decreased LOC places the patient at a significantly greater risk for aspiration.
Neurological Deficits Brain injured patients exhibit delayed gastric emptying and impaired lower esophageal sphincter as a result of increased intracranial pressure (16). Patients with chronic neurological disorders such as stroke, amyotrophic lateral sclerosis (ALS), and Parkinson’s disease may exhibit varying degrees of dysphagia and silent aspiration. Age >60 Years Advanced age may play a role in aspiration risk. Older patients may have decreased swallowing ability for a variety of reasons (2). They are more likely to have mul-tiple medical conditions, neurological deficits from an underlying disease process, and alterations in mental sta-tus from medications, or experience “sundowning,” towards the end of the day thus increasing their risk. Enteral Intubation The existence of a nasally inserted enteral access device may also place patients at aspiration risk. This is pre-sumed due to increased secretions from tube irritation, impairment of laryngeal function, and disruption of the esophageal sphincters during intubation. Gastric vs. Small Bowel Feeding Increased incidence of aspiration has not been clearly associated with any one enteral feeding method. Some clinicians hold the belief that transpyloric, small intestinal feedings protect the patient from reflux and subsequent aspiration. Studies to date have shown that this concept has not been clearly substantiated (14,18–20). Studies done in the ICU population concur with a recent prospective randomized trial by Neu-mann and Delegge (21). Sixty intensive care patients received either gastric or small bowel feedings with a 12-Fr nasal feeding tube. The gastric feedings demon-strated no increase in aspiration when compared to small bowel feedings. Unless a patient has a signifi-cant history of severe reflux, gastroparesis, refractory vomiting, or esophageal dysmotility, then initiation of gastric feedings is reasonable.
Malpositioned Tubes Regurgitation, coughing, or vomiting usually results when tubes are incorrectly placed or distal tip migration into the esophagus or gastroesophageal junction has occurred. A new procedure for confirming feeding tube placement using a CO2 monitoring device (thereby obvi-ating the need for x-ray confirmation) shows promising results (22). Bolus vs. Continuous Gastric Feedings Rapid infusion of formula using the syringe-bolus deliv-ery method may have a higher aspiration risk than lower volume continuous gastric feedings. Although there are no definitive studies to date, the bolus method of feeding may decrease lower esophageal sphincter pressure and increase the chance for reflux to occur (16).
Other Risk Factors Other risk factors that have been identified in the litera-ture include tracheal intubation, mechanical ventilation, seizures, poor oral health, and inadequate nurse to patient ratios (16). Despite multiple risk factors, enteral nutrition remains the safest and most cost effective means to reduce catabolism in hospitalized patients who cannot take nutrients orally (23). Implementation of pre-vention strategies is a key factor for improving the safety of enteral feeding delivery. See Table 3 for a list of pre-vention strategies.
PRACTICAL GASTROENTEROLOGY • APRIL 2003 93 NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Aspiration Risk and Enteral Feeding: A Clinical Approach PRACTICAL GASTROENTEROLOGY • APRIL 2003 94 PREVENTION Head of Bed Elevation There is good evidence that the 30 to 45 degree semi-recumbent body position minimizes reflux and poten-tial aspiration. Ibanez, et al showed in two studies that there is less gastroesophageal reflux and aspira-tion when nasogastrically-fed, mechanically venti-lated patients are maintained in the semi-recumbent position rather than supine (14,24). Drakulovic, et al also examined body position in 86 mechanically ven-tilated patients and found that nosocomial pneumonia was highest in enterally fed patients in the supine position (13). Head of bed elevation is an easy and economical nursing intervention for most hospitalized patients.
Grap, et al collected measurements of backrest eleva-tion in a medical intensive care unit and found that 86% of patients were maintained in the supine position despite enteral feedings (25). There was no change in hemodynamic status when semi-recumbent position-ing was maintained. The rationale for supine position-ing was attributed to convenience, patient comfort, and usual practice on the unit. The evidence overwhelm-ingly supports head of the bed elevation during feed-ing. This will significantly lessen the risk of aspiration and should be practiced without exception.
Verify Tube Placement Another nursing measure for prevention of aspiration is frequent monitoring of tube placement. Nasally or orally placed feeding tubes are generally anchored to the nose or face with tape. These tubes become easily dislodged into the esophagus with normal patient movement. Tube length and secure tape should be checked every four hours during tube feeding. Nasal and oral tubes are indicated for < 3–4 week regimens and should be removed as soon as feasible (26). Gastric Aspirates It is standard practice in many institutions to check aspirates in gastrically-fed patients to determine if for-mula is being retained in the stomach, theoretically placing the patient at risk for gastroesophageal reflux and potential aspiration. The general practice is to hold feedings for gastric residual volume of > 150–200 mL.
This amount is not standardized and there are no stud-ies to date that can predict an actual “safe” amount for gastric residual volume (27). Sometimes patients do not receive adequate nutritional support because feed-ings are being held for a predetermined, institution-specific high gastric residual. A more practical approach is to monitor the patient for complaints of fullness, abdominal distention and discomfort, ade-quate bowel function, and trends in elevated gastric residuals. Constipation can be the cause of these com-plaints, and a consistent bowel program may improve the residual volume. Unfortunately, there is no clear evidence that has established what volume constitutes an unsafe high residual and the assessment and inter-vention is left to the clinician at the bedside.
It has been suggested that promotility drugs such as metoclopramide and erythromycin may promote gastric emptying and improve high gastric residual volumes. In a prospective, randomized, controlled trial of 305 intensive care patients with nasogastric tubes it was determined that rates of nosocomial pneumonia and mortality were not different between the metoclo-pramide and the placebo group (28). In a review of randomized trials by Booth, et al, promotility agents demonstrated no positive effects on clinical outcomes (29). A trial period on one of these drugs may be ben-eficial for an individual patient with chronic high gas-tric residual trends, but ongoing, and, diligent assess-ment of clinical status is paramount.
NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Aspiration Risk and Enteral Feeding: A Clinical Approach Table 3 Prevention of Aspiration • Maintain head of bed > 30 degrees • Routinely verify tube placement • Clinical assessment of GI tolerance: – Abdominal distention – Fullness – Discomfort – Excessive residual trends • Remove naso/oro enteric tubes as soon as possible (continued on page 96) PRACTICAL GASTROENTEROLOGY • APRIL 2003 96 SUMMARY Delivery of enteral nutrition can be challenging in the hospitalized patient. Assessment of patient risk factors will help in the selection of the best enteral access route and method of feeding each individual patient. A team approach with the patient, nurses, dietitians and physi-cians will provide the best strategy for safe and success-ful nutritional support. Clinicians at the bedside are the front line of defense in preventing aspiration, the most serious complication of tube feeding. Simple nursing measures like elevation of the head of the bed have been shown to decrease aspi-ration risk. It is time for clinicians to abandon the prac-tices of glucose testing of pulmonary secretions and addition of blue dye to enteral feeding and rely more on clinical monitoring of abdominal and pulmonary assess-ment of tube fed patients. This article has reviewed the current understanding of aspiration risk in enterally-fed patients in the hospital setting. Prevention strategies to decrease that risk have also been provided. I References 1. Marik PE. Aspiration pneumonitis and aspiration pneumonia. N Engl J Med, 2001; 344:665-671.
2. Elpern EH. Pulmonary aspiration in hospitalized adults. Nutr Clin Pract, 1997;12:5-13.
3. Zaloga GP. Aspiration-related illnesses: Definitions and diagno-sis. J Parenter Enteral Nutr, 2002;26:S2-S7.
4. DeMeo MT, Bruninga K. Physiology of the aerodigestive system and aberrations in that system resulting in aspiration. J Parenter Enteral Nutr, 2002;26:S9-S18.
5. Leder SB. Videofluoroscopic evaluation of aspiration with visual examination of the gag reflex and velar movement. Dysphagia, 1997;12:21-23.
6. Metheny NA, St. John RE, Clouse RE. Measurement of glucose in tracheobronchial secretions to detect aspiration of enteral feed-ings. Heart Lung, 1998;27:285-292.
7. Kinsey GC, Murray MJ, Swensen SJ, Miles JM. Glucose content of tracheal aspirates: Implications for the detection of tube feed-ing aspiration. Crit Care Med, 1994;22:1557-1562.
8. Maloney JP, Ryan TA, Brasel KJ, Binion DG, et al. Food dye in enteral feedings: A review and a call for a moratorium. Nutr Clin Pract, 2002;17:169-181.
9. Metheny NA, Aud MA, Wunderlich RJ. A survey of bedside methods used to detect pulmonary aspiration of enteral formula in intubated tube-fed patients. Am J Crit Care, 1999;8:160-167.
10. Maloney JP, Halbower AC, Fouty BF, Fagan KA, et al. Systemic absorption of feed dye in patients with sepsis. N Engl J Med, 2000;343:14.
11. Maloney JP, Ryan TA. Detection of aspiration in enterally fed patients: A requiem for bedside monitors of aspiration. J Parenter Enteral Nutr, 2002;26:S34-S41.
12. Guenter P. Enteral nutrition complications. In Guenter P, Silkroski M.(Eds). Tube Feeding Practical Guidelines and Nurs-ing Protocols. Aspen Publishers, Gaithersburg, MD. 2001;114.
13. Drakulovic MB, Torres A, Bauer TT, Nicolas JM, et al. Supine body position as a risk factor for nosocomial pneumonia in mechanically ventilated patients: A randomized trial. Lancet, 1999; 354:1851-1857.
14. Ibanez J, Penafiel A, Marse P, Jorda R, et al. Incidence of gas-troesophageal reflux and aspiration in mechanically ventilated patients using small-bore nasogastric tubes. J Parenter Enteral Nutr, 2000;24:103-106.
15. Torres A, Serra-Batlles J, Ros E, Piera C, et al. Pulmonary aspi-ration of gastric contents in patients receiving mechanical venti-lation: The effect of body position. Ann Int Med, 1992;116:540-543.
16. Metheny NA. Risk factors of aspiration. J Parenter Enteral Nutr, 2002;26:S26-S31.
17. Murphy LM, Bickford V. Gastric residuals in tube feeding: How much is too much? Nutr Clin Pract, 1999;14:304-306.
18. Esparza J, Boivan MA, Hartshorne MF, Levy H. Equal aspiration rates in gastrically and transpylorically fed critically ill patients.
Intensive Care Med, 2001;27:660-664.
19. Heyland DK, Konopad E, Alberda C, Keefe L, et al. How well do critically ill patients tolerate early, intragastric enteral feeding?
Results of a prospective, multicenter trial. Nutr Clin Pract, 1999; 14:23-28.
20. Kearns PJ, Chin D, Mueller L, Wallace K, et al. The incidence of ventilator-associated pneumonia and success in nutrient delivery with gastric versus small intestinal feeding: A randomized clini-cal trial. Crit Care Med, 2000;28:1742-1746.
21. Neumann DA, DeLegge MH. Gastric versus small-bowel tube feeding in the intensive care unit: A prospective comparison of efficacy. Crit Care Med, 2002;30:1436-1438. 22. Burns S, Carpenter R, Truwit J. Report on the development of a procedure to prevent placement of feeding tubes into the lungs using end-tidal CO2 measurements. Crit Care Med, 2001;29:936-939.
23. Braunschweig CL, Levy P, Sheean PM, Wang X. Enteral com-pared with parenteral nutrition: A meta-analysis. Am J Clin Nutr, 2001;74:534-542.
24. Ibanez J, Penafiel A, Raurich JM, Marse P, et al. Gastroe-sophageal reflux in intubated patients receiving enteral nutrition: Effect of supine and semirecumbent positions. J Parenter Enteral Nutr, 1992;16:419-422.
25. Grap MJ, Cantley M, Munro CL, Corley MC. Use of backrest ele-vation in critical care: A pilot study. Am J Crit Care, 1999;8:475-480. 26. Guenter P. Enteral nutrition complications. In Guenter P, Sildroski M. (Eds). Tube Feeding Practical Guidelines and Nursing Protocols. Aspen Publishers, Gaithersburg, MD. 2001; 52-53.
27. McClave SA, Snidor HL. Clinical use of gastric residual volumes as a monitor for patients on enteral tube feeding. J Parenter Enteral Nutr, 2002;26:S43-S48.
28. Yavagal DR, Karnad DR, Oak JL. Metoclopramide for prevent-ing pneumonia in critically ill patients receiving enteral tube feeding: A randomized controlled trial. Crit Care Med, 2000; 28:1408-1411.
29. Booth CM, Heyland DK, Paterson WG: Gastrointestinal pro-motility drugs in the critical care setting: A systematic review of the evidence. Crit Care Med, 2002;30:1429-1435.
NUTRITION ISSUES IN GASTROENTEROLOGY, SERIES #4 Aspiration Risk and Enteral Feeding: A Clinical Approach (continued from page 94) V I S I T O U R W E B S I T E A T V I S I T O U R W E B S I T E A T W W W .
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Published Time: 2019-11-21T17:18:22+05:30
In the number system, each number is represented by its base. If the base is 2 it is a binary number, if the base is 8 it is an octal number, if the base is 10, then it is called decimal number system and if the base is 16, it is part of the hexadecimal number system. The conversion of decimal numbers to any other number system is an easy method. But to convert other base number systems into decimal numbers requires practice. In this article, let us learn more on the decimal number system and the conversion from a decimal number system to other systems here in detail.
Table of Contents:
Definition
Conversion from Other Bases to Decimal Number System
Binary to Decimal
Octal to Decimal
Hexadecimal to Decimal
Decimal Number System to Other Bases
Decimal to Binary
Decimal to Octal
Decimal to Hexadecimal
What is the Decimal Number System?
In the decimal number system, the numbers are represented with base 10. The way of denoting the decimal numbers with base 10 is also termed as decimal notation. This number system is widely used in computer applications. It is also called the base-10 number system which consists of 10 digits, such as, 0,1,2,3,4,5,6,7,8,9. Each digit in the decimal system has a position and every digit is ten times more significant than the previous digit. Suppose, 25 is a decimal number, then 2 is ten times more than 5. Some examples of decimal numbers are:-
(12)10, (345)10, (119)10, (200)10, (313.9)10
A number system which uses digits from 0 to 9 to represent a number with base 10 is the decimal system number. The number is expressed in base-10 where each value is denoted by 0 or first nine positive integers. Each value in this number system has the place value of power 10. It means the digit at the tens place is ten times greater than the digit at the unit place. Let us see some more examples:
(92)10 = 9×10 1+2×10 0
(200)10 = 2×10 2+0x10 1+0x10 0
The decimal numbers which have digits present on the right side of the decimal (.) denote each digit with decreasing power of 10. Some examples are:
(30.2)10= 30×10 1+0x10 0+2×10-1
(212.367)10 = 2×10 2+1×10 1+2×10 0+3×10-1+6×10-2+7×10-3
Also, read:
Binary Number System
Octal Number System
Number System Pdf
Number System For Class 9
Conversion From Other Bases to Decimal Number System
Let us see here how in number system conversion, we can convert any other base system such as binary, octal and hexadecimal to the equivalent decimal number.
Binary to Decimal
In this conversion, a number with base 2 is converted into number with base 10. Each binary digit here is multiplied by decreasing power of 2. Let us see one example:
Example: Convert (11011)2 to decimal number.
Solution: Given (11011)2 a binary number.
We need to multiply each binary digit with the decreasing power of 2. That is;
1×2 4+1×2 3+0x2 2+1×2 1+1×2 0
=16+8+0+2+1
=27
Therefore, (11011)2 = (27)10
Octal to Decimal
In this conversion, a number with base 8 is converted into number with base 10. Each digit of octal number here is multiplied by decreasing power of 8. Let us see one example:
Example: Convert 121 8 into the equivalent decimal number.
Solution: Given (121)8 is an octal number
Here, we have to multiply each octal digit with the decreasing power of 8, such as;
1×8 2+2×8 1+1×8 0
=64+16+1
=81
Hexadecimal to Decimal
In this conversion, a number with base 16 is converted into number with base 10. Each digit of hex number here is multiplied by decreasing power of 16. Let us understand with the help of an example:
Example: Convert 12 16 into a decimal number.
Solution: Given 12 16
Multiply each digit with decreasing power of 16 to obtain an equivalent decimal number.
1×16 1+2×16 0
=16+2
=18
Decimal Number System to Other Bases
Earlier we learned about converting other base number systems into a decimal number, Here we will learn how to convert a decimal number into different base numbers. Let us see one by one.
Decimal to Binary
To convert a decimal number into an equivalent binary number we have to divide the original number system by 2 until the quotient is 0, when no more division is possible. The remainder so obtained is counted for the required number in the order of LSB (Least significant bit) to MSB (most significant bit). Let us go through the example.
Example: Convert 26 10 into a binary number.
Solution: Given 26 10 is a decimal number.
Divide 26 by 2
26/2 = 13 Remainder →0 (MSB)
13/2 = 6 Remainder →1
6/2 = 3 Remainder →0
3/2 = 1 Remainder →1
½ = 0 Remainder →1 (LSB)
Hence, the equivalent binary number is (11010)2
Decimal to Octal
Here the decimal number is required to be divided by 8 until the quotient is 0. Then, in the same way, we count the remainder from LSB to MSB to get the equivalent octal number.
Example: Convert 65 10 into an octal number.
Solution: Given 65 10 is a decimal number.
Divide by 8
65/8 = 8 Remainder →1 (MSB)
8/8 = 1 Remainder →0
⅛ = 0 Remainder →1 (LSB)
Hence, the equivalent octal number is (101)8
Decimal to Hexadecimal
The given decimal number here is divided by 16 to get the equivalent hex. The division of the number continues until we get the quotient 0.
Example: Convert 127 10 to a hexadecimal number.
Solution: Given 127 10 is a decimal number.
Divide by 16
127/16 = 7 Remainder →15
7/16 = 0 Remainder → 7
In the hexadecimal number system, alphabet F is considered as 15.
Hence, 127 10 is equivalent to 7F 16
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Finding the number of subsets of a set such that an element divides the succeeding element.
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Hello everybody! The problem you see above is a combinatorial problem that I could not solve. :(
Since we are to choose distinct numbers from {1,2,3…K}{1,2,3…K}, the numbers must be in increasing order because they are all positive. I think that the difficulty of this problem is in the fact that the Special Sequence can be of any length. We can go about choosing say n n elements in (K n)(K n) ways but again how do we control that? Talking about a recursive solution, I think that is the way to go but I can't get how to do that (that is get a recursive formula for that).
Any help in solving this problem would be much appreciated. Thanks.
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asked Nov 4, 2019 at 11:06
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Find the number of special sequences of length one. Then find the number with length two. Then length three. And so on. Alternatively, your idea of using recursion is good. E.g., for K=19 K=19, the only special sequences that use 19 19 are (19)(19) and (1,19)(1,19), so just two more than the answer for K=18 K=18. But relating the number for K=18 K=18 to that for K=17 K=17 will be trickier.Gerry Myerson –Gerry Myerson 2019-11-04 12:16:18 +00:00 Commented Nov 4, 2019 at 12:16
maximum length is ⌈l o g 2(K)⌉⌈l o g 2(K)⌉ only if it contains 1.user645636 –user645636 2019-11-04 12:41:14 +00:00 Commented Nov 4, 2019 at 12:41
Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. Scanned pages from books are discouraged on SE network. Questions should contain sufficient context so that it is answerable with the text alone.GNUSupporter 8964民主女神 地下教會 –GNUSupporter 8964民主女神 地下教會 2020-12-16 07:51:57 +00:00 Commented Dec 16, 2020 at 7:51
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We denote with a[n],n≥1 a[n],n≥1 the number of special sequences and with b[n],n≥1 b[n],n≥1 the number of special sequences where each element contains n n as greatest element. We observe a[n]a[n] contains all special sequences of a[n−1]a[n−1] together with all special sequences b[n],(n>1)b[n],(n>1).
We have
a=b=|{(1)}|=1 a[n]=a[n−1]+b[n]n>1 a=b=|{(1)}|=1 a[n]=a[n−1]+b[n]n>1
In order to find b[n]b[n] we need to analyse the prime factor decomposition of n n. We create a small knowledge base of numbers which we need to factorise K=1,…,22 K=1,…,22.
Let p,q p,q be primes. We obtain
b[p]b[p 2]b[p 3]b[p 4]b[p q]b[p 2 q]=|{(p),(1,p)}|=2=∣∣{(p 2),(1,p 2),(p,p 2),(1,p,p 2)}∣∣=4=∣∣{(p 3),(1,p 3),(p,p 3),(p 2,p 3),(1,p,p 3),(1,p 2,p 3),(1,p,p 2,p 3)}∣∣=8=2 4=16=|{(p q),(1,p q),(p,p q),(q,p q),(1,p,p q),(1,q,p q)}|=6=∣∣{(p 2 q),(1,p 2 q),(p,p 2 q),(q,p 2 q),(p 2,p 2 q),(p q,p 2 q),(1,p,p 2 q),(1,q,p 2 q),(1,p 2,p 2 q),(1,p q,p 2 q),(p,p 2,p 2 q),(p,p q,p 2 q),(q,p q,p 2 q),(1,p,p 2,p 2 q),(1,p,p q,p 2 q),(1,q,p q,p 2 q))}∣∣=16 b[p]=|{(p),(1,p)}|=2 b[p 2]=|{(p 2),(1,p 2),(p,p 2),(1,p,p 2)}|=4 b[p 3]=|{(p 3),(1,p 3),(p,p 3),(p 2,p 3),(1,p,p 3),(1,p 2,p 3),(1,p,p 2,p 3)}|=8 b[p 4]=2 4=16 b[p q]=|{(p q),(1,p q),(p,p q),(q,p q),(1,p,p q),(1,q,p q)}|=6 b[p 2 q]=|{(p 2 q),(1,p 2 q),(p,p 2 q),(q,p 2 q),(p 2,p 2 q),(p q,p 2 q),(1,p,p 2 q),(1,q,p 2 q),(1,p 2,p 2 q),(1,p q,p 2 q),(p,p 2,p 2 q),(p,p q,p 2 q),(q,p q,p 2 q),(1,p,p 2,p 2 q),(1,p,p q,p 2 q),(1,q,p q,p 2 q))}|=16
Now it's time to harvest. We obtain
aaa=a+b=a+b+b=a+∑j=2 13 b[j]=1+∑j∈{2,3,5,7,11,13}b[j]+∑j∈{4,9}b[j]+∑j∈{6,10,14,15}b[j]+b+b=1+6 b[p]+2 b[p 2]+4 b[p q]+b[p 3]+b[p 2 q]=1+12+8+24+8+16=69=a+b+b+b+b=69+b[2 4]+b+b[3 2⋅2]+b=69+b[p 4]+2 b[p]+b[p 2 q]=69+16+4+16=105=a+b+b+b=105+b[2 2⋅5]+b[3⋅7]+b[2⋅11]=105+b[p 2 q]+2 b[p q]=105+16+12=133 a=a+b=a+b+b=a+∑j=2 13 b[j]=1+∑j∈{2,3,5,7,11,13}b[j]+∑j∈{4,9}b[j]+∑j∈{6,10,14,15}b[j]+b+b=1+6 b[p]+2 b[p 2]+4 b[p q]+b[p 3]+b[p 2 q]=1+12+8+24+8+16=69 a=a+b+b+b+b=69+b[2 4]+b+b[3 2⋅2]+b=69+b[p 4]+2 b[p]+b[p 2 q]=69+16+4+16=105 a=a+b+b+b=105+b[2 2⋅5]+b[3⋅7]+b[2⋅11]=105+b[p 2 q]+2 b[p q]=105+16+12=133
Note: Interestingly, the sequence (a[n])n≥1=(1,3,5,9,11,17,19,27,31,…)(a[n])n≥1=(1,3,5,9,11,17,19,27,31,…) doesn't seem to be archived in OEIS, but the sequence (b[n])n≥1=(1,2,2,4,2,6,2,8,4,6,2,16,2,…)(b[n])n≥1=(1,2,2,4,2,6,2,8,4,6,2,16,2,…) is archived in OEIS as A067824.
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edited Nov 4, 2019 at 15:32
answered Nov 4, 2019 at 13:54
Markus ScheuerMarkus Scheuer
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1 Program output (1,3,5,9,11,17,19,27,31,37,39,55,57,63,69,85,87,103,105,121,127,133,135,175)(1,3,5,9,11,17,19,27,31,37,39,55,57,63,69,85,87,103,105,121,127,133,135,175) disagrees with your results.Daniel Mathias –Daniel Mathias 2019-11-04 14:39:30 +00:00 Commented Nov 4, 2019 at 14:39
@DanielMathias: Thanks a lot for pointing at the mistake. I've missed four cases at b[p 2 q]b[p 2 q]. Corrected.Markus Scheuer –Markus Scheuer 2019-11-04 15:03:27 +00:00 Commented Nov 4, 2019 at 15:03
Thank you Markus for the brilliant solution!Vasu090 –Vasu090 2019-11-04 15:22:43 +00:00 Commented Nov 4, 2019 at 15:22
I just want to know how did you think of this solution? The solution is clear and easy to understand, but how might have I come up with this solution on my own? Also, how can I be sure to not have made any errors in all these computations?Vasu090 –Vasu090 2019-11-04 15:25:26 +00:00 Commented Nov 4, 2019 at 15:25
1 Many thanks for your nice comment. I started with small numbers and drew Hasse diagrams which are useful when looking at divisibility aspects and their partial ordering. This way I also recognized the recurrence relation between a[n] and b[n]. To not calculate each case from 1 to 22 it's a matter of experience to use some kind of repertoire method and reduce this way calculations. – Markus Scheuer 6 mins ago Markus Scheuer –Markus Scheuer 2019-11-04 15:47:43 +00:00 Commented Nov 4, 2019 at 15:47
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How to Navigating Complex Scenarios: Multi-step Word Problems with Remainders
Word problems often mirror real-life situations, requiring multiple steps to reach a solution.
When these problems involve division, remainders can add an extra layer of complexity. In this guide, we’ll tackle multi-step word problems that result in remainders, offering strategies to interpret and solve them effectively.
Step-by-step Guide to Solve Multi-step Word Problems with Remainders:
1. Understanding the Problem:
Begin by reading the problem carefully. Identify the information given and determine what you’re being asked to find.
2. Breaking Down the Steps:
Determine the sequence of operations required. This might involve addition, subtraction, multiplication, and division.
3. Handling Remainders:
When you encounter a division step that results in a remainder:
– Understand its significance in the context of the problem.
– Decide whether to round up, round down, or use the remainder as is.
4. Solving the Problem:
Execute the operations in the correct sequence, keeping track of any remainder.
5. Interpreting the Remainder:
In the context of the problem, decide how to represent the remainder. For instance, if you divide candies among children, the remainder represents the leftover candies.
Example 1:
Anna has 53 apples. She wants to pack them in bags with 8 apples in each bag. How many full bags will she have, and how many apples will be left over?
Solution:
Dividing 53 by 8 gives 6 with a remainder of 5.
Anna will have 6 full bags, with 5 apples left over.
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A school is organizing a trip for 89 students. Each bus can carry 30 students. How many buses are needed, and how many seats will be empty on the last bus?
Solution:
Dividing 89 by 30 gives 2 with a remainder of 29.
The school needs 3 buses (2 full buses and 1 more for the remaining 29 students). In the last bus, 1 seat will be empty.
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A bakery has 125 muffins and wants to pack them in boxes of 10. How many full boxes will they have, and how many muffins will be left unpacked?
There are 74 students in a competition. They are to be grouped into teams of 9. How many teams will there be, and how many students will not be in a team?
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Answers:
12 full boxes, 5 muffins left unpacked.
8 teams, 2 students will not be in a team.
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A Strongly Subcubic Combinatorial Algorithm for Triangle Detection with Applications
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Adrian Dumitrescu
Algoresearch L.L.C., Milwaukee, WI, USA.
Email ad.dumitrescu@algoresearch.org
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Abstract
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We revisit the algorithmic problem of finding a triangle in a graph:
We give a randomized combinatorial algorithm for triangle detection in a given -vertex graph with edges
running in time, or alternatively in time,
and whose probability of error is at most (or smaller than any fixed , as required,
by a modest runtime increase). This may come as a surprise since it invalidates several conjectures
in the literature. In particular,
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1.
the runtime surpasses the long-standing fastest algorithm for triangle detection
based on matrix multiplication running in time, due to Itai and Rodeh (1978).
Here is current state-of-the-art upper bound on , the exponent of matrix multiplication.
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2. 2.
the runtime surpasses the long-standing fastest algorithm for triangle detection
in sparse graphs based on matrix multiplication running in time
due to Alon, Yuster, and Zwick (1997).
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3. 3.
it is the first combinatorial algorithm for this problem running in truly subcubic time,
and the first combinatorial algorithm that surpasses a triangle detection algorithm based on
fast matrix multiplication.
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4. 4.
the time algorithm for triangle detection leads to a time
combinatorial algorithm for Boolean matrix multiplication, by a reduction of
V. V. Williams and R. R. Williams (2018).
This invalidates a conjecture of A. Abboud and V. V. Williams (FOCS 2014) that any combinatorial algorithm
for computing the Boolean product of two matrices requires time in expectation.
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5. 5.
the runtime invalidates a conjecture of A. Abboud and V. V. Williams (FOCS 2014) that
any combinatorial algorithm for triangle detection requires time.
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6. 6.
as a direct application of the triangle detection algorithm, we obtain a faster exact algorithm for the
-Clique problem, surpassing an almost years old algorithm of Nešetřil and Poljak (1985).
This gives positive answers to questions of (i) R. R. Williams (2005), (ii) G. Woeginger (2008), and
(iii) A. Abboud, A. Backurs, and V. V. Williams (2018). At the same time, it strongly disproves the
combinatorial -Clique conjecture.
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7. 7.
as another direct application of the triangle detection algorithm, we obtain a faster exact algorithm for the
Max-Cut problem, surpassing an almost years old algorithm of R. R. Williams (2005).
This gives a positive answer to a question of Woeginger (2008).
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Keywords: triangle detection problem, Boolean matrix multiplication,
witness matrix, combinatorial algorithm, randomized algorithm.
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1 Introduction
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Triangle detection.
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Consider the problem of deciding whether a given graph contains a complete subgraph
on vertices. If the subgraph size is part of the input, then the problem is
-complete — the well-known Clique problem, see — whereas for every fixed ,
it can be answered by a brute-force algorithm running in time, where .
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For , deciding whether a graph contains a triangle and finding one if it does
(or counting all triangles in a graph)
can be done in time by the algorithm of Itai and Rodeh :
the algorithm computes using Fast Matrix Multiplication (FMM), where is the graph adjacency matrix.
The existence of entries where , , and indicates the presence of
triangles(s) with edge . If at least one pair satisfies this condition, then contains a triangle.
An equivalent characterization is: contains a triangle iff has a non-zero entry on its main diagonal.
See [28, Ch. 10] for a short exposition of this elegant — though impractical — method.
Alternatively, this task can be done in time
by the algorithm of Alon, Yuster, and Zwick .
Here is the exponent of matrix multiplication, with being the current
bound [16, 40]. It is generally conjectured that , however, in spite of the ongoing
effort, the progress has been slow.
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It should be noted that the algorithms for fast matrix multiplication (FMM)
have mainly a theoretical importance and are otherwise largely impractical [27, Ch. 10.2.4].
See also for a more up to date discussion concerning their limitations in relation to generalizability,
elegance, and practical efficiency.
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Boolean matrix multiplication.
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Given two -matrices and , the Boolean product, , is an matrix
such that
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| | | |
---
| | | |
As a relative of matrix multiplication (MM) Boolean matrix multiplication (BMM) is one of the fundamental
problems in computer science, with direct applications to triangle finding [28, 39],
transitive closure [20, 19, 29],
grammar parsing [25, 32, 34],
and other problems ;
see also [24, Lect. 5 & 7], [27, Ch. 9.5], and .
One way to multiply two Boolean matrices is to treat them as integer matrices,
and apply a fast matrix multiplication algorithm over the integers (i.e., BMM via FMM).
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A different group of BMM algorithms, often called “combinatorial” algorithms,
generally have nice properties like simplicity that make them attractive.
The notion of “combinatorial” algorithm does not have a formal or precise definition.
Following , such an algorithm must be both theoretically and practically efficient.
The oldest and probably the best known combinatorial algorithm for BMM is the “Four Russians” algorithm by
Arlazarov, Dinic, Kronrod, and Faradzhev which can be implemented to run in time.
Subsequent improvements took about years and only affected the polylogarithmic factors:
Bansal and R. R. Williams gave a combinatorial algorithm for BMM running in time
relying on the weak regularity lemma for graphs, whereas Chan further improved the running time to
. Here the notation hides factors.
The fastest known combinatorial algorithm for BMM, due to Yu , runs in time.
Very recently, an even faster algorithm — achieving a super-poly-logarithmic saving — has been announced ;
it runs in time.
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In fact, Satta noted years ago that “the design of practical algorithms for Boolean matrix multiplication
that considerably improve the cubic time upper bound is regarded as a very difficult enterprise”.
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On a side note, it is worth noting that much better running times can be obtained for random Boolean matrices.
For example, under the assumption that all Boolean matrices are equally
likely choices for and , the expected number of operations to perform the
multiplication is . Moreover, for large , almost all pairs of matrices can
be multiplied using an algorithm of P. O’Neil and E. O’Neil in operations for any .
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In addition to the line of work trying to enhance the algebraic machinery aiming to reach
the conjectured bound , a parallel line of work aims to approach subcubic bounds
with combinatorial techniques. In this vein,
V. V. Williams and R. R. Williams showed [38, 39, Thm. 1.3 or Thm. 6.1] that either triangle detection
and Boolean matrix multiplication (BMM) both have truly subcubic “combinatorial” algorithms,
or none of them has. Their key lemma is next, see also from which we borrow
the following formulation:
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Theorem 1.
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()
If Triangle Detection is in time for some nondecreasing function , then
Boolean Matrix Multiplication is in time .
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Here we show that the former, subcubic option, materializes:
we obtain the first truly subcubic combinatorial algorithm for triangle detection running in time;
and via the above reduction the first truly subcubic combinatorial algorithm for BMM,
running in time, where notation hides factors.
As such, these results represent the greatest advance in triangle detection in over years
and the greatest advance in Boolean matrix multiplication in over years.
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Our results.
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(i)
Given a graph with vertices and edges,
there is a combinatorial randomized algorithm for triangle detection whose probability
of success can be made arbitrarily close to ,
running in time or in time
(Theorem 2 in Section 2).
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This is the first combinatorial algorithm for this problem running in truly subcubic time,
and the first combinatorial algorithm that surpasses a triangle detection algorithm based on
Fast Matrix Multiplication — the comparison is according to current bounds
on the exponent of FMM [16, 40].
The runtime invalidates a conjecture of A. Abboud and V. V. Williams that
any combinatorial algorithm for triangle detection requires time .
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2. (ii)
Given two Boolean square matrices of size ,
there is a randomized combinatorial algorithm for computing their Boolean product with high probability
running in time (Theorem 3 in Section 3).
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This result invalidates a conjecture of A. Abboud and V. V. Williams
(Conjecture 5 in the full version ), that any combinatorial algorithm
for computing the Boolean product of two matrices requires time in expectation.
The conjecture is also associated to some works in the ’s [25, 32]
and reappears (as Conjecture 1.1) in the recent preprints and
The conjecture was even “promoted” to the rank of “hypothesis” and used as a basis of proving
hardness of other problems, see .
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3. (iii)
As a direct application of the triangle detection algorithm, we obtain a faster exact algorithm for the
-Clique problem, running in (roughly) time, and whose probability of error is at most ,
surpassing an almost years old algorithm of
Nešetřil and Poljak (Theorem 4 in Section 3).
This provides a first solution to the longstanding open problem of improving this algorithm by more than
a polylogarithmic factor mentioned by A. Abboud, A. Backurs, and V. V. Williams ;
and gives a positive answer to a question of R. R. Williams ,
and to problem 4.3 (a) in Woeginger’s list of open problems around exact algorithms .
At the same time this result strongly invalidates — by a polynomial factor, —
the combinatorial -clique conjecture
according to which, combinatorial algorithms cannot solve -Clique in time
for any and positive constant ,
see [3, Conj. 2], [13, p. 48:3], [12, Hypothesis 1.1].
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4. (iv)
As another direct application of the triangle detection algorithm, we obtain a faster exact algorithm for the
Max-Cut problem, running in time, and whose probability of error
is at most , surpassing an almost years old algorithm of R. R. Williams (2005)
(Theorem 5 in Section 3).
This gives a positive answer to problem 4.7 (a) in Woeginger’s list of open problems around exact
algorithms .
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5. (v)
We obtain improved combinatorial algorithms for finding small complete subgraphs (Subsection 3.4
in Section 3). In particular we obtain a combinatorial randomized algorithm for detecting a
running in time; this gives a positive answer to a question of Alon, Yuster, and Zwick,
who asked for an improvement of the running time .
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Definitions and notations.
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Let be an undirected graph. The neighborhood of a vertex is
the set of all adjacent vertices, and its cardinality
is called the degree of in .
Similarly, the neighborhood of a vertex subset is
the set of all vertices
adjacent to vertices in . Define the closed neighborhood of as
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| | | | |
--- --- |
| | | | (1) |
We write for a time complexity of the form .
The notation hides factors, whereas
the notation hides factors.
Unless specified otherwise:
(i) all algorithms discussed are assumed to be deterministic;
(ii) all graphs mentioned are undirected;
(iii) all logarithms are in base .
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2 A fast combinatorial algorithm for triangle detection
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The idea in our algorithm for triangle detection is very simple, but apparently not so easy to find
(it took more than years to surpass the running time of the algorithm of Itai and Rodeh ).
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In this section we outline a randomized algorithm for triangle detection; it comes in two
variants, Triangle-Detection-1 and Triangle-Detection-2111Alternatively this
result can be presented as follows: There is a one-sided error randomized triangle detection algorithm
whose probability of success is at least , running in the specified time..
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Theorem 2.
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There is a randomized algorithm for triangle detection in a given -vertex graph with edges
running in time, and whose probability of error is at most .
Alternatively, this task can be handled in time, or in time.
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If the graph has no triangles, the algorithm is always correct; whereas if has at least one triangle,
the algorithms finds one with probability at least .
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Some intuition.
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The idea behind the algorithm is to get in a position to be able to detect a triangle during a
Breadth First Search (BFS) from a suitable vertex. Specifically we will use the following
slightly modified version of BFS with a start vertex .
For a standard version, the reader can refer for instance to [27, Ch. 7].
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Assume that contains at least one triangle, and fix one, say, .
For instance a vertex of is a suitable vertex. Another favorable scenario is when BFS discovers exactly
one vertex of on some level of the search tree and the other two vertices of appear on the next level.
In particular starting BFS from a vertex that is a “neighbor” of a triangle leads to successful triangle detection.
See Step of the modified BFS procedure and Fig. 1 (i).
In general, however, vertices could be unsuitable as start vertices for the task at hand as they might cause BFS
to discover two or all three vertices of on the same level of the search tree, making the detection of difficult.
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To achieve this privileged “neighbor” position, the algorithm distinguishes between high and low degree
vertices—also a key step in the seminal work on triangle detection of Alon, Yuster, and Zwick .
This distinction is exploited by the algorithm in a novel way by using random sampling.
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2.1 The first algorithm
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Analysis.
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The algorithm works with a parameter suitably set.
We say that a vertex in has low degree if .
Assume that is a triangle in , i.e., and .
If is incident to a low degree vertex, i.e.,
, , or has low degree, then is found in step .
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Let (recall (1)). Then clearly .
For each of the trials, let denote the bad event that the picked vertex is not in .
Let denote the bad overall event that no picked vertex is in .
We have
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| | | |
---
| | | |
Since the trials are independent from each other we have
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| | | |
---
| | | |
by applying the standard inequality for . Indeed, we have
for .
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It remains to show that if then BFS+ finds a triangle in (not necessarily ,
unless is the unique triangle in ). We distinguish two cases:
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Case 1: . By symmetry, we may assume that (the root, on level of the BFS tree).
Then and appear on the next level (level ) of the BFS tree. Assuming that is discovered before ,
the triangle is detected when scanning the adjacency list of . Refer to Fig. 1 (ii).
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Case 2: . We may assume without loss of generality that . Then
appears on level of the BFS tree. We distinguish two subcases:
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Case 2.1: Exactly one vertex, namely , appears on level . Then and appear on the next level
(level ) of the BFS tree. Assuming that is discovered before , the triangle is detected while
scanning the adjacency list of . Refer to Fig. 1 (iii).
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Case 2.2: At least two vertices of , say, and , appear on level . This means that
. Since we also have , is a triangle in that is detected while
scanning the adjacency list of , assuming that (and possibly ) are discovered in this order.
Refer to Fig. 1 (iv).
Note that in this subcase a triangle different from is detected by the algorithm.
(This case does not occur if is the only triangle present in .)
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This completes the case analysis and thereby shows that the algorithm performs as intended.
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Time analysis.
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Note that Phase I (lines 2–4) of Triangle-Detection-1
takes time, as there are at most vertices of degree at most , and
checking such a vertex amounts to edge tests; indeed, there are
two-edge paths whose middle vertex is and an edge test takes time using the adjacency matrix of .
Phase II (lines 6–9) takes time; indeed, each BFS+ takes time on a connected graph
with edges and there are such calls.
With the setting , the run-time is for each phase,
thus overall. In particular, this is , completing the proof of the first upper bound
in Theorem 2.
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2.2 A second algorithm
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We slightly modify the previous algorithm as follows.
In Phase I the algorithm generates all paths of length two in whose intermediate vertex is of low degree
(the previous algorithm achieves the same operation in a slightly different way).
There are at most such paths and they can be found in time.
For each such path, checking whether its endpoints are connected by an edge takes time
using the adjacency matrix of . Consequently, Phase I takes time.
Phase II is left unchanged; it executes BFS trials.
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Phase I takes time (see above) and Phase II takes time as explained earlier.
With the setting , the run-time is for each phase,
thus overall.
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Running time in terms of .
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Next we derive the upper bound on the running time in two different ways,
namely starting from Triangle-Detection-1 or starting from Triangle-Detection-2.
The two methods, based on the high-degree vs. low-degree distinction, have been distilled
in the following lemma of A. Abboud and V. V. Williams .
For concreteness we instantiate them below.
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Lemma 1.
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()
If there is an time algorithm for triangle detection for graphs on nodes
and edges, then there is an time algorithm for triangle detection for graphs
on edges (and arbitrary number of nodes). And similarly, if there is an time algorithm,
then there is an time algorithm as well.
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It can be assumed that the graph is connected, otherwise the algorithm is run on each connected component.
First, consider Triangle-Detection-1 and note that (by Theorem 2, first part)
it satisfies the conditions in the first part
of Lemma 1 with . The lemma yields a randomized algorithm
for triangle detection running in time and whose probability of error is as claimed.
Second, consider Triangle-Detection-2 and note that (by Theorem 2, second part)
it satisfies the conditions in the second part
of Lemma 1 with . Once again, the lemma yields a randomized algorithm
for triangle detection running in time and whose probability of error is as claimed.
The proof of Theorem 2 is now complete.
∎
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3 Applications
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3.1 Boolean matrix multiplication
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By a result of V. V. Williams and R. R. Williams [39, Thm. 6.1], the problems of triangle detection in an
-vertex graph and that of Boolean matrix multiplication of two matrices are subcubically equivalent
in the sense that both have truly subcubic combinatorial algorithms or none of them do.
Since prior to this work no truly subcubic combinatorial algorithms were known for either problem the above statement
had until now more of a speculative nature. Thus, by the above reduction, our first truly subcubic combinatorial algorithm
for triangle detection implies the first truly subcubic combinatorial algorithm for Boolean matrix multiplication.
The new faster combinatorial algorithm for BMM owes a lot to the work of V. V. Williams and R. R. Williams on
this reduction.
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Theorem 3.
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There is a randomized combinatorial algorithm for multiplying two Boolean matrices with high probability
running in time.
[Moreover, a Boolean product witness matrix can be computed by the algorithm.]
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Proof.
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According to V. V. Williams and R. R. Williams [39, Thm 6.3], if there exists a time algorithm
for triangle detection in an -vertex graph, where , then there is a randomized algorithm that computes
the Boolean product of two given matrices with high probability,
say, at least , in time .
Substituting and using random trials in Triangle-Detection-1
yields the claimed bound.
∎
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Given two Boolean matrices and , index is said to be a witness for the
index pair iff . An integer matrix is a Boolean product witness matrix
for and iff
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| | | |
---
| | | |
In many applications, in addition to computing the Boolean product it is also desired to
compute a matrix of witnesses. When there is more than one witness for a given pair ,
any witness usually suffices. An elegant randomized algorithm due to Seidel ,
also noticed by other researchers ,
computes a matrix of witnesses in time ; if , the runtime becomes .
Slightly slower deterministic variants have been obtained, among others, by Alon et al. and by
Alon and Naor .
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All previous algorithms that compute witnesses in subcubic time use FMM and are therefore impractical.
Moreover, computing witnesses is done by a separate algorithm independent of the one computing the matrix product.
In contrast, our new combinatorial algorithm for BMM outlined in Theorem 3 can compute witnesses
on its own by its intrinsic nature. More precisely, the algorithm of V. V. Williams and R. R. Williams
implementing the BMM to Triangle Detection reduction, can do this as follows, see [39, p. 16].
The algorithm repeatedly finds222Whereas algorithm in is phrased in terms of negative triangle detection,
it works equaly well with a triangle detection subroutine.
a triangle in a tripartite graph with parts . Whenever a triangle is found,
the algorithm simply records in the matrix as a witness for the pair , removes the edge from the graph
and attempts to find a new triangle. The algorithm ensures that a witness is found for each index pair ,
if it exists. The statement in square brackets in Theorem 3 is implied.
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3.2 The -CLIQUE problem
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As mentioned in the introduction, the -clique problem can be answered by a brute-force algorithm running in
time, where (and is fixed). The current fastest algorithm is due to
Nešetřil and Poljak .
It is based on the idea that triangle detection can be done in subcubic time using FMM (recall the algorithm of
Itai and Rodeh ). For , the algorithm of Nešetřil and Poljak runs
in time; see for instance [41, Sec. 4].
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Obtaining faster than trivial combinatorial algorithms, by more than polylogarithmic factors,
for -clique is a longstanding open question, as is the case of BMM, see . The fastest combinatorial algorithm
runs in time . A slightly faster combinatorial algorithm was just announced in .
Assume that .
Replacing the FMM algorithm in the algorithm of Nešetřil and Poljak
with that in Theorem 2 yields a faster randomized combinatorial algorithm for detecting a -clique,
running in time, thereby providing the first solution to the above open problem.
If , the runtime is , whereas if , the runtime is .
For example, if , the algorithm runs in time, whereas the previous record was
.
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Theorem 4.
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There is a randomized combinatorial algorithm for detecting a -clique in a given -vertex graph,
where , running in time, and whose probability of error is at most .
If , the runtime is , whereas if , the runtime is .
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We only give here a brief outline of the argument and refer the reader to , , or for the details.
For simplicity, assume that .
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Proof.
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For every -clique in , create a corresponding vertex in an auxiliary graph.
Two vertices and are connected by an edge in the auxiliary graph, if and only if
forms a –clique in . Note that the auxiliary graph has vertices.
Furthermore, the graph contains a -clique if and only if the auxiliary graph contains a triangle.
The latter graph is where the triangle detection algorithm is run. Its runtime is .
∎
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In the general context of constraint satisfaction optimization, R. R. Williams outlined a method in which
an optimum weight -clique corresponds to an optimum solution of the problem to be solved. In the conclusion of his paper,
and in direct relation to the technique used above he specifically asked:
“Is there a randomized -clique detection algorithm running in ? (Is there merely a good one that does not
use matrix multiplication?)” Our Theorem 2 gives a positive answer to both questions.
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3.3 The MAX-CUT problem
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Our presentation closely follows that of Woeginger regarding this problem.
In the Max-Cut problem, given an -vertex graph , the problem is to find a cut of
maximum cardinality, that is, a subset of the vertices that maximizes the number of edges between and
. The problem can be solved easily in time by enumerating all possible subsets .
R. R. Williams developed the following method effective in reducing the runtime of an exact algorithm;
his algorithm runs in time.
Here we offer a “black box” replacement that yields a faster randomized combinatorial algorithm.
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Partition the vertex set into three parts of roughly the same size, : .
Introduce a complete tripartite auxiliary graph that contains one vertex for every subset ,
one vertex for every subset , and one vertex for every subset .
Note that has vertices.
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For every subset and every with , introduce the directed edge
from to . This edge receives a weight that equals the number of edges in
between and plus the number of edges between and
plus the number of edges between and .
As a result, the cut cuts exactly edges in .
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Write . There are less than possible triples to consider, where
, , and ,
such that the auxiliary graph contains a triangle on vertices , ()
with these edge-weights. For each such triple, the algorithm computes a corresponding simplified version
of the auxiliary graph that only contains the edges of weight between vertices
and . It remains to find a triangle in the simplified auxiliary graph.
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Since the triangle detection algorithm we use is randomized, we slightly modify it so that given a graph
with vertices, it runs in time and its probability of success
is at least . By the union bound, the probability of error is at most
. We obtain the following result.
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Theorem 5.
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Given an -vertex graph , a maximum cut can be computed by a randomized combinatorial algorithm
in time with probability at least .
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3.4 Finding small complete subgraphs
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Triangle detection (or counting) is a key ingredient in many algorithms for detecting (or counting)
small complete subgraphs, i.e., for a fixed . See for instance and
some recent developments in . Our new triangle detection algorithm yields many
improvements, always in practicality and in most cases also in the running time.
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For example, the fastest known algorithm for detecting a , due to Alon, Yuster, and Zwick, runs
in and since it relies on fast matrix multiplication is largely impractical.
Using the “extension method”, the problem easily reduces to calls for triangle detection in a graph with
vertices; see or . Indeed detecting a in reduces to detecting a
in the graph induced by , for some . Our new algorithm for triangle detection immediately
improves the running time to with a simpler practical combinatorial randomized algorithm.
The bound on the error probability still holds (it is in fact better, one can check, since a can be detected
from any of its four vertices). Alternatively, consider the following algorithm. For a parameter ,
the algorithm first checks the existence of a incident to a vertex of low degree, , and then
attempts detecting a in the subgraph induced by vertices of high degree, , using the
randomized algorithm described above. The resulting running time is
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| | | |
---
| | | |
when setting . Whereas this runtime does not quite match the current record, , due to Kloks, Kratsch, and Müller ,
it gets close to it via a practical solution.
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As another example, the fastest known algorithm for detecting a running in
or time [18, 23] also relies on FMM thus is largely impractical.
Using the “triangle method”, the problem easily reduces to triangle detection in a (auxiliary) graph with
vertices, see or . Thus our Theorem 2 immediately gives a simpler
and faster combinatorial randomized algorithm running in time.
Similarly, we obtain randomized combinatorial algorithms for detecting a in time, or in time;
and for detecting a in time, or in time. These running time match or surpass previous ones
attainable via FMM algorithms, see . We recall that Woeginger asked whether a
can be detected in time, and so this particular challenge remains.
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More examples of a similar nature can be shown along the lines of , however, here we do not elaborate
further in this direction.
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3.5 Triangles in segment intersection graphs
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Clearly we have to miss many applications. Here is one in computational geometry that we did not want to miss,
where a “black box” replacement substantially improves the algorithm.
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Recently, Chan showed that detecting a triangle in the intersection graphs of segments in the plane
can be done in in time, where denotes the time complexity
of finding a triangle in a graph with edges. However, his algorithm is rather impractical
since it uses fast matrix multiplication (BMM via FMM).
As such, the algorithm runs in time.
Replacing the BMM algorithm with that in Theorem 2 yields
a practical algorithm for detecting a triangle in the intersection graphs of segments in the plane
that is also faster, running in time.
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3.6 Further applications
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Besides the subcubic equivalence between Triangle Detection and BMM we used here,
V. V. Williams and R. R. Williams [39, Thm. 6.1] have also shown that two other problems,
(i) Listing up to triangles in a graph, and (ii) Verifying the correctness of a matrix product
over the Boolean semiring are also subcubic equivalent in the same sense.
Since Theorem 2 shows that there is a randomized algorithm for triangle detection
in a given -vertex graph running in truly subcubic time, the same holds for the two problems mentioned above,
i.e., each admits a randomized combinatorial algorithm running in time , for some constant
. More details on these algorithms are expected to appear elsewhere.
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3.7 Remarks and open problems
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It is perfectly possible (even expected) that the running time of the Itai-Rodeh algorithm for triangle detection will
surpass the running time of our algorithm at some point in time, provided that sufficient advances in
FMM algorithms are made. When this would happen, it is hard to tell.
However, the improvement in practicality will remain, it seems, on the side of our algorithm.
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We conclude with two problems for further investigation.
The first was initially posed by Alon, Yuster, and Zwick , and later included by Woeginger in his survey
on exact algorithms : Is detection as difficult as Boolean matrix multiplication?
Further, one can ask: if there is a combinatorial algorithm for detection in an -vertex graph running in time,
is there a combinatorial algorithm for multiplying two Boolean matrices in time that is close to that?
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Another interesting question is whether the running times of our randomized combinatorial algorithms can be matched
or approached by deterministic ones.
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References
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↑
A. Abboud, A. Backurs, and V. V. Williams,
If the current clique algorithms are optimal, so is Valiant’s parser,
SIAM Journal of Computing
47(6) (2018), 2527–2555.
↑
A. Abboud, N. Fischer, Z. Kelley, S. Lovett, and R. Meka,
New graph decompositions and combinatorial Boolean matrix multiplication algorithms,
arXiv preprint , Nov. 2023.
↑
A. Abboud, N. Fischer, and Y. Schechter,
Faster combinatorial -clique algorithms,
arXiv preprint , Jan. 2024.
↑
A. Abboud and V. V. Williams,
Popular conjectures imply strong lower bounds for dynamic problems,
Proc. 2014 IEEE 55th Annual Symposium on Foundations of Computer Science (FOCS),
2014, pp. 434–443.
↑
A. Abboud and V. V. Williams,
Popular conjectures imply strong lower bounds for dynamic problems,
arXiv preprint , Feb. 2014.
↑
N. Alon, Z. Galil, O. Margalit, and M. Naor,
Witnesses for Boolean matrix multiplication and for shortest paths,
Proc. 33rd IEEE Annual Symposium on Foundations of Computer Science (FOCS),
1992, pp. 417–426.
↑
N. Alon and M. Naor,
Derandomization, witnesses for Boolean matrix multiplication and construction of perfect hash functions,
Algorithmica
16(4-5) (1996), 434–449.
↑
N. Alon, R. Yuster, and U. Zwick,
Color-coding,
Journal of the ACM
42(4) (1995), 844–856.
↑
N. Alon, R. Yuster, and U. Zwick,
Finding and counting given length cycles,
Algorithmica
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Vibrio vulnificus necrotizing fasciitis with sepsis presenting with pain in the lower legs in winter: a case report
BMC Infectious Diseases
volume 22, Article number: 670 (2022)
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Abstract
Background
Vibrio vulnificus infections develop rapidly and are associated with a high mortality rate. The rates of diagnosis and treatment are directly associated with mortality.
Case presentation
We describe an unusual case of a 61-year-old male patient with chronic liver disease and diabetes who presented with a chief complaint of pain in both lower legs due to V. vulnificus infection in winter. Within 12 h of arrival, typical skin lesions appeared, and the patient rapidly developed primary sepsis. Despite prompt appropriate antibiotic and surgical treatment, the patient died 16 days after admission.
Conclusion
Our case findings suggest that V. vulnificus infection should be suspected in patients with an unclear infection status experiencing pain of unknown origin in the lower legs, particularly in patients with liver disease or diabetes, immunocompromised status, and alcoholism.
Peer Review reports
Background
Vibrio vulnificus is a gram-negative, halophilic, alkaliphilic marine bacterial pathogen commonly found in plankton and shellfish, especially oysters, which was first described in 1976. [1, 2]. V. vulnificus favorably grows in warm water and low-salinity areas in coastal regions; therefore, V. vulnificus infections show seasonal and regional patterns . V. vulnificus infections are generally acquired by consuming contaminated seafood, particularly oysters. Wound infections can occur after skin lesions are exposed to contaminated seawater [4, 5].
In this report, we present an unusual case of necrotizing fasciitis with sepsis presenting with pain in the lower legs caused by V. vulnificus infection in winter in our hospital located in Binzhou, a coastal city in eastern China. The patient complained of pain in both lower legs without presenting with other prodromal symptoms upon admission. To the best of our knowledge, this clinical presentation is unusual and could alert clinicians to possible V. vulnificus infection in patients with an unclear infection status, who experience pain of unknown origin in the lower legs even before the appearance of typical skin lesions, especially if the patients have risk factors.
Case presentation
A 61-year-old male patient was transferred to our hospital on November 29, 2020, at 23:30 pm, with complaints of pain in both lower legs for 16 h. He had a medical history of hepatitis B, hepatitis B-decompensated cirrhosis, and liver cancer, along with hypertension, diabetes mellitus, and tobacco use (2 packs of cigarettes per day). Moreover, he had a history of esophageal variceal ligation under endoscopic surgery 5 years earlier. He had consumed seafood (including turtle, hairtail, and shellfish) for lunch 3 days before admission.
The patient was in a normal state until the morning of the 3rd day after consuming seafood with his family when he developed pain in both lower legs on awakening. He could recall no traumatic events or any history of recreational marine exposure. He did not consider the pain as a concern and ingested 800 mg of ibuprofen and sprayed diclofenac sodium on his lower legs. He experienced worsening of pain and arrived at our hospital at night. Color ultrasound examination of the legs showed normal findings.
Initial physical examination revealed the following: clear consciousness; body temperature, 36.5 °C; respiration rate, 17 breaths per minute; blood pressure, 92/50 mmHg; and heart rate, 68 beats per minute. Reddish patches were observed on the right ankle.
The initial laboratory investigations revealed the following: white blood cell count, 6.1 × 109/L (94.3% neutrophils, 2.9% lymphocytes, and 2.8% monocytes); platelet count, 35 × 109/L; hematocrit, 28%; serum glucose, 8.73 mmol/L; alanine aminotransferase, 22.84 U/L; ceramic oxaloacetic transaminase, 34.3 U/L; total bilirubin, 32.7 µmol/L (direct bilirubin, 9.1 µmol/L; indirect bilirubin, 23.6 µmol/L), serum γ-glutamyltransferase, 101.4 µmol/L; prothrombin time, 18.8 s; partial thromboplastin time, 43.5 s; internationally standardized ratio, 1.64; d-dimer, 3.07 mg/L; serum urea nitrogen, 14.88 mmol/L; myoglobin, 672.2 ng/mL; creatine kinase, 210.2 U/L; C-reactive protein, 223.90 mg/L; blood sodium, 124 mmol/L; blood chlorine, 93.90 mmol/L; and serum potassium, 5.52 mmol/L. Hemoglobin, hematocrit, creatinine, and routine urine test results were normal.
Blood test results indicated thrombocytopenia, unclear infection status, coagulation dysfunction, hypoalbuminemia, electrolyte disturbance, and renal dysfunction. Blood culture was performed, and empirical antibiotic therapy was initiated. We used levofloxacin (600 mg, qd) intravenously.
Within 12 h of arriving to the hospital, the patient experienced swelling and severe pain. Creatine kinase increased to 14,141 U/L, and his white blood cell count decreased to 1.5 × 109/L. Color ultrasound and magnetic resonance imaging of the legs revealed extensive swelling in the soft tissue of the lower legs, suggesting cellulitis and edema (Fig. 1). Several tense bullae, erythematous plaques, color changes in both lower legs, and hemorrhagic bullae in the right ankle were observed (Fig. 2). Subsequently, the patient continued to rapidly develop skin lesions, bullae, and vesicles even after 3 h (Fig. 3), and he further complained of decreased strength and paresthesia of the legs. Urgent surgical debridement and decompression were performed. During the operation, fishy odor was detected in the calf muscles and plantar compartments. Some of the muscle and fascia tissues were necrotic, and light-yellow fluid exudate was noted; no abscess was formed. Biopsy of the muscle and fascia tissue of both lower legs revealed swollen muscle fibers, dissolved and necrotic sarcoplasm, considerable bacterial concentration in the muscle and the surrounding adipose tissue, infiltration of extensive acute and chronic inflammatory cells, and dilated blood vessels congested with vasculitis (Fig. 4). The images were collected by cellSens image acquisition system of OLYBUS microscope BX51. The resolution of images is 1920 × 1440.
MRI showing extensive swelling in the soft tissue of the lower legs. MRI magnetic resonance imaging
Twelve hours after admission: Bullae and erythematous plaque in both lower legs
Fifteen hours after admission: Progression of the skin lesions in both legs
Histopathological findings: Swollen muscle fibers, dissolved necrotic sarcoplasm, high bacterial concentration in the muscle and surrounding adipose tissue, extensive infiltration of acute and chronic inflammatory cells, and dilated blood vessels with congested vasculitis. A Hematoxylin and eosin (H&E) stain, ×100. B H&E stain, ×400
Postoperatively, the patient’s condition deteriorated rapidly, with the development of multiple organ dysfunction syndrome (circulation, disseminated intravascular coagulation, respiratory, liver, and kidney), and he was immediately admitted to the intensive care unit. Intravenous therapy with imipenem/cilastatin sodium (2 g, q8h), cefoperazone sodium/sulbactam sodium (3 g, q6h), and levofloxacin was initiated. Four days after admission, bacterial cultures from the blood, bullae fluid, and tissues (matrix-assisted laser desorption/ionization–time of flight mass spectrometry) were all positive for V. vulnificus. Susceptibility was determined with Vitek2 Compact according to the recommendations of Clinical Laboratory Standards Institute, M45, third edition. The cultured microbes revealed sensitivity to all tested antibiotics, including piperacillin, ceftazidime, cefoperazone, cefepime, imipenem, meropenem, ciprofloxacin, levofloxacin, doxycycline, and tigecycline.
Despite prompt appropriate antibiotic and surgical treatment, the skin lesions spread to the right thigh on the 2nd day after surgery (Fig. 5). Amputation was required; however, considering the progressively deteriorating condition and his previously verbalized wishes, his family refused amputation. Despite vigorous treatment, the patient died 16 days after admission. Postmortem examination was not performed.
Day 4 after admission: progression of tissue necrosis to the thigh
Discussion and conclusions
Vibrio vulnificus multiplies in approximately 10 min at 20 °C, and its multiplication is controlled below 15 °C . V. vulnificus infections are common in warm climates, but our case occurred in winter. Although it is relatively unusual, there have been previous reports of cases in cold climates ; thus, we should pay attention to this disease even in winter.
Previous reports have shown that patients with liver disease are more susceptible to V. vulnificus infection, particularly those with hepatic cirrhosis. This may be due to the fact that patients with chronic liver disease often have portal hypertension, which can cause shunting of the bacteria around reticuloendothelial cells in the liver . Other risk factors include diabetes, immunocompromised status, and alcoholism [5, 9, 10]. Moreover, it is believed that approximately 1 million bacilli are required to cause infection through oral ingestion . The patient in our case was 61 years old, had an underlying liver disease and diabetes, and was at a high risk of infection. In addition, the man had a history of esophageal varix, which most likely indicates that he had portal hypertension, and that he may have had vascular ulceration in the gastrointestinal tract, making V. vulnificus more likely to enter the bloodstream and cause bacteremia on consumption of contaminated seafood. V. vulnificus disseminates from the gastrointestinal tract to the bloodstream, leading to the rapid development of primary sepsis. This may explain why only our patient developed primary sepsis 3 days after consuming seafood, although all his family members consumed the same diet.
Infection with this pathogen can result in three major discernible manifestations: primary sepsis, wound infections, and gastrointestinal diseases [1, 12]. Primary sepsis usually manifests as an acute symptom of systemic infection; it occurs within 24–48 h of ingesting the microorganism and often begins with prodromal symptoms, including watery diarrhea, fever, chills, nausea, vomiting, and abdominal pain, followed by skin lesions . Typical skin lesions include local or flaky erythema and ecchymosis, blood blisters with exudation, necrosis and cellulitis, necrotising fasciitis, and muscle inflammation . The wound infections due to V. vulnificus can progress to necrotizing soft-tissue infections . The contact history of patients with a rapid onset of cellulitis can alert clinicians to a differential diagnosis of soft-tissue infection with V. vulnificus (contact with seawater or raw seafood) or Aeromonas species (contact with fresh or brackish water, soil, or wood) . In addition, sepsis progresses rapidly, and most patients experience shock with hypotension at the time of hospital arrival.
In the present case, the patient presented with pain in both lower legs as the chief complaint. We did not detect any other prodromal symptoms; therefore, diagnosis was challenging until the development of the characteristic bullae. Despite the inapparent or atypical presentation, there were some indicators of infection. His blood pressure was slightly low, and laboratory examination findings suggested infection; these may indicate septic shock. Reddish patches on the right ankle, elevated d-dimer, and coagulation dysfunction may indicate progression to disseminated intravascular coagulation. Moreover, elevated creatine kinase may signify rhabdomyolysis. We should obtain critical information quickly when treating patients with suspected limb infections, which can help identify patients with life-threatening limb infections at an early stage.
Antibiotic therapy and debridement are the most effective methods for the treatment of necrotizing myofascitis caused by V. vulnificus infection. A recent study suggested that for patients presenting with septic shock with a recent history of raw seafood consumption, treatment with doxycycline or ciprofloxacin for appropriate coverage of V. vulnificus and gram-negative resistant organisms should commence while awaiting microbiological cultures. Once a diagnosis of V. vulnificus septicemia is confirmed, treatment can be safely changed to ceftriaxone combined with doxycycline or ciprofloxacin . Early debridement plays a vital role in improving the prognosis of patients with V. vulnificus infection . In the present case, after the occurrence of bullae, even with sensitive antibiotic treatment and surgical intervention, the patient died. The patient may have died owing to not only delayed diagnosis and treatment but also the patient’s medical condition.
In conclusion, V. vulnificus is a deadly and opportunistic human pathogen that usually infects humans through seafood consumption or direct contact with open wounds. Infection with this pathogen can rapidly progress to septic shock and even death. Therefore, early culture and diagnosis are crucial. Our case alerts clinicians to a possible V. vulnificus infection in patients with an unclear infection status, who experience pain of unknown origin in the lower legs, especially if the patient has hypotensive shock, leukopenia, hypoalbuminemia, coagulation dysfunction, increased creatine kinase, or underlying liver disease and diabetes.
Availability of data and materials
The data that support the findings of this study are available from the corresponding author (Yong Gao).
Abbreviations
Vibrio vulnificus
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Acknowledgements
We would like to thank Editage (www.editage.cn) for English language editing.
Funding
The present study was supported by the Research Foundation of Binzhou Medical University (BY2015KYQD23 and BY2015KJ01), Shandong Province Traditional Chinese Medicine Development Plan (2019-0515).
Author information
Authors and Affiliations
Department of Pain, Binzhou Medical University Hospital, Binzhou, 256603, China
Weihua Di, Hui Yu, Xiao Cui, Huanlan Sa, Cuijie Shao & Yong Gao
Department of Rehabilitation Medicine, Binzhou Medical University Hospital, Binzhou, 256603, China
Jing Cui
Department of Intensive Care Unit, Binzhou Medical University Hospital, Binzhou, 256603, China
Zhong Fu
Department of Foot and Ankle Surgery, Binzhou Medical University Hospital, Binzhou, 256603, China
Bingjin Fu, Guofeng Guan & Rui Du
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Contributions
WD and YG participated in the design of the study, contributed to the acquisition of data, and wrote the final draft of the manuscript. JC and HY edited the first draft of the manuscript. XC and HS participated in the investigation process, data curation and formal analysis of the work. ZF, BF, GG, RD, and CS participated in the investigation process and data curation. All authors read and approved the final manuscript.
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Di, W., Cui, J., Yu, H. et al. Vibrio vulnificus necrotizing fasciitis with sepsis presenting with pain in the lower legs in winter: a case report.
BMC Infect Dis 22, 670 (2022).
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Complex-number method to minimize equilateral-triangle area inside right triangle of side lengths 2 3–√2 3, 5 5, and 37−−√37
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The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths 2 3–√2 3, 5 5, and 37−−√37, as shown, is m p√n m p n, where m m, n n, and p p are positive integers, m m and n n are relatively prime, and p p is not divisible by the square of any prime. Find m+n+p m+n+p.
I am trying to solve this problem just with complex numbers. I let the vertices of the large triangle to be a,b,c a,b,c for b=0,a=2 3–√i,c=5.b=0,a=2 3 i,c=5. I let the vertices of the equilateral triangle to be z 1,z 2,z 3 z 1,z 2,z 3 for z 1 z 1 along A B A B, z 2 z 2 along A C A C, and z 3 z 3 along B C B C. In the first place, it is well-known that z 2 1+z 2 2+z 2 3=z 1 z 2+z 2 z 3+z 3 z 1.z 1 2+z 2 2+z 3 2=z 1 z 2+z 2 z 3+z 3 z 1.
It is futile to consider the equations for A,Z 1,B A,Z 1,B and B,Z 3,C B,Z 3,C to be collinear because we set those as the imaginary and real axes. With the equation for A,Z 2,C A,Z 2,C to be collinear, we have
z 2−a z 2−c=z 2¯¯¯¯¯−a¯¯¯z 2¯¯¯¯¯−c¯¯=z 2¯¯¯¯¯+a z 2¯¯¯¯¯−c,z 2−a z 2−c=z 2¯−a¯z 2¯−c¯=z 2¯+a z 2¯−c,
with that step from a¯¯¯=−a a¯=−a and c¯¯=c,c¯=c, as these are along the imaginary and real axes respectively. With some manipulation, we get to
a(z 2+z 2¯¯¯¯¯)+c(z 2−z 2¯¯¯¯¯)−2 a c=0 a(z 2+z 2¯)+c(z 2−z 2¯)−2 a c=0
⟺a R(z 2)+c I(z 2)i=10 3–√⟺a ℜ(z 2)+c ℑ(z 2)i=10 3
⟺2 3–√x 2+5 y 2=10 3–√,⟺2 3 x 2+5 y 2=10 3,
for R(z)=x,I(z)=y.ℜ(z)=x,ℑ(z)=y.
But I'm not quite sure what to do now, because I can't think of any synthetic simplifications.
geometry
complex-numbers
contest-math
euclidean-geometry
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edited Dec 25, 2020 at 16:57
Quanto
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asked Dec 23, 2020 at 19:10
explicitEllipticGroupActionexplicitEllipticGroupAction
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Solution (#6) on AOPS : artofproblemsolving.com/wiki/index.php/2017_AIME_I_Problems/…cosmo5 –cosmo5 2020-12-23 19:59:26 +00:00 Commented Dec 23, 2020 at 19:59
Can you check your definitions of a,b,c a,b,c? you defined b b twice.Calvin Lin –Calvin Lin 2020-12-24 00:53:12 +00:00 Commented Dec 24, 2020 at 0:53
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Rewrite the equilateral condition z 2 1+z 2 2+z 2 3=z 1 z 2+z 2 z 3+z 3 z 1 z 1 2+z 2 2+z 3 2=z 1 z 2+z 2 z 3+z 3 z 1 as
z 2 3−(z 1+z 2)z 3+z 2 2+z 2 3−z 1 z 2=0 z 3 2−(z 1+z 2)z 3+z 2 2+z 3 2−z 1 z 2=0
which is quadratic in z 3 z 3, yielding
z 3=e i π 3 z 1+e−i π 3 z 2 z 3=e i π 3 z 1+e−i π 3 z 2
Substitute z 3 z 3 into the line equation below for the hypotenuse
z+z¯10+z−z¯4 3–√i=1 z+z¯10+z−z¯4 3 i=1
obtained from the given side lengths, leading to
7 20|z 1|+11 3–√60|z 2|=1 7 20|z 1|+11 3 60|z 2|=1
where z 1=z¯1 z 1=z¯1 and z 2=−z¯2 z 2=−z¯2 are used. Then, the area of the equilateral triangle is
A=3–√4|z 2−z 1|2=3–√4(|z 2|2+|z 1|2)≥3–√4(7 20|z 1|+11 3√60|z 2|)2(7 20)2+(11 3√60)2=75 3–√67 A=3 4|z 2−z 1|2=3 4(|z 2|2+|z 1|2)≥3 4(7 20|z 1|+11 3 60|z 2|)2(7 20)2+(11 3 60)2=75 3 67
where the C-S inequality is applied.
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edited Dec 24, 2020 at 1:16
answered Dec 24, 2020 at 1:01
QuantoQuanto
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1 Great answer, I realized that instead of using the criterion for an equilateral triangle, we can perform a spiral similarity and arrive at the same expression regarding the point on the hypotenuse: z 3↦e i π/3(z 3−z 1)+z 1=z 2,z 3↦e i π/3(z 3−z 1)+z 1=z 2, where we take z 2 z 2 on the hypotenuse and z 1 z 1 on the imaginary axis.explicitEllipticGroupAction –explicitEllipticGroupAction 2020-12-24 04:59:42 +00:00 Commented Dec 24, 2020 at 4:59
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https://www.youtube.com/watch?v=637mYMLUf8s
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Solve the Last Layer / Third Layer - 3x3 Cube Tutorial - Only 4 moves to learn - Easy Instructions
Zman872
48700 subscribers
108028 likes
Description
8552787 views
Posted: 20 Jun 2019
Simple, easy-to-follow, step-by-step explanation to solve the TOP, LAST, or THIRD layer. No complicated terms. All 3rd layer situations covered.
Visit our Cool Cube Merch store for an AWESOME selection of Fun Cube-themed Shirts, Pillows, Blankets, and other items. Use coupon code ZMAN872 to save 10%.
Table of contents:
0:00 - Intro
0:26 - Yellow CROSS:
1:14 - Yellow Cross - Dot
1:49 - Yellow Cross - Hook
2:19 - Yellow Cross - Bar
2:49 - Yellow EDGES
4:30 - Yellow Edges - Opposite Edges Match
5:06 - Yellow Edges - Adjacent Edges Match
6:10 - Yellow CORNERS
8:59 - Finishing the Cube
I worked hard to find the most simple method. I hope it helps you a lot!
If you want to learn an easier way to solve the SECOND layer, you can find that video here:
Now that you can solve your cube, if you're looking for a smoother/faster cube, here are some of my favorites:
GAN 356:
GAN 11 M Pro:
Best Under $10:
Highest-rated cubes:
Best mirror cube ever!:
Thank you!
-Zman872
Transcript:
Intro Welcome to my video for the easiest way
to solve the last layer of your 3x3 cube. If you're still trying to figure out the
best way to do the first two layers, you can check out my other videos, with links
in the description. This video won't be of much use to you unless you already
have the first two layers solved. The last layer is not nearly as difficult as other videos make it seem. Other videos make it much more complicated
than the approach I'll teach you. Alright. Let's start with the yellow cross. Once
you've solved the second layer, take a Yellow CROSS look at your yellow side. If you look at
just the center and edge pieces, not the corners, you'll see one of these four
situations. (And once again we do not care about the corners for this step) So
you'll either have the yellow dot, which is just the center with none of the four
yellow pieces yellow edge pieces around it, a yellow hook or L (whichever you want
to call it), which has two adjacent edge pieces, the yellow bar or
line, which has two opposite yellow edge pieces, or the yellow cross, with all
four edge pieces already solved. For this step, we're going to be doing the same set
of moves regardless of which of these scenarios
you have, to progress through each of these stages, to get to the yellow cross.
So if you have no yellow edges around Yellow Cross - Dot your yellow Center (meaning you have the
dot) and then once again we're ignoring our corners so it doesn't matter if
they're diagonal, next to each other, if all four are on the top -- we don't care.
Then we want to do this move: We want to turn the front of the cube clockwise,
turn the right side up, the top to the left, right side down, top back to the
right, and the front counterclockwise. Now we should have an L or a hook in the
bottom right corner. So the next step is Yellow Cross - Hook to get from the hook to the bar. So for
this, we want to put the hook so it's in the top left corner, meaning we have a yellow
edge on the top, and yellow edge on the left. So now we want to do the same set
of moves with the yellow on top and the hook in the top left. So front clockwise,
right side up, top to the left, right side down, top back to the right,
and the front counterclockwise. Now you should have the bar horizontally. So next,
to get from the bar to the cross, Yellow Cross - Bar you want to make sure that our bar is
horizontal. If the bar is vertical, just rotate it once so that it's horizontal.
Now we want to do the same set of moves as the other two steps, with the yellow
on top and the bar horizontal. So front clockwise, right side up, top to the left,
right side down, top back to the right, and front counterclockwise. Now you
should have the yellow cross. So now that Yellow EDGES we have our yellow cross, we need to see
if the edges match up with the colors of the sides. We're looking at our yellow
edge pieces matching up with the center right below it. So if all four already do,
you got lucky. But usually when you first get your yellow cross, only two of them
will match up, but we need to rotate the top layer until we can find those two. So
if you look around the cube you can see that this blue and the green don't
match, the green and the red don't match, the red and the blue don't match, and the
orange and the orange does match, but because there's only one that matches, we
need to rotate the top again. So now you want to look around again: so red and
orange doesn't match, orange and green doesn't match,
blue and red doesn't match, green and blue doesn't match. We need to rotate it
again: blue and blue does match, green and orange doesn't match, red and green
doesn't match, and orange and red doesn't match, meaning once again we only had one.
And if you rotate it again, we can see that orange and blue don't match, blue
and orange don't match, but green and green, and red and red DO both match, so
this is the correct way that we need to have our top side rotated. Once you've
rotated your yellow sides so that two match, the question is: are the two that match
next to each other or across from each other? So in this example, we have the
orange and the blue that both match and they're right next to each other. In this
example, we have the red and the orange that match and they're opposite sides
from each other. When we do this next set of moves, it's going to progress
the edges from across from each other, to next to each other, to all four
matching. So first, when the two edges Yellow Edges - Opposite Edges Match that match are across from each other,
then you want to rotate the cube so that the edges are on the front and the back.
And then you want to do this set of moves: You want to turn the right side
up, top to the left, right side down, top to the left, right side up, top two times,
and the right side back down. And now you should have two yellow edge pieces that
match their corresponding centers and are right next to each other. Once you've
gotten this (or if you already had it) Yellow Edges - Adjacent Edges Match then you want to put those edges so that
they're on the back and on the right side, and then do that same set of moves
again. So once again, that's the right side up, top to the left, right side down,
top to the left, right side up, top two times, and the right side back down. And
now, if you rotate your top one more time to the left, all four edge pieces
should all match. That set of moves may seem a little bit confusing, so one way
to help you remember, and knowing that we're working with the right side, is to
think of it as up, over, down, over, up, over, over, down. This might just help you
remember this set of moves a little bit better. So once again that's up-over-down-over-up-over-over-down. And that should help you to remember this step when
trying to solve your yellow edges. Now Yellow CORNERS it's the first time we actually pay any
attention to the yellow corners since they are the last pieces we're going to
put in place. First, we're going to put them into the correct corner, and then
the next step will be to make them face the right way to complete the cube. Okay,
let's get them into the right places. First, look to see if you have any corners
that are exactly right (like this) where it's both in the right place and rotated
the correct way. You might have this, but you also might not. So if you DO have
this, wait a minute - we'll get there. Next, look
to see if you have any yellow corners that are in the right place for that
corner, but not facing the right way (like this). You can see that these three colors
are the same as these three colors around it, but this isn't facing the
right way. We need to rotate this corner for it to be correct, but that's okay -
that's our goal for this step - to get all the corners at least in the right
places, even if they aren't facing the right way.
Once you've got all four in the right place, then you'll be ready to go to the next
step, but for now, we're going to assume that you have none in the right place. So
if you have no corner pieces in the right place (like this), where this one
doesn't match up, this corner piece doesn't match up, this corner piece
doesn't match up, and this one doesn't match up, then it doesn't matter how your
top layer is rotated, but you want to do this set of moves: it's the top to the
left, the right side up, top back to the right so it knocks off this white piece,
the left side up, top back to the left so knocks this white piece off, the right
side back down, top back to the right, and the left side back down. Now you want to
look around your cube again to look for a yellow corner piece that matches up.
This one doesn't, this one doesn't, this one doesn't, but this one DOES right
here. There's also a chance that you still might not have one corner piece
that matches up, in which case you just need to do that set of moves again, and
then you should have at least one corner piece that matches up. So once you have
one corner piece that is in the correct place, then you want to put it so it's on
the bottom right of the yellow side - so right here. Then you want to do that set
of moves again. So once again: that's top to the left, the right side up, top to the
right knocking off this white piece, left side up, top back to the left knocking
this white piece off, right side down, top back to the right, and then the left
side down. Now you want to look around your cube again and see if all of
your yellow corner pieces are in their correct spot. So this one is, this one is,
this one is, and this one also is, meaning that we're now done with this step.
And similarly to trying to get one corner piece in the correct spot, you may
have to do that set of moves twice. At Finishing the Cube this point, you should have all four
corners in the right place. You might have none of them facing the right way
(like this), or might have some of them facing the right way - for example this
one is facing the right way, and this one's not, but that's still fine.
It doesn't matter, as long as all four are in the right places, you're ready for
the final step. Before we start this step, make sure you've got your yellow cross
all matched up like this, and all the yellow corner pieces are in their
correct corners, but not necessarily turned the correct way - like this. In this step,
we're going to get all the yellow corners to face the correct way, which
will solve the cube. Put one of the yellow corners that needs to be
corrected into the front right of the yellow side, then hold your cube so you've
got these four pieces locked into place like this, because for this set of moves,
we're only going to be turning the right side and the bottom. So now we're going to
do this next set of moves twice, and then check to see if it's solved, and if not,
we're going to do this set of moves another two times. So the set of moves is:
the right side down, the bottom to the left, the right side up, and the bottom to
the right. Now we're going to do that a second time: the right side down, bottom
to the left, right side up, and bottom to the right. And now we're going to check that
piece and see if it's been solved yet. As you can see, mine is still not, so I'm
going to do that set of four moves another two times. It's the right side down,
bottom to the left, right side up, bottom to the right. And again: right side down,
bottom to the left, right side up, bottom to the right. So once you've done that
full set of four moves however many times you need, then you want to rotate
your top layer so you have another piece that needs to be solved in that spot. So
once again I want to do: right side down, bottom to the left, right side up, bottom
to the right. And then again: right side down, bottom to the left, right side up,
bottom to the right. Now we want to check that piece and see it
has also been solved, so we want to rotate our top again until we get
another piece that needs to be solved. So this one. Then once again, we want to do:
right side down, bottom to the left, right side up, bottom to the right. And then
again: right side down, bottom to the left, right side up, bottom to the right. And
see that one now has been solved. Then rotate our top again - for our last
piece that needs to be solved, and do that same set of moves again. So right
side down, bottom to the left, right side up, bottom to the right. And again: right
side down, bottom to the left, right side up, bottom to the right. And then check
our piece to see if it has been solved. It has not yet been solved, so we have to do
that set of moves again. It's: right side down, bottom to the left, right side
up, bottom to the right. And then again: right side down, bottom to the left, right
side up, bottom to the right. And now, if we just rotate our top . . . CONGRATULATIONS!
You've just solved the Rubik's Cube! If your two pieces that need to be solved
are opposite corners from each other, then you need to do the same thing where
you put this piece in this position to be solved. Then you need to turn: right side
down, bottom to the left, right side up, bottom to the right. And then again: right
side down bottom to the left, right side up, bottom to the right. And now check to
see if it's been solved. It is not, so you have to do that again. So right side down,
bottom to the left, right side up, bottom to the right. And then again: right side
down, bottom to the left, right side up, bottom to the right. And now, if we rotate our
top, you can see that this piece has already been solved, so we just want to
keep on rotating past that piece - to another one that hasn't been solved,
which is over here, and then we can do that set of moves again. So it's right
side down, bottom to the left, right side up, bottom to the right. And again: right
side down, bottom to the left, right side up, bottom to the right. Now we just need
to rotate our top . . . and you've solved the Rubik's Cube!!
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Approximating square roots walk through (article) | Khan Academy
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8th grade (Eureka Math/EngageNY)
Course: 8th grade (Eureka Math/EngageNY)>Unit 7
Lesson 1: Topic A: Square and cube roots
Intro to square roots
Square roots of perfect squares
Square roots
Intro to cube roots
Cube roots
Worked example: Cube root of a negative number
Equations with square roots & cube roots
Square root of decimal
Roots of decimals & fractions
Equations with square roots: decimals & fractions
Dimensions of a cube from its volume
Square and cube challenge
Square roots review
Cube roots review
Intro to rational & irrational numbers
Classifying numbers: rational & irrational
Classify numbers: rational & irrational
Classifying numbers
Classify numbers
Classifying numbers review
Approximating square roots
Approximating square roots walk through
Approximating square roots
Approximating square roots to hundredths
Math>
8th grade (Eureka Math/EngageNY)>
Module 7: Introduction to irrational numbers using geometry>
Topic A: Square and cube roots
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Approximating square roots walk through
CCSS.Math: 8.NS.A.2
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Walk through a series of questions and examples that will help you learn how to approximate square roots.
In this article, you'll learn how to approximate square roots! Let's dive in.
Example
Spot the pattern:
(25)2=25
(6)2=6
(9482)2=9482
Reflection question
When you square the square root of a number...
Choose 1 answer:
Choose 1 answer:
(Choice A) the answer is less than the original number. A the answer is less than the original number.
(Choice B) the answer is the original number. B the answer is the original number.
(Choice C) the answer is greater than the original number. C the answer is greater than the original number.
Check
Practice Set 1:
Problem 1A
Current
(16)2=
Check Explain
Example
Without using a calculator, order the following numbers from least to greatest.
30, 5, 6
Step 1: Square each of the numbers:
| n: | 30 | 5 | 6 |
--- --- |
| n 2: | 30 | 25 | 36 |
Step 2: Order the numbers from least to greatest:
| n: | 5 | 30 | 6 |
--- --- |
| n 2: | 25 | 30 | 36 |
The answer:
| | | | |
--- --- |
| 5 | 30 | 6 |
Reflection question
When ordering square roots and integers, which of the following is the best strategy?
For example: 6,28,5.
Choose 1 answer:
Choose 1 answer:
(Choice A) Square just the square roots. For example: 6,(28)2,5. A Square just the square roots. For example: 6,(28)2,5.
(Choice B) Square both the square roots and the integers. For example: 6 2,(28)2,5 2. B Square both the square roots and the integers. For example: 6 2,(28)2,5 2.
Check
Practice Problem 2:
Without using a calculator, order the following numbers from least to greatest.
4
14
5
Check Explain
Example
Without using a calculator, find two consecutive integers to complete the following inequality.
?<97<?
Step 1: Create a table of perfect squares.
| n: | 7 | 8 | 9 | 10 | 11 |
--- --- --- |
| n 2: | 49 | 64 | 81 | 100 | 121 |
Step 2: Ask yourself, "Where does n=97 fit into the table?"
| n: | 7 | 8 | 9 | 97 | 10 | 11 |
--- --- ---
| n 2: | 49 | 64 | 81 | 97 | 100 | 121 |
The answer:
9<97<10
Practice Problem 3:
Without using a calculator, fill in the blanks with two consecutive integers to complete the following inequality.
<56<
Check Explain
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Shaurya_Patel 8 years ago Posted 8 years ago. Direct link to Shaurya_Patel's post “Is this the easiest way o...” more Is this the easiest way of doing this?
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Coolman 3 years ago Posted 3 years ago. Direct link to Coolman's post “yes it is” more yes it is
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SuperDavido05 2 years ago Posted 2 years ago. Direct link to SuperDavido05's post “is they a way to solve it...” more is they a way to solve it easily
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look4livesunsh1ne 2 years ago Posted 2 years ago. Direct link to look4livesunsh1ne's post “I think that if you want ...” more I think that if you want to see the easiest way. It's to find the square root of it; (ex 155). And find its perfect square root and find the closest whole number below and above it
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1203197 7 years ago Posted 7 years ago. Direct link to 1203197's post “Does it make sense if I f...” more Does it make sense if I for example have the equation-14.5 (looking for the squared root). And i see what perfect square is less than 14 and what perf square is greater than 14 than order them like you do from least to greater. I than take those 2 other numbers and combine there squared roots.EX)
1- 3=9 ?=14 4=16
2- 3.9 approximately =‘s. 14.5
3- I used as squared and my answer might not be exact but in estimation it’s not always perfect.
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gordonmalik2020 2 years ago Posted 2 years ago. Direct link to gordonmalik2020's post “yes it is” more yes it is
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Elle B. 2 years ago Posted 2 years ago. Direct link to Elle B.'s post “is there a simpler way to...” more is there a simpler way to do this
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Emily Kayy 2 years ago Posted 2 years ago. Direct link to Emily Kayy's post “Yeah, if there is a root ...” more Yeah, if there is a root and you square it, then the square and the root cancel out.
As for finding the consecutive integers, use the process of elimination. and square
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🧠 2 years ago Posted 2 years ago. Direct link to 🧠's post “I wonder why in school th...” more I wonder why in school they do not teach how to do it without a calculator
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mohmdsaleh800 9 months ago Posted 9 months ago. Direct link to mohmdsaleh800's post “Hi it's very good explan...” more Hi
it's very good explanation, but there little bet confusion with greater and less then sign they all look the same while they supposed be in different direction as we know greater sign > and less then sign <
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OceanBreeze2010 9 months ago Posted 9 months ago. Direct link to OceanBreeze2010's post “So mohmdsaleh800, if you ...” more So mohmdsaleh800, if you read the equation, using both less than signs works when defining a parameter for the certain number.
For example: 4 <x < 8
This means that x is greater than for but it is less than 8, so all possible values could be 5, 6, 7 and any decimals between that. Using the greater than value is not needed when writing these equations.
Hopefully this helps!
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LexaBudexa 2 years ago Posted 2 years ago. Direct link to LexaBudexa's post “why does it say 30 square...” more why does it say 30 squared is the square root of 30 in Step 1 of the first example?
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A.J. a year ago Posted a year ago. Direct link to A.J.'s post “because you are squaring ...” more because you are squaring each number so the order is still the same, but now you can easily compare them.
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Ryan!) a year ago Posted a year ago. Direct link to Ryan!)'s post “you cant use the calculat...” more you cant use the calculator to square number, at least for me it doesn't work
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FierceDragonEm🐉 a year ago Posted a year ago. Direct link to FierceDragonEm🐉's post “Actullay, you can use a c...” more Actullay, you can use a calculator. a basic calculator can only do the basic functions. But, a graphing calculator has a whole lot more stuff, including things you will almost never even need. (it has buttons that make the buttons it already has be able to do multiple different things too)
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lucadoc2029 a year ago Posted a year ago. Direct link to lucadoc2029's post “this is hard, can u give ...” more this is hard, can u give me more examples
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AlexisP 23 days ago Posted 23 days ago. Direct link to AlexisP's post “Is 67 a number?” more Is 67 a number?
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Jerry Nilsson 22 days ago Posted 22 days ago. Direct link to Jerry Nilsson's post “Yup, 67 is a number. It s...” more Yup, 67 is a number. It sits exactly 67 units to the right of zero on the number line.
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New: More links in the Sidebar, including to the Lexeis project; enhancements to morphology.
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Parsed as a form of: λόγος,
See λόγος in Μορφώ
λόγος
Short Definition
λόγος, λόγος, λόγου, ὁ, word, speech, story
Frequency
λόγος is the 57th most frequent word LSJBrill-MontanariBailly 2024Grieks NederlandsPapeCunliffe HomerSlater PindarAbbott-Smith NTLMPGAutenrieth HomerMiddle LiddellExamples from the corpus LSJBrill-MontanariBailly 2024Grieks NederlandsPapeCunliffe HomerSlater PindarAbbott-Smith NTLMPGAutenrieth HomerMiddle LiddellExamples from the corpus
λόγος, ὁ, verbal noun of λέγω (B), with senses corresponding to λέγω (B) II and III (on the various senses of the word v. Theo Sm. pp. 72,73 H., An.Ox. 4.327): common in all periods in Prose and Verse, exc. Epic, in which it is found in signf. derived from λέγω (B) III, cf.infr. VI. 1 a:
I computation, reckoning (cf. λέγω (B) II).
1 account of money handled, σανίδες εἰς ἃς τὸν λ. ἀναγράφομεν IG 1(2).374.191; ἐδίδοσαν τὸν λ. ib. 232.2; λ. δώσεις τῶν μετεχείρισας χρημάτων Hdt. 3.142, cf. 143; οὔτε χρήματα διαχειρίσας τῆς πόλεως δίδωμι λ. αὐτῶν οὔτε ἀρχὴν ἄρξας οὐδεμίαν εὐθύνας ὑπέχω νῦν αὐτῆς Lys. 24.26; λ. ἀπενεγκεῖν Arist. Ath. 54.1; ἐν ταῖς εὐθύναις τοῦ τοιούτου λ. ὑπεχέτω Pl. Lg. 774b; τὸν τῶν χρημάτων λ. παρὰ τούτων λαμβάνειν D. 8.47; ἀδικήματα εἰς ἀργυρίου λ. ἀνήκοντα Din. 1.60; συνᾶραι λόγον μετά τινος settle accounts with, Ev.Matt. 18.23, etc.; δεύτεροι λ. a second audit, Cod.Just. 1.4.26.1; ὁ τραπεζιτικὸς λ. banking account, Theo Sm. p.73 H.: metaph., οὐκ ἂν πριαίμην οὐδενὸς λ. βροτόν S. Aj. 477.
b public accounts, i. e. branch of treasury, ἴδιος λ., in Egypt, OGI 188.2, 189.3, 669.38; also as title of treasurer, ib. 408.4, Str. 17.1.12; ὁ ἐπὶ τῶν λ. IPE 2.29 A (Panticapaeum); δημόσιος λ., = Lat. fiscus, OGI 669.21 (Egypt, i A.D.), etc. (but later, = aerarium, Cod.Just. 1.5.15); also Καίσαρος λ. OGI 669.30; κυριακὸς λ. ib. 18.
2 generally, account, reckoning, μὴ φῦναι τὸν ἅπαντα νικᾷ λ. excels the whole account, i.e. is best of all, S. OC 1225 (lyr.); δόντας λ. τῶν ἐποίησαν accounting for, i.e. paying the penalty for their doings, Hdt. 8.100; λ. αἰτεῖν Pl. Plt. 285e; λ. δοῦναι καὶ δέξασθαι Id. Prt. 336c, al.; λαμβάνειν λ. καὶ ἐλέγχειν Id. Men. 75d; παρασχεῖν τῶν εἰρημένων λ. Id. R. 344d; λ. ἀπαιτεῖν D. 30.15, cf. Arist. EN 1104a3; λ. ὑπέχειν, δοῦναι, D. 19.95; λ. ἐγγράψαι Id. 24.199, al.; λ. ἀποφέρειν τῇ πόλει Aeschin. 3.22, cf. Eu. Luc. 16.2, Ep.Hebr. 13.17; τὸ παράδοξον τῶν συμβεβηκότων ὑπὸ λόγον ἄγειν Plb. 15.34.2; λ. ἡ ἐπιστήμη, πολλὰ δὲ ὁ λ. the account is manifold, Plot. 6.9.4; ἔχων λόγον τοῦ διὰ τί an account of the cause, Arist. APo. 74b27; ἐς λ. τινός on account of, ἐς χρημάτων λ. Th. 3.46, cf. Plb. 5.89.6, LXX 2 Ma 1.14, JRS 18.152 (Jerash); λόγῳ c. gen., by way of, Cod.Just. 3.2.5. al.; κατὰ λόγον τοῦ μεγέθους if we take into account his size, Arist. HA 517b27; πρὸς ὃν ἡμῖν ὁ λ. Ep.Hebr. 4.13, cf. D.Chr. 31.123.
3 measure, tale (cf. infr. II.1), θάλασσα . . μετρέεται ἐς τὸν αὐτὸν λ. ὁκοῖος πρόσθεν Heraclit. 31; ψυχῆς ἐστι λ. ἑαυτὸν αὔξων Id. 115; ἐς τούτου (sc. γήραος) λ. οὐ πολλοί τινες ἀπικνέονται to the point of old age, Hdt. 3.99, cf. 7.9.β; ὁ ξύμπας λ. the full tale, Th. 7.56, cf. Ep.Phil. 4.15; κοινῷ λ. νομίσαντα common measure, Pl. Lg. 746e; sum, total of expenditure, IG 42(1).103.151 (Epid., iv B.C.); ὁ τῆς οὐσίας λ., = Lat. patrimonii modus, Cod.Just. 1.5.12.20.
4 esteem, consideration, value put on a person or thing (cf. infr. VI. 2 d), οὗ πλείων λ. ἢ τῶν ἄλλων who is of more worth than all the rest, Heraclit. 39; βροτῶν λ. οὐκ ἔσχεν οὐδένʼ A. Pr. 233; οὐ σμικροῦ λ. S. OC 1163: freq. in Hdt., Μαρδονίου λ. οὐδεὶς γίνεται 8.102; τῶν ἦν ἐλάχιστος ἀπολλυμένων λ. 4.135, cf. E. Fr. 94; περὶ ἐμοῦ οὐδεὶς λ. Ar. Ra. 87; λόγου οὐδενὸς γίνεσθαι πρός τινος to be of no account, repute with.., Hdt. 1.120, cf. 4.138; λόγου ποιήσασθαί τινα make one of account, Id. 1.33; ἐλαχίστου, πλείστου λ. εἶναι, to be highly, lowly esteemed, Id. 1.143, 3.146; but also λόγον τινὸς ποιεῖσθαι, like Lat. rationem habere alicujus, make account of, set a value on, Democr. 187, etc.: usu. in neg. statements, οὐδένα λ. ποιήσασθαί τινος Hdt. 1.4, cf. 13, Plb. 21.14.9, etc.; λ. ἔχειν Hdt. 1.62, 115; λ. ἴσχειν περί τινος Pl. Ti. 87c; λ. ἔχειν περὶ τοὺς ποιητάς Lycurg. 107; λ. ἔχειν τινός D. 18.199, Arist. EN 1102b32, Plu. Phil. 18 (but also, have the reputation of . ., v. infr. VI. 2 e); ἐν οὐδενὶ λ. ποιήσασθαί τι Hdt. 3.50; ἐν οὐδενὶ λ. ἀπώλοντο without regard, Id. 9.70; ἐν σμικρῷ λ. εἶναι Pl. R. 550a; ὑμεῖς οὔτʼ ἐν λ. οὔτʼ ἐν ἀριθμῷ Orac. ap. Sch. Theoc. 14.48; ἐν ἀνδρῶν λ. [εἶναι] to be reckoned, count as a man, Hdt. 3.120; ἐν ἰδιώτεω λόγῳ καὶ ἀτίμου reckoned as . ., Eus.Mynd. Fr. 59; σεμνὸς εἰς ἀρετῆς λ. καὶ δόξης D. 19.142.
II relation, correspondence, proportion,
1 generally, ὑπερτερίης λ. relation (of gold to lead), Thgn. 418 = 1164; πρὸς λόγον τοῦ σήματος A. Th. 519; κατὰ λόγον προβαίνοντες τιμῶσι in inverse ratio, Hdt. 1.134, cf. 7.36; κατὰ λ. τῆς ἀποφορῆς Id. 2.109; τἄλλα κατὰ λ. in like fashion, Hp. VM 16, Prog. 17: c. gen., κατὰ λ. τῶν πρόσθεν ib. 24; κατὰ λ. τῶν ἡμερῶν Ar. Nu. 619; κατὰ λ. τῆς δυνάμεως X. Cyr. 8.6.11; ἐλάττω ἢ κατὰ λ. Arist. HA 508a2, cf. PA 671a18; ἐκ ταύτης ἐγένετο ἐκείνη κατὰ λ. Id. Pol. 1257a31; cf. εὔλογος: sts. with ὁ αὐτός added, κατὰ τὸν αὐτὸν λ. τῷ τείχεϊ in fashion like to . ., Hdt. 1.186; περὶ τῶν νόσων ὁ αὐτὸς λ. analogously, Pl. Tht. 158d, cf. Prm. 136b, al.; εἰς τὸν αὐτὸν λ. similarly, Id. R. 353d; κατὰ τὸν αὐτὸν λ. in the same ratio, IG 1(2).76.8; by parity of reasoning, Pl. Cra. 393c, R. 610a, al.; ἀνὰ λόγον τινός, τινί, Id. Ti. 29c, Alc. 2.145d; τοῦτον ἔχει τὸν λ. πρὸς . . ὃν ἡ παιδεία πρὸς τὴν ἀρετήν is related to . . as . ., Procl. in Euc. p.20 F., al.
2 Math., ratio, proportion (ὁ κατʼ ἀνάλογον λ., λ. τῆς ἀναλογίας, Theo Sm. p.73 H.), Pythag. 2; ἰσότης λόγων Arist. EN 113a31; λ. ἐστὶ δύο μεγεθῶν ἡ κατὰ πηλικότητα ποιὰ σχέσις Euc. 5 Def. 3; τῶν ἁρμονιῶν τοὺς λ. Arist. Metaph. 985b32, cf. 1092b14; λόγοι ἀριθμῶν numerical ratios, Aristox. Harm. p.32 M.; τοὺς φθόγγους ἀναγκαῖον ἐν ἀριθμοῦ λ. λέγεσθαι πρὸς ἀλλήλους to be expressed in numerical ratios, Euc. Sect.Can. Proëm.: in Metre, ratio between arsis and thesis, by which the rhythm is defined, Aristox. Harm. p.34 M.; ἐὰν ᾖ ἰσχυροτέρα τοῦ αἰσθητηρίου ἡ κίνησις, λύεται ὁ λ. Arist. de An. 424a31; ἀνὰ λόγον analogically, Archyt. 2; ἀνὰ λ. μερισθεῖσα [ἡ ψυχή] proportionally, Pl. Ti. 37a; so κατὰ λ. Men. 319.6; πρὸς λόγον in proportion, Plb. 6.30.3, 9.15.3 (but πρὸς λόγον ἐπὶ στενὸν συνάγεται narrows uniformly, Sor. 1.9, cf. Diocl. Fr. 171); ἐπὶ λόγον IG 5(1).1428 (Messene).
3 Gramm., analogy, rule, τῷ λ. τῶν μετοχικῶν, τῆς συγκοπῆς, by the rule of the participles, of syncope, Choerob. in Theod. 1.75 Gaisf., 1.377 H.; εἰπέ μοι τὸν λ. τοῦ Αἴας Αἴαντος, τουτέστι τὸν κανόνα An.Ox. 4.328.
III explanation,
1 plea, pretext, ground, ἐκ τίνος λ.; A. Ch. 515; ἐξ οὐδενὸς λ. S. Ph. 731; ἀπὸ παντὸς λ. Id. OC 762; χὠ λ. καλὸς προσῆν Id. Ph. 352; σὺν ἀφανεῖ λ. Id. OT 657 (lyr., v.l. λόγων) ; ἐν ἀφανεῖ λ. Antipho 5.59; ἐπὶ τοιούτῳ λ. Hdt. 6.124; κατὰ τίνα λ.; on what ground? Pl. R. 366b; οὐδὲ πρὸς ἕνα λ. to no purpose, Id. Prt. 343d; ἐπὶ τίνι λ.; for what reason? X. HG 2.2.19; τὸν λ. τοῦτον this ground of complaint, Aeschin. 3.228; τίνι δικαίῳ λ.; what just cause is there? Pl. Grg. 512c; τίνι λ.; on what account? Act. Ap. 10.29; κατὰ λόγον ἂν ἠνεσχόμην ὑμῶν reason would that . ., ib. 18.14; λ. ἔχειν, with personal subject, εἶχον ἄν τινα λ. I (i.e. my conduct) would have admitted of an explanation, Pl. Ap. 31b; τὸν ὀρθὸν λ. the true explanation, ib. 34b.
b plea, case, in Law or argument (cf. VIII. I), τὸν ἥττω λ. κρείττω ποιεῖν to make the weaker case prevail, ib. 18b, al., Arist. Rh. 1402a24, cf. Ar. Nu. 1042 (pl.); personified, ib. 886, al.; ἀμύνεις τῷ τῆς ἡδονῆς λ. Pl. Phlb. 38a; ἀνοίσεις τοὺς λ. αὐτῶν πρὸς τὸν θεόν LXX Ex. 18.19; ἔχειν λ. πρός τινα to have a case, ground of action against . ., Act.Ap. 19.38.
2 statement of a theory, argument, οὐκ ἐμεῦ ἀλλὰ τοῦ λ. ἀκούσαντας prob. in Heraclit. 50; λόγον ἠδὲ νόημα ἀμφὶς ἀληθείης discourse and reflection on reality, Parm. 8.50; δηλοῖ οὗτος ὁ λ. ὅτι . . Democr. 7; οὐκ ἔχει λόγον it is not arguable, i.e. reasonable, S. El. 466, Pl. Phd. 62d, etc.; ἔχει λ. D. 44.32; οὐδεὶς αὐτὰ καταβαλεῖ λ. E. Ba. 202; δίκασον . . τὸν λ. ἀκούσας Pl. Lg. 696b; personified, φησὶ οὗτος ὁ λ. ib. 714d, cf. Sph. 238b, Phlb. 50a; ὡς ὁ λ. (sc. λέγει) Arist. EN 1115b12; ὡς ὁ λ. ὁ ὀρθὸς λέγει ib. 1138b20, cf. 29; ὁ λ. θέλει προσβιβάζειν Phld. Rh. 1.41, cf.1.19 S.; οὐ γὰρ ἂν ἀκούσειε λόγου ἀποτρέποντος Arist. EN 1179b27; λ. καθαίρων Aristo Stoic. 1.88; λόγου τυγχάνειν to be explained, Phld. Mus. p.77 K.; ὁ τὸν λ. μου ἀκούων my teaching, Ev.Jo. 5.24; ὁ προφητικὸς λ., collect., of VT prophecy, 2 Ep.Pet. 1.19: pl., ὁκόσων λόγους ἤκουσα Heraclit. 108; οὐκ ἐπίθετο τοῖς ἐμοῖς λ. Ar. Nu. 73; of arguments leading to a conclusion (ὁ λ.), Pl. Cri. 46b; τὰ Ἀναξαγόρου βιβλία γέμει τούτων τῶν λ. Id. Ap. 26d; λ. ἀπὸ τῶν ἀρχῶν, ἐπὶ τὰς ἀρχάς, Arist. EN 1095a31; συλλογισμός ἐστι λ. ἐν ᾧ τεθέντων τινῶν κτλ. Id. APr. 24b18; λ. ἀντίτυπός τε καὶ ἄπορος, of a self-contradictory theory, Plot. 6.8.7.
b ὁ περὶ θεῶν λ., title of a discourse by Protagoras, D.L. 9.54; ὁ Ἀχιλλεὺς λ., name of an argument, ib. 23; ὁ αὐξόμενος λ. Plu. Vind. 2.559b; καταβάλλοντες (sc. λόγοι), title of work by Protagoras, S.E. M. 7.60; λ. σοφιστικοί Arist. SE 165a34, al.; οἱ μαθηματικοὶ λ. Id. Rh. 1417a19, etc.; οἱ ἐξωτερικοὶ λ., current outside the Lyceum, Id. Ph. 217b31, al.; Δισσοὶ λ., title of a philosophical treatise (= Dialex.); Λ. καὶ Λογίνα, name of play of Epicharmus, quibble, argument, personified, Ath. 8.338d.
c in Logic, proposition, whether as premiss or conclusion, πρότασίς ἐστι λ. καταφατικὸς ἢ ἀποφατικός τινος κατά τινος Arist. APr. 24a16.
d rule, principle, law, as embodying the result of λογισμός, Pi. O. 2.22, P. 1.35, N. 4.31; πείθεσθαι τῷ λ. ὃς ἄν μοι λογιζομένῳ βέλτιστος φαίνηται Pl. Cri. 46b, cf. c; ἡδονὰς τοῖς ὀρθοῖς λ. ἑπομένας obeying right principles, Id. Lg. 696c; προαιρέσεως [ἀρχὴ] ὄρεξις καὶ λ. ὁ ἕνεκά τινος principle directed to an end, Arist. EN 1139a32; of the final cause, ἀρχὴ ὁ λ. ἔν τε τοῖς κατὰ τέχνην καὶ ἐν τοῖς φύσει συνεστηκόσιν Id. PA 639b15; ἀποδιδόασι τοὺς λ. καὶ τὰς αἰτίας οὗ ποιοῦσι ἑκάστου ib. 18; [τέχνη] ἕξις μετὰ λ. ἀληθοῦς ποιητική Id. EN 1140a10; ὀρθὸς λ. true principle, right rule, ib. 1144b27, 1147b3, al.; κατὰ λόγον by rule, consistently, ὁ κατὰ λ. ζῶν Pl. Lg. 689d, cf. Ti. 89d; τὸ κατὰ λ. ζῆν, opp. κατὰ πάθος, Arist. EN 1169a5; κατὰ λ. προχωρεῖν according to plan, Plb. 1.20.3.
3 law, rule of conduct, ᾧ μάλιστα διηνεκῶς ὁμιλοῦσι λόγῳ Heraclit. 72; πολλοὶ λόγον μὴ μαθόντες ζῶσι κατὰ λόγον Democr. 53; δεῖ ὑπάρχειν τὸν λ. τὸν καθόλου τοῖς ἄρχουσιν universal principle, Arist. Pol. 1286a17; ὁ νόμος . . λ. ὢν ἀπό τινος φρονήσεως καὶ νοῦ Id. EN 1180a21; ὁ νόμος . . ἔμψυχος ὢν ἑαυτῷ λ. conscience, Plu. Ad. princ. ind. 2.780c; τὸν λ. πρόχειρον ἔχειν precept, Phld. Piet. 30, cf. 102; ὁ προστακτικὸς τῶν ποιητέων ἢ μὴ λ. κοινός M.Ant. 4.4.
4 thesis, hypothesis, provisional ground, ὡς ἂν εἰ λέγοι λόγον maintain a thesis, Pl. Prt. 344b; ὑποθέμενος ἑκάστοτε λ. provisionally assuming a proposition, Id. Phd. 100a; τὸν τῆς ὁμοιότητος λ. hypothesis of equivalence, Arist. Cael. 296a20.
5 reason, ground, πάντων γινομένων κατὰ τὸν λ. τόνδε Heraclit. 1; οὕτω βαθὺν λ. ἔχει Id. 45; ἐκ λόγου, opp. μάτην, Leucipp. 2; μέγιστον σημεῖον οὗτος ὁ λ. Meliss. 8; [ἐμπειρία] οὐκ ἔχει λ. οὐδένα ὧν προσφέρει has no grounds for . ., Pl. Grg. 465a; μετὰ λόγου τε καὶ ἐπιστήμης θείας Id. Sph. 265c; ἡ μετα λόγου ἀληθὴς δόξα ( ἐπιστήμη) Id. Tht. 201c; λόγον ζητοῦσιν ὧν οὐκ ἔστι λ. proof, Arist. Metaph. 1011a12; οἱ ἁπάντων ζητοῦντες λ. ἀναιροῦσι λ. Thphr. Metaph. 26.
6 formula (wider than definition, but freq. equivalent thereto), term expressing reason, λ. τῆς πολιτείας Pl. R. 497c; ψυχῆς οὐσία τε καὶ λ. essential definition, Id. Phdr. 245e; ὁ τοῦ δικαίου λ. Id. R. 343a; τὸν λ. τῆς οὐσίας ib. 534b, cf. Phd. 78d; τὰς πολλὰς ἐπιστήμας ἑνὶ λ. προσειπεῖν Id. Tht. 148d; ὁ τῆς οἰκοδομήσεως λ. ἔχει τὸν τῆς οἰκίας Arist. PA 646b3; τεθείη ἂν ἴδιον ὄνομα καθʼ ἕκαστον τῶν λ. Id. Metaph. 1006b5, cf. 1035b4; πᾶς ὁρισμὸς λ. τίς ἐστι Id. Top. 102a5; ἐπὶ τῶν σχημάτων λ. κοινός generic definition, Id. de An. 414b23; ἀκριβέστατος λ. specific definition, Id. Pol. 1276b24; πηγῆς λ. ἔχον Ph. 2.477; τὸ ᾠὸν οὔτε ἀρχῆς ἔχει λ. fulfils the function of . ., Plu. QConv. 2.637d; λ. τῆς μίξεως formula, i.e. ratio (cf. supr. II) of combination, Arist. PA 642a22, cf. Metaph. 993a17.
7 reason, law exhibited in the world-process, κατὰ λόγον by law, κόσμῳ πάντα καὶ κατὰ λ. ἔχοντα Pl. R. 500c; κατ τὸν <αὐτὸν αὖ> λ. by the same law, Epich. 170.18; ψυχῆς τὸ πᾶν τόδε διοικούσης κατὰ λ. Plot. 2.3.13; esp. in Stoic Philos., the divine order, τὸν τοῦ παντὸς λ. ὃν ἔνιοι εἱμαρμένην καλοῦσιν Zeno Stoic. 1.24; τὸ ποιοῦν τὸν ἐν [τῇ ὕλῃ] λ. τὸν θεόν ibid., cf. 42; ὁ τοῦ κόσμου λ. Chrysipp.Stoic. 2.264; λόγος, = φύσει νόμος, Stoic. 2.169; κατὰ τὸν κοινὸν θεοῖς καὶ ἀνθρώποις λ. M.Ant. 7.53; ὁ ὀρθὸς λ. διὰ πάντων ἐρχόμενος Chrysipp.Stoic. 3.4: so in Plot., τὴν φύσιν εἶναι λόγον, ὃς ποιεῖ λ. ἄλλον γέννημα αὑτοῦ 3.8.2.
b σπερματικὸς λ. generative principle in organisms, ὁ θεὸς σπ. λ. τοῦ κόσμου Zeno Stoic. 1.28: usu. in pl., Stoic. 2.205,314, al.; γίνεται τὰ ἐν τῷ παντὶ οὐ κατὰ σπερματικούς, ἀλλὰ κατὰ λ. περιληπτικούς Plot. 3.1.7, cf. 4.4.39: so without σπερματικός, ὥσπερ τινὲς λ. τῶν μερῶν Cleanth.Stoic. 1.111; οἱ λ. τῶν ὅλων Ph. 1.9.
c in Neo-Platonic Philos., of regulative and formative forces, derived from the intelligible and operative in the sensible universe, ὄντων μειζόνων λ. καὶ θεωρούντων αὑτοὺς ἐγὼ γεγέννημαι Plot. 3.8.4; οἱ ἐν σπέρματι λ. πλάττουσι . . τὰ ζῷα οἷον μικρούς τινας κόσμους Id. 4.3.10, cf. 3.2.16, 3.5.7; opp. ὅρος, Id. 6.7.4; ἀφανεῖς λ. τῆς φύσεως Procl. in R. 1.18 K.; τεχνικοὶ λ. ib. 142 K., al.
IV inward debate of the soul (cf. λ. ὃν αὐτὴ πρὸς αὑτὴν ἡ ψυχὴ διεξέρχεται Pl. Tht. 189e (διάλογος in Sph. 263e); ὁ ἐν τῇ ψυχῇ, ὁ ἔσω λ. (opp. ὁ ἔξω λ.), Arist. APo. 76b25, 27; ὁ ἐνδιάθετος, opp. ὁ προφορικὸς λ., Stoic. 2.43, Ph. 2.154),
1 thinking, reasoning, τοῦ λ. ἐόντος ξυνοῦ, opp. ἰδία φρόνησις, Heraclit. 2; κρῖναι δὲ λόγῳ . . ἔλεγχον test by reflection, Parm. 1.36; reflection, deliberation (cf. VI.3), ἐδίδου λόγον ἑωυτῷ περὶ τῆς ὄψιος Hdt. 1.209, cf. 34, S. OT 583, D. 45.7; μὴ εἰδέναι . . μήτε λόγῳ μήτε ἔργῳ neither by reasoning nor by experience, Anaxag. 7; ἃ δὴ λόγῳ μὲν καὶ διανοίᾳ ληπτά, ὄψει δʼ οὔ Pl. R. 529d, cf. Prm. 135e; ὁ λ. ἢ ἡ αἴσθησις Arist. EN 1149a35, al.; αὐτῷ μόνον τῷ λ. πιστεύειν (opp. αἰσθήσεις), of Parmenides and his school, Aristocl. ap. Eus. PE 14.17: hence λόγῳ or τῷ λ. in idea, in thought, τῷ λ. τέμνειν Pl. R. 525e; τῷ λ. δύο ἐστίν, ἀχώριστα πεφυκότα two in idea, though indistinguishable in fact, Arist. EN 1102a30, cf. GC 320b14, al.; λόγῳ θεωρητά mentally conceived, opp. sensibly perceived, Placit. 1.3.5, cf. Demetr.Lac. Herc. 1055.20; τοὺς λ. θεωρητοὺς χρόνους Epicur. Ep. 1p.19U.; διὰ λόγου θ. χ. ib. p.10 U.; λόγῳ καταληπτός Phld. Po. 5.20, etc.; ὁ λ. οὕτω αἱρέει analogy proves, Hdt. 2.33; ὁ λ. or λ. αἱρέει reasoning convinces, Id. 3.45, 6.124, cf. Pl. Cri. 48c (but, our argument shows, Lg. 663d): also c. acc. pers., χρᾶται ὅ τι μιν λ. αἱρέει as the whim took him, Hdt. 1.132; ἢν μὴ ἡμέας λ. αἱρῇ unless we see fit, Id. 4.127, cf. Pl. R. 607b; later ὁ αἱρῶν λ. ordaining reason, Zeno Stoic. 1.50, M.Ant. 2.5, cf. 4.24, Arr. Epict. 2.2.20, etc.: coupled or contrasted with other functions, καθʼ ὕπνον ἐπειδὴ λόγου καὶ φρονήσεως οὐ μετεῖχε since reason and understanding are in abeyance, Pl. Ti. 71d; μετὰ λόγου τε καὶ ἐπιστήμης, opp. αἰτία αὐτομάτη, of Natureʼs processes of production, Id. Sph. 265c; τὸ μὲν δὴ νοήσει μετὰ λόγου περιληπτόν embraced by thought with reflection, opp. μετʼ αἰσθήσεως ἀλόγου, Id. Ti. 28a; τὸ μὲν ἀεὶ μετʼ ἀληθοῦς λ., opp. τὸ δὲ ἄλογον, ib. 51e, cf. 70d, al.; λ. ἔχων ἑπόμενον τῷ νοεῖν Id. Phlb. 62a; ἐπιστήμη ἐνοῦσα καὶ ὀρθὸς λ. scientific knowledge and right process of thought, Id. Phd. 73a; πᾶς λ. καὶ πᾶσα ἐπιστήμη τῶν καθόλου Arist. Metaph. 1059b26; τὸ λόγον ἔχον Id. EN 1102b15, 1138b9, al.: in sg. and pl., contrasted by Pl. and Arist. as theory, abstract reasoning with outward experience, sts. with depreciatory emphasis on the former, εἰς τοὺς λ. καταφυγόντα Pl. Phd. 99e; τὸν ἐν λόγοις σκοπούμενον τὰ ὄντα, opp. τὸν ἐν ἔργοις (realities), ib. 100a; τῇ αἰσθήσει μᾶλλον τῶν λ. πιστευτέον Arist. GA 760b31; γνωριμώτερα κατὰ τὸν λ., opp. κατὰ τὴν αἴσθησιν, Id. Ph. 189a4; ἐκ τῶν λ. δῆλον, opp. ἐκ τῆς ἐπαγωγῆς, Id. Mete. 378b20; ἡ τῶν λ. πίστις, opp. ἐκ τῶν ἔργων φανερόν, Id. Pol. 1326a29; ἡ πίστις οὐ μόνον ἐπὶ τῆς αἰσθήσεως ἀλλὰ καὶ ἐπὶ τοῦ λ. Id. Ph. 262a19; μαρτυρεῖ τὰ γιγνόμενα τοῖς λ. Id. Pol. 1334a6; ὁ μὲν λ. τοῦ καθόλου, ἡ δὲ αἴσθησις τοῦ κατὰ μέρος explanation, opp. perception, Id. Ph. 189a7; ἔσονται τοῖς λ. αἱ πράξεις ἀκόλουθοι theory, opp. practice, Epicur. Sent. 25; in Logic, of discursive reasoning, opp. intuition, Arist. EN 1142a26, 1143b1; reasoning in general, ib. 1149a26; πᾶς λ. καὶ πᾶσα ἀπόδειξις all reasoning and demonstration, Id. Metaph. 1063b10; λ. καὶ φρόνησιν Phld. Mus. p.105 K.; ὁ λ. ἢ λογισμός ibid.; τὸ ἰδεῖν οὐκέτι λ., ἀλλὰ μεῖζον λόγου καὶ πρὸ λόγου, of mystical vision, opp. reasoning, Plot. 6.9.10.—Phrases, κατὰ λ. τὸν εἰκότα by probable reasoning, Pl. Ti. 30b; οὔκουν τόν γʼ εἰκότα λ. ἂν ἔχοι Id. Lg. 647d; παρὰ λόγον, opp. κατὰ λ., Arist. Rh.Al. 1429a29, cf. EN 1167b19; cf. παράλογος (but παρὰ λ. unexpectedly, E. Ba. 940).
2 reason as a faculty, ὁ λ. ἀνθρώπους κυβερνᾷ [Epich.] 256; [θυμοειδὲς] τοῦ λ. κατήκοον Pl. Ti. 70a; [θυμὸς] ὑπὸ τοῦ λ. ἀνακληθείς Id. R. 440d; σύμμαχον τῷ λ. τὸν θυμόν ib. b; πειθαρχεῖ τῷ λ. τὸ τοῦ ἐγκρατοῦς Arist. EN 1102b26; ἄλλο τι παρὰ τὸν λ. πεφυκός, ὃ μάχεται τῷ λ. ib. 17; ἐναντίωσις λόγου πρὸς ἐπιθυμίας Plot. 4.7.13(8); οὐ θυμός, οὐκ ἐπιθυμία, οὐδὲ λ. οὐδέ τις νόησις Id. 6.9.11: freq. in Stoic. Philos. of human Reason, opp. φαντασία, Zeno Stoic. 1.39; opp. φύσις, Stoic. 2.206; οὐ σοφία οὐδὲ λ. ἐστὶν ἐν [τοῖς ζῴοις] ibid.; τοῖς ἀλόγοις ζῴοις ὡς λ. ἔχων λ. μὴ ἔχουσι χρῶ M.Ant. 6.23; ὁ λ. κοινὸν πρὸς τοὺς θεούς Arr. Epict. 1.3.3; οἷον [εἰκὼν] λ. ὁ ἐν προφορᾷ λόγου τοῦ ἐν ψυχῇ, οὕτω καὶ αὐτὴ λ. νοῦ Plot. 5.1.3; τὸ τὸν λ. σχεῖν τὴν οἰκείαν ἀρετήν (sc. εὐδαιμονίαν) Procl. in Ti. 3.334 D.; also of the reason which pervades the universe, θεῖος λ. [Epich.] 257; τὸν θεῖον λ. καθʼ Ἡράκλειτον διʼ ἀναπνοῆς σπάσαντες νοεροὶ γινόμεθα S.E. M. 7.129 (cf. infr. x).
b creative reason, ἀδύνατον ἦν λόγον μὴ οὐκ ἐπὶ πάντα ἐλθεῖν Plot. 3.2.14; ἀρχὴ οὖν λ. καὶ πάντα λ. καὶ τὰ γινόμενα κατʼ αὐτόν Id. 3.2.15; οἱ λ. πάντες ψυχαί Id. 3.2.18.
V continuous statement, narrative (whether fact or fiction), oration, etc. (cf. λέγω (B) II.2),
1 fable, Hdt. 1.141; Αἰσώπου λόγοι Pl. Phd. 60d, cf. Arist. Rh. 1393b8; ὁ τοῦ κυνὸς λ. X. Mem. 2.7.13.
2 legend, ἱρὸς λ. Hdt. 2.62, cf. 47, Pi. P. 3.80 (pl.); συνθέντες λ. E. Ba. 297; λ. θεῖος Pl. Phd. 85d; ἱεροὶ λ., of Orphic rhapsodies, Suid. S.V. Ὀρφεύς.
3 tale, story, ἄλλον ἔπειμι λ. Xenoph. 7.1, cf. Th. 1.97, etc.; συνθέτους λ. A. Pr. 686; σπουδὴν λόγου urgent tidings, E. Ba. 663; ἄλλος λ. ‘another story’, Pl. Ap. 34e; ὁμολογούμενος ὁ λ. ἐστίν the story is consistent, Isoc. 3.27: pl., histories, ἐν τοῖσι Ἀσσυρίοισι λ. Hdt. 1.184, cf. 106, 2.99; so in sg., a historical work, Id. 2.123, 6.19, 7.152: also in sg., one section of such a work (like later βίβλος), Id. 2.38, 6.39, cf. VI.3d; so in pl., ἐν τοῖσι Λιβυκοῖσι λ. Id. 2.161, cf. 1.75, 5.22, 7.93, 213; ἐν τῷ πρώτῳ τῶν λ. Id. 5.36; ὁ πρῶτος λ., of St. Lukeʼs gospel, Act.Ap. 1.1: in Pl., opp. μῦθος, as history to legend, Ti. 26e; ποιεῖν μύθους ἀλλʼ οὐ λόγους Phd. 61b, cf. Grg. 523a (but μῦθον λέγειν, opp. λόγῳ (argument) διεξελθεῖν Prt. 320c, cf. 324d); περὶ λόγων καὶ μύθων Arist. Pol. 1336a30; ὁ λ . . . μῦθός ἐστι Ael. NA 4.34.
4 speech, delivered in court, assembly, etc., χρήσομαι τῇ τοῦ λ. τάξει ταύτῃ Aeschin. 3.57, cf. Arist. Rh. 1358a38; δικανικοὶ λ. Id. EN 1181a4; τρία γένη τῶν λ. τῶν ῥητορικῶν, συμβουλευτικόν, δικανικόν, ἐπιδεικτικόν Id. Rh. 1358b7; τῷ γράψαντι τὸν λ. Thphr. Char. 17.8, cf. λογογράφος II; ἐπιτάφιος λ. funeral oration, Pl. Mx. 236b; esp. of the body of a speech, opp. ἐπίλογος, Arist. Rh. 1420b3; opp. προοίμιον, ib. 1415a12; body of a law, opp. proem, Pl. Lg. 723b; spoken, opp. written word, τὸν τοῦ εἰδότος λ. ζῶντα καὶ ἔμψυχον οὗ ὁ γεγραμμένος εἴδωλόν τι Id. Phdr. 276a; ὁ ἐκ τοῦ βιβλίου ῥηθεὶς [λ.] speech read from a roll, ib. 243c; published speech, D.C. 40.54; rarely of the speeches in Tragedy (ῥήσεις), Arist. Po. 1450b6, 9.
VI verbal expression or utterance (cf. λέγω (B) ΙΙΙ), rarely a single word, v. infr. b, never in Gramm. signf. of vocable (ἔπος, λέξις, ὄνομα, ῥῆμα), usu. of a phrase, cf. IX.3 (the only sense found in Ep.).
a pl., without Art., talk, τὸν ἔτερπε λόγοις Il. 15.393; αἱμύλιοι λ. Od. 1.56, h.Merc. 317, Hes. Th. 890, Op. 78, 789, Thgn. 704, A.R. 3.1141; ψευδεῖς Λ., personified, Hes. Th. 229; ἀφροδίσιοι λ. Semon. 7.91; ἀγανοῖσι λ. Pi. P. 4.101; ὄψον δὲ λ. φθονεροῖσιν tales, Id. N. 8.21; σμικροὶ λ. brief words, S. Aj. 1268 (s.v.l.), El. 415; δόκησις ἀγνὼς λόγων bred of talk, Id. OT 681 (lyr.): also in sg., λέγʼ εἴ σοι τῷ λ. τις ἡδονή speak if thou delightest in talking, Id. El. 891.
b sg., expression, phrase, πρὶν εἰπεῖν ἐσθλὸν ἢ κακὸν λ. Id. Ant. 1245, cf. E. Hipp. 514; μυρίας ὡς εἰπεῖν λόγῳ Hdt. 2.37; μακρὸς λ. rigmarole, Simon. 189, Arist. Metaph. 1091a8; λ. ἠρέμα λεχθεὶς διέθηκε τὸ πόρρω a whispered message, Plot. 4.9.3; ἑνὶ λόγῳ to sum up, in brief phrase, Pl. Phdr. 241e, Phd. 65d; concisely, Arist. EN 1103b21 (but also, = ἁπλῶς, περὶ πάντων ἑνὶ λ. Id. GC 325a1): pl., λ. θελκτήριοι magic words, E. Hipp. 478; rarely of single words, λ. εὐσύνθετος οἷον τὸ χρονοτριβεῖν Arist. Rh. 1406a36; οὐκ ἀπεκρίθη αὐτῇ λ. answered her not a word, Ev.Matt. 15.23.
c coupled or contrasted with words expressed or understood signifying act, fact, truth, etc., mostly in a depreciatory sense, λ. ἔργου σκιή Democr. 145; ὥσπερ μικρὸν παῖδα λόγοις μʼ ἀπατᾷς Thgn. 254; λόγῳ, opp. ἔργῳ, Democr. 82, etc.; νηπίοισι οὐ λ. ἀλλὰ ξυμφορὴ διδάσκαλος Id. 76; ἔργῳ κοὐ λόγῳ τεκμαίρομαι A. Pr. 338, cf. S. El. 59, OC 782; λόγῳ μὲν λέγουσι . . ἔργῳ δὲ οὐκ ἀποδεικνῦσι Hdt. 4.8; οὐ λόγων, φασίν, ἡ ἀγορὴ δεῖται, χαλκῶν δέ Herod. 7.49; οὔτε λ. οὔτε ἔργῳ Lys. 9.14; λόγοις, opp. ψήφῳ, Aeschin. 2.33; opp. νόῳ, Hdt. 2.100; οὐ λόγῳ μαθών E. Heracl. 5; ἐκ λόγων, κούφου πράγματος Pl. Lg. 935a; λόγοισι εἰς τὸ πιθανὸν περιπεπεμμένα ib. 886e, cf. Luc. Anach. 19; ἵνα μὴ λ. οἴησθε εἶναι, ἀλλʼ εἰδῆτε τὴν ἀλήθειαν Lycurg. 23, cf. D. 30.34; opp. πρᾶγμα, Arist. Top. 146a4; opp. βία, Id. EN 1179b29, cf. 1180a5; opp. ὄντα, Pl. Phd. 100a; opp. γνῶσις, 2 Ep.Cor. 11.6; λόγῳ in pretence, Hdt. 1.205, Pl. R. 361b, 376d, Ti. 27a, al.; λόγου ἕνεκα merely as a matter of words, ἄλλως ἕνεκα λ. ἐλέγετο Id. Cri. 46d; λόγου χάριν, opp. ὡς ἀληθῶς, Arist. Pol. 1280b8; but also, let us say, for instance, Id. EN 1144a33, Plb. 10.46.4, Phld. Sign. 29, M.Ant. 4.32; λόγου ἕνεκα let us suppose, Pl. Tht. 191c; ἕως λόγου, μέχρι λ., = Lat. verbo tenus, Plb. 10.24.7, Epict. Ench. 16: sts. without depreciatory force, the antithesis or parallelism being verbal (cf. ‘word and deedʼ), λόγῳ τε καὶ σθένει S. OC 68; ἔν τε ἔργῳ καὶ λ. Pl. R. 382e, cf. D.S. 13.101, Ev.Luc. 24.19, Act.Ap. 7.22, Paus. 2.16.2; ὅσα μὲν λόγῳ εἶπον, opp. τὰ ἔργα τῶν πραχθέντων, Th. 1.22.
2 common talk, report, tradition, ὡς λ. ἐν θνητοῖσιν ἔην Batr. 8; λ. ἐκ πατέρων Alc. 71; οὐκ ἔστʼ ἔτυμος λ. οὗτος Stesich. 32; διξὸς λέγεται λ. Hdt. 3.32; λ. ὑπʼ Αἰγυπτίων λεγόμενος Id. 2.47; νέον [λ.] tidings, S. Ant. 1289 (lyr.); τὰ μὲν αὐτοὶ ὡρῶμεν, τὰ δὲ λόγοισι ἐπυνθανόμεθα by hearsay, Hdt. 2.148: also in pl., ἐν γράμμασιν λόγοι κείμενοι traditions, Pl. Lg. 886b.
b rumour, ἐπὶ παντὶ λ. ἐπτοῆσθαι Heraclit. 87; αὐδάεις λ. voice of rumour, B. 14.44; περὶ θεῶν διῆλθεν ὁ λ. ὅτι . . Th. 6.46; λ. παρεῖχεν ὡς . . Plb. 3.89.3; ἐξῆλθεν ὁ λ. οὗτος εῖς τινας ὅτι . . Ev.Jo. 21.23, cf. Act.Ap. 11.22; fiction, Ev.Matt. 28.15.
c mention, notice, description, οὐκ ὕει λόγου ἄξιον οὐδέν worth mentioning, Hdt. 4.28, cf. Plb. 1.24.8, etc.; ἔργα λόγου μέζω beyond expression, Hdt. 2.35; κρεῖσσον λόγου τὸ εἶδος τῆς νόσου beyond description, Th. 2.50; μείζω ἔργα ἢ ὡς τῷ λ. τις ἂν εἴποι D. 6.11.
d the talk one occasions, repute, mostly in good sense, good report, praise, honour (cf. supr. I.4), πολλὰ φέρειν εἴωθε λ. . . πταίσματα Thgn. 1221; λ. ἐσλὸν ἀκοῦσαι Pi. I. 5(4).13; πλέονα . . λ. Ὀδυσσέος ἢ πάθαν Id. N. 7.21; ἵνα λ. σε ἔχῃ πρὸς ἀνθρώπων ἀγαθός Hdt. 7.5, cf. 9.78; Τροίαν . . ἧς ἁπανταχοῦ λ. whose fame, story fills the world, E. IT 517; οὐκ ἂν ἦν λ. σέθεν Id. Med. 541: less freq. in bad sense, evil report, λ. κακόθρους, κακός, S. Aj. 138 (anap.), E. Heracl. 165: pl., λόγους ψιθύρους πλάσσων slanders, S. Aj. 148 (anap.).
e λ. ἐστί, ἔχει, κατέχει, the story goes, c. acc. et inf., ἔστι τις λ. τὰν Ἀρετὰν ναίειν Simon. 58.1, cf. S. El. 417; λ. μὲν ἔστʼ ἀρχαῖος ὡς . . Id. Tr. 1; λ. alone, E. Heracl. 35; ὡς λ. A. Supp. 230, Pl. Phlb. 65c, etc.; λ. ἐστί Hdt. 7.129, 9.26, al.; λ. αἰὲν ἔχει S. OC 1573 (lyr.); ὅσον ὁ λ. κατέχει tradition prevails, Th. 1.10: also with a personal subject in the reverse construction. Κλεισθένης λ. ἔχει τὴν Πυθίην ἀναπεῖσαι has the credit of . ., Hdt. 5.66, cf. Pl. Epin. 987b, 988b; λ. ἔχοντα σοφίας Ep.Col. 2.23, v. supr. I.4.
3 discussion, debate, deliberation, πολλὸς ἦν ἐν τοῖσι λ. Hdt. 8.59; συνελέχθησαν οἱ Μῆδοι ἐς τὠυτὸ καὶ ἐδίδοσαν σφίσι λόγον, λέγοντες περὶ τῶν κατηκόντων Id. 1.97; οἱ Πελασγοὶ ἑωυτοῖσι λόγους ἐδίδοσαν Id. 6.138; πολέμῳ μᾶλλον ἢ λόγοις τὰ ἐγκλήματα διαλύεσθαι Th. 1.140; οἱ περὶ τῆς εἰρήνης λ. Aeschin. 2.74; τοῖς ἔξωθεν λ. πεπλήρωκε τὸν λ. [Plato] has filled his dialogue with extraneous discussions, Arist. Pol. 1264b39; τὸ μῆκος τῶν λ. D.Chr. 7.131; μεταβαίνων ὁ λ. εἰς ταὐτὸν ἀφῖκται our debate, Arist. EN 1097a24; ὁ παρὼν λ. ib. 1104a11; θεῶν ὧν νῦν ὁ λ. ἐστί discussion, Pl. Ap. 26b, cf. Tht. 184a, M.Ant. 8.32; τῷ λ. διελθεῖν, διϊέναι, Pl. Prt. 329c, Grg. 506a, etc.; τὸν λ. διεξελθεῖν conduct the debate, Id. Lg. 893a; ξυνελθεῖν ἐς λόγον confer, Ar. Eq. 1300: freq. in pl., ἐς λόγους συνελθόντες parley, Hdt. 1.82; ἐς λ. ἐλθεῖν τινι have speech with, ib. 86; ἐς λ. ἀπικέσθαι τινί Id. 2.32; διὰ λόγων ἰέναι E. Tr. 916; ἐμαυτῇ διὰ λ. ἀφικόμην Id. Med. 872; ἐς λ. ἄγειν τινά X. HG 4.1.2; κοινωνεῖν λόγων καὶ διανοίας Arist. EN 1170b12.
b right of discussion or speech, ἢ ʼπὶ τῷ πλήθει λ.; S. OC 66; λ. αἰτήσασθαι ask leave to speak, Th. 3.53; λ. διδόναι X. HG 5.2.20; οὐ προυτέθη σφίσιν λ. κατὰ τὸν νόμον ib. 1.7.5; λόγου τυχεῖν D. 18.13, cf. Arist. EN 1095b21, Plb. 18.52.1; οἱ λόγου τοὺς δούλους ἀποστεροῦντες Arist. Pol. 1260b5; δοῦλος πέφυκας, οὐ μέτεστί σοι λόγου Trag.Adesp. 304; διδόντας λ. καὶ δεχομένους ἐν τῷ μέρει Luc. Pisc. 8: hence, time allowed for a speech, ἐν τῷ ἐμῷ λ. And. 1.26, al.; ἐν τῷ ἑαυτοῦ λ. Pl. Ap. 34a; οὐκ ἐλάττω λ. ἀνήλωκε D. 18.9.
c dialogue, as a form of philosophical debate, ἵνα μὴ μαχώμεθα ἐν τοῖς λ. ἐγώ τε καὶ σύ Pl. Cra. 430d; πρὸς ἀλλήλους τοὺς λ. ποιεῖσθαι Id. Prt. 348a: hence, dialogue as a form of literature, οἱ Σωκρατικοὶ λ. Arist. Po. 1447b11, Rh. 1417a20; cf. διάλογος.
d section, division of a dialogue or treatise (cf. V.3), ὁ πρῶτος λ. Pl. Prm. 127d; ὁ πρόσθεν, ὁ παρελθὼν λ., Id. Phlb. 18e, 19b; ἐν τοῖς πρώτοις λ. Arist. PA 682a3; ἐν τοῖς περὶ κινήσεως λ. in the discussion of motion (i. e. Ph. bk. 8), Id. GC 318a4; ἐν τῷ περὶ ἐπαίνου λ. Phld. Rh. 1.219; branch, department, division of a system of philosophy, τὴν φρόνησιν ἐκ τριῶν συνεστηκέναι λ., τῶν φυσικῶν καὶ τῶν ἠθικῶν καὶ τῶν λογικῶν Chrysipp.Stoic. 2.258.
e in pl., literature, letters, Pl. Ax. 365b, Epin. 975d, D.H. Comp. 1, 21 (but, also in pl., treatises, Plu. Aud.poet. 2.16c); οἱ ἐπὶ λόγοις εὐδοκιμώτατοι Hdn. 6.1.4; Λόγοι, personified, AP 9.171 (Pall.).
VII a particular utterance, saying:
1 divine utterance, oracle, Pi. P. 4.59; λ. μαντικοί Pl. Phdr. 275b; οὐ γὰρ ἐμὸν ἐρῶ τὸν λ. Pl. Ap. 20e; ὁ λ. τοῦ θεοῦ Apoc. 1.2, 9.
2 proverb, maxim, saying, Pi. N. 9.6, A. Th. 218; ὧδʼ ἔχει λ. ib. 225; τόνδʼ ἐκαίνισεν λ. ὡς . . Critias 21, cf. Pl. R. 330a, Ev.Jo. 4.37; ὁ παλαιὸς λ. Pl. Phdr. 240c, cf. Smp. 195b, Grg. 499c, Lg. 757a, 1 Ep.Ti. 1.15, Plu. Comm.not. 2.1082e, Luc. Alex. 9, etc.; τὸ τοῦ λόγου δὴ τοῦτο Herod. 2.45, cf. D.Chr. 66.24, Luc. JTr. 3, Alciphr. 3.56, etc.: pl., Arist. EN 1147a21.
3 assertion, opp. oath, S. OC 651; ψιλῷ λ. bare word, opp. μαρτυρία, D. 27.54.
4 express resolution, κοινῷ λ. by common consent, Hdt. 1.141, al.; ἐπὶ λ. τοιῷδε, ἐπʼ ᾧ τε . . on the following terms, Id. 7.158, cf. 9.26; ἐνδέξασθαι τὸν λ. Id. 1.60, cf. 9.5; λ. ἔχοντες πλεονέκτην a greedy proposal, Id. 7.158: freq. in pl., terms, conditions, Id. 9.33, etc.
5 word of command, behest, A. Pr. 17, 40 (both pl.), Pers. 363; ἀνθρώπους πιθανωτέρους ποιεῖν λόγῳ X. Oec. 13.9; ἐξέβαλε τὰ πνεύματα λόγῳ Ev.Matt. 8.16; οἱ δέκα λ. the ten Commandments, LXX Ex. 34.28, Ph. 1.496.
VIII thing spoken of, subject-matter (cf. III.1 b and 2), λ. τοῦτον ἐάσομεν Thgn. 1055; προπεπυσμένος πάντα λ. the whole matter, Hdt. 1.21, cf. 111; τὸν ἐόντα λ. the truth of the matter, ib. 95, 116; μετασχεῖν τοῦ λ. to be in the secret, ib. 127; μηδενὶ ἄλλῳ τὸν λ. τοῦτον εἴπῃς Id. 8.65; τίς ἦν λ.; S. OT 684 ( = πρᾶγμα, 699); περί τινος λ. διελεγόμεθα subject, question, Pl. Prt. 314c; [τὸ προοίμιον] δεῖγμα τοῦ λ. case, Arist. Rh. 1415a12, cf. III.1b; τέλος δὲ παντὸς τοῦ λ. ψηφίζονται the end of the matter was that . ., Aeschin. 3.124; οὐκ ἔστεξε τὸν λ. Plb. 8.12.5; οὐκ ἔστι σοι μερὶς οὐδὲ κλῆρος ἐν τῷ λ. τούτῳ Act.Ap. 8.21; ἱκανὸς αὐτῷ ὁ λ. Pl. Grg. 512c; οὐχ ὑπολείπει [Γοργίαν] ὁ λ. matter for talk, Arist. Rh. 1418a35; μηδένα λ. ὑπολιπεῖν Isoc. 4.146; πρὸς λόγον to the point, apposite, οὐδὲν πρὸς λ. Pl. Phlb. 42e, cf. Prt. 344a; ἐὰν πρὸς λ. τι ᾖ Id. Phlb. 33c; also πρὸς λόγου Id. Grg. 459c (s. v.l.).
2 plot of a narrative or dramatic poem, = μῦθος, Arist. Po. 1455b17, al.
b in Art, subject of a painting, ζωγραφίας λόγοι Philostr. VA 6.10; λ. τῆς γραφῆς Id. Im. 1.25.
3 thing talked of, event, μετὰ τοὺς λ. τούτους LXX 1 Ma. 7.33, cf. Act.Ap. 15.6.
IX expression, utterance, speech regarded formally, τὸ ἀπὸ [ψυχῆς] ῥεῦμα διὰ τοῦ στόματος ἰὸν μετὰ φθόγγου λ., opp. διάνοια, Pl. Sph. 263e; intelligent utterance, opp. φωνή, Arist. Pol. 1253a14; λ. ἐστὶ φωνὴ σημαντικὴ κατὰ συνθήκην Id. Int. 16b26, cf. Diog.Bab.Stoic. 3.213; ὅθεν (from the heart) ὁ λ. ἀναπέμπεται Stoic. 2.228, cf. 244; Protagoras was nicknamed λόγος, Hsch. ap. Sch. Pl. R. 600c, Suid.; λόγου πειθοῖ Democr. 181: in pl., eloquence, Isoc. 3.3, 9.11; τὴν ἐν λόγοις εὐρυθμίαν Epicur. Sent.Pal. 5p.69 v. d. M.; λ. ἀκριβής precise language, Ar. Nu. 130 (pl.), cf. Arist. Rh. 1418b1; τοῦ μὴ ᾀδομένου λ. Pl. R. 398d; ἡδυσμένος λ., of rhythmical language set to music, Arist. Po. 1449b25; ἐν παντὶ λ. in all manner of utterance, 1 Ep.Cor. 1.5; ἐν λόγοις in orations, Arist. Po. 1459a13; λ. γελοῖοι, ἀσχήμονες, ludicrous, improper speech, Id. SE 182b15, Pol. 1336b14.
2 of various modes of expression, esp. artistic and literary, ἔν τε ᾠδαῖς καὶ μύθοις καὶ λόγοις Pl. Lg. 664a; ἐν λόγῳ καὶ ἐν ᾠδαῖς X. Cyr. 1.4.25, cf. Pl. Lg. 835b; prose, opp. ποίησις, Id. R. 390a; opp. ψιλομετρία, Arist. Po. 1448a11; opp. ἔμμετρα, ib. 1450b15 (pl.); τῷ λ. τοῦτο τῶν μέτρων (sc. τὸ ἰαμβεῖον) ὁμοιότατον εἶναι Id. Rh. 1404a31; in full, ψιλοὶ λ. prose, ib. b33 (but ψιλοὶ λ., = arguments without diagrams, Pl. Tht. 165a); λ. πεζοί, opp. ποιητική, D.H. Comp. 6; opp. ποιήματα, ib. 15; κοινὰ καὶ ποιημάτων καὶ λόγων Phld. Po. 5.7; πεζὸς λ. ib. 27, al.
b of the constituents of lyric or dramatic poetry, words, τὸ μέλος ἐκ τριῶν . . λόγου τε καὶ ἁρμονίας καὶ ῥυθμοῦ Pl. R. 398d; opp. πρᾶξις, Arist. Po. 1454a18; dramatic dialogue, opp. τὰ τοῦ χοροῦ, ib. 1449a17.
3 Gramm., phrase, complex term, opp. ὄνομα, Id. SE 165a13; λ. ὀνοματώδης noun-phrase, Id. APo. 93b30, cf. Rh. 1407b27; expression, D.H. Th. 2, Demetr. Eloc. 92.
b sentence, complete statement, "ἄνθρωπος μανθάνει λόγον εἶναί φῃς . . ἐλάχιστόν τε καὶ πρῶτον Pl. Sph. 262c; λ. αὐτοτελής A.D. Synt. 3.6, D.T. 634.1; ῥηθῆναι λόγῳ to be expressed in a sentence, Pl. Tht. 202b; λ. ἔχειν to be capable of being so expressed, ib. 201e, cf. Arist. Rh. 1404b26.
c language, τὰ τοῦ λ. μέρη parts of speech, Chrysipp.Stoic. 2.31, S.E. M. 9.350, etc.; τὰ μόρια τοῦ λ. D.H. Comp. 6; μέρος λ. D.T. 633.26, A.D. Pron. 4.6, al. (but ἓν μέρος <τοῦ cod.> λόγου one word, Id. Synt. 340.10, cf. 334.22); περὶ τῶν στοιχείων τοῦ λ., title of work by Chrysippus.
X the Word or Wisdom of God, personified as his agent in creation and world-government, ὁ παντοδύναμός σου λ. LXX Wi. 18.15; ὁ ἐκ νοὸς φωτεινὸς λ. υἱὸς θεοῦ Corp.Herm. 1.6, cf. Plu. Isid. 2.376c; λ. θεοῦ διʼ οὗ κατεσκευάσθη [ὁ κόσμος] Ph. 1.162; τῆς τοῦ θεοῦ σοφίας· ἡ δέ ἐστιν ὁ θεοῦ λ. ib. 56; λ. θεῖος . . εἰκὼν θεοῦ ib. 561, cf. 501; τὸν τομέα τῶν συμπάντων [θεοῦ] λ. ib. 492; τὸν ἄγγελον ὅς ἐστι λ. ib. 122: in NT identified with the person of Christ, ἐν ἀρχῇ ἦν ὁ λ. Ev.Jo. 1.1, cf. 14, 1 Ep.Jo. 2.7, Apoc. 19.13; ὁ λ. τῆς ζωῆς 1 Ep.Jo. 1.1.
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λόγος
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| Author: | Christopher J. Wells |
| Website: | www.technologyuk.net |
| Page title: | Matter - Electron Shells and Orbitals |
| URL: | |
| Published: | 13-03-2017 |
| Last revised: | 30-06-2025 |
| Accessed: | 16-09-2025 |
Matter
States of matter
Discovering the atom - a brief history
Inside the atom
Electron shells and orbitals
Chemical bonding
Electronegativity and the Pauling Scale
Lewis structures
VSEPR and molecular geometry
Solids - an overview
Minerals and organic solids
Crystalline solids
Network covalent solids
Ionic solids
Metallic solids
Alkali and alkaline earth metals
Transition metals
The lanthanides
An Introduction to Liquids
Electron Shells and Orbitals
Overview
According to the Rutherford-Bohr model, electrons were thought to occupy fixed, circular orbits around the nucleus of an atom. The electrons with the lowest energy levels occupied the lowest orbits. Electrons with higher energy levels would occupy higher orbits. Whereas the planetary orbits in our solar system all lie on (or very close to) a two-dimensional orbital plane, electron orbits were believed to occupy a number of different orbital planes, spawning the concept of three-dimensional electron shells.
We know now that the Rutherford-Bohr model does not accurately represent the way in which electrons behave. They certainly do not occupy neat circular orbits - the reality is far more complex. Nevertheless, the Rutherford-Bohr model is still taught in schools and colleges because it gives us a good conceptual framework for thinking about electrons and their energy levels. For that reason, we will start by by exploring the concept of electron shells, and then go on to examine the concept of orbitals.
Before we do anything else, let's try and get a little perspective. At the time of writing, the periodic table contains one hundred and eighteen (118) elements. The first element in the table is hydrogen (H), which has the atomic number one (1) because it has one proton and one electron. The last element in the table is oganesson (Og), which has the atomic number one hundred and eighteen (118) because it has one hundred and eighteen protons and one hundred and eighteen electrons.
When dealing with the complexity of an atom's electron configuration, therefore, the worst-case scenario is that we have one hundred and eighteen electrons to consider. Bear in mind, however, that element numbers 95 to 118 in the periodic table are all synthetic radioisotopes. They have been produced in a laboratory and are not naturally occurring, often having a very brief existence. Oganesson, for example, has a half-life of just seven-tenths of a millisecond (0.7 ms).
Most of the elements that physicists and chemists deal with in the normal course of their work are far less exotic. That said, the subject of electron configuration is complex enough to require a reasonable amount of study if we are to get a good understanding of how it all works. And it is important to achieve that understanding, because the way in which electrons are organised within atoms determines the properties of the different elements and how they interact with one another.
To digress for a moment, it is fascinating to consider that everything you are, and everything you can see, hear or touch - the entire world around you, in fact - all of that is made up from less than a hundred different kinds of atomic building block!
Electron shells in the Rutherford-Bohr model
In the Rutherford-Bohr model of the atom, electrons occupy electron shells, each of which is located at a certain distance from the nucleus. Each of the electrons in a particular electron shell possesses a discrete amount of energy, designated by a quantum number (n). Those with the least energy are in the electron shell closest to the nucleus. The electrons in the shell furthest away from the nucleus (called the valence shell) possess the greatest amount of energy.
Of the one hundred and eighteen known elements, electron configurations for the first one hundred and eight have so far been determined. The electron configurations suggested for the remaining elements are based on what we would expect to see rather than on actual observational data (these elements have extremely short half-lives, making it difficult to obtain accurate measurements).
The electron configuration of the known elements is represented using seven main electron shells, each of which is represented by a concentric circle with the nucleus at its centre, as shown below. The shells are designated as quantum numbers n=1 to n=7, with n=1 being the shell nearest to the nucleus (you may also see them identified with the uppercase letters K, L, M, N, O, P and Q, with K being the shell nearest to the nucleus).
The element Hassium (Hs) has 108 electrons in 7 electron shells
Electrons and the periodic table
The periodic table orders the elements according to the number of protons in the nucleus of each element. Since, in a non-ionised atom, the number of electrons will be the same as the number of protons, we can infer the number of electrons in an atom of a particular element from that element's position in the periodic table (i.e. its atomic number).
Atoms tend to assume the most stable configuration and the lowest energy level possible. For that reason, the electron shells in an atom are (usually) filled in strict order, starting with the lowest energy level. Each of the electron shells (n1 to n7, or K to Q) can hold a maximum number of electrons determined by the formula: electron capacity = 2n2 (where n is the electron shell number), as shown below. Note, however, that the maximum number of electrons actually seen in an electron shell to date is thirty-two (32).
| Electron shell | 2n2 |
--- |
| n1 (K) | 2 |
| n2 (L) | 8 |
| n3 (M) | 18 |
| n4 (N) | 32 |
| n5 (O) | 50 |
| n6 (P) | 72 |
| n7 (Q) | 98 |
The position of an element in the periodic table tells us a lot about its electron configuration. The table itself is divided into rows (or periods), and columns (or groups). There are seven periods in the table, numbered from 1 to 7. Elements in each period have the same number of electron shells as the number of that period. Hydrogen and helium are in period 1, and both have one electron shell. Carbon, nitrogen and oxygen are all in period 2, and all have two electron shells. And so on.
There are eighteen groups in the periodic table. Elements in the same group tend to have similar chemical properties. This is related to the fact that elements in the same group tend to have the same number of electrons in their outer electron shell (the valence shell). Elements in groups 1, 2, 13, 14, 15, 16, 17 and 18 have 1, 2, 3, 4, 5, 6, 7 and 8 electrons in their valence shells respectively.
Group numbers 3 through 12 contain three special groups of metals called transition metals, lanthanoids and actinoids. We will look at the significance of these special groups elsewhere. For the purposes of this discussion, the important point to note is that most of the sixty-eight elements in these ten groups have two electrons in their outer shell. A small number of the elements have only one electron in their outer shell, and one element (lawrencium) has three.
The periodic table organises elements into periods (rows 1-7) and groups (columns 1-18)
More detailed versions of the periodic table (you can find an excellent example here) often show the electron configuration as a comma-separated list of values showing the number of electrons in each shell. For example, silicon (Si) would have the electron configuration 2, 8, 4. Electron shells 1n and 2n are full, containing two and eight electrons respectively, while electron shell 3n contains four electrons for a total of fourteen electrons.
We said earlier that electron shells in an atom are usually filled in strict order, starting with the lowest energy level. This is clearly the case for the elements in first three periods in the periodic table. After that, things get a little more complicated. In the fourth period, for example, we see electrons appearing in shell 4n before shell 3n is completely full. The reasons for this apparent anomaly will hopefully become clear in due course, when we start to look at orbitals.
The illustration below shows the first ten elements in the periodic table. Note that helium and neon both have full outer shells. Both of these elements can be found in Group 18 in the periodic table. All Group 18 elements (except for helium) have eight electrons in their valence shell. This makes them very stable and non-reactive. The Group 18 elements helium, neon, argon, krypton, xenon and radon, once known as the inert gases, are today more often referred to as the noble gases because they tend not to interact with other elements.
The first ten elements in the periodic table
The electron configuration of an element, and in particular the number of electrons it has in its valence (outer) shell, has a huge influence on how atoms of that element interact with atoms of other elements. For example, the Group 17 elements fluorine, chlorine, bromine and iodine (collectively known as the halogens) all need one additional electron in order to achieve a full valence shell.
These elements react vigorously with the so-called alkali metals (lithium, sodium, potassium, rubidium and caesium) in Group 1 of the periodic table, all of which have just one electron in their valence shells. When sodium reacts with chlorine, these elements form the highly stable ionic compound sodium chloride (NaCl) - better known as common salt. By combining in this way, both elements effectively achieve complete outer shells.
Orbitals
We know now of course that electrons do not move around the nucleus of an atom in the neat circular orbits described by the Rutherford-Bohr model. In fact, thanks to the work of Heisenberg and Schrödinger, we know that at any given time, an electron could be anywhere inside the electron cloud. Which is where orbitals come in.
We have stated that we cannot know with any certainty where an electron will be at any given time. We can, however, identify a three-dimensional region within the space occupied by the atom where a particular electron can be found for ninety percent (90%) of the time. We call this region an orbital. The name is rather misleading, since an electron orbital does not have anything to do with an "orbit".
Orbitals can have different shapes, and the geometry can get quite complex. The simplest kind of orbital is the s orbital, which is spherical in shape, with the nucleus of the atom at its centre. Other orbitals include the p, d and f orbitals. For historical reasons the designated letters stand for sharp, principal, diffuse, and fundamental, based on the spectral line observations associated with them, but the association is no longer of any significance. Orbits beyond the f orbitals are designated as g, h, i and so on (i.e. in alphabetical order).
The p orbital is shaped like a dumbell, again centred on the nucleus of the atom. Orbitals of this kind can co-exist within an electron shell, but each is aligned on a different perpendicular axis (imagine an x, y or z axis passing through the nucleus of the atom). It is not practical to try and describe the geometry of all of the different kinds of electron orbital, but the following graphic shows some of the possible permutations.
Electron orbitals can have complex geometries (image: chem.libretexts.org)
Each of the orbitals shown in the illustration above, even the ones that look quite complex, can only contain a maximum of two electrons. The periodic table now contains more than a hundred elements, so you can probably imagine how complicated the orbital geometry can become! Fortunately, you are not usually required to describe anything too advanced in terms of orbital geometry in an exam. It is perhaps more important to understand how orbitals relate to electron shells.
Even though we know that electron shells do not actually consist of electrons travelling in neat circular orbits around the nucleus, they are real enough in the sense that each electron shell (sometimes called a principal energy level) represents a specific quantum energy level, represented by the principal quantum number (n). Each electron shell consists of one or more subshells, and each subshell consists of one or more orbitals.
It is often stated that electrons in the same electron shell have the same energy levels. This is at best a generalisation. All electrons in the same subshell will have exactly the same energy level. However, whist the overall energy levels in electron shells increases with the principal quantum number n, energy levels between subshells in the same electron shell will be different. Electrons in subshells that fill later will have more energy than those in subshells that fill earlier. In some cases, the energy ranges associated with two electron shells overlap, as we shall see.
Note that as well as being identified by the designated letter (s, p, d, f, etc.), orbitals can also be associated with a particular electron shell using the appropriate number. For example, the s orbitals in the first three electron shells will be refered to as the 1s, 2s and 3s orbitals. Likewise, the p orbitals in electron shells 2n, 3n and 4n will be referred to as the 2p, 3p and 4p orbitals respectively.
Secondary, magnetic, and spin quantum numbers
Before we progress further, we would like to introduce some additional quantum numbers. The first of these is the secondary (or angular momentum, or azimuthal) quantum number, ℓ. This number identifies the shape of the orbital. We normally use the letters s, p, d, f, etc. for this purpose to avoid confusion with the principal quantum number (n). Values of ℓ relate to orbital types as follows:
| Secondary quantum number (ℓ ) | Orbital type |
--- |
| 0 | s |
| 1 | p |
| 2 | d |
| 3 | f |
| 4 | g |
| 5 | h |
| . . . | . . . |
The secondary quantum number identifies the subshell, since each subshell consists of the set of orbitals of a particular type within a shell. It also has significance with respect to the energy level of the subshell, which will increase slightly as ℓ increases.
The next quantum number we want to talk about is the magnetic quantum number, m. This number identifies the orientation of an orbital in space. All of the orbitals in a subshell will have the same energy values, but will have different spatial orientations. For any given subshell, the values that m can take will be integer values in the range -ℓ . . . 0 . . . ℓ. The number of orbitals in any given subshell will therefore be 2ℓ + 1.
The last quantum number of interest in this context is the spin quantum number, s. This number gives the intrinsic angular momentum (or spin angular momentum or just spin) of an electron, and is designated by the letter s (the term intrinsic angular momentum simply means that the electron is spinning around its own axis).
For reasons that are beyond the scope of this page (and because a straightforward explanation has eluded even Nobel Prize winning physicists on occasion), the spin quantum number (s) of an electron can take values of s = +1/2 or s = -1/2. These two spin orientations are sometimes referred to as spin-up and spin down, respectively. The important thing to note here is that, when two electrons occupy the same orbital, they must have opposite spin orientations.
An electron has a spin quantum number of -1/2 or +1/2
In 1925, the German physicist Friedrich Hermann Hund (1896-1997) established a set of rules governing electron configuration in atoms. The one that applies here (and the one commonly referred to as Hund's rule) is that, if two or more orbitals in the same subshell are available, electrons will occupy an empty orbital first, if one is available. Only after all of the orbitals in a subshell have one electron each will the electrons start to pair up. The first electron in an orbital has a spin of +1/2, and the second has a spin of -1/2.
The four quantum numbers (the principal quantum number n, the secondary quantum number ℓ, the magnetic quantum number m, and the spin quantum number s) together describe the unique quantum state of a single electron. They completely describe the movement and trajectories of each electron within an atom.
No two electrons can share the same combination of quantum numbers (a principle known as the Pauli Exclusion Principle, named after the Austrian physicist Wolfgang Ernst Pauli (1900-1958) who discovered it, and who made significant contributions in the field of spin theory (the study of the intrinsic angular momentum of elementary particles).
Quantum numbers are important for a number of reasons. They can be used to determine the electron configuration of an atom and the probable location of the electrons within an atom. They are also factors in determining characteristics such as the ionisation energy and atomic radius of an atom, how the atom will behave in a magnetic field, and whether or not the atom can generate its own magnetic field.
Nodes
Most orbitals have one or more nodes. A node is a point at which the probablitity of an electron occurring is zero. There are two kinds of node - radial nodes, and angular nodes. A radial node is essentially the set of points that lie at some specified radial distance from the nucleus. An angular node is typically represented as a plane of symmetry bisecting the orbital.
With all types of orbital, the number of nodes increases with the principal quantum number (n). There are also some simple formulas that you can use if you need to determine the number and type of nodes for a particular orbital. These are as follows:
number of nodes = n - 1
number of angular nodes = ℓ
number of radial nodes = n - ℓ - 1
where n is the principal quantum number and ℓ is the secondary quantum number. Let's look at a couple of examples to see how this works. Suppose we want to determine the number and type of the nodes in a p type orbital in the 3n electron shell. We know that the value of n is 3 (given), and we saw earlier that the value of ℓ for a p type orbital is 1. We therefore have:
number of nodes = n - 1 = 3 - 1 = 2
number of angular nodes = ℓ = 1
number of radial nodes = n - ℓ - 1 = 3 - 1 - 1 = 1
Now suppose that we want to find the number and type of the nodes in a d type orbital in the 5n electron shell. The numbers are getting bigger, but it's still a fairly trivial problem. The value of n is 5, and the value of ℓ for a d type orbital is 2. We therefore have:
number of nodes = n - 1 = 5 - 1 = 4
number of angular nodes = ℓ = 2
number of radial nodes = n - ℓ - 1 = 5 - 2 - 1 = 2
The s orbitals
The simplest electronic configuration, unsurprisingly, can be seen in the hydrogen atom, which has just one electron. The hydrogen atom therefore has a single electron shell consisting of a single subshell containing a single s orbital. The s orbitals are spherical, and are the only orbitals that occur in every electron shell. Like all orbitals, the s orbitals have a number associated with them; the principal quantum number (n) of the shell in which they appear. The illustration below shows the first three s orbitals.
The 1s, 2s and 3s orbitals
The s orbitals only have radial nodes (the 1s orbital has no nodes at all - it is simply a sphere of electron density). The following illustration should hopefully help you to visualise the electron distribution within the s orbitals.
Electron distribution in the 1s, 2s and 3s orbitals
With all types of orbital - as we mentioned earlier - the number of nodes increases with the principal quantum number (n). The s orbitals are no exception - the 2s orbital has one radial node, the 3s orbital has two radial nodes, and so on (note that an electron shell may only contain one s orbital).
The p orbitals
The 1n electron shell can contain only two electrons, both of which will occupy a single s orbital. The 2n electron shell, on the other hand, can contain up to eight electrons. Of these eight electrons, the first two will occupy the 2s orbital. Any additional electrons will occupy one of three p orbitals. The illustration below shows all of the orbitals found in the 1n and 2n electron shells.
As you can see, the three dumbbell-shaped p orbitals lie at right angles to one another along imaginary x, y and z axes. We identify individual p orbitals in an electron shell using the labels px, py, and pz.
The 2n electron shell is the first to contain p orbitals
As we have seen, the value of the secondary quantum number (ℓ) for p orbitals is one (1). This means that each p orbital has one angular node. You can imagine the angular node as a two-dimensional plane bisecting the nuclueus, with one lobe of the p orbital on either side of it. To give an example, the angular node for the px orbital (that's the one that lies along the x axis) would be the plane on which both the y and z axes lie.
Since the probability of finding an electron at a node is zero, you might well be wondering how electrons manage to get from one lobe of a p orbital to the other. Whilst a detailed explanation is perhaps beyond the scope of this page, suffice it to say that it has to do with the fact that electrons can behave like both particles and waves.
Possible values for the magnetic quantum number m for a p orbital are -1, 0 and +1 (since ℓ is equal to one), which means that there can be three p orbitals in any of the electron shells except 1n. Once the s orbital in each electron shell has its complement of two electrons, the next six electrons will find a home in one of the p orbitals.
The d orbitals
For the first eighteen elements in the periodic table up to and including argon (i.e. the first three periods of the table), all of the electrons will be found in either s orbitals or p orbitals. In fact, that continues to be the case for the first two elements of period four - potassium and calcium. In these two elements, we see electrons appearing in the s orbital of electron shell 4n, despite the fact that electron shell 3n is not yet full.
The reason for this seeming anomaly is that the energy level of the s orbital in electron shell 4n (designated as the 4s orbital) is actually slightly lower than the energy level of the d orbitals in electron shell 3n, and we have already seen that electrons will (usually) occupy the orbitals with the lowest energy levels first (just to make things more complicated, the energy level of the 4s orbital increases when the 3d orbitals start to fill up, but we're not going to get into that here).
The element scandium - one of the so-called transition metals - follows calcium in the periodic table. Scandium has nine (9) electrons in the 3n electron shell. Of these, two are accommodated in the 3s orbital and six more in the 3p orbitals. The ninth electron occupies a d orbital. This is where things start to get really interesting.
The d orbitals have a secondary quantum number (ℓ) of two (2). Each d orbital therefore has two angular nodes, and possible values for the magnetic quantum number m of -2, -1, 0, +1 and +2, which means that there can be five d orbitals in any of the electron shells from 3n onwards. Most of the d orbitals look a bit like a four-leaf clover made from two pairs of dumbbells, although one looks like a dumbbell encircled by a doughnut! The illustration below should help you to visualise them.
Electron shells 3n and upwards contain d orbitals
The five d orbitals in the 3n electron shell, together with the single s orbital and the three p orbitals, allow it to contain a total of eighteen (18) electrons (remember that each orbital can contain only two electrons, regardless of the complexity of its geometry).
As with the p orbitals, individual d orbitals are identified according to their relationship with arbitrary x, y and z axes. The dyz, dxz, and dxy orbitals lie on the planes formed by the y and z, x and z, and x and y axes respectively, with each lobe forming a forty-five-degree angle with each axis on that plane. In the dx2-y2 orbital, one pair of lobes lies on the x axis and the other lies on the y axis.
The dz2 orbital has a somewhat novel configuration, in which two lobes lie on the z axis and are encircled by a doughnut-shaped orbital centred on the z axis. The doughnut is bisected through its circumference by the plane formed by the x and y axes.
The f orbitals
Electron shell 4n contains thirty-two (32) electrons. The first eighteen of these electrons occupy the s, p and d orbitals. The last fourteen electrons occupy f orbitals. In fact, the f orbitals do not begin to fill until we get to the third element in period five of the periodic table, yttrium (Y) - after both the 5p and the 6s orbital have been populated (because these orbitals have lower energy levels than the 4f orbitals).
The f orbitals have a secondary quantum number (ℓ) of three (3). Each f orbital therefore has three angular nodes, and possible values for the magnetic quantum number m of -3, -2, -1, 0, +1, +2 and +3, which means that there can be seven f orbitals in any of the electron shells from 4n onwards.
The seven f orbitals in the 4n electron shell, together with the single s orbital, three p orbitals and four d orbitals, allow it to contain a total of thirty-two (32) electrons. The geometry of the f orbitals is more complex than any of the previous orbitals we have looked at. We will not attempt to describe them here, but the illustration below should help you to visualise them.
Electron shells 4n and upwards contain f orbitals
The number of electron shells could in theory increase indefinitely, and we could expect to see electrons populating g, h and i orbitals (and beyond). However, although the periodic table contains elements with partially filled 5n, 6n and 7n electron shells, no naturally occurring or man-made element has ever been observed to have more than thirty-two electrons in any of its electron shells. From a practical viewpoint, therefore, our interest in orbitals ends with the f orbital.
The aufbau principle
The aufbau principle takes its name from the German word Aufbauprinzip, which roughly translated means building-up principle. The principle, which dates from the 1920s, has been attributed to Niels Bohr and Wolfgang Pauli. It states the following:
"The orbitals of lower energy are filled in first with the electrons and only then the orbitals of high energy are filled."
The principle follows from the natural tendency for atoms to assume the most stable electron configuration possible. We have already seen that an electron shell does not need to be completely filled before the next electron shell starts to fill with electrons, because the energy levels of the suborbitals in two electron shells can overlap. So, are there any rules that we can apply here?
Actually, there is a relatively simple rule that can be used to determine - at least in the majority of cases - the order in which electrons fill orbitals. The German physicist Erwin Madelung (1881-1972) developed the Madelung rule (also known by several other names), which states that atomic orbitals are filled in order of increasing values of n + ℓ (i.e. the sum of the principal and secondary quantum numbers). The rule is often represented graphically using the following diagram:
A diagrammatic representation of the Madelung rule
Hopefully, the diagram is relatively easy to follow. To find the order in which the subshells fill with electrons, simply follow the arrows from top to bottom and from right to left. Each entry gives a principal quantum number (which identifies the electron shell) and a letter that represents one of the orbital types (s, p, d or f). Note that if two subshells have the same value of n + ℓ, the subshell with the lower value of n will fill first. Here is the complete sequence:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
Another way of representing this progression graphically is shown below. Each subshell is represented by a number of boxes corresponding to the number of orbitals it contains of a particular type. Electron shells are shown in a strict left-to-right order, but the sequence in which the subshells are populated is vertical (starting from the bottom of the diagram).
This diagram shows the (approximate) order in which the electron subshells are filled
The aufbau principle, which was arrived at through extensive experimentation, is successful in predicting the electron configuration of all but a handful of elements (copper, chromium and palladium are examples of exceptions to the rule). Be aware, however, that it is not always so successful in predicting the precise order in which the subshells are filled, and anomalies do occur.
The element scandium (Sc) is a good example. Scandium has the atomic number twenty-one (21), and so has twenty-one electrons in its neutral (non-ionised) state. It is the first of the transition metals, and lies in period four of the periodic table. The Madelung rule successfully predicts the final electron configuration, which is as follows (note that the superscripted numbers represent the number of electrons in each orbital):
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d1
Is this the order in which the subshells are populated, however? According to experimental results obtained so far, the answer is no. These experiments indicate that the single electron occupying the 3d orbital is present before the 4s orbital is filled.
A full examination of why this apparent anomaly occurs is beyond the scope of this page. Note, however, that it is important to be aware of such inconsistencies, especially when considering what happens during ionisation. In the case of scandium, a positive Sc+ ion is formed when an electron is lost from the 4s orbital. This makes sense, since the 4s orbital is the last orbital to be filled.
Electron configuration shorthand
As we saw above with the element scandium, the electron configuration for any element can be written out using notation that tells us exactly what we need to know in terms of both the orbital type and the number of electrons in each subshell. For the lighter elements, this form of notation is relatively easy going. For example, the electron configuration for hydrogen is written as 1s1; helium is written as 1s2; and lithium is written as 1s2, 2s1.
As we can see from the scandium example, however, the longhand form of the notation becomes increasingly cumbersome as the atomic number increases. When we get to the heavier elements, it becomes clear that a shorthand version of the notation would certainly make life a little easier. Fortunately, an abbreviated form of the notation can be used.
The most frequently used shorthand version of electron configuration seems to be the noble gas configuration, so-called because it uses the noble gas occupying the last column of the periodic table in the row immediately preceding the element of interest. The notation essentially consists of the symbol for this noble gas, followed by the standard notation for the remaining subshell(s).
To find the noble gas configuration for an element, use the following steps:
Find the element you wish to find the configuration for in the periodic table.
Determine which noble gas to use (this will be the last element in the row above the element you wish to find the configuration for).
Write the symbol for the noble gas within square brackets, e.g. [He], [Ne], [Ar] etc., and make a note of both the atomic number of the noble gas and the row number in which it occurs.
Determine how many electrons remain (this will be the atomic number of the element you are interested in, minus the atomic number of the noble gas).
Begin the remainder of the configuration with the number of the row in which the element of interest resides, followed by orbital type s (we always start with orbital type s).
Continue writing your electron configuration according to the aufbau diagram we saw earlier until the remaining electrons have all been accounted for.
An example should help to clarify the procedure. Let's find the noble gas configuration for selenium (Se). The image below shows the relevant part of the periodic table.
Part of the periodic table, with selenium and argon highlighted
Selenium is in row four of the periodic table, so the noble gas we must use is argon, which is in the last column of row three (3) and has the atomic number eighteen (18). We can now start our configuration by writing the symbol for argon in square brackets, thus:
The atomic number of selenium is thirty-four (34). Subtracting the atomic number of argon gives us a total of sixteen (16) electrons that still need to be accounted for. So far so good. We know that selenium is in row four, so we can write:
[Ar] 4s . . .
We still need to account for sixteen electrons. The 4s subshell accounts for two of them. Here is the sequence in which the subshells (from 4s onwards) are filled according to the aufbau principle:
4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
The complete 3d subshell will add ten (10) more electrons to the total, leaving just four electrons to find. Since the 4p subshell can hold a total of six electrons, we need look no further than that. Our final (abbreviated) configuration looks like this:
[Ar] 4s2, 3d10, 4p4
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Section 9.6 Counting Principles 691 Simple Counting Problems This section and Section 9.7 present a brief introduction to some of the basic counting principles and their application to probability. In Section 9.7, you will see that much of probability has to do with counting the number of ways an event can occur. The following two examples describe simple counting problems.
Selecting Pairs of Numbers at Random Eight pieces of paper are numbered from 1 to 8 and placed in a box. One piece of paper is drawn from the box, its number is written down, and the piece of paper is replaced in the box. Then, a second piece of paper is drawn from the box, and its number is written down. Finally, the two numbers are added togeth-er. How many different ways can a sum of 12 be obtained?
Solution To solve this problem, count the different ways that a sum of 12 can be obtained using two numbers from 1 to 8.
First number 4 5 6 7 8 Second number 8 7 6 5 4 From this list, you can see that a sum of 12 can occur in five different ways.
Now try Exercise 5.
Selecting Pairs of Numbers at Random Eight pieces of paper are numbered from 1 to 8 and placed in a box. Two pieces of paper are drawn from the box at the same time, and the numbers on the pieces of paper are written down and totaled. How many different ways can a sum of 12 be obtained?
Solution To solve this problem, count the different ways that a sum of 12 can be obtained using two different numbers from 1 to 8.
First number 4 5 7 8 Second number 8 7 5 4 So, a sum of 12 can be obtained in four different ways.
Now try Exercise 7.
The difference between the counting problems in Examples 1 and 2 can be described by saying that the random selection in Example 1 occurs with replacement, whereas the random selection in Example 2 occurs without replacement, which eliminates the possibility of choosing two 6’s.
What you should learn • Solve simple counting problems.
• Use the Fundamental Counting Principle to solve counting problems.
• Use permutations to solve counting problems.
• Use combinations to solve counting problems.
Why you should learn it You can use counting principles to solve counting problems that occur in real life. For instance, in Exercise 65 on page 700, you are asked to use counting principles to determine the number of possible ways of selecting the winning numbers in the Powerball lottery.
Counting Principles © Michael Simpson/FPG/Getty Images 9.6 Example 1 Example 2 333202_0906.qxd 12/5/05 11:39 AM Page 691 The Fundamental Counting Principle Examples 1 and 2 describe simple counting problems in which you can list each possible way that an event can occur. When it is possible, this is always the best way to solve a counting problem. However, some events can occur in so many different ways that it is not feasible to write out the entire list. In such cases, you must rely on formulas and counting principles. The most important of these is the Fundamental Counting Principle.
The Fundamental Counting Principle can be extended to three or more events. For instance, the number of ways that three events and can occur is Using the Fundamental Counting Principle How many different pairs of letters from the English alphabet are possible?
Solution There are two events in this situation. The first event is the choice of the first letter, and the second event is the choice of the second letter. Because the English alphabet contains 26 letters, it follows that the number of two-letter pairs is Now try Exercise 13.
Using the Fundamental Counting Principle Telephone numbers in the United States currently have 10 digits. The first three are the area code and the next seven are the local telephone number. How many different telephone numbers are possible within each area code? (Note that at this time, a local telephone number cannot begin with 0 or 1.) Solution Because the first digit of a local telephone number cannot be 0 or 1, there are only eight choices for the first digit. For each of the other six digits, there are 10 choices.
So, the number of local telephone numbers that are possible within each area code is Now try Exercise 19.
8 10 10 10 10 10 10 8,000,000.
Local Number 10 10 10 10 10 10 8 Area Code 26 26 676.
m1 m2 m3.
E3 E1, E2, 692 Chapter 9 Sequences, Series, and Probability Fundamental Counting Principle Let and be two events. The first event can occur in different ways. After has occurred, can occur in different ways. The number of ways that the two events can occur is m1 m2.
m2 E2 E1 m1 E1 E2 E1 You may want to consider opening class by asking students if they can predict which offers more choices for license plates: (a) a plate with three different letters of the alphabet in any order or (b) a plate with four different nonzero digits in any order.You may want to end class by verifying the answers: (a) offers choices and (b) offers choices.
9 8 7 6 3024 26 25 24 15,600 Example 3 Example 4 333202_0906.qxd 12/5/05 11:39 AM Page 692 Permutations One important application of the Fundamental Counting Principle is in determining the number of ways that elements can be arranged (in order). An ordering of elements is called a permutation of the elements.
Finding the Number of Permutations of n Elements How many permutations are possible for the letters A, B, C, D, E, and F?
Solution Consider the following reasoning.
First position: Any of the six letters Second position: Any of the remaining five letters Third position: Any of the remaining four letters Fourth position: Any of the remaining three letters Fifth position: Any of the remaining two letters Sixth position: The one remaining letter So, the numbers of choices for the six positions are as follows.
The total number of permutations of the six letters is Now try Exercise 39. 720.
6! 6 5 4 3 2 1 Permutations of six letters 6 5 4 3 2 1 n n Section 9.6 Counting Principles 693 Additional Examples a. How many permutations are possible for the numbers 0, 1, 2, and 3?
Solution b. Write out the possible permutations of the letters A, B, and C.
Solution ABC, ACB, BAC, BCA, CAB, CBA 4! 24 Definition of Permutation A permutation of different elements is an ordering of the elements such that one element is first, one is second, one is third, and so on.
n Number of Permutations of n Elements The number of permutations of elements is In other words, there are different ways that elements can be ordered.
n n!
n n 1 . . . 4 3 2 1 n!.
n Example 5 333202_0906.qxd 12/5/05 11:39 AM Page 693 Counting Horse Race Finishes Eight horses are running in a race. In how many different ways can these horses come in first, second, and third? (Assume that there are no ties.) Solution Here are the different possibilities.
Win (first position): Eight choices Place (second position): Seven choices Show (third position): Six choices Using the Fundamental Counting Principle, multiply these three numbers together to obtain the following.
So, there are different orders.
Now try Exercise 43.
It is useful, on occasion, to order a subset of a collection of elements rather than the entire collection. For example, you might want to choose and order elements out of a collection of elements. Such an ordering is called a permutation of n elements taken r at a time.
Using this formula, you can rework Example 6 to find that the number of permutations of eight horses taken three at a time is which is the same answer obtained in the example. 336 8 7 6 5!
5! 8!
5!
8P3 8!
8 3!
n r 8 7 6 336 Different orders of horses 8 7 6 694 Chapter 9 Sequences, Series, and Probability Permutations of n Elements Taken r at a Time The number of permutations of elements taken at a time is nn 1n 2 . . . n r 1.
nP r n!
n r!
r n Example 6 Most graphing calculators are programmed to evaluate Consult the user’s guide for your calculator and then evaluate You should get an answer of 6720.
8P5.
nPr.
Technology Eleven thoroughbred racehorses hold the title of Triple Crown winner for winning the Kentucky Derby, the Preakness, and the Belmont Stakes in the same year.
Forty-nine horses have won two out of the three races.
Vaughn Youtz/Newsmakers/Getty Images 333202_0906.qxd 12/5/05 11:39 AM Page 694 Remember that for permutations, order is important. So, if you are looking at the possible permutations of the letters A, B, C, and D taken three at a time, the permutations (A, B, D) and (B, A, D) are counted as different because the order of the elements is different.
Suppose, however, that you are asked to find the possible permutations of the letters A, A, B, and C. The total number of permutations of the four letters would be However, not all of these arrangements would be distinguishable because there are two A’s in the list. To find the number of distinguishable permutations, you can use the following formula.
Distinguishable Permutations In how many distinguishable ways can the letters in BANANA be written?
Solution This word has six letters, of which three are A’s, two are N’s, and one is a B. So, the number of distinguishable ways the letters can be written is The 60 different distinguishable permutations are as follows.
Now try Exercise 45.
AAABNN AANABN ABAANN ANAABN ANBAAN BAAANN BNAAAN NAABAN NABNAA NBANAA AAANBN AANANB ABANAN ANAANB ANBANA BAANAN BNAANA NAABNA NANAAB NBNAAA AAANNB AANBAN ABANNA ANABAN ANBNAA BAANNA BNANAA NAANAB NANABA NNAAAB AABANN AANBNA ABNAAN ANABNA ANNAAB BANAAN BNNAAA NAANBA NANBAA NNAABA AABNAN AANNAB ABNANA ANANAB ANNABA BANANA NAAABN NABAAN NBAAAN NNABAA AABNNA AANNBA ABNNAA ANANBA ANNBAA BANNAA NAAANB NABANA NBAANA NNBAAA 60. 6 5 4 3!
3! 2!
n!
n1! n2! n3! 6!
3! 2! 1!
4P4 4!.
Section 9.6 Counting Principles 695 Additional Example In how many different ways can the letters in INDIANA be written?
Solution 630 Distinguishable Permutations Suppose a set of objects has of one kind of object, of a second kind, of a third kind, and so on, with Then the number of distinguishable permutations of the objects is n!
n1! n2! n3! . . . nk !.
n n3 . . . nk .
n n1 n2 n3 n2 n1 n Example 7 333202_0906.qxd 12/5/05 11:39 AM Page 695 Combinations When you count the number of possible permutations of a set of elements, order is important. As a final topic in this section, you will look at a method of selecting subsets of a larger set in which order is not important. Such subsets are called combinations of n elements taken r at a time. For instance, the combinations and are equivalent because both sets contain the same three elements, and the order in which the elements are listed is not important. So, you would count only one of the two sets. A common example of how a combination occurs is a card game in which the player is free to reorder the cards after they have been dealt.
Combinations of n Elements Taken r at a Time In how many different ways can three letters be chosen from the letters A, B, C, D, and E? (The order of the three letters is not important.) Solution The following subsets represent the different combinations of three letters that can be chosen from the five letters.
From this list, you can conclude that there are 10 different ways that three letters can be chosen from five letters.
Now try Exercise 55.
Note that the formula for is the same one given for binomial coefficients.
To see how this formula is used, solve the counting problem in Example 8. In that problem, you are asked to find the number of combinations of five elements taken three at a time. So, and the number of combinations is which is the same answer obtained in Example 8.
5C3 5!
2!3! 5 4 3!
2 1 3! 10 2 n 5, r 3, nCr C, D, E B, D, E B, C, E B, C, D A, D, E A, C, E A, C, D A, B, E A, B, D A, B, C B, A, C A, B, C 696 Chapter 9 Sequences, Series, and Probability Careful attention to detail and numerous examples will help students understand when to count different orders and when not to.You may want to discuss the misnaming of the “combination lock”according to the definitions presented in this section.
Combinations of n Elements Taken r at a Time The number of combinations of elements taken at a time is which is equivalent to nCr nPr r! .
nCr n!
n r!r!
r n Example 8 333202_0906.qxd 12/5/05 11:39 AM Page 696 Counting Card Hands A standard poker hand consists of five cards dealt from a deck of 52 (see Figure 9.7). How many different poker hands are possible? (After the cards are dealt, the player may reorder them, and so order is not important.) Solution You can find the number of different poker hands by using the formula for the number of combinations of 52 elements taken five at a time, as follows.
Now try Exercise 63.
Forming a Team You are forming a 12-member swim team from 10 girls and 15 boys. The team must consist of five girls and seven boys. How many different 12-member teams are possible?
Solution There are ways of choosing five girls. The are ways of choosing seven boys. By the Fundamental Counting Principal, there are ways of choosing five girls and seven boys.
So, there are 1,621,620 12-member swim teams possible.
Now try Exercise 65.
When solving problems involving counting principles, you need to be able to distinguish among the various counting principles in order to determine which is necessary to solve the problem correctly. To do this, ask yourself the follow-ing questions.
1. Is the order of the elements important? Permutation 2. Are the chosen elements a subset of a larger set in which order is not important? Combination 3. Does the problem involve two or more separate events? Fundamental Counting Principle 1,621,620 252 6435 10C5 15C7 10!
5! 5! 15!
8! 7!
10C5 15C7 15C7 10C5 2,598,960 52 51 50 49 48 47!
5 4 3 2 1 47! 52!
47!5!
52C5 52!
52 5!5!
Section 9.6 Counting Principles 697 Activities 1. Evaluate (a) and (b) Answers: (a) 30 and (b) 15 2. A local building supply company is hiring extra summer help.They need four additional employees to work outside in the lumber yard and three more to work inside the store. In how many ways can these positions be filled if there are 10 applicants for outside work and 5 for inside work?
Answer: 2100 3. In how many distinguishable ways can the letters C A L C U L U S be written?
Answer: 5040 6C2.
6P 2 A 2 3 4 5 6 7 8 9 10 J Q K A 2 3 4 5 6 7 8 9 10 J Q K A 2 3 4 5 6 7 8 9 10 J Q K A 2 3 4 5 6 7 8 9 10 J Q K Example 9 Example 10 FIGURE 9.7 Standard deck of playing cards 333202_0906.qxd 12/5/05 11:39 AM Page 697 Random Selection In Exercises 1–8, determine the number of ways a computer can randomly generate one or more such integers from 1 through 12.
1. An odd integer 2. An even integer 3. A prime integer 4. An integer that is greater than 9 5. An integer that is divisible by 4 6. An integer that is divisible by 3 7. Two distinct integers whose sum is 9 8. Two distinct integers whose sum is 8 9. Entertainment Systems A customer can choose one of three amplifiers, one of two compact disc players, and one of five speaker models for an entertainment system.
Determine the number of possible system configurations.
10. Job Applicants A college needs two additional faculty members: a chemist and a statistician. In how many ways can these positions be filled if there are five applicants for the chemistry position and three applicants for the statistics position?
11. Course Schedule A college student is preparing a course schedule for the next semester. The student may select one of two mathematics courses, one of three science courses, and one of five courses from the social sciences and humanities. How many schedules are possible?
12. Aircraft Boarding Eight people are boarding an aircraft.
Two have tickets for first class and board before those in the economy class. In how many ways can the eight people board the aircraft?
13. True-False Exam In how many ways can a six-question true-false exam be answered? (Assume that no questions are omitted.) 14. True-False Exam In how many ways can a 12-question true-false exam be answered? (Assume that no questions are omitted.) 15. License Plate Numbers In the state of Pennsylvania, each standard automobile license plate number consists of three letters followed by a four-digit number. How many distinct license plate numbers can be formed in Pennsylvania?
16. License Plate Numbers In a certain state, each automo-bile license plate number consists of two letters followed by a four-digit number. To avoid confusion between “O” and “zero” and between “I” and “one,” the letters “O” and “I” are not used. How many distinct license plate numbers can be formed in this state?
17. Three-Digit Numbers How many three-digit numbers can be formed under each condition?
(a) The leading digit cannot be zero.
(b) The leading digit cannot be zero and no repetition of digits is allowed.
(c) The leading digit cannot be zero and the number must be a multiple of 5.
(d) The number is at least 400.
18. Four-Digit Numbers How many four-digit numbers can be formed under each condition?
(a) The leading digit cannot be zero.
(b) The leading digit cannot be zero and no repetition of digits is allowed.
(c) The leading digit cannot be zero and the number must be less than 5000.
(d) The leading digit cannot be zero and the number must be even.
19. Combination Lock A combination lock will open when the right choice of three numbers (from 1 to 40, inclusive) is selected. How many different lock combinations are possible?
698 Chapter 9 Sequences, Series, and Probability Exercises 9.6 VOCABULARY CHECK: Fill in the blanks.
1. The _ _ _ states that if there are ways for one event to occur and ways for a second event to occur, there are ways for both events to occur.
2. An ordering of elements is called a _ of the elements.
3. The number of permutations of elements taken at a time is given by the formula _.
4. The number of _ _ of objects is given by 5. When selecting subsets of a larger set in which order is not important, you are finding the number of _ of elements taken at a time.
PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.
r n n!
n1!n2!n3! . . . nk!.
n r n n m1 m2 m2 m1 333202_0906.qxd 12/5/05 11:39 AM Page 698 Section 9.6 Counting Principles 699 20. Combination Lock A combination lock will open when the right choice of three numbers (from 1 to 50, inclusive) is selected. How many different lock combinations are possible?
21. Concert Seats Four couples have reserved seats in a row for a concert. In how many different ways can they be seated if (a) there are no seating restrictions?
(b) the two members of each couple wish to sit together?
22. Single File In how many orders can four girls and four boys walk through a doorway single file if (a) there are no restrictions?
(b) the girls walk through before the boys?
In Exercises 23–28, evaluate 23.
24.
25.
26.
27.
28.
In Exercises 29 and 30, solve for 29.
30.
In Exercises 31–36, evaluate using a graphing utility.
31.
32.
33.
34.
35.
36.
37. Posing for a Photograph In how many ways can five children posing for a photograph line up in a row?
38. Riding in a Car In how many ways can six people sit in a six-passenger car?
39. Choosing Officers From a pool of 12 candidates, the offices of president, vice-president, secretary, and treasurer will be filled. In how many different ways can the offices be filled?
40. Assembly Line Production There are four processes involved in assembling a product, and these processes can be performed in any order. The management wants to test each order to determine which is the least time-consuming.
How many different orders will have to be tested?
In Exercises 41–44, find the number of distinguishable permutations of the group of letters.
41. A, A, G, E, E, E, M 42. B, B, B, T, T, T, T, T 43. A, L, G, E, B, R, A 44. M, I, S, S, I, S, S, I, P, P, I 45. Write all permutations of the letters A, B, C, and D.
46. Write all permutations of the letters A, B, C, and D if the letters B and C must remain between the letters A and D.
47. Batting Order A baseball coach is creating a nine-player batting order by selecting from a team of 15 players. How many different batting orders are possible?
48. Athletics Six sprinters have qualified for the finals in the 100-meter dash at the NCAA national track meet. In how many ways can the sprinters come in first, second, and third? (Assume there are no ties.) 49. Jury Selection From a group of 40 people, a jury of 12 people is to be selected. In how many different ways can the jury be selected?
50. Committee Members As of January 2005, the U.S.
Senate Committee on Indian Affairs had 14 members.
Assuming party affiliation was not a factor in selection, how many different committees were possible from the 100 U.S. senators?
51. Write all possible selections of two letters that can be formed from the letters A, B, C, D, E, and F. (The order of the two letters is not important.) 52. Forming an Experimental Group In order to conduct an experiment, five students are randomly selected from a class of 20. How many different groups of five students are possible?
53. Lottery Choices In the Massachusetts Mass Cash game, a player chooses five distinct numbers from 1 to 35. In how many ways can a player select the five numbers?
54. Lottery Choices In the Louisiana Lotto game, a player chooses six distinct numbers from 1 to 40. In how many ways can a player select the six numbers?
55. Defective Units A shipment of 10 microwave ovens contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?
56. Interpersonal Relationships The complexity of interper-sonal relationships increases dramatically as the size of a group increases. Determine the numbers of different two-person relationships in groups of people of sizes (a) 3, (b) 8, (c) 12, and (d) 20.
57. Poker Hand You are dealt five cards from an ordinary deck of 52 playing cards. In how many ways can you get (a) a full house and (b) a five-card combination containing two jacks and three aces? (A full house consists of three of one kind and two of another. For example, A-A-A-5-5 and K-K-K-10-10 are full houses.) 58. Job Applicants A toy manufacturer interviews eight people for four openings in the research and development department of the company. Three of the eight people are women. If all eight are qualified, in how many ways can the employer fill the four positions if (a) the selection is random and (b) exactly two selections are women?
10C7 20C5 10P8 100P3 100P5 20P5 nP5 18 n2P4 14 nP3 n2P4 n.
7P4 5P4 20P2 8P3 5P5 4P4 nPr.
333202_0906.qxd 12/5/05 11:39 AM Page 699 59. Forming a Committee A six-member research committee at a local college is to be formed having one administrator, three faculty members, and two students.
There are seven administrators, 12 faculty members, and 20 students in contention for the committee. How many six-member committees are possible?
60. Law Enforcement A police department uses computer imaging to create digital photographs of alleged perpetra-tors from eyewitness accounts. One software package contains 195 hairlines, 99 sets of eyes and eyebrows, 89 noses, 105 mouths, and 74 chins and cheek structures.
(a) Find the possible number of different faces that the software could create.
(b) A eyewitness can clearly recall the hairline and eyes and eyebrows of a suspect. How many different faces can be produced with this information?
Geometry In Exercises 61–64, find the number of diago-nals of the polygon. (A line segment connecting any two nonadjacent vertices is called a diagonal of the polygon.) 61. Pentagon 62. Hexagon 63. Octagon 64. Decagon (10 sides) 66. Permutations or Combinations?
Decide whether each scenario should be counted using permutations or combinations. Explain your reasoning.
(a) Number of ways 10 people can line up in a row for con-cert tickets (b) Number of different arrangements of three types of flowers from an array of 20 types (c) Number of three-digit pin numbers for a debit card (d) Number of two-scoop ice cream cones created from 31 different flavors Synthesis True or False?
In Exercises 67 and 68,determine whether the statement is true or false. Justify your answer.
67. The number of letter pairs that can be formed in any order from any of the first 13 letters in the alphabet (A–M) is an example of a permutation.
68. The number of permutations of elements can be deter-mined by using the Fundamental Counting Principle.
69. What is the relationship between and ?
70. Without calculating the numbers, determine which of the following is greater. Explain.
(a) The number of combinations of 10 elements taken six at a time (b) The number of permutations of 10 elements taken six at a time Proof In Exercises 71–74, prove the identity.
71.
72.
73.
74.
75. Think About It Can your calculator evaluate If not, explain why.
76. Writing Explain in words the meaning of Skills Review In Exercises 77–80, evaluate the function at each specified value of the independent variable and simplify.
77.
(a) (b) (c) 78.
(a) (b) (c) 79.
(a) (b) (c) 80.
(a) (b) (c) In Exercises 81–84, solve the equation. Round your answer to two decimal places, if necessary.
81.
82.
83.
84. e x3 16 log2x 3 5 4 t 3 2t 1 x 3 x 6 f 20 f 1 f 4 x2 2x 5, x2 2, x ≤4 x > 4 f x f 11 f 1 f 5 f x x 5 6 gx 1 g7 g3 gx x 3 2 f 5 f 0 f 3 f x 3x2 8 nP r .
100P80?
nC r nP r r!
nC n 1 nC1 nC n nC0 nP n 1 nP n nCnr nCr n 700 Chapter 9 Sequences, Series, and Probability 65. Lottery Powerball is a lottery game that is operated by the Multi-State Lottery Association and is played in 27 states, Washington D.C., and the U.S. Virgin Islands.
The game is played by drawing five white balls out of a drum of 53 white balls (numbered 1–53) and one red powerball out of a drum of 42 red balls (numbered 1–42). The jackpot is won by matching all five white balls in any order and the red powerball.
(a) Find the possible number of winning Powerball numbers.
(b) Find the possible number of winning Powerball numbers if the jackpot is won by matching all five white balls in order and the red power ball.
(c) Compare the results of part (a) with a state lottery in which a jackpot is won by matching six balls from a drum of 53 balls.
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Published April 20, 2007
Open Your Eyes to Cycloplegia
An often-overlooked treatment can provide pain relief for patients with acute inflammation.
Joseph W. Sowka, O.D., and Alan G. Kabat O.D.
A 27-year-old female nursing student presents urgently for evaluation following an accident at work. She was reaching for a stack of papers high on a shelf when they fell down upon her; one sheet hit her in the face, giving her a paper cut across her right cornea. She went immediately to the emergency room, where she was diagnosed with a corneal abrasion and given a topical antibiotic solution to instill four times per day.
She did so dutifully, but she remained in agony through the rest of the day and into the night. She had so much pain that she couldnt sleep; she had to sequester herself to a darkened room with the shades drawn. The pain became so unbearable that she could not function. Now, she presents urgently for ophthalmic consultation.
Her best-corrected visual acuity is 20/200 O.D. She has a large central corneal abrasion, profuse corneal edema, folds in Descemets membrane and a modest anterior chamber reaction. There is no mucopurulent discharge or stromal infiltration to indicate corneal infection. The prescribed antibiotic was effective in preventing a secondary infection.
Why is the patient in such agony? Could it have been avoided? Clearly, the emergency room physician knew to provide prophylaxis against infection; however, the job doesnt stop there. What was overlooked: use of a cycloplegic agent.
When we were in training, our professors imparted a very important axiom that, to this day, we still teach to our students and residents: You cant go wrong to cycloplege. This simple saying has saved many patients from unnecessary suffering.
If a patient is in ocular pain for almost any reason (barring acute angle-closure glaucoma), and you arent sure how to manage the pain, the use of a cycloplegic agent would likely help and, at the very worst, slightly blur your patients vision. In this column, we examine the science behind cycloplegia to see if the old saying still holds true.
Mechanism of Action
Cycloplegics block the action of acetylcholine, a stimulatory neurotransmitter of the autonomic nervous system. So, they are known as anticholinergic or antimuscarinic drugs.1 In the eye, acetylcholine receptors are located within the iris sphincter muscle as well as the ciliary body. Activity of these receptors results in contraction of the iris and ciliary body. Cycloplegics temporarily inhibit this activity, causing ciliary body paralysis and pupillary mydriasis.
Cycloplegics are extremely effective in relieving pain caused by ocular inflammation; they relax ciliary spasm by paralyzing the muscle. Further, they help prevent posterior synechiae formation by decreasing the area of the posterior iris that contacts the anterior lens capsule when the pupil is dilated.2 Cycloplegics can also stabilize the blood-aqueous barrier, thus reducing the amount of cell and flare reaction in the anterior chamber. For these reasons, cycloplegics have long been used in the management of patients with corneal injury and uveitis, for example.3-6
Review of Cycloplegics
Atropine: This medication first was derived from the belladonna plant, Atropa belladonna, in 1831.1 Atropine is the most potent cycloplegic agent available clinically, with a duration of action lasting up to 12 days in a healthy eye. Atropine is available in 0.5%, 1%, and 2% ophthalmic solutions and a 1% ophthalmic ointment. A recommended regimen for atropine cycloplegia has been seven to 10 applications within three to four days; however, cycloplegia obtained after eight instillations was not greater than after four instillations in healthy eyes.7 Typically, dosing of atropine is b.i.d. in the affected eye. While atropine can usually be discontinued after a few days, more extensive usage may be required in highly inflamed eyes.
Scopolamine: Also known as hyoscine, this medication comes in 0.25% ophthalmic solution. Though scopolamine has a shorter duration of cycloplegia than atropine, its antimuscarinic activity is greater than that of atropine on a weight basis. Cycloplegia (as measured by accommodative ability) generally wears off within three days of treatment.8 Typical dosing is b.i.d. to t.i.d. in the affected eye.
Homatropine: This cycloplegic comes in 2% and 5% ophthalmic solutions and is typically dosed b.i.d. to t.i.d. Homatropine is only about one-tenth as potent as
atropine, and cycloplegic recovery occurs in one to three days.
While tropicamide and cyclopentolate are considered mydriatic or cycloplegic agents, we do not consider these to be therapeutic drugs and limit their use to diagnostic testing. Tropicamide is best used for routine pupillary dilation, and we have seen many practitioners fail to properly manage patients with uveitis because they used cyclopentolate as the cycloplegic agent. Cyclopentolate is not a strong enough cycloplegic agent when managing significant ocular inflammation, so it is best used for cycloplegic refractions on children. We like to say that, as far as cycloplegia is concerned, Cyclogyl is for kids.
Risks of Cycloplegia
The most feared complication of cycloplegic use is the potential for inducing acute angle-closure glaucoma from the mydriatic effect of these agents. A shallow anterior chamber is a risk factor that should be evaluated before inducing mydriasis and/or cycloplegia. The risk of precipitating an attack after evaluation should be minimal.9,10 One large population-based study saw the occurrence of acute angle-closure glaucoma following diagnostic mydriasis to be only 0.03%.11
Though uncommon, cycloplegics may induce mental and neurotoxic effects, and can even cause death in rare instances.12-15 The myriad of neurogenic symptoms induced by cycloplegic toxicity includes confusion, vivid visual hallucinations, restlessness, muscular incoordination, emotional lability, acute psychotic reactions, restlessness, excitement, euphoria, disorientation, stupor, coma and respiratory depression.12-15 But, these are rare, anecdotal reports. Exposure to cycloplegics should be taken into account in the differential diagnosis of acute confusional syndromes, and patients should be made aware of the rare possibility of these occurrences.
So, what became of our patient? The corneal abrasion appeared to be healing well, and the prescribed antibiotic was appropriate. So, we instilled two drops of scopolamine in-office and sent the patient home. Six hours later, we followed up with the patient by telephone, and she reported feeling remarkably well. She thanked us profusely for using the magic drop.
While in rare instances there can be untoward effects from cycloplegic use, this class of medication is extremely safe overall. Remember, when patients have ocular pain, you really cant go wrong to cycloplege.
Brown JH. Atropine, scopolamine and related antimuscarinic drugs. In: Gilman AG, Rall TW, Nies AS, et al, eds. Goodman and Gilmans The pharmacological basis of therapeutics. New York: McGraw and Hill, 1993, chapter 8.
Wang T, Liu L, Li Z, et al. Ultrasound biomicroscopic study on changes of ocular anterior segment structure after topical application of cycloplegia. Chin Med J (Engl) 1999 Mar;112(3):217-20.
Janda AM. Ocular trauma. Triage and treatment. Postgrad Med 1991 Nov 15;90(7):51-2,55-60.
Torok PG, Mader TH. Corneal abrasions: diagnosis and management. Am Fam Physician 1996 Jun;53(8):2521-9,2532.
Wilson SA, Last A. Management of corneal abrasions. Am Fam Physician 2004 Jul 1;70(1):123-8.
Kaiser PK. A comparison of pressure patching versus no patching for corneal abrasions due to trauma or foreign body removal. Corneal Abrasion Patching Study Group. Ophthalmology 1995 Dec;102(12):1936-42.
Stolovitch C, Loewenstein A, Nemmet P, et al. Atropine cycloplegia: how many instillations does one need? J Pediatr Ophthalmol Strabismus 1992 May-Jun;29(3):175-6.
Marron J. Cycloplegia and mydriasis by use of atropine, scopolamine, and homatropine-paredrine. Arch Ophthalmol 1940;23:340-50.
Terry JE. Mydriatic angle-closure glaucomamechanism, evaluation and reversal. J Am Optom Assoc 1977 Feb; 48(2):159-68.
Brooks AM, West RH, Gillies WE. The risks of precipitating acute angle-closure glaucoma with the clinical use of mydriatic agents. Med J Aust 1986 Jul 7;145(1):34-6.
Wolfs RC, Grobbee DE, Hofman A, et al. Risk of acute angle-closure glaucoma after diagnostic mydriasis in nonselected subjects: the Rotterdam Study. Invest Ophthalmol Vis Sci 1997 Nov;38(12):2683-7.
Jimenez-Jimenez FJ, Alonso-Navarro H, Fernandez-Diaz A, et al. Neurotoxic effects induced by the topical administration of cycloplegics. A case report and review of the literature. Rev Neurol 2006 Nov 16-30;43(10):603-9.
Kortabarria RP, Duran JA, Chacon JR, et al. Toxic psychosis following cycloplegic eyedrops. DICP 1990 Jul-Aug;24(7-8):708-9.
Hamborg-Petersen B, Nielsen MM, Thordal C. Toxic effect of scopolamine eye drops in children. Acta Ophthalmol (Copenh) 1984 Jun;62(3):485-8.
Muller J, Wanke K. Toxic psychoses from atropine and scopolamine. Fortschr Neurol Psychiatr 1998 Jul;66(7):289-95.
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[Solved] In an equilateral triangle ABC, if AD ⊥ BC, then which
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In an equilateral triangle ABC, if AD ⊥ BC, then which of the following is true?
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NVS TGT Mathematics held on 29th Nov 2022 shift 1
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2 AB 2 = 3 AD 2
4 AB 2 = 3 AD 2
3 AB 2 = 4 AD 2
3 AB 2 = 2 AD 2
Answer (Detailed Solution Below)
Option 3 : 3 AB 2 = 4 AD 2
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Detailed Solution
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Given:
An equilateral triangle ABC, and AD ⊥ BC.
Concept:
In an equilateral triangle the median is also the altitude.
Solution:
According to the question,
An equilateral triangle ABC, and AD ⊥ BC.
In triangle ADB,
Angle ADB = 90 o
Using Pythagoras theorem,
AB 2 = AD 2 + BD 2
Since, AD is a median, so
BD = DC = BC/2 = AB/2 , since the given triangle is an equilateral triangle.
AB 2= AD 2+ (BC/2)2
AB 2= AD 2+ (AB/2)2
AB 2= AD 2+ (AB 2/4)
3AB 2= 4AD 2
Hence, option 3 is correct.
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Permutation Pattern -- from Wolfram MathWorld
TOPICS
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Discrete Mathematics
Combinatorics
Permutations
Permutation Pattern
Download Wolfram Notebook
Let denote the number of permutations on the symmetric group which avoid as a subpattern, where " contains as a subpattern" is interpreted to mean that there exist such that for ,
iff.
For example, a permutation contains the pattern (123) iff it has an ascending subsequence of length three. Here, note that members need not actually be consecutive, merely ascending (Wilf 1997). Therefore, of the partitions of , all but (i.e., , , , , and ) contain the pattern (12) (i.e., an increasing subsequence of length two).
The following table gives the numbers of pattern-matching permutations of , , ..., numbers for various patterns of length .
pattern OEIS number of pattern-matching permutations
1A0001421, 2, 6, 24, 120, 720, 5040, ...
12A0333121, 5, 23, 119, 719, 5039, 40319, ...
A0569861, 10, 78, 588, 4611, 38890, ...
1234A1580051, 17, 207, 2279, 24553, ...
1324A1580091, 17, 207, 2278, 24527, ...
1342A1580061, 17, 208, 2300, 24835, ...
The following table gives the numbers of pattern-avoiding permutations of for various sets of patterns.
Wilf class OEIS number of pattern-avoiding permutations
A0001081, 2, 5, 14, 42, 132, ...
123, 132, 213A0000271, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
132, 231, 321A0000271, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
123, 132, 3214A0000731, 2, 4, 7, 13, 24, 44, 81, 149, ...
123, 132, 3241A0000711, 2, 7, 12, 20, 33, 54, 88, 143, ...
123, 132, 3412A0001241, 2, 4, 7, 11, 16, 22, 29, 37, 46, ...
123, 231, A0042751, 2, 4, 6, 8, 10, 12, 14, 16, 18, ...
123, 231, A0001241, 2, 4, 7, 11, 16, 22, 29, 37, 46, ...
123, 231, 4321 1, 2, 4, 6, 3, 1, 0, ...
132, 213, 1234A0000731, 2, 4, 7, 13, 24, 44, 81, 149, ...
213, 231, A0001241, 2, 4, 7, 11, 16, 22, 29, 37, 46, ...
Abbreviations used in the above table are summarized below.
abbreviation patterns in class
123, 132, 213, 232, 312, 321
1432, 2143, 3214, 4132, 4213, 4312
1234, 1243, 1324, 1342, 1423, 2134, 2314, 2341, 2413, 2431, 3124, 3142, 3241, 3412, 3421, 4123, 4231
1234, 1243, 1423, 1432
See also
Contained Pattern, Order Isomorphic, Permutation, Stanley-Wilf Conjecture, Wilf Class, Wilf Equivalent
Explore with Wolfram|Alpha
More things to try:
1/2 + 1/4 + 1/8 + 1/16 + ...
Does the set of perfect numbers contain 18?
logic circuit (p or ~q) and (r xor s)
References
Arratia, R. "On the Stanley-Wilf Conjecture for the Number of Permutations Avoiding a Given Pattern." Electronic J. Combinatorics6, No.1, N1, 1-4, 1999. S.; Jockusch, W.; and Stanley, R.P. "Some Combinatorial Properties of Schubert Polynomials." J. Alg. Combin.2, 345-374, 1993.Guibert, O. "Permutations sans sous séquence interdite." Mémoire de Diplô me d'Etudes Approfondies de L'Université Bordeaux I. 1992.Mansour, T. "Permutations Avoiding a Pattern from and at Least Two Patterns from ." 31 Jul 2000. R. and Schmidt, F.W. "Restricted Permutations." Europ. J. Combin.6, 383-406, 1985.Sloane, N.J.A. Sequences A000027/M0472, A000071/M1056, A000073/M1074, A000108/M1459, A000124/M1041, A000142/M1675, A004275, A033312, and A056986, A158005, and A158006 in "The On-Line Encyclopedia of Integer Sequences."Stankova, Z.E. "Forbidden Subsequences." Disc. Math.132, 291-316, 1994.West, J. "Generating Trees and Forbidden Subsequences." Disc. Math.157, 363-372, 1996.Wilf, H. "On Crossing Numbers, and Some Unsolved Problems." In Combinatorics, Geometry, and Probability: A Tribute to Paul Erdős. Papers from the Conference in Honor of Erdős' 80th Birthday Held at Trinity College, Cambridge, March 1993 (Ed. B.Bollobás and A.Thomason). Cambridge, England: Cambridge University Press, pp.557-562, 1997.
Referenced on Wolfram|Alpha
Permutation Pattern
Cite this as:
Weisstein, Eric W. "Permutation Pattern." From MathWorld--A Wolfram Resource.
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| 1 | Resistance and propulsion (propulsors) (PDF) | |
| 2 | Actuator disk Propeller testing - B series (PDF) | Actuator disk (PDF) |
| 3 | Design using Kt (Kq) curves (PDF) Detail design (PDF) | Propeller note book (PDF) |
| 4 | Cavitation (PDF) Waterjet notes (PDF) | |
| 5 | First law (PDF) | Summary of thermo (PDF) |
| 6 | Second law (PDF) Availability (PDF) | Keenan availabilty (PDF) |
| 7 | Propeller lifting line theory (Dr. Rich Kimball) | Propeller design plot example (PDF) Units (propulsors) (PDF) |
| 8 | Propeller lifting line theory (Dr. Rich Kimball) (cont.) | |
| 9 | Propeller lifting line theory (Dr. Rich Kimball) (cont.) | |
| 10 | Water properties (Prof. Doug Carmichael) (PDF) Rankine cycle (Prof. Doug Carmichael) (PDF) | |
| 11 | Rankine cycle vs. pressure and temperature (Prof. Doug Carmichael) (PDF) Practical Rankine cycle (Prof. Doug Carmichael) (PDF) Rankine cycle with regeneration (PDF) Rankine cycle vs. pressure with reheat (PDF) | |
| 12 | Combustion (PDF) | |
| 13 | Relationships for gases (PDF) | Properties of gases (PDF) |
| 14 | Basic dual cycle diesel notes (PDF) | |
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| 17 | Polytropic efficiency (PDF) Brayton cycle summary 2005 (PDF) | |
| 18 | Brayton cycle - irreversible examples (PDF) Open Brayton cycle (PDF) Creep (PDF) | |
| 19 | Electrical theory overview (PDF) | |
| 20 | Motors and generators overview (PDF) | |
| 21 | Electric propulsion presentation, guest lecturer Prof. Harbour | |
| 22 | Reliability and availability (PDF) Repairable systems supplement (PDF) | |
| 23 | Reduction gears notes (PDF) | Marine engineering (PDF) (Courtesy of SNAME. Used with permission.) |
| 24 | Gear geometry (PDF - 5.3 MB) Helical gear geometry (PDF) | |
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5
How do I prove that Ln(x+1) >x/(x+1), x>0?
Alexey Godin
Ph.D. in Mathematics & Economics, Moscow State University (Graduated 1998)
·
Author has 2.7K answers and 3.9M answer views
·
8y
Consider a function f(x)=ln(x+1)−xx+1.
So basically we need to prove that f(x)>0∀x>0.
f′(x)=1x+1−1(x+1)2=x(x+1)2>0∀x>0
Thus f(x) is a strongly increasing function.
As f(0)=0⇒f(x)>0∀x>0
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Sep 6
Consider f(x) = ln(1+x) − x/(1+x) for x > 0. Show f(x) > 0.
Compute derivative:
f′(x) = 1/(1+x) − [ (1+x) − x ]/(1+x)^2 = 1/(1+x) − 1/(1+x)^2 = x/(1+x)^2.
Since x > 0, f′(x) > 0 for all x > 0. Thus f is strictly increasing on (0, ∞).
Evaluate the limit at 0:
lim_{x→0+} f(x) = ln(1) − 0/(1) = 0.
Conclusion:
A strictly increasing function on (0, ∞) that has limit 0 at 0+ must satisfy f(x) > 0 for every x > 0. Therefore ln(1+x) − x/(1+x) > 0, i.e. ln(1+x) > x/(1+x) for all x > 0.
Alternate quick justification (Taylor / convexity perspective):
ln(1+x) has derivative 1/(1+x) which is decreasing, so
Consider f(x) = ln(1+x) − x/(1+x) for x > 0. Show f(x) > 0.
Compute derivative:
f′(x) = 1/(1+x) − [ (1+x) − x ]/(1+x)^2 = 1/(1+x) − 1/(1+x)^2 = x/(1+x)^2.
Since x > 0, f′(x) > 0 for all x > 0. Thus f is strictly increasing on (0, ∞).
Evaluate the limit at 0:
lim_{x→0+} f(x) = ln(1) − 0/(1) = 0.
Conclusion:
A strictly increasing function on (0, ∞) that has limit 0 at 0+ must satisfy f(x) > 0 for every x > 0. Therefore ln(1+x) − x/(1+x) > 0, i.e. ln(1+x) > x/(1+x) for all x > 0.
Alternate quick justification (Taylor / convexity perspective):
ln(1+x) has derivative 1/(1+x) which is decreasing, so ln(1+x) is concave. By the mean-value theorem or the integral representation ln(1+x) = ∫_0^x 1/(1+t) dt and comparing the integrand to the constant 1/(1+x) gives ln(1+x) = ∫_0^x 1/(1+t) dt > ∫_0^x 1/(1+x) dt = x/(1+x).
Reuven Harmelin
Lecturer at Technion - Israel Institute of Technology (1982–present)
·
Author has 2.3K answers and 1.9M answer views
·
3y
In fact, the more general inequality holds
with equality for and only for x=0, where both sides are equal to 0.
In the drawing below, the green graph is the graph of ln(x+1) and the violet is the graph of x/(x+1)
In order to prove that, let’s rewrite it as follows
Next substitute
and note that this inequality is equivalent to
with equality for and only for s=1. (see the attached drawing below)
Now make another substitution s=t+1, for t>-1 and observe that we now need to prove the inequality
instead of the original inequality.
Finally exponentiate both sides of that inequality and use the fact that the
In fact, the more general inequality holds
with equality for and only for x=0, where both sides are equal to 0.
In the drawing below, the green graph is the graph of ln(x+1) and the violet is the graph of x/(x+1)
In order to prove that, let’s rewrite it as follows
Next substitute
and note that this inequality is equivalent to
with equality for and only for s=1. (see the attached drawing below)
Now make another substitution s=t+1, for t>-1 and observe that we now need to prove the inequality
instead of the original inequality.
Finally exponentiate both sides of that inequality and use the fact that the exponential function is strictly increasing everywhere along the real axis, and therefore the original inequality is equivalent to the next basic exponential inequality
which holds for all real values of t, with equality for and just for
t=s-1=-x/(x+1)=0
The inequality exp(x)>=1+x is expressing the fact the the exponential function is convex and the line y=1+x is a tangent line to the graph of exp(x) at x=0, and therefore, this tangent must lie below the graph of exp(x) (see the attached drawing below)
Alexandru Carausu
Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010)
·
Author has 3K answers and 874.7K answer views
·
May 6
The two functions involved in the above inequality are defined (
i
) for
x
-
1 (the natural logarithm) , and (
ii
) , for the rational function
x
/ (
x
+ 1 ) , if
x
≠
-
1 ; the intersection of the two domains is the smaller one :
x
∈ (
-
1 ,
+∞
) . The sender of this question has taken into account an even smaller interval, ( 0 ,
+∞
) . Let us denote the two functions as
f
(
x
) = ln (
x
1 ) &
g
(
x
) =
x
/ (
x
1 ) . (1)
(1) ==>
f
( 0 ) =
g
( 0 ) = 0 ==>
O
( 0 , 0 ) ∈
G
_
f
∩ G
_
g
. (2)
The inequality to be proved means that, at right of 0 ,
f
(
x
) >
g
(
x
) : the graph
G
_
f
lies
above
G
_
g
.
The two functions involved in the above inequality are defined (
i
) for
x
-
1 (the natural logarithm) , and (
ii
) , for the rational function
x
/ (
x
+ 1 ) , if
x
≠
-
1 ; the intersection of the two domains is the smaller one :
x
∈ (
-
1 ,
+∞
) . The sender of this question has taken into account an even smaller interval, ( 0 ,
+∞
) . Let us denote the two functions as
f
(
x
) = ln (
x
1 ) &
g
(
x
) =
x
/ (
x
1 ) . (1)
(1) ==>
f
( 0 ) =
g
( 0 ) = 0 ==>
O
( 0 , 0 ) ∈
G
_
f
∩ G
_
g
. (2)
The inequality to be proved means that, at right of 0 ,
f
(
x
) >
g
(
x
) : the graph
G
_
f
lies
above
G
_
g
. The appropriate way to prove this consists in studying the variation (monotony) of the difference function
h
=
f
– g , and the proper “tool” to be used implies the (first) derivatives of the three functions:
f
ʹ
(
x
) = [ ln (
x
+ 1 ) ]
ʹ
= 1/(
x
+ 1 ) ;
g
ʹ
(
x
) = [
x
/(
x
+ 1 ) ]
ʹ
= . . . = 1/(
x
+ 1 )^2 . (3)
(3) ==>
h
ʹ
(
x
) =
f
ʹ
(
x
) –
g
ʹ
(
x
) = 1/(
x
+ 1 )
– 1/(
x
+ 1 )^2 = (
x
+ 1 – 1 ) / (
x
+ 1 )^2 . (4)
(4) ==>
h
ʹ
(
x
) = 1 / (
x
+ 1 )^2 > 0 over ( 0 ,
+∞
) ==>
( ∀
x
0 )
f
(
x
) >
g
(
x
)
. (5)
Final Comments: 1)
The inequality in the above question is thus proved. But let me add that it was possible to deduce it if the two graphs were plotted on the screen of a computer with a program of graphics. I have drawn the two graphs on a sheet of a copybook for Mathematics, an I have installed a program
SWP 5.5
(
Scientific WorkPlace
) which plots instantly the graphs of functions given by their analytic epxressions; but I cannot attach figures, mathematical equations, properly written matrices-determinants to my answers, edited by the TEXT format.
2)
It is easy to see that both functions
f
(
x
)
and
g
( x
) have a
common vertical asymptote
of equation (
x
=
-
1) ; their graphs are (very) close inside the strip (
-
1, 0 ) x (
-∞
, 0 ) .
G
_
f
&
G
_
g
meet at the origin
O
, but (then)
g
increases
at a lower rate
than
f
over the interval ( 0 ,
+∞
) ; this behaviour analytically follows from the expressions of the two derivatives in (3), but also from the limits
lim_( x
➛ +∞
) ln (
x
1 ) =
+∞
while lim_( x
➛ +∞
)
x
/ (
x
1 ) = 1 . (6)
3)
I’m closing my answer with the appreciation on this question to be a rather simple problem of
Calculus
, yet an interesting one. That’s why I have given a detailed answer. And I prefer to receive such a question instead of a dozen of
too simple
/
strange
/
improperly formulated
ones.
Related questions
How do you prove the inequality
x
x
+
1
<
ln
(
x
+
1
)
<
x
for
x
0
?
How do I prove that
l
n
(
1
+
1
/
x
)
1
/
(
1
+
x
)
(
1
+
x
)
,
x
0
?
How do I solve
x
−
ln
(
x
2
−
1
)
=
0
?
How do you prove the inequality
l
n
(
1
+
x
)
a
r
c
t
a
n
(
x
)
1
+
x
,
x
0
?
How do I evaluate
lim
x
→
∞
(
1
−
x
)
x
(
1
+
x
)
x
?
Prosun Ghosh
MSc from Vidyasagar University
·
7y
let us consider a function f(x)=ln(x+1)-x/(x+1)
now f(0)=0
differentiate w.r.t. x,we get,d/dx f(x)=1/x+1 +1/(x+1)
for every x>0,f'(x)>0.
thus f(x) is a monotone increasing function.
and hence f(x)>f(0)
ln(x+1)- x/x+1>0 for all x>0
ln(x+1)>x/x+1 for all x>0
Sponsored by Grammarly
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Vu Nguyen
Lecturer, Biostatistics
·
Author has 151 answers and 57.4K answer views
·
Updated 2y
Let u(t)=1+(t−1)et. Since t−1 and et are on t, then u is also increasing. Moreover, u is strictly positive for all positive t as u(0)=0.
Now take t=Ln(1+x) being positive with positive x , we get u=1+(1+x)(Ln(1+x)−1) strictly positive showing that Ln(x+1)>x/(x+1).
Pramesh Bhaila
Guide at Suryabinyak Academy (2018–present)
·
Author has 58 answers and 104K answer views
·
6y
Originally Answered: How do I prove that ln(1+1/x)>1/(1+x) (1+x),x>0 ?
·
Sol’n
We knew
lim ( x tend to infinity ) ln ( 1+ 1/x) = e^ x
Also,
Lim ( x ~> Infinity ) 1/ ( 1 + x)^2= 0
IF the above result was true, then the results obtain after takin limit to both should be correct.
Given that x > 0,
Then,
e^ x > 0 for all the value of x
Hence, the above expression is true.
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Updated Sep 16
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Ronald Deep
Worked at The University of Dayton
·
Author has 1.9K answers and 1.2M answer views
·
7y
Originally Answered: How do I prove that ln(1+1/x)>1/(1+x) (1+x),x>0 ?
·
Exponentiate both sides and compare exponents
(x + 1)/ x vs. 1/(1 + x) =>(1 + x) ^2 > x for x > 0
Note that 1 + x > 0 and thus can be multiplied across an inequality without changing signs.
Enrico Gregorio
Associate professor in Algebra
·
Author has 18.4K answers and 16M answer views
·
6y
For t>0, we have
1t+1>1(t+1)2
and therefore, for x>0,
∫x01t+1dt>∫x01(t+1)2dt
Computing the integrals
[log(t+1)]x0>[−1t+1]x0
which is exactly
log(x+1)>−1x+1+1=xx+1
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Imad Zghaib
BA in Mathematics & Engineering, Free University of Brussels (ULB) (Graduated 1989)
·
Author has 2.6K answers and 1.6M answer views
·
Apr 24
Related
How do I integrate between 0 and 1 in ln(x+1) /x²+1?
Simon Tsai
Lives in Taiwan
·
Author has 4.9K answers and 2M answer views
·
2y
Related
How do you prove
ln
(
x
+
1
)
<
x
√
x
2
+
1
<
sin
(
x
)
when
0
<
x
<
1
?
In this post, we are proving the following inequalities:
for [math]x \in \left( 0, 1 \right)[/math]. Proof 1 is trigonometric and Proof 2 is hyperbolic. All the diagrammes are made with Microsoft Paint.
[Proof 1]
Draw a trapezium [math]\text {ABOP}[/math] where [math]\overline {\text {AB}} = x[/math], [math]\overline {\text {AB}} \perp\overline {\text {BO}}[/math], [math]\overline {\text {BO}} = \overline {\text {OP}} = 1[/math], and [math]\overline {\text {AP}} \perp \overline {\text {AB}}[/math]. Draw a sector centred at [math]\text {O}[/math] passing through [math]\text {B}[/math] a
In this post, we are proving the following inequalities:
[math]\boxed {\displaystyle \ln \left( x + 1 \right) < \frac {x}{\sqrt {1 + {x}^{2}}} < \sin \left( x \right)} \tag{}[/math]
for [math]x \in \left( 0, 1 \right)[/math]. Proof 1 is trigonometric and Proof 2 is hyperbolic. All the diagrammes are made with Microsoft Paint.
[Proof 1]
Draw a trapezium [math]\text {ABOP}[/math] where [math]\overline {\text {AB}} = x[/math], [math]\overline {\text {AB}} \perp\overline {\text {BO}}[/math], [math]\overline {\text {BO}} = \overline {\text {OP}} = 1[/math], and [math]\overline {\text {AP}} \perp \overline {\text {AB}}[/math]. Draw a sector centred at [math]\text {O}[/math] passing through [math]\text {B}[/math] and [math]\text {P}[/math].
[math]\stackrel{\frown}{\text {BP}}[/math] cuts [math]\overline {\text {AO}}[/math] at [math]\text {C}[/math]. The sector [math]\text {BOC}[/math] being smaller than the triangle [math]\text {ABO}[/math] implies that [math]\stackrel{\frown}{\text {BC}} \; < \overline {\text {AB}}[/math]. Also, [math]\overline {\text {AB}} < \overline {\text {BP}} < \; \stackrel{\frown}{\text {BP}}[/math]. We therefore know that there exists such a point [math]\text {E}[/math] on [math]\stackrel{\frown}{\text {CP}}[/math] that [math]\stackrel{\frown}{\text {BE}} \; = \overline {\text {AB}}[/math]. Now, let [math]\text {D}[/math] and [math]\text {F}[/math] be on [math]\overline {\text {BO}}[/math] such that [math]\overline {\text {CD}}, \overline {\text {EF}} \perp \overline {\text {BO}}[/math]. Then,
[math]\displaystyle \frac {x}{\sqrt {1 + {x}^{2}}} = \overline {\text {CD}} < \overline {\text {EF}} = \sin \left( x \right) \tag{}[/math]
Hence, the rightmost inequality is proved.
[Proof 2]
Let [math]\text {O} \left( 0, 0 \right)[/math], [math]\text {A} \left( 1, 0 \right)[/math], and [math]\text {B} \left( \cosh \left( x \right), \sinh \left( x \right) \right)[/math]. (We have assumed the Cartesian coordinate system.) Then the area bounded by [math]\overline {\text {AO}}[/math], [math]\overline {\text {BO}}[/math], and [math]{x}^{2} - {y}^{2} = 1[/math] is [math]x / 2[/math], and the area of the triangle [math]\text {AOB}[/math] is [math]\sinh \left( x \right) / 2[/math], so [math]x < \sinh \left( x \right)[/math]. Let [math]x \mapsto \ln \left( 1 + x \right) / 2[/math] and see
[math]\displaystyle \ln \left( 1 + x \right) < 2 \sinh \Big( \frac {1}{2} \ln \left( 1 + x \right) \Big) = \sqrt {1 + x} - \frac {1}{\sqrt {1 + x}} = \frac {x}{\sqrt {1 + x}} \tag{}[/math]
The leftmost inequality follows if [math]x \in \left( 0, 1 \right)[/math].
Dhiraj Kumar
KNOWLEDGE & SHARP USE OF MIND IS THE POWER
·
Author has 107 answers and 272.9K answer views
·
7y
Related
How do you prove the inequality
[math]\frac{x}{x+1}< \ln(x+1)< x[/math]
for
[math]x > 0[/math]
?
Vijaykumar N. Joshi
Former Prof and Head , Dept. Of Computer Science , RETD (1980–2002)
·
Author has 1.9K answers and 1.7M answer views
·
5y
Related
How do I prove that limx->0 log a (1+x) /x=1/log a?
Related questions
How do I prove that
x
x
+
1
≤
ln
(
x
+
1
)
≤
x
?
How do I prove that
√
1
−
x
1
+
x
<
ln
(
1
+
x
)
arcsin
x
<
1
if
0
<
x
<
1
?
How can I solve
x
x
+
1
=
ln
(
x
)
ln
(
x
+
1
)
?
If
x
1
+
x
1
=
x
what is x ?
How do I solve
∫
1
0
2
x
1
+
x
ln
1
−
x
1
+
x
d
x
?
How do you prove the inequality
x
x
+
1
<
ln
(
x
+
1
)
<
x
for
x
0
?
How do I prove that
l
n
(
1
+
1
/
x
)
1
/
(
1
+
x
)
(
1
+
x
)
,
x
0
?
How do I solve
x
−
ln
(
x
2
−
1
)
=
0
?
How do you prove the inequality
l
n
(
1
+
x
)
a
r
c
t
a
n
(
x
)
1
+
x
,
x
0
?
How do I evaluate
lim
x
→
∞
(
1
−
x
)
x
(
1
+
x
)
x
?
If
1
x
= 1 what is x ?
Can you calculate
∫
1
0
(
x
x
)
(
x
x
)
(
x
x
)
(
x
x
)
…
d
x
?
If (x+1) +(x-1) =0, what is x?
Isn't ln(1+x) x could be anything, always 0 because of ln(1+x) =ln(1) ln(x) =0ln(x) =0?
If
−
x
x
=
−
1
,
what is x?
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Mechanical Engineering questions and answers
A model of a submarine, 1 : 15 scale, is to be tested at 195 ft/s in a wind tunnel with standard sea-level air, while the prototype will be operated in seawater. Determine the speed of the prototype to ensure Reynolds number similarity.
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Question: A model of a submarine, 1 : 15 scale, is to be tested at 195 ft/s in a wind tunnel with standard sea-level air, while the prototype will be operated in seawater. Determine the speed of the prototype to ensure Reynolds number similarity.
A model of a submarine, 1 : 15 scale, is to be tested at 195 ft/s in a wind tunnel with standard sea-level air, while the prototype will be operated in seawater. Determine the speed of the prototype to ensure Reynolds number similarity.
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To start with ensuring Reynolds number similarity between the model and the prototype, utilize the equation (d V m ν m)d l m V m=(d V p ν p)d l p V p and gather all known quantities for (V_m), (\nu_m), (l_m), and (l_p). We wish to match the Reynolds number between the model test in the wind tunnel (with air) and the pr... View the full answer Previous questionNext question
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(x^(-1) - y^(-1))/((xy)^(-1)(x - y)) = : Problem Solving (PS)
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(x^(-1) - y^(-1))/((xy)^(-1)(x - y)) =
Russ19
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Post URL Updated on: 11 Feb 2020, 02:34
Originally posted by Russ19 on 11 Feb 2020, 02:31.
Last edited by Bunuel on 11 Feb 2020, 02:34, edited 1 time in total.
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x−1−y−1(x y)−1(x−y)x−1−y−1(x y)−1(x−y) =
(A) - 1
(B) 1
(C) - xy
(D) xy
(E) 1 x y 1 x y
Source: GMAT Quantum
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BrentGMATPrepNow
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Post URL11 Feb 2020, 07:21
sjuniv32
x−1−y−1(x y)−1(x−y)x−1−y−1(x y)−1(x−y) =
(A) - 1
(B) 1
(C) - xy
(D) xy
(E) 1 x y 1 x y
Source: GMAT Quantum
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Rather than deal with performing a variety of operations involving many fractions, a fast approach is to plug some values into the original expression.
Let's see what happens when x = 2 and y = 4
We get: x−1−y−1(x y)−1(x−y)=2−1−4−1(2×4)−1(2−4)x−1−y−1(x y)−1(x−y)=2−1−4−1(2×4)−1(2−4)
Aside: 2−1=1 2=0.5 2−1=1 2=0.5, 4−1=1 4=0.25 4−1=1 4=0.25 and 8−1=1 8=0.125 8−1=1 8=0.125
So we get: =0.5−0.25(8)−1(−2)=0.5−0.25(8)−1(−2)
=0.5−0.25(0.125)(−2)=0.5−0.25(0.125)(−2)
=0.25−0.25=−1=0.25−0.25=−1
Since answer choice A is the only answer choice that evaluates to be -1 when x = 2 and y = 4, the correct answer is A
Cheers,
Brent
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BrentGMATPrepNow
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Post URL17 Jun 2020, 09:06
sjuniv32
x−1−y−1(x y)−1(x−y)x−1−y−1(x y)−1(x−y) =
(A) - 1
(B) 1
(C) - xy
(D) xy
(E) 1 x y 1 x y
Source: GMAT Quantum
Show more
A student asked me to solve the question without testing values. So here's an algebraic solution...
Let's examine and simplify the numerator and denominator individually
NUMERATOR: x−1−y−1=1 x−1 y=y x y−x x y=y−x x y x−1−y−1=1 x−1 y=y x y−x x y=y−x x y
DENOMINATOR: (x y)−1(x−y)=(1 x y)(x−y)=x−y x y=(−1)(−x+y)x y=(−1)(y−x)x y(x y)−1(x−y)=(1 x y)(x−y)=x−y x y=(−1)(−x+y)x y=(−1)(y−x)x y
Replace numerator and denominator with equivalent values to get: x−1−y−1(x y)−1(x−y)=y−x x y(−1)(y−x)x y x−1−y−1(x y)−1(x−y)=y−x x y(−1)(y−x)x y
Rewrite as: (y−x x y)(x y(−1)(y−x))(y−x x y)(x y(−1)(y−x))
Combine to get: (y−x)(x y)(x y)(−1)(y−x)(y−x)(x y)(x y)(−1)(y−x)
Simplify to get: 1−1 1−1, which equals −1−1
Answer: A
Cheers,
Brent
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Post URL24 Jun 2020, 19:39
Hi BrentGMATPrepNow
Can you explain how you got from (1/x)-(1/y) to (y/xy)-(x/xy) in the numerator?
Thanks in advance!
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Post URL25 Jun 2020, 07:12
LaurenGol
Hi BrentGMATPrepNow
Can you explain how you got from (1/x)-(1/y) to (y/xy)-(x/xy) in the numerator?
Thanks in advance!
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Take: 1 x 1 x
Create an equivalent fraction my multiplying top and bottom by y y to get: y x y y x y
Likewise, we can take: 1 y 1 y
And create an equivalent fraction my multiplying top and bottom by x x to get: x x y x x y
So, 1 x−1 y=y x y−x x y 1 x−1 y=y x y−x x y
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Post URL25 Jun 2020, 18:19
Got it. So essentially you manipulated the expression by multiplying the variables by xy/xy in order to have a common denominator.
Thank you!
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Post URL02 Jul 2020, 12:02
x^−1 − y^−1 / (xy)^−1(x−y) = [x^−1 / (xy)^−1(x−y)] - [y^−1 / (xy)^−1(x−y)]
= y/(x−y) - x/(x−y)
=(y-x)/(x-y)
= -1
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Post URL13 Sep 2023, 08:33
Here's the approach I took for future users if they find current solutions confusing:
x−1 y−1(x y)−1(x−y)x−1 y−1(x y)−1(x−y)
=x−1(1−y−1 x−1)(x y)−1(x−y)=x−1(1−y−1 x−1)(x y)−1(x−y) --> Factor out x−1 x−1
=x−1(1−x y−1)(x y)−1(x−y)=x−1(1−x y−1)(x y)−1(x−y) --> This part becomes x y−1 x y−1
=x−1(1−x y−1)(x y)−1(x−y)=x−1(1−x y−1)(x y)−1(x−y) --> Take the same approach to factor out y−1 y−1
=x−1 y−1(y−x)(x y)−1(x−y)=x−1 y−1(y−x)(x y)−1(x−y)
=(x y)−1(y−x)(x y)−1(x−y)=(x y)−1(y−x)(x y)−1(x−y) --> Exponent rules allow us to make this change and then cancel num and denom
=(y−x)(x−y)=(y−x)(x−y)
=−(x−y)(x−y)=−(x−y)(x−y)
=−1=−1
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proof of $\log(xy) =\log (x) + \log (y)$ [closed]
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11
$\begingroup$
$\log(xy) = \log (x) + \log ( y)$ and its division counter part were mentioned in an axiomatic way which I failed to proof.
I noted the correlation with the exponential rules of adding powers in case of multiplication of a single base and subtracting them in case of division but if we do proceed with that, how do we take log of terms linked with arithmetic operators ?
it doesn't have to be base $10$ but I picked it for convenience.
another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?
algebra-precalculus
logarithms
inverse-function
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edited Nov 7, 2016 at 21:39
sarah sarah
asked Nov 6, 2016 at 23:41
sarah sarah
47011 gold badge55 silver badges1313 bronze badges
$\endgroup$
5
1
$\begingroup$ Yes, do 10^ on both sides $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-11-06 23:44:50 +00:00
Commented Nov 6, 2016 at 23:44
12
$\begingroup$ Before you can prove anything about $\log$ you need a definition of $\log$. $\endgroup$
GEdgar
– GEdgar
2016-11-06 23:54:47 +00:00
Commented Nov 6, 2016 at 23:54
$\begingroup$ What's up with close votes? The question is (1) clear (2) has context. There may be a duplicate somewhere but that's not why people seem to be voting. $\endgroup$
MathematicsStudent1122
– MathematicsStudent1122
2016-11-07 07:12:06 +00:00
Commented Nov 7, 2016 at 7:12
$\begingroup$ This is direct result of the definition of $\log_b z$ being defined as the number, $y$ so that $b^y = z$. $\log_b xy$ is the $a$ so that $b^a = xy$ $\log_b x$ is the $c$ so that $b^c = x$ and $\log_b y$ is the $d$ so that $b^d = y$ so $b^a = xy = b^cb^d = b^{c+d}$. The number require to raise b so that it equals xy, is the sum of the numbers required to raise b to get x and to get y. $\endgroup$
fleablood
– fleablood
2016-11-07 21:47:55 +00:00
Commented Nov 7, 2016 at 21:47
$\begingroup$ "another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?" Syntactically, I don't think that sentence makes any sense. But if you mean, what I think you mean the yes. $10^{\log x} = x$ so $10^{\log x + \log y} = 10^{\log x}10^{\log y} = xy = 10^{\log xy}$. $\endgroup$
fleablood
– fleablood
2016-11-07 21:52:02 +00:00
Commented Nov 7, 2016 at 21:52
Add a comment |
6 Answers 6
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41
$\begingroup$
Let $r=\log(x), s=\log(y)$ and $t=\log(xy)$.
Then $x=10^r,y=10^s$ and $xy=10^t=10^r10^s=10^{r+s}$.
Therefore, $t=r+s$, that is, $\log(xy)=\log(x)+\log(y)$.
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answered Nov 6, 2016 at 23:53
John Wayland BalesJohn Wayland Bales
22.4k22 gold badges2222 silver badges5656 bronze badges
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8
$\begingroup$ +1 for no calculus. Plain and simple is all it needs to be. $\endgroup$
Fine Man
– Fine Man
2016-11-07 06:22:29 +00:00
Commented Nov 7, 2016 at 6:22
Add a comment |
19
$\begingroup$
Using a different definition of the logarithm, $\ln(x) := \int_1^x \frac{dt}{t}$, we can prove this without a definition of exponentiation.
$$ \ln(ab) = \int_1^{ab}\frac{dt}{t} = \int_1^a\frac{dt}{t} + \int_a^{ab} \frac{dt}{t} = \ln(a) + \int_a^{ab} \frac{dt}{t} $$ Next we can do a u-substitution with $u= \frac ta$, meaning the limits of the remaining integral can be re-expressed as $1$ and $b$. Also, note that $a du = dt$ as well as $au = t$, meaning we find that $ \int_a^{ab} \frac{dt}{t} = \int_1^b \frac{a\,du}{au} = \int_1^b \frac{du}{u} = \ln(b)$.
Now we find that $\ln(ab) = \ln(a) + \ln(b)$, using just basic properties of integrals.
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answered Nov 7, 2016 at 2:07
PTNobelPTNobel
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10
$\begingroup$
Considering that $\log(x)$ is the inverse function of $e^x$ you have
$$e^{\log(xy)} = xy = e^{\log(x)} \cdot e^{\log(y)} = e^{\log(x)+\log(y)}$$
But since the exponential function is injective you get $\log(xy) = \log(x)+\log(y)$
Share
answered Nov 6, 2016 at 23:51
cronos2cronos2
1,9671313 silver badges1919 bronze badges
$\endgroup$
5
$\begingroup$ Personally, I'd just go with log base $b$ for those who haven't seen $e$. $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-11-07 00:04:54 +00:00
Commented Nov 7, 2016 at 0:04
$\begingroup$ @SimpleArt I suppose if someone hasn't seen $e$, it could just as well be an arbitrary base. $\endgroup$
David Z
– David Z
2016-11-07 04:40:51 +00:00
Commented Nov 7, 2016 at 4:40
2
$\begingroup$ @SimpleArt Isn't it unlikely that someone has seen enough of $log$ to know that $log(xy)=log(x)+log(y)$, but has not seen $e$ yet? $\endgroup$
Right Leg
– Right Leg
2016-11-07 05:02:07 +00:00
Commented Nov 7, 2016 at 5:02
$\begingroup$ i m sorry , i used log in the maths way where it stands for logarithm of base ten not as the natural logarithm , as to the answer , i will check if i understand it . $\endgroup$
sarah
– sarah
2016-11-07 21:24:01 +00:00
Commented Nov 7, 2016 at 21:24
$\begingroup$ @sarah well, then $\log(x)$ is just the inverse of $10^x$ $\endgroup$
cronos2
– cronos2
2016-11-07 21:25:04 +00:00
Commented Nov 7, 2016 at 21:25
Add a comment |
4
$\begingroup$
$\log xy = \log x + \log y\ 10^{\log xy} = 10^{(\log x + \log y)}\ 10^{\log xy} = (10^{\log x})(10^{\log y})\ xy = (x)(y)$
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answered Nov 6, 2016 at 23:47
Doug MDoug M
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$\begingroup$
I want to take a geometric view here. I promise it won't be too bad! Let's start with the fact that, with a base $b$,
$b^{n}b^{m} = b^{n+m}$
Which makes sense just from thinking about groupings of numbers,
$(b\cdot b \cdots b)_{ntot}\cdot(b \cdot b \cdots b)_{mtot} = (b\cdot b \cdots b)_{ntot+mtot}$.
Let's define a function $f(x)$ to be,
$f(x)=b^x$.
Then we have the properties,
$f(n+m) = b^{n+m} = b^n b^m = f(n)f(m)$,
from what I discussed in the first sentence. But what does this actually mean? If we look at a graph (which I've crudely drawn because I am not good with drawing software),
This shows graphically what is happening with $n$ and $m$ and how the function responds (sorry for high quality?). But what about $log_b (x)$? Let's define a new function
$g(x) = log_b (x)$.
It turns out that $log_b(x)$ and $b^x$ are related by a reflection about the line $ y = x $ on a Cartesian grid. This is shown very nicely in this picture:
If you imagine the line $ y = x $ going straight through the origin you'll see that they are just reflections of each other! What does this have to do with your question? Well what if we imagine taking the $x$ axis in my original plot, and treating that as our new '$y$' axis (vertical axis), with the $f(x)$ axis being the $g(x)$ axis (horizontal axis). I've drawn another sketch (rotated by 90 degrees counterclockwise on right):
Now usually we like our positive direction to be to the right, so we'll flip it and I'll replot $n$ and $m$ on this graph:
Since it was just a reflection, we're seeing another exponentially increasing function! $n$ and $m$ are on our vertical 'x' axis and $g(n)$ and $g(m)$ are on our horizontal '$g(x)$' axis. But we know on the vertical axis the two values multiply to give a new value, then on the horizontal axis those two numbers add to give the final value for exponentially increasing functions! All we've done is swapped the axis and used the properties of $f(x)=b^x$ to show that,
$g(n)+g(m) = g(nm)$
For $g(x)=\log_b(x)$! Plugging in,
$\log_b(n) + \log_b(m) = \log_b(nm).$
Share
edited Dec 30, 2016 at 16:29
grg
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answered Nov 7, 2016 at 6:19
DougDoug
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$\begingroup$ appreciate this answer even tho it took time to be intuitive to +1 for using plain functions $\endgroup$
sarah
– sarah
2016-11-07 21:30:51 +00:00
Commented Nov 7, 2016 at 21:30
Add a comment |
3
$\begingroup$
Let $b, x, y > 0$, with $b \neq 1$. Suppose that $log_b x = u$ and that $\log_b y = v$. Then, by definition of the logarithm, $b^u = x$ and $b^v = y$. Therefore, $$\log_b{xy} = \log_b(b^ub^v) = \log_b b^{u + v} = u + v = \log_b x + \log_b y$$ and $$\log_b\left(\frac{x}{y}\right) = \log_b \left(\frac{b^u}{b^v}\right) = \log_b b^{u - v} = u - v = \log_b x - \log_b y$$
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answered Nov 6, 2016 at 23:53
N. F. TaussigN. F. Taussig
79.3k1414 gold badges6262 silver badges7777 bronze badges
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ResourcesAops Wiki 2023 AMC 12B Problems/Problem 6
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2023 AMC 12B Problems/Problem 6
The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5: Graphing
7 Solution 6: Solution 3 with powers
8 Video Solution 1 by OmegaLearn
9 Video Solution
10 Video Solution by Interstigation
11 See Also
Problem
When the roots of the polynomial
are removed from the number line, what remains is the union of disjoint open intervals. On how many of these intervals is positive?
Solution 1
The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about . We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are intervals.
~Aopsthedude
Solution 2
The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on , , , , and . The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is .
~darrenn.cp ~DarkPheonix
Solution 3
We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
First, we evaluate any value on the interval . Since the degree of is = = , and every term in is negative, multiplying negatives gives a negative value. So is a negative interval.
We know that the roots of are at . When the degree of the term of each root is odd, the graph of will pass through the graph and change signs, and vice versa. So at , the graph will change signs; at , the graph will not, and so on.
This tells us that the interval is positive, is also positive, is negative, is also negative, and so on, with the pattern being .
The positive intervals are therefore , , , , , and , for a total of .
~nm1728 ~ESAOPS (minor edits)
Solution 4
Denote by the interval for and the interval .
Therefore, the number of intervals that is positive is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5: Graphing
Recall two key facts about the roots of a polynomial: - If a root has an odd multiplicity (e.g. the root appears an odd number of times), then the graph will cross the x-axis. - If a root has an even multiplicity (e.g. the root appears an even number of times), then the graph will bounce off the x-axis.
Sketching the graph and noting the multiplicity of the roots, we see that there are positive intervals.
Solution 6: Solution 3 with powers
When the range of P(x) is from −∞ to 1, the sign is negative because in
, the terms corresponding to x=1,3,5,7, and 9 are all negative, and since they have odd exponents, they contribute a negative sign. The other terms with even exponents (like (x−2)2, (x−4)4, etc.) are positive regardless of the sign inside, so they don't affect the overall sign. However, in the range 1 to 2, the power of flips, causing this range to be positive. For the range - , the sign of any even inside the exponent doesn't matter, so this range stays positive. Continueing this pattern, We see the pattern is and so on. there are 3 complete cycles, so ~BlueKite
Video Solution 1 by OmegaLearn
~OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
~Interstigation
See Also
2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11Followed by
Problem 13
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 10 Problems and Solutions
2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5Followed by
Problem 7
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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Which is more efficient for iterating over an array of objects in Javascript: using a 'for' loop or 'forEach'? Why?
Full-Stack Software Developer & SEO Specialist
·
4y
The short answer is for loop.
But it depends on you how you use the loop in your code.
For Loop: An iteration structure that’s useful when the number of iterations is known. A for loop’s declaration has three components: initialization of a variable, expression to determine if the loop will end, and a statement that’s executed at the end of each iteration.
.forEach() Method: A method of the array class that takes a single argument, which is a function, and performs that function with every element within the array.
To test the speed of each loop, I created arrays of 10, 1000, and 100000 integers.
The short answer is for loop.
But it depends on you how you use the loop in your code.
For Loop: An iteration structure that’s useful when the number of iterations is known. A for loop’s declaration has three components: initialization of a variable, expression to determine if the loop will end, and a statement that’s executed at the end of each iteration.
.forEach() Method: A method of the array class that takes a single argument, which is a function, and performs that function with every element within the array.
To test the speed of each loop, I created arrays of 10, 1000, and 100000 integers. I used performance.now() method to test the speed.
All the times are in milliseconds.
Now you have to choose which one you want to use.
Related questions
When should I use each, forEach or a simple for loop in Javascript?
How do you loop in arrangement of objects in JavaScript (Javascript, array, for loop, object, development)?
How do I make a for loop in an array in JavaScript for beginners?
How do you iterate through a JavaScript array with foreach and while?
How do you handle a for loop applied to an array in JavaScript (JavaScript development)?
Logan R. Kearsley
BS in CS and 12 years of work experience in programming.
·
Author has 8.7K answers and 8.1M answer views
·
5y
Originally Answered: Is forEach faster than for Javascript?
·
That depends.
forEach has function call overhead on every iteration, and may require allocating a new function closure whenever it is run. A basic C-style for loop will not have these issues, and should be faster in most cases.
A for-of loop, however, also has function call overhead, as it must call the next() method of the iterator protocol on every iteration. It will never allocate a closure, but does have it’s own start-up costs associated with creating an iterator.
So, a C-style loop will almost always be the fastest, but whether forEach or for-of are significantly slower, and, if so, which o
That depends.
forEach has function call overhead on every iteration, and may require allocating a new function closure whenever it is run. A basic C-style for loop will not have these issues, and should be faster in most cases.
A for-of loop, however, also has function call overhead, as it must call the next() method of the iterator protocol on every iteration. It will never allocate a closure, but does have it’s own start-up costs associated with creating an iterator.
So, a C-style loop will almost always be the fastest, but whether forEach or for-of are significantly slower, and, if so, which of those two is faster, will depend on individual circumstances. If it matters, use a profiler.
Mike Dougherty
Sr.Developer at UHS (2014–present)
·
Author has 65 answers and 27.8K answer views
·
1y
This question strikes me like asking which is sharper: a sugeon's scalpel or a lumberjack's ax?
You won't manage to cut firewood with a scalpel or perform surgery with an ax. Use the proper tool for the job.
If clearly readable code is important:
for(i=0;i<10;i++){ / use i / }
boxes.forEach(box => { / use box / })
The first one only tells us that something happens ten times… but that's not even guaranteed because I could arbitrarily increment or decrement that variable inside the body, so ‘something' might actually happen fewer or more than ten times. Sure, this is an unlikely scenario but illu
This question strikes me like asking which is sharper: a sugeon's scalpel or a lumberjack's ax?
You won't manage to cut firewood with a scalpel or perform surgery with an ax. Use the proper tool for the job.
If clearly readable code is important:
for(i=0;i<10;i++){ / use i / }
boxes.forEach(box => { / use box / })
The first one only tells us that something happens ten times… but that's not even guaranteed because I could arbitrarily increment or decrement that variable inside the body, so ‘something' might actually happen fewer or more than ten times. Sure, this is an unlikely scenario but illustrates that it is more likely to “oops" if the loop control variable is available to the program. Compare with .forEach() - no chance to stumble over a loop control variable because there is none, and you also know that function will run for each box element in the boxes array.
I suppose next you could ask about for() vs .some() or .every() but my answer would be very similar. 🙂
Shiv
Working professionally on JS
·
6y
Originally Answered: Which is faster 'for' or 'forEach' in JavaScript?
·
In general the for loop is faster than forEach. If you check the polyfill of forEach it does a lot of things that you may not require in general.
Comparison here for vs forEach · jsPerf
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Related questions
How can I use a for loop to iterate over an Array in JavaScript?
How do you loop through multiple objects and get similar key values in a new object (JavaScript, arrays, object, development)?
How do you do a double foreach loop with JavaScript condition (JavaScript development)?
What is the equivalent of a for-each loop in JavaScript? How would you implement it?
Why doesn't JavaScript have a normal foreach loop?
Subramanyam.D
Former Frontend React Trainee at DCT Academy (2021–2021)
·
Author has 1.4K answers and 450.7K answer views
·
4y
Originally Answered: Which is faster, foreach or for loop JavaScript?
·
In terms of faster execution.
The hierachy is,
for…loop>forEach>for of loop(ES6).
You can use ‘break’ loop control for…loop not forEach.
Traditional for…loop is so fast.
Rafał Klinowski
Freelance .NET Developer
·
4y
Originally Answered: Which is faster, foreach or for loop?
·
For arrays, the difference between for and foreach is so minimal it can be omitted. The performance is very much the same.
For lists, and generally other non-array collections, foreach is significantly slower than for. The reason is because it doesn’t naturally go over elements one after one, but rather reference elements using MoveNext or Current.
Generally, for lists, the hierarchy is as follows:
for
foreach (loop)
ForEach (LINQ)
ForEach has the worst performance because it uses a delegate (Action).
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Alan Mellor
Started programming 8 bit computers in 1981
·
Author has 13.5K answers and 102.3M answer views
·
Updated Jan 8
Related
Do you think the .forEach() method is better than using a for loop?
I do; forEach() is better by far.
It expresses the intent of the code much more clearly. forEach says “I don’t care how many items there are, or even if there are any - but I want to do this to all of them”.
A traditional counting loop like for ( int i=0; i < thing.size(); i++ ) says something else to me. It jams mechanics right up in your face.
Does it start from one or zero? Is the end condition less than or less than or equal? What is the maximum size limit for an int and would we ever exceed it?
I’m thinking all that stuff when we could have chosen a construct that more clearly says “I just wa
I do; forEach() is better by far.
It expresses the intent of the code much more clearly. forEach says “I don’t care how many items there are, or even if there are any - but I want to do this to all of them”.
A traditional counting loop like for ( int i=0; i < thing.size(); i++ ) says something else to me. It jams mechanics right up in your face.
Does it start from one or zero? Is the end condition less than or less than or equal? What is the maximum size limit for an int and would we ever exceed it?
I’m thinking all that stuff when we could have chosen a construct that more clearly says “I just want to do this thing for each item”.
Trausti Thor Johannsson
Computer programmer
·
Author has 19.6K answers and 101.9M answer views
·
Jan 28
Related
Do you think the .forEach() method is better than using a for loop?
I take it that you are talking about Javascript.
I like .forEach, .map, .filter a lot. I use them, a lot. There are however things you should consider.
Each .forEach is significantly slower than doing a for loop.
If you do a for loop like this
for (let i = 0, c = list.length; i < c; i++) { ...
It is significantly faster than doing a forEach. Why? Because each forEach call is a new function, also, it calls the length of the list only once. Pretty smart stuff.
The Dark Side of 'Foreach()' : Why You Should Think Twice Before Using It
The other day, I was writing some code, and it just wasn't working for me.
"community.appsmith.com")
The speed difference is reallt dramatic, as in that it is at least around twice as slow, you can’t exit the
I take it that you are talking about Javascript.
I like .forEach, .map, .filter a lot. I use them, a lot. There are however things you should consider.
Each .forEach is significantly slower than doing a for loop.
If you do a for loop like this
for (let i = 0, c = list.length; i < c; i++) { ...
It is significantly faster than doing a forEach. Why? Because each forEach call is a new function, also, it calls the length of the list only once. Pretty smart stuff.
The Dark Side of 'Foreach()' : Why You Should Think Twice Before Using It
The other day, I was writing some code, and it just wasn't working for me.
"community.appsmith.com")
The speed difference is reallt dramatic, as in that it is at least around twice as slow, you can’t exit the loop and there are other penalties as well. The amount of memory use is also noticable.
For a small list of less than 100 items, guaranteed, why not, it makes your code more readable. But anything above that and your for loop is taking up real time, you don’t need a large list to run in a high amount of milliseoncds. It adds up quickly. Do you have 4–5 drop down boxes or lists on the same page? Your users might even notice how slow it feels.
There is a time and place for everything and with experience you learn when to use what.
A short story I often tell people. My father in law was extremely good handy man, there was nothing the guy did not know, he was super good with electricity, plumbing, carpentry, you name it. One of the hardest fields in house building is getting doors and windows correct and he was extremely good. So one time I was fixing stuff at home and he was staying with us for a while. In my view he was behaving like a toddler complaining I did not have the right tools. What do you mean, I have a small hammer, larger hammer and a sledge I said. He looked at me like I was speaking gibberish, never forget this look. If the man ever though I was stupid, this was the proof, look. So, I caved and went shopping with the guy. I wasn’t extremely happy when I came back, with 6–7 new hammers. Not happy at all because this stuff costs actual money, it isn’t cheap, times 6–7.
But as soon as we got going, putting together stuff, I needed to swallow my pride many times. His hammer purchases were great. When you are dealing with window frames for example, a tiny hammer, as far as I know it is even called window frame hammer (who would have known?), allowed you to use tiny thin nails and it did not leave a mark on the window frame.
This was extreme, I could have done the job with my 2–3 hammers I already owned, but it would have been crap work which I would have regretted.
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Wim ten Brink
Studied at Hogere Beroepsopleiding (Graduated 1987)
·
Author has 2.2K answers and 8.4M answer views
·
Updated 5y
Related
Do you think the .forEach() method is better than using a for loop?
Many already say “Yes”. Which makes sense. Wrong, but it makes sense! They just don’t get the difference between for() and foreach(), though. In reality, both methods have completely different purposes!
Both numbers are generally used for enumerating data. And the for() method is used when you need some fine control over how the data gets enumerated. You’re basically defining the initial state, the end condition and the stepping method for any loop. The most common usage would be a simple loop like for(int i = 0; i < 10; i++) {…} is just the simplest way to do loops. But what many don’t realize
Many already say “Yes”. Which makes sense. Wrong, but it makes sense! They just don’t get the difference between for() and foreach(), though. In reality, both methods have completely different purposes!
Both numbers are generally used for enumerating data. And the for() method is used when you need some fine control over how the data gets enumerated. You’re basically defining the initial state, the end condition and the stepping method for any loop. The most common usage would be a simple loop like for(int i = 0; i < 10; i++) {…} is just the simplest way to do loops. But what many don’t realize is that the for() loop actually uses three methods and can thus walk through any collection of data in various ways.
The foreach() method differs in that it gives no control over the way you enumerate the data. This is defined by the enumeration class itself. A foreach() method will call the GetEnumerator() method of any enumerating object and this method will return a single record every time until it’s out of data. Then it returns null and the foreach() loop knows it has ended.
This actually means that the for() method gives you far more control over walking through data than the foreach() method. But in most situation, the default enumerator is just fine for your purposes.
A lot of developers underestimate how powerful the for() method can be. And how it actually serves a different purpose than the foreach() method. They consider the for() loop to be some simple counter and generally use it that way. Thus they will answer that the foreach() method is better…
But they are wrong! For() and ForEach() cannot be compared! They have very different, yet similar purposes. I think that in general, the ForEach() method is good enough so you will rarely find the for() method more practical. But don’t focus on foreach() as if it’s a silver bullet! It’s not!
Alan Mellor
Self taught programmer starting in 1981
·
Author has 13.5K answers and 102.3M answer views
·
3y
Related
Isn’t forEach() an iterating method in JavaScript?
Yes. That’s a great way of looking at it from an imperative perspective.
It is a fancy - and preferred - way of saying “here are a bunch of things. Call the function I’ve given you on each of those things”
You can also consider it as part of a chain of transforms in a more functional or declarative sense: “Transform things using this function, given this set of many things”
That second way hints at some declarative programming power.
An iterator implies that your group of things gets processed one after the other. A forEach does not imply that. It says that your function will be applied for each t
Yes. That’s a great way of looking at it from an imperative perspective.
It is a fancy - and preferred - way of saying “here are a bunch of things. Call the function I’ve given you on each of those things”
You can also consider it as part of a chain of transforms in a more functional or declarative sense: “Transform things using this function, given this set of many things”
That second way hints at some declarative programming power.
An iterator implies that your group of things gets processed one after the other. A forEach does not imply that. It says that your function will be applied for each thing. That could happen in parallel, with each thing being allocated to a different CPU or GPU core. Maybe even a different computer across a network (in theory; practically, not so much).
Like all source code constructions, it’s more about what this tells our fellow programmer as a reader of the code. For me, I use an iterator to say “do this stuff in sequence” and forEach to say “Get all this work done, series or parallel, it doesn’t matter”
The concepts of map and reduce in functional programming take that idea a step further.
Mohit
IT and Web Services at Freelancing
·
Author has 109 answers and 128.5K answer views
·
3y
Related
How does a for loop know which array to count in JavaScript?
This is How does a for loop know which array to count in JavaScript
Description. JavaScript array length property returns an unsigned, 32-bit integer that specifies the number of elements in an array.
Syntax. Its syntax is as follows − array.length.
Return Value. Returns the length of the array.
Example. Try the following example. ...
Output. arr.length is : 3.
This is How does a for loop know which array to count in JavaScript
Description. JavaScript array length property returns an unsigned, 32-bit integer that specifies the number of elements in an array.
Syntax. Its syntax is as follows − array.length.
Return Value. Returns the length of the array.
Example. Try the following example. ...
Output. arr.length is : 3.
Rick Viscomi
web dev @ google
·
Author has 243 answers and 2.7M answer views
·
9y
Related
When should I use each, forEach or a simple for loop in Javascript?
each is a jQuery method, non-native to JavaScript. Use that only if jQuery is included on the page.
All other times, the question is really whether to use the forEach method or the old school for loop. But it’s important to point out that there are different variations of loops with the for keyword. Here’s how I’d distinguish each use case:
If the data I’m iterating over is an object, I would use a for loop. Specifically, the for..in loop. There’s no better way to iterate over the properties of an object.
If I’m writing ES6, I would use a for..of loop if possible. This allows me to declare the
each is a jQuery method, non-native to JavaScript. Use that only if jQuery is included on the page.
All other times, the question is really whether to use the forEach method or the old school for loop. But it’s important to point out that there are different variations of loops with the for keyword. Here’s how I’d distinguish each use case:
If the data I’m iterating over is an object, I would use a for loop. Specifically, the for..in loop. There’s no better way to iterate over the properties of an object.
If I’m writing ES6, I would use a for..of loop if possible. This allows me to declare the local variable representing each item in the iterable quite easily without the need for additional functions.
If I’m stuck in ES5 and I needed to do something like return or break early from the loop, I would use a standard for loop. You know, the kind with i++. You can write any loop with this syntax, so it gives you the most control. If you need to return early from the loop, this is the best way to do it.
If I’m stuck in ES5 and I’m in no rush to short-circuit the loop, I’d use forEach.
Ganesh Pandey
Studied at Institute of Engineering, Pulchowk Campus
·
8y
Related
In JavaScript, how can I use a "for" loop to cycle through each item in an array?
You can use forEach() or map() ES5 iterative method for arrays. Each of the methods accepts two arguments.
A function to run on each item and
Optional scope object in which to run the function (affecting the value of this).
The function passed into one of these methods will receive three arguments. The array item value, the position of the item in the array, and the array object itself.
forEach(): Runs the given function on every item in the array. This method has no return value.
map(): Runs the given function on every item in the array and returns the result of each function call in an array.
You can use forEach() or map() ES5 iterative method for arrays. Each of the methods accepts two arguments.
A function to run on each item and
Optional scope object in which to run the function (affecting the value of this).
The function passed into one of these methods will receive three arguments. The array item value, the position of the item in the array, and the array object itself.
forEach(): Runs the given function on every item in the array. This method has no return value.
map(): Runs the given function on every item in the array and returns the result of each function call in an array.
var numbers = [1,2,3,4,5,4,3,2,1];
var mapResult = numbers.map(function(item, index, array){
return item 2;
alert(mapResult); //[2,4,6,8,10,8,6,4,2]
numbers.forEach(function(item, index, array){//do something here});
Related questions
When should I use each, forEach or a simple for loop in Javascript?
How do you loop in arrangement of objects in JavaScript (Javascript, array, for loop, object, development)?
How do I make a for loop in an array in JavaScript for beginners?
How do you iterate through a JavaScript array with foreach and while?
How do you handle a for loop applied to an array in JavaScript (JavaScript development)?
How can I use a for loop to iterate over an Array in JavaScript?
How do you loop through multiple objects and get similar key values in a new object (JavaScript, arrays, object, development)?
How do you do a double foreach loop with JavaScript condition (JavaScript development)?
What is the equivalent of a for-each loop in JavaScript? How would you implement it?
Why doesn't JavaScript have a normal foreach loop?
Why might you prefer to use for loop instead of built-in functions that loop through arrays and objects in JavaScript?
Which is faster 'for' or 'forEach' in JavaScript?
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From Simple English Wikipedia, the free encyclopedia
In algebra, a montonic function is any function whose gradient never changes sign. In simple words, it is a function which is either always increasing or decreasing. For example, sin x=f(x) isn't a monotonous function or simply, non monotonous function. That is to say there are no turning points but there may be stationary points where the gradient is momentarily zero. The derivative function of a monotonic function which describes its gradient will never change sign.
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↑ Monotonic Function Mathworld , Wolfram alpha , Accessed March 2017
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What Is the Density of Air at STP?
How the Density of Air Works
photo-graphe/Pixabay
What is the density of air at STP? In order to answer the question, you need to understand what density is and how STP, or standard temperature and pressure, is defined.
Key Takeaways: Density of Air at STP
The density of air is the mass per unit volume of atmospheric gases. It is denoted by the Greek letter rho, ρ. The density of air, or how light it is, depends on the temperature and pressure of the air. Typically, the value given for the density of air is at STP.
STP is one atmosphere of pressure at 0 degrees C. Since this would be a freezing temperature at sea level, dry air is less dense than the cited value most of the time. However, air typically contains a lot of water vapor, which would make it denser than the cited value.
The Density of Air Values
The density of dry air is 1.29 grams per liter (0.07967 pounds per cubic foot) at 32 degrees Fahrenheit (0 degrees Celsius) at average sea-level barometric pressure (29.92 inches of mercury or 760 millimeters).
Affect of Altitude on Density
The density of air decreases as you gain altitude. For example, the air is less dense in Denver than in Miami. The density of air decreases as you increase temperature, providing the volume of the gas is allowed to change. As an example, air would be expected to be less dense on a hot summer day versus a cold winter day, providing other factors remain the same. Another example of this would be a hot air balloon rising into a cooler atmosphere.
STP Versus NTP
While STP is standard temperature and pressure, not many measured processes occur when it's freezing. For ordinary temperatures, another common value is NTP, which stands for normal temperature and pressure. NTP is defined as air at 20 degrees C (293.15 K, 68 degrees F) and 1 atm (101.325 kN/m2, 101.325 kPa) of pressure. The average density of air at NTP is 1.204 kg/m3 (0.075 pounds per cubic foot).
Calculate the Density of Air
If you need to calculate the density of dry air, you can apply the ideal gas law. This law expresses density as a function of temperature and pressure. Like all gas laws, it is an approximation where real gases are concerned but is very good at low (ordinary) pressures and temperatures. Increasing temperature and pressure adds error to the calculation.
The equation is:
ρ = p / RT
where:
Sources
“Air Density and Specific Weight Table, Equations and Calculator.” Engineers Edge, LLC.
Dry Air Density a IUPAC, www.vcalc.com.
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Fundamentals of Chemical Reaction Engineering Fundal11entals of Chel11ical Reaction Engineering Mark E. Davis California Institute of Technology Robert J. Davis University ofVirginia Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto McGraw-Hill Higher Education 'ZZ A Division of The MGraw-Hill Companies FUNDAMENTALS OF CHEMICAL REACTION ENGINEERING Published by McGraw-Hili, a business unit of The McGraw-Hili Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2003 by The McGraw-Hili Companies, Inc. All rights reserved.
No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hili Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the United States.
This book is printed on acid-free paper.
International Domestic 1234567890DOCroOC098765432 1234567890DOCroOC098765432 ISBN 0-07-245007-X ISBN 0-07-119260-3 (ISE) Publisher: Elizabeth A. Jones Sponsoring editor: Suzanne Jeans Developmental editor: Maja Lorkovic Marketing manager: Sarah Martin Project manager: Jane Mohr Production supervisor: Sherry L. Kane Senior media project manager: Tammy Juran Coordinator of freelance design: Rick D. Noel Cover designer: Maureen McCutcheon Compositor: TECHBOOKS Typeface: 10/12 Times Roman Printer: R. R. Donnelley/Crawfordsville, IN Cover image: Adaptedfrom artwork provided courtesy ofProfessor Ahmed Zewail's group at Caltech.
In 1999, Professor Zewail received the Nobel Prize in Chemistry for studies on the transition states of chemical reactions using femtosecond spectroscopy.
Library of Congress Cataloging-in-Publication Data Davis, Mark E.
Fundamentals of chemical reaction engineering / Mark E. Davis, Robert J. Davis. -1st ed.
p.
em. -(McGraw-Hili chemical engineering series) Includes index.
ISBN 0-07-245007-X (acid-free paper) -ISBN 0-07-119260-3 (acid-free paper: ISE) I. Chemical processes.
I. Davis, Robert J.
II. Title.
III. Series.
TP155.7 .D38 2003 660'.28-dc21 2002025525 CIP INTERNATIONAL EDITION ISBN 0-07-119260-3 Copyright © 2003. Exclusive rights by The McGraw-Hill Companies, Inc., for manufacture and export.
This book cannot be re-exported from the country to which it is sold by McGraw-HilI. The International Edition is not available in North America.
www.mhhe.com McGraw.Hili Chemical Engineering Series Editorial Advisory Board Eduardo D. Glandt, Dean, School of Engineering and Applied Science, University of Pennsylvania Michael T. Klein, Dean, School of Engineering, Rutgers University Thomas F. Edgar, Professor of Chemical Engineering, University of Texas at Austin Bailey and Ollis Biochemical Engineering Fundamentals Bennett and Myers Momentum, Heat and Mass Transfer Coughanowr Process Systems Analysis and Control de Nevers Air Pollution Control Engineering de Nevers Fluid Mechanics for Chemical Engineers Douglas Conceptual Design of Chemical Processes Edgar and Himmelblau Optimization of Chemical Processes Gates, Katzer, and Schuit Chemistry of Catalytic Processes King Separation Processes Luyben Process Modeling, Simulation, and Control for Chemical Engineers Marlin Process Control: Designing Processes and Control Systems for Dynamic Performance McCabe, Smith, and Harriott Unit Operations of Chemical Engineering Middleman and Hochberg Process Engineering Analysis in Semiconductor Device Fabrication Perry and Green Perry's Chemical Engineers' Handbook Peters and Timmerhaus Plant Design and Economics for Chemical Engineers Reid, Prausnitz, and Poling Properties of Gases and Liquids Smith, Van Ness, and Abbott Introduction to Chemical Engineering Thermodynamics Treybal Mass Transfer Operations To Mary, Kathleen, and our parents Ruth and Ted.
__C.Ott:rEHl:S Preface xi Nomenclature xii Chapter 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 1 1.1 The Scope of Chemical Reaction Engineering I 1.2 The Extent of Reaction 8 1.3 The Rate of Reaction 16 1.4 General Properties of the Rate Function for a Single Reaction 19 1.5 Examples of Reaction Rates 24 Chapter 2 Rate Constants of Elementary Reactions 53 2.1 Elementary Reactions 53 2.2 Arrhenius Temperature Dependence of the Rate Constant 54 2.3 Transition-State Theory 56 Chapter 3 Reactors for Measuring Reaction Rates 64 3.1 Ideal Reactors 64 3.2 Batch and Semibatch Reactors 65 3.3 Stirred-Flow Reactors 70 3.4 Ideal Tubular Reactors 76 3.5 Measurement of Reaction Rates 82 3.5.1 Batch Reactors 84 3.5.2 Flow Reactors 87 Chapter 4 The Steady-State Approximation: Catalysis 100 4.1 Single Reactions 100 4.2 The Steady-State Approximation 105 4.3 Relaxation Methods 124 Chapter 5 Heterogeneous Catalysis 133 5.1 Introduction 133 5.2 Kinetics of Elementary Steps: Adsorption, Desorption, and Surface Reaction 140 5.3 Kinetics of Overall Reactions 157 5.4 Evaluation of Kinetic Parameters 171 Chapter 6 Effects of Transport Limitations on Rates of Solid-Catalyzed Reactions 184 6.1 Introduction 184 6.2 External Transport Effects 185 6.3 Internal Transport Effects 190 6.4 Combined Internal and External Transport Effects 218 6.5 Analysis of Rate Data 228 Chapter 7 Microkinetic Analysis of Catalytic Reactions 240 7.1 Introduction 240 7.2 Asymmetric Hydrogenation of Prochiral Olefins 240 ix x Contents 7.3 Ammonia Synthesis on Transition Metal Catalysts 246 7.4 Ethylene Hydrogenation on Transition Metals 252 7.5 Concluding Remarks 257 Chapter 8 Nonideal Flow in Reactors 260 8.1 Introduction 260 8.2 Residence Time Distribution (RTD) 262 8.3 Application of RTD Functions to the Prediction of Reactor Conversion 269 804 Dispersion Models for Nonideal Reactors 272 8.5 Prediction of Conversion with an Axially-Dispersed PFR 277 8.6 Radial Dispersion 282 8.7 Dispersion Models for Nonideal Flow in Reactors 282 Chapter 9 Nonisothermal Reactors 286 9.1 The Nature of the Problem 286 9.2 Energy Balances 286 9.3 Nonisothermal Batch Reactor 288 9.4 Nonisothermal Plug Flow Reactor 297 9.5 Temperature Effects in a CSTR 303 9.6 Stability and Sensitivity of Reactors Accomplishing Exothermic Reactions 305 Chapter 10 Reactors Accomplishing Heterogeneous Reactions 315 10.1 Homogeneous Versus Heterogeneous Reactions in Tubular Reactors 315 10.2 One-Dimensional Models for Fixed-Bed Reactors 317 10.3 Two-Dimensional Models for Fixed-Bed Reactors 325 lOA Reactor Configurations 328 10.5 Fluidized Beds with Recirculating Solids 331 Appendix A Review of Chemical Equilibria 339 A.1 Basic Criteria for Chemical Equilibrium of Reacting Systems 339 A.2 Determination of Equilibrium Compositions 341 Appendix B Regression Analysis 343 B.1 Method of Least Squares 343 B.2 Linear Correlation Coefficient 344 B.3 Correlation Probability with a Zero Y-Intercept 345 BA Nonlinear Regression 347 Appendix C Transport in Porous Media 349 C.1 Derivation of Flux Relationships in One-Dimension 349 C.2 Flux Relationships in Porous Media 351 Index 355 T his book is an introduction to the quantitative treatment of chemical reaction en-gineering. The level of the presentation is what we consider appropriate for a one-semester course. The text provides a balanced approach to the understanding of: (1) both homogeneous and heterogeneous reacting systems and (2) both chemical reaction engineering and chemical reactor engineering. We have emulated the teach-ings of Prof. Michel Boudart in numerous sections of this text. For example, much of Chapters 1 and 4 are modeled after his superb text that is now out of print (Kinetics a/Chemical Processes), but they have been expanded and updated. Each chapter con-tains numerous worked problems and vignettes. We use the vignettes to provide the reader with discussions on real, commercial processes and/or uses of the molecules and/or analyses described in the text. Thus, the vignettes relate the material presented to what happens in the world around us so that the reader gains appreciation for how chemical reaction engineering and its principles affect everyday life. Many problems in this text require numerical solution. The reader should seek appropriate software for proper solution of these problems. Since this software is abundant and continually improving, the reader should be able to easily find the necessary software. This exer-cise is useful for students since they will need to do this upon leaving their academic institutions. Completion of the entire text will give the reader a good introduction to the fundamentals of chemical reaction engineering and provide a basis for extensions into other nontraditional uses of these analyses, for example, behavior of biological systems, processing of electronic materials, and prediction of global atmospheric phe-nomena. We believe that the emphasis on chemical reaction engineering as opposed to chemical reactor engineering is the appropriate context for training future chemi-cal engineers who will confront issues in diverse sectors of employment.
We gratefully acknowledge Prof. Michel Boudart who encouraged us to write this text and who has provided intellectual guidance to both of us. MED also thanks Martha Hepworth for her efforts in converting a pile of handwritten notes into a final prod-uct. In addition, Stacey Siporin, John Murphy, and Kyle Bishop are acknowledged for their excellent assistance in compiling the solutions manual. The cover artwork was provided courtesy of Professor Ahmed Zewail's group at Caitech, and we gratefully thank them for their contribution. We acknowledge with appreciation the people who reviewed our project, especially A. Brad Anton of Cornell University, who provided extensive comments on content and accuracy. Finally, we thank and apologize to the many students who suffered through the early drafts as course notes.
We dedicate this book to our wives and to our parents for their constant support.
Mark E. Davis Pasadena, CA Robert J. Davis Charlottesville. VA Nomenclature xii Ci or [Ai] CB CiS C p C p de dp dt D a De Dij DKi Dr DTA Da Da E ED E(t) E It activity of species i external catalyst particle surface area per unit reactor volume representation of species i cross sectional area of tubular reactor cross sectional area of a pore heat transfer area pre-exponential factor dimensionless group analogous to the axial Peclet number for the energy balance concentration of species i concentration of species i in the bulk fluid concentration of species i at the solid surface heat capacity per mole heat capacity per unit mass effective diameter particle diameter diameter of tube axial dispersion coefficient effective diffusivity molecular diffusion coefficient Knudsen diffusivity of species i radial dispersion coefficient transition diffusivity from the Bosanquet equation Damkohler number dimensionless group activation energy activation energy for diffusion E(t)-curve; residence time distribution total energy in closed system friction factor in Ergun equation and modified Ergun equation fractional conversion based on species i fractional conversion at equilibrium hi ht H t:.H t:.Hr Hw Hw I I Ii k k kc Ka Kc Kp Kx K¢ L m,.
Mi M MS ni Nomeoclatllre fugacity of species i fugacity at standard state of pure species i frictional force molar flow rate of species i gravitational acceleration gravitational potential energy per unit mass gravitational constant mass of catalyst change in Gibbs function ("free energy") Planck's constant enthalpy per mass of stream i heat transfer coefficient enthalpy change in enthalpy enthalpy of the reaction (often called heat of reaction) dimensionless group dimensionless group ionic strength Colburn I factor flux of species i with respect to a coordinate system rate constant Boltzmann's constant mass transfer coefficient equilibrium constant expressed in terms of activities portion of equilibrium constant involving concentration portion of equilibrium constant involving total pressure portion of equilibrium constant involving mole fractions portion of equilibrium constant involving activity coefficients length of tubular reactor length of microcavity in Vignette 6.4.2 generalized length parameter length in a catalyst particle mass of stream i mass flow rate of stream i molecular weight of species i ratio of concentrations or moles of two species total mass of system number of moles of species i xiii xiv Nomenclatl J[e Ni NCOMP NRXN P Pea Per PP q Q Q r LlS Sc Si Sp S SA Sc SE Sh (t) t T TB Ts TB u flux of species i number of components number of independent reactions pressure axial Peelet number radial Peelet number probability heat flux heat transferred rate of heat transfer reaction rate turnover frequency or rate of turnover radial coordinate radius of tubular reactor recyele ratio universal gas constant radius of pellet radius of pore dimensionless radial coordinate in tubular reactor correlation coefficient Reynolds number instantaneous selectivity to species i change in entropy sticking coefficient overall selectivity to species i surface area of catalyst particle number of active sites on catalyst surface area Schmidt number standard error on parameters Sherwood number time mean residence time student t-test value temperature temperature of bulk fluid temperature of solid surface third body in a collision process linear fluid velocity (superficial velocity) v Vi V p VR Vtotal We X -Z z "Ii r r 8(t) 8 -e Nomeoclatl ire laminar flow velocity profile overall heat transfer coefficient internal energy volumetric flow rate volume mean velocity of gas-phase species i volume of catalyst particle volume of reactor average velocity of all gas-phase species width of microcavity in Vignette 6.4.2 length variable half the thickness of a slab catalyst particle mole fraction of species i defined by Equation (B.1.5) dimensionless concentration yield of species i axial coordinate height above a reference point dimensionless axial coordinate charge of species i when used as a superscript is the order of reaction with respect to species i coefficients; from linear regression analysis, from integration, etc.
parameter groupings in Section 9.6 parameter groupings in Section 9.6 Prater number dimensionless group dimensionless groups Arrhenius number activity coefficient of species i dimensionless temperature in catalyst particle dimensionless temperature Dirac delta function thickness of boundary layer molar expansion factor based on species i deviation of concentration from steady-state value porosity of bed porosity of catalyst pellet xv xvi Nomenclat! j[e YJo YJ e p., ~ P PB P p T -T Vi w intraphase effectiveness factor overall effectiveness factor interphase effectiveness factor dimensionless time fractional surface coverage of species i dimensionless temperature universal frequency factor effective thermal conductivity in catalyst particle parameter groupings in Section 9.6 effective thermal conductivity in the radial direction chemical potential of species i viscosity number of moles of species reacted density (either mass or mole basis) bed density density of catalyst pellet standard deviation stoichiometric number of elementary step i space time tortuosity stoichiometric coefficient of species i Thiele modulus Thiele modulus based on generalized length parameter fugacity coefficient of species i extent of reaction dimensionless length variable in catalyst particle dimensionless concentration in catalyst particle for irreversible reaction dimensionless concentration in catalyst particle for reversible reaction dimensionless concentration dimensionless distance in catalyst particle Notation used for stoichiometric reactions and elementary steps Irreversible (one-way) Reversible (two-way) Equilibrated Rate-determining ~_1~ The Basics of Reaction Kinetics for Chemical Reaction Engineering 1.1 I The Scope of Chemical Reaction Engineering The subject of chemical reaction engineering initiated and evolved primarily to accomplish the task of describing how to choose, size, and determine the optimal operating conditions for a reactor whose purpose is to produce a given set of chem-icals in a petrochemical application. However, the principles developed for chemi-cal reactors can be applied to most if not all chemically reacting systems (e.g., at-mospheric chemistry, metabolic processes in living organisms, etc.). In this text, the principles of chemical reaction engineering are presented in such rigor to make possible a comprehensive understanding of the subject. Mastery of these concepts will allow for generalizations to reacting systems independent of their origin and will furnish strategies for attacking such problems.
The two questions that must be answered for a chemically reacting system are: (1) what changes are expected to occur and (2) how fast will they occur? The initial task in approaching the description of a chemically reacting system is to understand the answer to the first question by elucidating the thermodynamics of the process.
For example, dinitrogen (N2) and dihydrogen (H2) are reacted over an iron catalyst to produce ammonia (NH3): N2 + 3H2 = 2NH3, - b.H, = 109 kllmol (at 773 K) where b.H, is the enthalpy of the reaction (normally referred to as the heat of reac-tion). This reaction proceeds in an industrial ammonia synthesis reactor such that at the reactor exit approximately 50 percent of the dinitrogen is converted to ammo-nia. At first glance, one might expect to make dramatic improvements on the production of ammonia if, for example, a new catalyst (a substance that increases 2 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering the rate of reaction without being consumed) could be developed. However, a quick inspection of the thermodynamics of this process reveals that significant enhance-ments in the production of ammonia are not possible unless the temperature and pressure of the reaction are altered. Thus, the constraints placed on a reacting sys-tem by thermodynamics should always be identified first.
EXAMPLE 1.1.1 I In order to obtain a reasonable level of conversion at a commercially acceptable rate, am-monia synthesis reactors operate at pressures of 150 to 300 atm and temperatures of 700 to 750 K. Calculate the equilibrium mole fraction of dinitrogen at 300 atm and 723 K starting from an initial composition of XN2 = 0.25, XHz = 0.75 (Xi is the mole fraction of species i).
At 300 atm and 723 K, the equilibrium constant, Ka , is 6.6 X 10-3. (K. Denbigh, The Prin-ciples of Chemical Equilibrium, Cambridge Press, 1971, p. 153).
• Answer (See Appendix A for a brief overview of equilibria involving chemical reactions): CHAPTER 1 The Basics of Rear.tion Kinetics for Chemical Reaction Engineering The definition of the activity of species i is: fugacity at the standard state, that is, 1 atm for gases and thus 3 K = [_lN~3 ] [(]~,)I/2(]~Y/2] a fI/2 f3/2 (t'O ) N, H, JNH] Use of the Lewis and Randall rule gives: [ fNH; ] [ ] ]J;2]J;2 I atm /; = Xj cPj P, cPj = fugacity coefficient of pure component i at T and P of system then [ XNH; ] [ cPNH;] I K = K K-K = ---P-1 atm a X (p P XlI2 X3 /2 -:1,1(2-:1,3/2 [ ] [ ] N, H, 'VN, 'VH, Upon obtaining each cPj from correlations or tables of data (available in numerous ref-erences that contain thermodynamic information): If a basis of 100 mol is used (g is the number of moles of N2reacted): then N 2 Hz NH3 total 25 75 o 100 (2g)(100 - 2g) ------- = 2.64 (25 - g)l/2(75 - 3g)3/2 Thus, g = 13.1 and XN, (25 -13.1)/(100 26.2) = 0.16. At 300 atm, the equilibrium mole fraction of ammonia is 0.36 while at 100 atm it falls to approximately 0.16. Thus, the equilibrium amount of ammonia increases with the total pressure of the system at a constant temperature.
4 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering The next task in describing a chemically reacting system is the identifica-tion of the reactions and their arrangement in a network. The kinetic analysis of the network is then necessary for obtaining information on the rates of individ-ual reactions and answering the question of how fast the chemical conversions occur. Each reaction of the network is stoichiometrically simple in the sense that it can be described by the single parameter called the extent of reaction (see Sec-tion 1.2). Here, a stoichiometrically simple reaction will just be called a reaction for short. The expression "simple reaction" should be avoided since a stoichio-metrically simple reaction does not occur in a simple manner. In fact, most chem-ical reactions proceed through complicated sequences of steps involving reactive intermediates that do not appear in the stoichiometries of the reactions. The iden-tification of these intermediates and the sequence of steps are the core problems of the kinetic analysis.
If a step of the sequence can be written as it proceeds at the molecular level, it is denoted as an elementary step (or an elementary reaction), and it represents an ir-reducible molecular event. Here, elementary steps will be called steps for short. The hydrogenation of dibromine is an example of a stoichiometrically simple reaction: If this reaction would occur by Hz interacting directly with Brz to yield two mole-cules of HBr, the step would be elementary. However, it does not proceed as writ-ten. It is known that the hydrogenation of dibromine takes place in a sequence of two steps involving hydrogen and bromine atoms that do not appear in the stoi-chiometry of the reaction but exist in the reacting system in very small concentra-tions as shown below (an initiator is necessary to start the reaction, for example, a photon: Brz + light -+ 2Br, and the reaction is terminated by Br + Br + TB -+ Brz where TB is a third body that is involved in the recombination process-see below for further examples): Br + Hz -+ HBr + H H + Brz -+ HBr + Br In this text, stoichiometric reactions and elementary steps are distinguished by the notation provided in Table 1.1.1.
Table 1.1.1 I Notation used for stoichiometric reactions and elementary steps.
Irreversible (one-way) Reversible (two-way) Equilibrated Rate-determining CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction EnginAering 5 In discussions on chemical kinetics, the terms mechanism or model fre-quently appear and are used to mean an assumed reaction network or a plausi-ble sequence of steps for a given reaction. Since the levels of detail in investi-gating reaction networks, sequences and steps are so different, the words mechanism and model have to date largely acquired bad connotations because they have been associated with much speculation. Thus, they will be used care-fully in this text.
As a chemically reacting system proceeds from reactants to products, a number of species called intermediates appear, reach a certain concentration, and ultimately vanish. Three different types of intermediates can be identified that correspond to the distinction among networks, reactions, and steps. The first type of intermediates has reactivity, concentration, and lifetime compara-ble to those of stable reactants and products. These intermediates are the ones that appear in the reactions of the network. For example, consider the follow-ing proposal for how the oxidation of methane at conditions near 700 K and atmospheric pressure may proceed (see Scheme l.l.l). The reacting system may evolve from two stable reactants, CH4 and° 2, to two stable products, CO2 and H20, through a network of four reactions. The intermediates are formaldehyde, CH20; hydrogen peroxide, H2 0 2; and carbon monoxide, CO. The second type of intermediate appears in the sequence of steps for an individual reaction of the network. These species (e.g., free radicals in the gas phase) are usually pres-ent in very small concentrations and have short lifetimes when compared to those of reactants and products. These intermediates will be called reactive in-termediates to distinguish them from the more stable species that are the ones that appear in the reactions of the network. Referring to Scheme 1.1.1, for the oxidation of CH2 0 to give CO and H2 0 2, the reaction may proceed through a postulated sequence of two steps that involve two reactive intermediates, CHO and H02 . The third type of intermediate is called a transition state, which by definition cannot be isolated and is considered a species in transit. Each ele-mentary step proceeds from reactants to products through a transition state.
Thus, for each of the two elementary steps in the oxidation of CH20, there is a transition state. Although the nature of the transition state for the elementary step involving CHO, 02' CO, and H02 is unknown, other elementary steps have transition states that have been elucidated in greater detail. For example, the configuration shown in Fig. 1.1.1 is reached for an instant in the transition state of the step: The study of elementary steps focuses on transition states, and the kinetics of these steps represent the foundation of chemical kinetics and the highest level of understanding of chemical reactivity. In fact, the use of lasers that can gen-erate femtosecond pulses has now allowed for the "viewing" of the real-time transition from reactants through the transition-state to products (A. Zewail, The 6 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Br 7 Br I C H/ I "'CH H 3 ~OW ) ..
H OH Figure 1.1.1 I The transition state (trigonal bipyramid) of the elementary step: OH- + C2HsBr ~ HOC2Hs + Br-The nucleophilic substituent OH- displaces the leaving group Br-.
Br-) H I H '" C/ CH3 I OH • J 8 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Chemical Bond: Structure and Dynamics, Academic Press, 1992). However, in the vast majority of cases, chemically reacting systems are investigated in much less detail. The level of sophistication that is conducted is normally dictated by the purpose of the work and the state of development of the system.
1.2 I The Extent of Reaction The changes in a chemically reacting system can frequently, but not always (e.g., complex fermentation reactions), be characterized by a stoichiometric equation. The stoichiometric equation for a simple reaction can be written as: NCOMP 0= L: viA; i=1 (1.2.1) where NCOMP is the number of components, A;, of the system. The stoichiomet-ric coefficients, Vi' are positive for products, negative for reactants, and zero for inert components that do not participate in the reaction. For example, many gas-phase oxidation reactions use air as the oxidant and the dinitrogen in the air does not par-ticipate in the reaction (serves only as a diluent). In the case of ammonia synthesis the stoichiometric relationship is: Application of Equation (1.2.1) to the ammonia synthesis, stoichiometric relation-ship gives: For stoichiometric relationships, the coefficients can be ratioed differently, e.g., the relationship: can be written also as: since they are just mole balances. However, for an elementary reaction, the stoi-chiometry is written as the reaction should proceed. Therefore, an elementary re-action such as: 2NO + O2 -+ 2N02 CANNOT be written as: (correct) (not correct) CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 9 EXAMPLE 1.2.1 I If there are several simultaneous reactions taking place, generalize Equation (1.2.1) to a sys-tem of NRXN different reactions. For the methane oxidation network shown in Scheme 1.1.1, write out the relationships from the generalized equation.
• Answer If there are NRXN reactions and NCOMP species in the system, the generalized form of Equa-tion (1.2.1) is: NCOMP o = 2: vi,jAi, j 1; ",NRXN i (1.2.2) For the methane oxidation network shown in Scheme 1.1.1: 0= OCOz + IHzO lOz + OCO + OHzOz + lCHzO -ICH4 0= OCOz + OHp -lOz + lCO + 1HzOz -lCHzO + OCH4 o = ICOz + ORzO ! Oz -ICO + OHzOz + OCHzO + OCH4 0= OCOz + IHzO +! Oz + OCO -I HzOz + OCHp + OCH4 or in matrix form: I -I o o o 1 o -1 o 1 -1 o 1 o o I -o~l HzOz CHp CH4 Note that the sum of the coefficients of a column in the matrix is zero if the component is an intermediate.
Consider a closed system, that is, a system that exchanges no mass with its sur-roundings. Initially, there are n? moles of component Ai present in the system. If a single reaction takes place that can be described by a relationship defined by Equa-tion (1.2.1), then the number of moles of component Ai at any time t will be given by the equation: ni (t) = n? + Vi <p(t) (1.2.3) (1.2.4) that is an expression of the Law of Definitive Proportions (or more simply, a mole balance) and defines the parameter, <P, called the extent of reaction. The extent of reaction is a function of time and is a natural reaction variable.
Equation (1.2.3) can be written as: <p(t) = ni (t) -n?
Vi 10 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Since there is only one <P for each reaction: (1.2.5) or (1.2.6) EXAMPLE 1.2.2 I Thus, if ni is known or measured as a function of time, then the number of moles of all of the other reacting components can be calculated using Equation (1.2.6).
If there are numerous, simultaneous reactions occurring in a closed system, each one has an extent of reaction. Generalize Equation (1.2.3) to a system with NRXN reactions.
• Answer EXAMPLE 1.2.3 I NRXN nj = n? + 2: Vj,j<Pj j~l (1.2.7) Carbon monoxide is oxidized with the stoichiometric amount of air. Because of the high temperature, the equilibrium: Nz + Oz =@:: 2NO has to be taken into account in addition to: (1) (2) The total pressure is one atmosphere and the equilibrium constants of reactions (l) and (2) are: Kx, (XNO)Z (XN,)(Xo,)' whereKx, = 8.26 X 1O-3,Kx, = 0.7, andXjis the mole fraction of species i (assuming ideal gas behavior). Calculate the equilibrium composition.
• Answer Assume a basis of 1 mol of CO with a stoichiometric amount of air ({; I and tz are the num-ber of moles of Nz and CO reacted, respectively): CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Nz 1.88 1.88 gj Oz 0.5 0.5 ~g2 gj CO 1 1 gz COZ 0 gz NO 0 2gj total 3.38 3.38 ~gz The simultaneous solution of these two equations gives: 11 Therefore, gj = 0.037, gz = 0.190 EXAMPLE 1.2.4 I Nz Oz CO COz NO 0.561 0.112 0.247 0.058 0.022 1.000 VIGNETTE 1.2.1 Using the results from Example 1.2.3, calculate the two equilibrium extents of reaction.
• Answer <P1q g1q = 0.037 <p2 Q ~Q = 0.190 12 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering CHAPTER 1 The Basics of Reaction Kinetics~hemical Reaction Engineering 13 significantly contributed to pollution reduction and are one of the major success stories for chemical reaction en~sim,ering.
Insulation cover The drawback of is that it is an extensive variable, that is, it is dependent upon the mass of the system. The fractional conversion, f, does not suffer from this problem and can be related to <1>. In general, reactants are not initially present in stoichiometric amounts and the reactant in the least amount determines the maxi-mum value for the extent of reaction, max. This component, called the limiting component (subscript f) can be totally consumed when reaches max. Thus, The fractional conversion is defined as: f(t) = (t) <P max and can be calculated from Equations (1.2.3) and (1.2.8): Equation (1.2.10) can be rearranged to give: (1.2.8) (1.2.9) (1.2.10) (1.2.11) where 0 :::; fi :::; 1. When the thermodynamics of the system limit such that it can-not reach max (where n/ 0), will approach its equilibrium value eg (n/ =1= 0 value of n/ determined by the equilibrium constant). When a reaction is limited by thermodynamic equilibrium in this fashion, the reaction has historically been called 14 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Rear,tion Engineering reversible. Alternatively, the reaction can be denoted as two-way. When <peg is equal to 0 + 4H2, 50 percent of feed is n-hexane SA = 0.5;~oTAL [4 + 1 - 1] = 2 nTOTAL 1-11 EXAMPLE 1.2.6 I If the decomposition of N20 S into N20 4 and O2 were to proceed to completion in a closed volume of size V, what would the pressure rise be if the starting composition is 50 percent N20 S and 50 percent N2?
• Answer The ideal gas law is: PV = nTOTALRgT (Rg : universal gas constant) At fixed T and V, the ideal gas law gives: 16 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering The reaction proceeds to completion so fA = 1 at the end of the reaction. Thus, with ~A - 0.50n~OTAL [1 + 0.5 1] o = 0.25 nTOTAL 1-11 Therefore, p pO = 1.25 1.3 I The Rate of Reaction For a homogeneous, closed system at uniform pressure, temperature, and composition in which a single chemical reaction occurs and is represented by the stoichiometric Equation (1.2.1), the extent of reaction as given in Equation (1.2.3) increases with time, t. For this situation, the reaction rate is defmed in the most general way by: d<p (mOl) dt time (1.3.1) This expression is either positive or equal to zero if the system has reached equi-librium. The reaction rate, like <P, is an extensive property of the system. A specific rate called the volumic rate is obtained by dividing the reaction rate by the volume of the system, V: 1 d<p r= V dt ( mol ) time-volume (1.3.2) (1.3.4) (1.3.3) Differentiation of Equation (1.2.3) gives: dni = Vid The specific rates for 4.88 wt. % Pd on Al20 3 and 3.75 wt. % Pd on Si02 were 7.66 X 10-4 and 1.26 X 10-3 mo!/(gcat . s), respectively. Using a technique called titration. the percent-age of Pd metal atoms on the surface of the Pd metal particles on the Al20 3 and Si02 was 21 percent and 55 percent, respectively. Since the specific rates for Pd on Al20 3 and Si02 are different, does the metal oxide really influence the reaction rate?
CHAPTER 1 The Basics of Reac1lollKLoetics for Chemical Reaction Engineering 19 Titration is a technique that can be used to measure the number of surface metal atoms.
The procedure involves first chemisorbing (chemical bonds formed between adsorbing species and surface atoms) molecules onto the metal atoms exposed to the reaction environment.
Second, the chemisorbed species are reacted with a second component in order to recover and count the number of atoms chemisorbed. By knowing the stoichiometry of these two steps, the number of surface atoms can be calculated from the amount of the recovered chemisorbed atoms. The technique is illustrated for the problem at hand: Pd metal particle \ G2J Metal oxide Step I Step II +°z (Step I) • Metal oxide 2Pd, + Oz -+ 2PdP PdP + ~ Hz =? Pd,H + HzO Oxygen atoms ~ chemisorbed on ~ surface Pd atoms (only surface Pd) (not illustrated) By counting the number of Hz molecules consumed in Step II, the number of surface Pd atoms (PdJ can be ascertained. Thus, the percentage of Pd atoms on the surface can be cal-culated since the total number of Pd atoms is known from the mass of Pd.
• Answer The best way to determine if the reaction rates are really different for these two catalysts is to compare their values of the turnover frequency. Assume that each surface Pd atom is an active site. Thus, to convert a specific rate to a turnover frequency: _ I ( mOl) ( gcat ) (mOleCUlar weight of metal) r (s ) = r --.
.
t gcat. s mass metal fraction of surface atoms = (7.66 X 10-4). (oo~88)(106.4)(o.21)-1 8.0s- 1 Likewise for Pd on SiOz, Since the turnover frequencies are approximately the same for these two catalysts, the metal oxide support does not appear to influence the reaction rate.
1.4 I General Properties of the Rate Function for a Single Reaction The rate of reaction is generally a function of temperature and composition, and the development of mathematical models to describe the form of the reaction rate is a central problem of applied chemical kinetics. Once the reaction rate is known, 20 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering infonnation can often be derived for rates of individual steps, and reactors can be designed for carrying out the reaction at optimum conditions.
Below are listed general rules on the fonn of the reaction rate function (M.
Boudart, Kinetics of Chemical Processes, Butterworth-Heinemann, 1991, pp.
13-16). The rules are of an approximate nature but are sufficiently general that ex-ceptions to them usually reveal something of interest. It must be stressed that the utility of these rules is their applicability to many single reactions.
CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 21 The coefficient k does not depend on the c~mposition of the system or time. For this reason, k is called the rate constant. If F is not a function of the temperature, r = k(T)F(C;) then the reaction rate is called sepa~able since the temperature and composition de-pendencies are separated in k and F, respectively.
In Equation (1.4.2), the pre-exponential factor, A, does not depend appreciably on temperature, and E is called the activation energy. Figure 1.4.2 is an example of a typical Arrhenius plot.
The product n is taken over all components of the system. The exponents Ui are small integers or fractions that are positive, negative, or zero and are temperature independent at least over a restricted interval (see Table 1.4.1 for an example).
Consider the general reaction: aA + bB =} wW If the reaction rate follows Rule IV then it can be written as: r = kC'AAC!3B (1.4.3) The exponent Ui is called the order of reaction with respect to the corresponding component of the system and the sum of the exponents is called the total order of 22 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 80 70 T (OC) 60 50 40 10 8 6 4 ~ 2 .S 9 N E-1.0 "0 E- 0.8 '" 0 0.6 x "'" 0.4 0.2 0.1 2.8 2.9 3.0 3.1 lX103 (Krl T 3.2 Figure 1.4.2 I A typical Arrhenius plot, In k vs liT. The slope corresponds to -EIRg- Adapted from D. E. Mears and M.
Boudart, AlChE J. 12 (1966) 313, with permission of the American Institute of Chemical Engineers. Copyright © 1966 AIChE. All rights reserved.
the reaction. In general ai -=1= IViI and is rarely larger than two in absolute value. If a i = IViI for reactants and equal to zero for all other components of the system, the expression: (for reactants only) would be of the form first suggested by Guldberg and Waage (1867) in their Law of Mass Action. Thus, a rate function of the form: (1.4.4) CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Table 1.4.1 I Kinetic parameters for the simultaneous hydroformylation 1 and hydrogenation2 of propene over a rhodium zeolite catalyst at T 423 K and 1 atm total pressure [from Rode et aI., J. Gatal., 96 (1985) 563].
( mal ) Rate gRh' hr 23 A E (kcal/mol), (393-423 K) £Xl £Xz £X3 4.12 X 106 15.8 1.01 0.35 -0.71 1.36 X 103 11.0 1.00 0040 -0.70 1.86 X 103 10.8 0.97 0.48 -0.72 Table 1.4.21 I Examples of rate functions of the type: r kII Gii.
CH3CHO =;. CH4 + CO CZ H6 + Hz =;. 2CH4 (catalytic) SbH3 =;. Sb + ~ Hz Nz + 3Hz=;. 2NH3 (catalytic) k(CH3CHOt5 k(CzH6r (Hzr07 k(SbH3 )o.6 k(Nz)(Hz?Z5(NH3)-15 IProm M. Boudart, Kinetics alChemical Processes, Butterworth-Heinemaun, 1991, p. 17.
which is normally referred to as "pseudo mass action" or "power law" is really the Guldberg-Waage form only when (Xi = 1Vi I for reactants and zero for all other com-ponents (note that orders may be negative for catalytic reactions, as will be discussed in a later chapter). For the reaction described by Equation (1.4.3), the Guldberg-Waage rate expression is: Examples of power law rate expressions are shown in Table 1.4.2.
For elementary steps, (Xi = IVi I. Consider again the gas-phase reaction: Hz + Brz =} 2HBr If this reaction would occur by Hz interacting directly with Brz to yield two mole-cules of HBr, the step would be elementary and the rate could be written as: r = kCH,CBr2 24 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering However, it is known that this is not how the reaction proceeds (Section 1.1) and the real rate expression is: For elementary steps the number of molecules that partcipate in the reaction is called the molecularity of the reaction (see Table 1.4.3).
When Rule II applies to the rate functions r+ and r ~ so that: r+ = k+F+(Ci) L = LF~(CJ both rate constants k+ and k~ are related to the equilibrium constant, Kc. For example, the reaction A ( ) B at ideal conditions gives (see Chapter 5 for a more rigorous explanation of the relationships between rates and equilibrium expressions): (1.4.6) 1.5 I Examples of Reaction Rates Consider the unimolecular reaction: A -+ products (1.5.1) Using the Guldberg-Waage form of the reaction rate to describe this reaction gives: r = kCA From Equations (1.3.4) and (1.5.2): (1.5.2) 1 dni r= vY dt or -knA djA = k(1 - f) dt A -1 dnA -=kCA V dt (variable V) [using Equation (1.2.11)] (1.5.3) (1.5.4) Table 1.4.3 I Molecularity and rates of elementary steps.
Unimolecular Bimolecular Trimolecular (rare) 2 3 A ~ products 2A ~ products A + B ~ products 3A ~ products 2A + B ~ products A + B + C ~ products NPs ~ N02 + N03 NO + N03 ~ 2N02 2NO + O2 ~ 2N02 NO + N02 + H20 ~ 2HN02 1.96 X 1014 exp[ -10660/T], S-1 2.0 X 1011, cm3/ s/ molecule (2) 3.3 X 10- 39 exp(530IT), cm61sl molecule2 (2) ~ 4.4 X 10- 4°, cm6/s/molecuIe2 (2) (I) From J. H. Seinfeld, Atmospheric Chemistry and Physics ofAir Pollution, Wiley, 1986, p. 175.
(2) Concentrations are in moleculeslcm3 .
26 CHAPTER 1 The Basics of Reaction KinAtics for Chemical Reaction Engineering Table 1.5.1 I Examples of reactions that can be described using first-order reaction rates.
Isomerizations Decompositions Radioactive decay (each decay can be described by a first-order reaction rate) N20 S ===> N02 + N03 CH2 CH2 ===> CH4 + CO ~/ ° dCA dt dPA -= -kP dt A (constant V) [constant V: Ci = P;j(RgT) ] (1.5.5) (1.5.6) EXAMPLE 1 .5.1 I Thus, for first-order systems, the rate, r, is proportional (via k) to the amount present, ni' in the system at any particular time. Although at first glance, first-order reaction rates may appear too simple to describe real reactions, such is not the case (see Table 1.5.1). Additionally, first-order processes are many times used to approximate complex systems, for example, lumping groups of hydrocarbons into a generic hypothetical component so that phenomenological behavior can be described.
In this text, concentrations will be written in either of two notations. The nota-tions Ci and [AJ are equivalent in terms of representing the concentration of species i or Ai, respectively. These notations are used widely and the reader should become comfortable with both.
The natural abundance of 235U in uranium is 0.79 atom %. If a sample of uranium is enriched to 3 at. % and then is stored in salt mines under the ground, how long will it take the sam-ple to reach the natural abundance level of 235U (assuming no other processes form 235U; this is not the case if 238U is present since it can decay to form 235U)? The half-life of 235U is 7.13 X 108 years.
• Answer Radioactive decay can be described as a first-order process. Thus, for any first-order decay process, the amount of material present declines in an exponential fashion with time. This is easy to see by integrating Equation (1.5.3) to give: 11; 11? exp(- kt), where 11? is the amount of l1i present at t O.
CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 27 The half-life, tj, is defined as the time necessary to reduce the amount of material in half. For a first-order process tj can be obtained as follows: ! n? = n? exp(-ktj) or Given tj, a value of k can be calculated. Thus, for the radioactive decay of 235U, the first-order rate constant is: In(2) k = --9.7 X 10- 10 years-I t1 2 To calculate the time required to have 3 at. % 235U decay to 0.79 at. %, the first-order expression: can be used. Thus, n· , = exp(-kt) or t n?
(nO) In ~i k EXAMPLE 1.5.2 I or a very long time.
In(_3) 0.79 t = = 1.4 X 10 9 years 9.7 X 10- 10 NzOs decomposes into NOz and N03 with a rate constant of 1.96 X 1014 exp [-10,660/T]s-l.
At t = 0, pure NzOs is admitted into a constant temperature and volume reactor with an initial pressure of 2 atm. After 1 min, what is the total pressure of the reactor? T = 273 K .
• Answer Let n be the number of moles of NzOs such that: dn -= -kn dt Since n nO (1 - I): df dt k(l - f),f = O@ t = 0 Integration of this first-order, initial-value problem yields: In(ft) kt fort~O 28 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering or f I - exp( -kt) for t 2:: 0 At 273 K, k = 2.16 X 10-3 S-I. After reaction for I min: f= I exp[ -(60)(2.16 X 10-3)) = 0.12 EXAMPLE 1.5.3 I From the ideal gas law at constant T and V: P n n°(1 + sf) pO nO nO For this decomposition reaction: Thus, p = pO(1 +f) 2(1 + 0.12) = 2.24 atm Often isomerization reactions are highly two-way (reversible). For example, the isomeriza-tion of I-butene to isobutene is an important step in the production of methyl tertiary butyl ether (MTBE), a common oxygenated additive in gasoline used to lower emissions. MTBE is produced by reacting isobutene with methanol: In order to make isobutene, n-butane (an abundant, cheap C4 hydrocarbon) can be dehydro-genated to I-butene then isomerized to isobutene. Derive an expression for the concentration of isobutene formed as a function of time by the isomerization of I-butene: k, CHCH2CH3 ( ) CH3CCH3 k2 II CH2 • Answer Let isobutene be denoted as component J and I-butene as B. If the system is at constant T and V, then: dfB1 , J dt CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Since [B] = [BJO(l - Is): = 0 + [B]Ols [B]O(M + Is), ;W = [I]°/[BJO 0 Thus, A 'l'b' dis 0 t eqUl 1 flum -= , so: dt 29 [B]O(M +nq) [B J0(1 - IEq) Insertion of the equilibrium relationship into the rate expression yields: dis dt or after rearrangement: dis kl(M + 1) dt = (M +n q) (tE q - Is), Is = 0 @ t = 0 Integration of this equation gives: In[_ll = [kl(M + 1)]t, M 0 1 _ Is M + IE q n q or { [ ( kl(M+ 1))]} Is = IE q 1 - exp -M + IEq t Using this expression for Is: Consider the bimolecular reaction: A + B --+ products (1.5,7) Using the Guldberg-Waage form of the reaction rate to describe this reaction gives: (1.5.8) 30 C H APT ER 1 The Basics Qf Reaction Kinetics for Chemical ReactiQn Engineering From Equations (1.3.4) and (1.5.8): 1 dni r= vY dt or (variable V) (constant V) dnA V- = -knAnB dt dCA -= -kCACB dt dPA k dt = - R r PAPB [constant V: Ci = pJ(R~T)] g (1.5.9) (1.5.10) (1.5.11) For second-order kinetic processes, the limiting reactant is always the appropriate species to follow (let species denoted as A be the limiting reactant). Equations (1.5.9-1.5.11) cannot be integrated unless Cs is related to CA' Clearly, this can be done via Equation (1.2.5) or Equation (1.2.6). Thus, or if the volume is constant: -n~ c2 P~ If M = -= -= -then: n~ c1 P~ nB = nA + n~(M - 1) CB = CA + C~(M -1) PB = PA + P~(M -1) (variable V)} (constant V) (constant V) (1.5.12) Inserting Equation (1.5.12) into Equations (1.5.9-1.5.11) gives: (variable V) (1.5.13) (1.5.14) (1.5.15) If V is not constant, then V = VO (1 + SAJA) by using Equation (1.2.15) and the ideal gas law. Substitution of this expression into Equation (1.5.13) gives: CHAPTER 1 The-.BBS.ics of Reaction Kinetics fOLChemical Reaction Engineering k(~) (1 - JA)[M - JA] (1 + SAJA) EXAMPLE 1.5.4 I 31 (1.5.16) Equal volumes of 0.2 M trimethylamine and 0.2 M n-propylbromine (both in benzene) were mixed, sealed in glass tubes, and placed into a constant temperature bath at 412 K.
After various times, the tubes were removed and quickly cooled to room temperature to stop the reaction: + N(CH3)3 + C3H7Br =? C3H7 N(CH3)3Br-The quaternization of a tertiary amine gives a quaternary ammonium salt that is not soluble in nonpolar solvents such as benzene. Thus, the salt can easily be filtered from the remain-ing reactants and the benzene. From the amount of salt collected, the conversion can be cal-culated and the data are: 5 13 25 34 45 59 80 100 120 4.9 11.2 20.4 25.6 31.6 36.7 45.3 50.7 55.2 Are these data consistent with a first- or second-order reaction rate?
• Answer The reaction occurs in the liquid phase and the concentrations are dilute. Thus, a good as-sumption is that the volume of the system is constant. Since C~ = C~: (first-order) In[_1] = kt 1 - fA (second-order) fA = kC~t I fA In order to test the first-order model, the In[ 1 -is plotted versus t while for the second-order model, [,) is plotted versus t (see Figures 1.5.1 and 1.5.2). Notice that both models conform to the equation y ex It + az. Thus, the data can be fitted via linear 32 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 1.0 -,----------------...., 0.8 0.6 0.4 0.2 0.0 o 20 40 60 80 100 120 140 Figure 1.5.1 I Reaction rate data for first-order kinetic model.
regression to both models (see Appendix B). From visual inspection of Figures 1.5.1 and 1.5.2, the second-order model appears to give a better fit. However, the results from the lin-ear regression are (5£ is the standard error): 5£(aIl = 2:51 X 10-4 5£(a2) 1.63 X 10-2 first-order (Xl = 6.54 X 10- 3 a2 = 5.55 X 10- 2 Ree 0.995 second-order (Xl = 1.03 X 10- 2 5£(al) = 8.81 X 10-5 (X2 -5.18 X 10- 3 5£(a2) 5.74 X 10- 3 Ree 0.999 Both models give high correlation coefficients (Reel, and this problem shows how the correlation coefficient may not be useful in determining "goodness of fit." An appropriate way to determine "goodness of fit" is to see if the models give a2 that is not statistically different from zero. This is the reason for manipulating the rate expressions into forms that have zero intercepts (i.e., a known point from which to check statistical significance). If a student 1-test is used to test significance (see Appendix B), then: (X2 01 5£(a2) CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 33 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 o 20 40 60 80 100 120 140 t~ = Figure 1.5.2 I Reaction rate data for second-order kinetic model.
The values of t for the first- and second-order models are: 15.55 X 10-2 - 01 = 3.39 1.63 X 10- 2 -5.18 X 10- 3 - 01 t; = --------:-- = 0.96 5.74 X For 95 percent confidence with 9 data points or 7 degrees of freedom (from table of student t values): expected deviation t;xp = = 1.895 standard error M . min Since t~ > t;xp and t; < is accepted. Thus, and the first-order model is rejected while the second-order model kC~ = 1.030 X 10- 2 1.030 X 10- 2 k = = 0.1030 ---O.IM 34 EXAMPLE 1.5.5 I CHAPTER 1 The Basics Qf BeactiQn Kinetics for Chemical Beaction Engineering When the standard error is known, it is best to report the value of the correlated parameters: The following data were obtained from an initial solution containing methyl iodide (MI) and dimethyl-p-toludine (PT) both in concentrations of 0.050 mol/L. The equilibrium constant for the conditions where the rate data were collected is 1.43. Do second-order kinetics ade-quately describe the data and if so what are the rate constants?
Data: • Answer IO 26 36 78 0.18 0.34 0.40 0.52 CH3I + (CH3hN-0- CH3 (:~) (CH3h~-0- CH3 + 1-(MI) (PT) (NQ) (I) At constant volume, dCPT --;jt = -k1CPTCMI + k2 CNQCI C~ = C~l1 = 0.05, C~Q = C? = 0, CPT = C~(l fPT)' CMI = clPT(l - fPT) CNQ = CI = C~fPT Therefore, dE ~ = k CO (I -E )2 - k CO f2 dt 1 PT JPT 2 PT PT At equilibrium, K - ~ -(fM-?
c -k2 -(l - fM-?
Substitution of the equilibrium expression into the rate expression gives: dfPT = k CO (I _feq)2[( I - fPT)2 _ ([PT)2] dt 1 PT PT I f::i-v::i-CHAPTER 1 The Basics Qf ReactiQn Kinetics for Chemical Reaction Engineering Upon integration with lIT = 0 at t 0: 35 Note that this equation is again in a form that gives a zero intercept. Thus, a plot and linear least squares analysis of: In[ /~~ - (2/;;} l)/IT] versus t (f;;} lIT) will show if the model can adequately describe the data. To do this, I;;} is calculated from Kc and it is I;;} = 0.545. Next, from the linear least squares analysis, the model does fit the data and the slope is 0.0415. Thus, _ [1 ] 0 0.0415 - 2k] eq -1 CIT IPT giving k] = 0.50 Llmol/min. From Kc and kj, Consider the trimolecular reaction: 0.35 L/mol/min.
A + B + C ~ products (1.5.17) Using the Guldberg-Waage form of the reaction rate to describe this reaction gives: (constant V) r = kCACBCC From Equations (1.3.4) and (1.5.18): 1 dni 1 dnA r=-=--=kCACBCC vy dt V dt 2 dnA V -= -knAnBnC (variable V) dt dCA dt (1.5.18) (1.5.19) (1.5.20) (constant V) (1.5.21) Trimolecular reactions are very rare. If viewed from the statistics of collisions, the probability of three objects colliding with sufficient energy and in the correct configuration for reaction to occur is very small. Additionally, only a small amount of these collisions would successfully lead to reaction (see Chapter 2, for a de-tailed discussion). Note the magnitudes of the reaction rates for unimolecular and 36 CHAPTER 1 The Rasics of Reaction Kinetics for Chemical Reaction Engineering bimolecular reactions as compared to trimolecular reactions (see Table 1.4.3). How-ever, trimolecular reactions do occur, for example: ° + O2 + TB ~ 0 3 + TB where the third body TB is critical to the success of the reaction since it is neces-sary for it to absorb energy to complete the reaction (see Vignette 1.2.1).
In order to integrate Equation (1.5.20), Cs and Cc must be related to CA and this can be done by the use of Equation (1.2.5). Therefore, analysis of a trimolecular process is a straightforward extension of bimolecular processes.
If trimolecular processes are rare and give slow rates, then the question arises as to how reactions like hydroformylations (Table 1.4.1) can be accomplished on a commercial scale (Vignette 1.5.1). The hydroformylation reaction is for example (see Table 1.4.1): ° II CH2 CH-R+CO+H2 ==> R CH2-CH2 C-H H / co o C I Rh o II Rh -C -CHzCHzR H I H-Rh \ H-Rh o " II H-C CHzCHzR I/Hz o II C-CHzCHzR Figure 1.5.3 I Simplified version of the hydroformylation mechanism. Note that other ligands on the Rh are not shown for ease in illustrating what happens with reactants and products.
CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 37 This reaction involves three reactants and the reason that it proceeds so efficiently is that a catalyst is used. Referring to Figure 1.5.3, note that the rhodium catalyst coordinates and combines the three reactants in a closed cycle, thus breaking the "statistical odds" of having all three reactants collide together simultaneously. With-out a catalyst the reaction proceeds only at nonsignificant rates. This is generally true of reactions where catalysts are used. More about catalysts and their functions will be described later in this text.
When conducting a reaction to give a desired product, it is common that other reactions proceed simultaneously. Thus, more than a single reaction must be considered (i.e., a reaction network), and the issue of selectivity becomes important.
In order to illustrate the challenges presented by reaction networks, small reaction networks are examined next. Generalizations of these concepts to larger networks are only a matter of patience.
Consider the reaction network of two irreversible (one-way), first-order reac-tions in series: k k A-4B-4C (1.5.22) This network can represent a wide variety of important classes of reactions. For ex-ample, oxidation reactions occurring in excess oxidant adhere to this reaction net-work, where B represents the partial oxidation product and C denotes the complete oxidation product CO2: 38 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering ° excess air II excess air CH3CH20H > CH3C -H > 2C02 + 2H20 ethanol acetaldehyde For this situation the desired product is typically B, and the difficulty arises in how to obtain the maximum concentration of B given a particular k j and k2• Using the Guldberg-Waage form of the reaction rates to describe the network in Equation (1.5.22) gives for constant volume: with dCA dt dCB ---;It = kjCA - k2CB (1.5.23) C~ + C~ + C& = CO = CA + CB + Cc Integration of the differential equation for CA with CA = C~ at t = 0 yields: (1.5.24) Substitution of Equation (1.5.24) into the differential equation for Cs gives: dCB _ ° -+ k1CB - kjCAexp[ -kjt] dt This equation is in the proper form for solution by the integrating factor method, that is: : + p(t)y = g(t), I = exp [J p(t)dt] (d(IY) = (hdt)dt ) .' ) u ..
Now, for the equation concerning CB, p(t) = k1 so that: Integration of the above equation gives: CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering or 39 [ k]C~ ] Cs = exp( -kIt) + l' exp(-k2t) k2 ki where "I is the integration constant. Since Cs = C~ at t = 0, "I can be written as a function of C~, C~, ki and k2• Upon evaluation of "I, the following expression is found for Cs (t): k]C~ Cs = k 2 _ k i [exp(-kIt) - exp(-k2 t)] + C~exp(-k2t) (1.5.25) By knowing CB(t) and CA(t), Cdt) is easily obtained from the equation for the con-servation of mass: (1.5.26) (1.5.27) For cg = cg = 0 and ki = k2 , the normalized concentrations of CA , CB , and Cc are plotted in Figure 1.5.4.
Notice that the concentration of species B initially increases, reaches a maximum, and then declines. Often it is important to ascertain the maximum amount of species B and at what time of reaction the maximum occurs. To find these quantities, notice that at CIJ'ax, dCs/dt = O. Thus, if the derivative of Equation (1.5.25) is set equal to zero then tmax can be found as: 1 [(k2 )( cg (k2 ) Cg)] t = In -1+----max (k2 kI) ki C~ k] C~ Using the expression for tmax (Equation (1.5.27)) in Equation (1.5.25) yields CIJ'ax.
1.0 o Time Figure 1.5.4 I Nonnalized concentration of species i as a function of time for kl k2• 40 EXAMPLE 1.5.6 I CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering For C~ = C~ = 0, find the maximum concentration of Cs for k] = 2k2• • Answer From Equation (1.5.27) with C~ = 0, tmax is: Substitution of tmax into Equation (1.5.25) with C~ = ° gives: or This equation can be simplified as follows: or C/lax = 0.5C~ When dealing with multiple reactions that lead to various products, issues of selectivity and yield arise. The instantaneous selectivity, Si, is a function of the local conditions and is defined as the production rate of species i divided by the produc-tion rates of all products of interest: ri Si = --, j = I,. .., all products of interest 2:rj (1.5.28) where ri is the rate of production of the species i. An overall selectivity, Si, can be defined as: total amount of species i S=------------I total amount of products of interest (1.5.29) CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering The yield, Yi , is denoted as below: total amount of product i fonned y. = -------"---------I initial amount of reactant fed 41 (1.5.30) where the initial amount of reactant fed is for the limiting component. For the net-work given by Equation (1.5.22): amount of B fonned amount of B fonned SB = = --------amount of Band C fonned amount of A reacted or and amount of B fonned CB Y ---------B -initial amount of A fed C ~ (1.5.31) (1.5.32) The selectivity and yield should, of course, correctly account for the stoichiometry of the reaction in all cases.
EXAMPLE 1.5.7 I Plot the percent selectivity and the yield of B [Equation (1.5.31) multiplied by 100 percent and Equation (1.5.32), respectively] as a function of time. Does the time required to reach Ce ax give the maximum percent selectivity to B and/or the maximum fractional yield of B?
Let kl = 2k2, C~ = C~ = O .
• Answer From the plot shown below the answer is obvious. For practical purposes, what is important is the maximum yield.
0.5 100 "Q 0.4 80 4-0 if) 0 CZl '1;l 0.3 60 " " 1> '>, () " ~: c 0.2 40 q .S a t) " tl:l ci: 0.1 20 0 0 Time~ 42 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Consider the reaction network of two irreversible (one-way), first-order reac-tions in parallel: yDP A ~SP (1.5.33) Again, like the series network shown in Equation (1.5.22), the parallel network of Equation (1.5.33) can represent a variety of important reactions. For example, de-hydrogenation of alkanes can adhere to this reaction network where the desired product DP is the alkene and the undesired side-product SP is a hydrogenolysis (C -C bond-breaking reaction) product: ~ CHz=CHz + Hz CH3 ~ CH4 + carbonaceous residue on catalyst Using the Guldberg-Waage form of the reaction rates to describe the network in Equation (1.5.33) gives for constant volume: (1.5.34) with c2 + c2p + cgp= CO = CA + CDP + CSP Integration of the differential equation for CA with CA = c2 at t = 0 gives: (1.5.35) Substitution of Equation (1.5.35) into the differential equation for CDP yields: The solution to this differential equation with CDP = cgp at t = 0 is: (1.5.36) CHAPTER 1 The Basics Qf ReactiQn Kinetics for Chemical Reaction Engineering Likewise, the equation for Csp can be obtained and it is: The percent selectivity and yield of DP for this reaction network are: 43 (1.5.37) (1.5.38) and CD?
y=-C~ EXAMPLE 1.5.8 I The following reactions are observed when an olefin is epoxidized with dioxygen: alkene + O2 ===> epoxide epoxide + O2 ===> CO2 + H20 alkene + O2 ===> CO2 + H20 (1.5.39) Derive the rate expression for this mixed-parallel series-reaction network and the expression for the percent selectivity to the epoxide.
• Answer The reaction network is assumed to be: k, A+O ~ EP+O CD CD where A: alkene, 0: dioxygen, EP: epoxide, and CD: carbon dioxide. The rate expressions for this network are: dt 44 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering The percent selectivity to EP is: EXAMPLE 1.5.9 I EXAMPLE 1.5.10 In Example 1.5.6, the expression for the maximum concentration in a series reaction network was illustrated. Example 1.5.8 showed how to determine the selectivity in a mixed-parallel series-reaction network. Calculate the maximum epoxide selectivity attained from the reac-tion network illustrated in Example 1.5.8 assuming an excess of dioxygen.
• Answer If there is an excess of dioxygen then Co can be held constant. Therefore, From this expression it is clear that the selectivity for any CA will decline as k2CEP increases.
Thus, the maximum selectivity will be: and that this would occur at t = 0, as was illustrated in Example 1.5.7. Here, the maximum selectivity is not 100 percent at t = ° but rather the fraction kI/(k2 + k3) due to the parallel portion of the network.
Find the maximum yield of the epoxide using the conditions listed for Example 1.5.9.
• Answer The maximum yield, CEpx/C~ will occur at CEPx. If k I = kICO, k 2 = k2CO' k 3 = k3CO' y CA/C~ and x = CEP/C~, the rate expressions for this network can be written as: dy dt dx dt Note the analogy to Equation (1.5.23). Solving the differential equation for y and substitut-ing this expression into the equation for x gives: dx + dt CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 45 o~ v 0.5 0-0,< v 0.5 0-0 1Z 0 0,< v 0.5 --G 0 1Z 0 propane Figure 1.5.5 I Normalized concentration of species i as a function of time for k) = k 2• Solution of this differential equation by methods employed for the solution of Equation (1.5.23) gives: x = _ _ [exp[-(k) + k3)t] - exp(-k2t)] k 2 (k) + k3) If k 3 = 0, then the expression is analogous to Equation (1.5.25). In Figure 1.5.5, the nor-malized concentration profiles for various ratios ofkik3 are plotted. The maximum yields of epoxide are located at the xuax for each ratio of kik3 • Note how increased reaction rate to deep oxidation of alkane decreases the yield to the epoxide.
Exercises for Chapter 1 1.
Propylene can be produced from the dehydrogenation of propane over a catalyst. The reaction is: Hz H /C"" =9= /C~ + Hz H3C CH3 H3C CHz propylene At atmospheric pressure, what is the fraction ofpropane converted to propylene at 400, 500, and 600°C if equilibrium is reached at each temperature? Assume ideal behavior.
Temperature (0C) 400 0.000521 500 0.0104 600 0.104 propane 46 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 2.
An alternative route to the production of propylene from propane would be through oxydehydrogenation: H2 H /C"" + 1/202:::§:: /C~ + H20 H3C CH3 H3C CH2 propylene At atmospheric pressure, what is the fraction ofpropane converted to propylene at 400, 500, and 600°C if equilibrium is reached at each temperature? Compare the results to those from Exercise 1. What do you think is the major impediment to this route of olefin formation versus dehydrogenation? Assume ideal behavior.
Temperature (0C) 400 500 5.34 X 1023 600 8.31 X 1021 3.
The following reaction network represents the isomerization of I-butene to cis- and trans-2-butene: I-butene CH3 / HC=C / H H3C trans-2-butene CH \ CH3 cis-2-butene The equilibrium constants (Ka's) for steps 1 and 2 at 400 K are 4.30 and 2.15, respectively. Consider the fugacity coefficients to be unity.
(a) Calculate the equilibrium constant of reaction 3.
(b) Assuming pure I-butene is initially present at atmospheric pressure, calculate the equilibrium conversion of I-butene and the equilibrium composition of the butene mixture at 400 K. (Hint: only two of the three reactions are independent.) 4.
Xylene can be produced from toluene as written schematically: CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 47 CH3 c5 nOo= CH3 ~ I 14.85 kJ maC' ()CH 3 (' + ::§:: I + U ~ / toluene ortho-xylene benzene c5 c5 CH, nOo= 10.42 kJ mo!-l 6, 0 + ::§:: I / CH3 toluene meta-xylene benzene C CH3 nOo= Q 6 15.06 kJ moC' 0 + ::§:: + 1/ toluene CH3 benzene para-xylene The values of ~Go were determined at 700 K. What is the equilibrium composition (including all xylene isomers) at 700 K and 1.0 atm pressure?
Propose a method to manufacture para-xylene without producing significant amounts of either ortho- or meta-xylene.
5.
Vinyl chloride can be synthesized by reaction of acetylene with hydrochloric acid over a mercuric chloride catalyst at 500 K and 5.0 atm total pressure. An undesirable side reaction is the subsequent reaction of vinyl chloride with HCl.
These reactions are illustrated below.
HC==CH + HCl acetylene H2C = CHCl + HCl H2C= CHCl vinyl chloride H 3C-CHC1 2 1,2 dichloroethane (1) (2) The equilibrium constants at 500 K are 6.6 X 103 and 0.88 for reaction 1 and 2, respectively. Assume ideal behavior.
(a) Find the equilibrium composition at 5.0 atm and 500 K for the case when acetylene and HCl are present initially as an equimolar mixture. What is the equilibrium conversion of acetylene?
(b) Redo part (a) with a large excess of inert gas. Assume the inert gas constitutes 90 vol. % of the initial gas mixture.
6.
Acetone is produced from 2-propanol in the presence of dioxygen and the photocatalyst Ti02 when the reactor is irradiated with ultraviolet light. For a 48 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering reaction carried out at room temperature in 1.0 mol of liquid 2-propano1 containing 0.125 g of catalyst, the following product concentrations were measured as a function of irradiation time (J. D. Lee, M.S. Thesis, Univ. of Virginia, 1993.) Calculate the first-order rate constant.
Reaction time (min) 20 Acetone produced 1.9 (gacetone!c1?2.propano!) X 10 4 40 60 80 3.9 5.0 6.2 100 8.2 160 13.2 180 14.0 7.
The Diels-Alder reaction of 2,3-dimethyl-1,3-butadiene (DMB) and acrolein produces 3,4-dimethyl-~3-tetrahydro-benzaldehyde.
+ Acrolein 3,4-Dimethyl-tl.3-tetrahydro-benzaldehyde This overall second-order reaction was performed in methanol solvent with equimolar amounts of DMB and acrolein (C. R. Clontz, Jr., M.S. Thesis, Univ.
of Virginia, 1997.) Use the data shown below to evaluate the rate constant at each temperature.
323 0 0.097 323 20 0.079 323 40 0.069 323 45 0.068 298 0 0.098 298 74 0.081 298 98 0.078 298 125 0.074 298 170 0.066 278 0 0.093 278 75 0.091 278 110 0.090 278 176 0.088 278 230 0.087 8.
Some effluent streams, especially those from textile manufacturing facilities using dying processes, can be highly colored even though they are considered to be fairly nontoxic. Due to the stability of modem dyes, conventional CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 49 biological treatment methods are ineffective for decolorizing such streams.
Davis et al. studied the photocatalytic decomposition of wastewater dyes as a possible option for decolorization [R. J. Davis, J. L. Gainer, G. O'Neal, and L-W. Wu, Water Environ. Res. 66 (1994) 50]. The effluent from a municipal water treatment facility whose influent contained a high proportion of dyeing wastewater was mixed with Ti02 photocatalyst (0.40 wt. %), sparged with air, and irradiated with UV light. The deep purple color of the original wastewater lightened with reaction time. Since the absolute concentration of dye was not known, the progress of the reaction was monitored colorimetrically by measuring the relative absorbance of the solution at various wavelengths.
From the relative absorbance data collected at 438 nm (shown below), calculate the apparent order of the decolorization reaction and the rate constant. The relative absorbance is the absorbance at any time t divided by the value at t = O.
Reaction time (min) Relative absorbance 210 0.17 9.
The following reaction is investigated in a constant density batch reactor: NOz N,o,+,o=z6+ H'o (A) (B) The reaction rate is: The reactor is initially charged with c1 and c2 and c2/c1 = 3.0. Find the value of kif c1 = 0.01 mol/L and after 10 min of reaction the conversion of NzOs is 50 percent.
10.
As discussed in Vignette 1.1.1, ammonia is synthesized from dinitrogen and dihydrogen in the presence of a metal catalyst. Fishel et al. used a constant volume reactor system that circulated the reactants over a heated ruthenium metal catalyst and then immediately condensed the product ammonia in a cryogenic trap [c. T. Fishel, R. J. Davis, and J. M. Garces, J. Catal. 163 (1996) 148]. A schematic diagram of the system is: 50 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering Catalytic reactor Ammonia trap From the data presented in the following tables, determine the rates of ammonia synthesis (moles NH3 produced per min per gcat) at 350°C over a supported ruthenium catalyst (0.20 g) and the orders of reaction with respect to dinitrogen and dihydrogen. Pressures are referenced to 298 K and the total volume of the system is 0.315 L. Assume that no ammonia is present in the gas phase.
Pressure (torr) Time (min) Pressure (torr) Time (min) Pressure (torr) Time (min) 661.5 54 700.3 60 675.5 55 11.
In Example 1.5.6, the series reaction: A~B~C was analyzed to determine the time (tmax) and the concentration (c;ax) associated with the maximum concentration of the intermediate B when k] = 2k2 • What are tmax and c;ax when k] = k2?
CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 51 12.
The Lotka-Volterra model is often used to characterize predator-prey interactions.
For example, ifR is the population ofrabbits (which reproduce autocatalytically), G is the amount of grass available for rabbit food (assumed to be constant), L is the population oflynxes that feeds on the rabbits, and D represents dead lynxes, the following equations represent the dynamic behavior of the populations of rabbits and lynxes: R + G ~ 2R (1) L + R ~ 2L (2) L ~ D (3) Each step is irreversible since, for example, rabbits cannot tum back into grass.
(a) Write down the differential equations that describe how the populations of rabbits (R) and lynxes (L) change with time.
(b) Assuming G and all of the rate constants are unity, solve the equations for the evolution of the animal populations with time. Let the initial values of Rand L be 20 and I, respectively. Plot your results and discuss how the two populations are related.
13.
Diethylamine (DEA) reacts with l-bromobutane (BB) to form diethylbutylamine (DEBA) according to: H2 C2HS Br C CH3 I + ~ / ~ / NH C C H2 H2 + HBr diethylamine I-bromobutane C2HS I C2HS- N I C4H9 diethylbutylamine From the data given below (provided by N. Leininger, Univ. of Virginia), find the effect of solvent on the second-order rate constant.
o 35 115 222 455 0.000 0.002 0.006 0.012 0.023 o 31 58 108 190 0.000 0.004 0.007 0.013 0.026 Solvent = IA-butanediol T 22°C [DEA]o = [BB]o = O.SO mol L- 1 Solvent = acetonitrile T = 22°C [DEA]o 1.0 mol L- 1 [BB]O 0.10 mol L- 1 52 CHAPTER 1 The Basics of Reaction Kinetics for Chemical Reaction Engineering 14.
A first-order homogeneous reaction ofA going to 3B is carried out in a constant pressure batch reactor. It is found that starting with pure A the volume after 12 min is increased by 70 percent at a pressure of 1.8 atm. If the same reaction is to be carried out in a constant volume reactor and the initial pressure is 1.8 atm, calculate the time required to bring the pressure to 2.5 atm.
15.
Consider the reversible, elementary, gas phase reaction shown below that occurs at 300 K in a constant volume reactor of 1.0 L.
k] A + B ( ) C, k1 = 6.0Lmol-1h-1,k2 = 3.0h-1 k2 For an initial charge to the reactor of 1.0 mol of A, 2.0 mol of B, and no C, find the equilibrium conversion of A and the final pressure of the system. Plot the composition in the reactor as a function of time.
16.
As an extension of Exercise 15, consider the reversible, elementary, gas phase reaction ofA and B to form C occurring at 300 K in a variable volume (constant pressure) reactor with an initial volume of 1.0 L. For a reactant charge to the reactor of 1.0 mol ofA, 2.0 mol ofB, and no C, find the equilibrium conversion of A. Plot the composition in the reactor and the reactor volume as a function of time.
17.
As an extension of Exercise 16, consider the effect of adding an inert gas, I, to the reacting mixture. For a reactant charge to the reactor of 1.0 mol of A, 2.0 mol of B, no C, and 3.0 mol of I, find the equilibrium conversion of A. Plot the composition in the reactor and the reactor volume as a function of time.
_~~....2...,f~ Rate Constants of Elementary Reactions 2.1 I Elementary Reactions Recall from the discussion of reaction networks in Chapter 1 that an elementary re-action must be written as it proceeds at the molecular level and represents an irre-ducible molecular event. An elementary reaction normally involves the breaking or making of a single chemical bond, although more rarely, two bonds are broken and two bonds are formed in what is denoted a four-center reaction. For example, the reaction: is a good candidate for possibly being an elementary reaction, while the reaction: is not. Whether or not a reaction is elementary must be determined by experimentation.
As stated in Chapter 1, an elementary reaction cannot be written arbitrarily and must be written the way it takes place. For example (see Table 1.4.3), the reaction: (2.1.1) cannot be written as: (2.1.2) since clearly there is no such entity as half a molecule of dioxygen. It is important to note the distinction between stoichiometric equations and elementary reactions (see Chapter 1) is that for the stoichiometric relation: (2.1.3) S4 CHAPTER 2 Rate Constants of Elementary Reactions one can write (although not preferred): NO + ~02 = N02 (2.1.4) The remainder of this chapter describes methods to determine the rate and tem-perature dependence of the rate of elementary reactions. This information is used to describe how reaction rates in general are appraised.
2.2 I Arrhenius Temperature Dependence of the Rate Constant The rate constant normally depends on the absolute temperature, and the functional form of this relationship was first proposed by Arrhenius in 1889 (see Rule III in Chapter 1) to be: k = Ii exp[ -E/(RgT)J (2.2.1) EXAMPLE 2.2.1 I where the activation energy, E, and the pre-exponential factor, A, both do not de-pend on the absolute temperature. The Arrhenius form of the reaction rate constant is an empirical relationship. However, transition-state theory provides a justification for the Arrhenius formulation, as will be shown below. Note that the Arrhenius law (Equation 2.2.1) gives a linear relationship between In k and T~ 1• The decomposition reaction: can proceed at temperatures below 100°C and the temperature dependence of the first-order rate constant has been measured. The data are: 288 298 313 323 338 1.04 X 1O~5 3.38 X 1O~5 2.47 X 1O~4 7.59 X 1O~4 4.87 X 10 3 Suggest an experimental approach to obtain these rate constant data and calculate the acti-vation energy and pre-exponential factor. (Adapted from C. G. Hill, An Introduction to Chem-ical Engineering Kinetics & Reactor Design. Wiley, New York, 1977.) • Answer Note that the rate constants are for a first-order reaction. The material balance for a closed system at constant temperature is: dt CHAPTER 2 Bate Constants of Elementary Reactions 55 where nN,o, is the number of moles of NzOs. If the system is at constant volume (a closed vessel), then as the reaction proceeds the pressure will rise because there is a positive mole change with reaction. That is to say that the pressure will increase as NzOs is reacted because the molar expansion factor is equal to 0.5. An expression for the total moles in the closed system can be written as: where n is the total number of moles in the system. The material balance on the closed sys-tem can be formulated in terms of the fractional conversion and integrated (see Example 1.5.2) to give: fN,o, = 1 - exp(-kt) Since the closed system is at constant T and V (PV = nRgT): P n -=-=(1 +0.5fNo) Po no 2 , and the pressure can therefore be written as: P = Po[1.5 - 0.5exp(-kt)] If the pressure rise in the closed system is monitored as a function of time, it is clear from the above expression how the rate constant can be obtained at each temperature.
In order to determine the pre-exponential factor and the activation energy, the In k is plotted against as shown below: From a linear regression analysis of the data, the slope and intercept can be obtained, and they are 1.21 X 104 and 30.4, respectively. Thus, slope = -EjRg intercept = In A E = 24 kcal/mol A 1.54 X 1013 S-1 56 CHAPTER 2 Rate Constants of Elementary Reactions Consider the following elementary reaction: k j A+B( )S+W k2 At equilibrium, and (2.2.2) (2.2.3) (2.2.4) (2.2.5) k1 (CsCw) Kc = k2 = CACB eq If the Arrhenius law is used for the reaction rate constants, Equation (2.2.4) can be written: Kc = k2 = (~Jexp(E2R~/I) = (~:~:)eq It is easy to see from Equation (2.2.5) that if (E2 -E1) > 0 then the reaction is exothermic and likewise if (E2 -E1) < 0 it is endothermic (refer to Appendix A for temperature dependence of Kc). In a typical situation, the highest yields of prod-ucts are desired. That is, the ratio (CSCW/CACB)eq will be as large as possible. If the reaction is endothermic, Equation (2.2.5) suggests that in order to maximize the product yield, the reaction should be accomplished at the highest possible temper-ature. To do so, it is necessary to make exp[(E2 - E1)/(RgT) ] as large as possible by maximizing (RgT), since (E2 -E 1) is negative. Note that as the temperature in-creases, so do both the rates (forward and reverse). Thus, for endothermic reactions, the rate and yield must both increase with temperature. For exothermic reactions, there is always a trade-off between the equilibrium yield of products and the reac-tion rate. Therefore, a balance between rate and yield is used and the T chosen is dependent upon the situation.
2.3 I Transition-State Theory For most elementary reactions, the rearrangement of atoms in going from reactants to products via a transition state (see, for example, Figure 1.1.1) proceeds through the movements of atomic nuclei that experience a potential energy field that is gen-erated by the rapid motions of the electrons in the system. On this potential energy surface there will be a path of minimum energy expenditure for the reaction to pro-ceed from reactants to products (reaction coordinate). The low energy positions of reactants and products on the potential energy surface will be separated by a higher energy region. The highest energy along the minimum energy pathway in going from reactants to products defines the transition state. As stated in Chapter I, the transition state is not a reaction intermediate but rather a high energy configuration of a system in transit from one state to another.
CHAPTER 2 Rate Constants of Elementary Reactions 57 Transition state s+w A+f"i""~ \ Energy difference related to tJ.H,.
Transition state VIGNETTE 2.3.1 Reaction coordinate Reaction coordinate (a) (b) Figure 2.3.1 I Potential energy profiles for the elementary reaction A + B -+ S + W for (a) an endothermic reaction and (b) an exothermic reaction.
The difference in energies of the reactants and products is related to the heat of reaction-a thermodynamic quantity. Figure 2.3.1 shows potential energy profiles for endothermic and exothermic elementary reactions.
Transition-state theory is used to calculate an elementary reaction that con-forms to the energetic picture illustrated in Figure 2.3.1. How this is done is de-scribed next.
58 C H APT E R 2 Rate Constants oLElemen1a.yrYLJRU'ewa",c.u..til.lollns~ ~ 8 '" i 0 I' E ~ >, 4 W CH3CCH3 + H2 over a series of solid catalysts called hydrotalcites, and obtained the following data: 1 2 3 4 4.3 X 1012 2.3 X 1011 2.2 X 1010 1.6 X 109 172 159 146 134 60 CHAPTER 2 Rate Constants of Elementary Reactions If Equation (2.3.9) is compared to Equation (2.2.1), then: A= (~) exp[ A:gt ] E= AHt (2.3.10) (2.3.11) The data of McKenzie et al. clearly show that as the energy barrier increases (higher E), the entropy of activation becomes more positive (larger ASt implies more favorable configurational driving force for the reaction, since entropy will always attempt to be maximized). Thus, energy and entropy compensateJn this example.
The Arrhenius form of the rate constant specifies that both A and E are inde-pendent of T. Note that when the formulation derived from transition-state theorx is compared to the Arrhenius formulation [Equations (2.3.10) and (2.3.11)], both A and Edo have some dependence on T. However, AHt is very weakly dependent on T and the temperature dependence of: .(kT) [ASt] -exp--h Rg (2.3.12) can normally be neglected as compared to the strong exponential dependence of: [ -AHt] exp R T g (2.3.13) Thus, the Arrhenius form is an excellent approximation to that obtained from transition-state theory, provided the temperature range does not become too large.
Previously, the concentration CTS was expressed in simple terms assuming ideal conditions. However, for nonideal systems, K4" must be written as: (2.3.14) (2.3.15) where ai is the activity of species i. If activity coefficients, 'Yi' are used such that ai = YiCi, then Equation (2.3.14) can be formulated as: K4" - [~~J [~~J Following the substitutions and equation rearrangements used for ideal conditions, the nonideal system yields: k =(kT) Y~YB K4" h 'Yrs (2.3.16) At infinite dilution where the activity coefficients are one (ideal conditions), ko can be written as: (2.3.17) CHAPTER 2 Rate Constants of Elementary Reactions Thus, the rate constant at nonideal conditions relative to the ideal system is: 61 EXAMPLE 2.3.1 I k YAYS -=--Predict how the rate constant of the reaction: A+B--+S (2.3.18) would vary as a function of ionic strength, J, if A and B are both ions in aqueous solutions at 25°C. The Debye-Hlickel theory predicts that: -log(Yi) = 0.5z1VI where Zi is the charge of species i and - 12:-2 1=-Zc 2 i I I (i=A,B,S) where Ci is the concentration of species i in units of molality.
• Answer Although the structure of the transition state is not given, it must have a charge of ZA + ZB' Using the Debye-Hlickel equation in Equation (2.3.18) gives: (k) (YAYB) log ko = log YTS or After simplification, the above equation reduces to: This relationship has been experimentally verified. Note that if one of the reactants is un-charged, for example, B, then ZB = 0 and k = ko. If the rate is mistakenly written: application of the Debye-Hlickel equation gives: 10g(~) IOgeO~:YB) = -0.5[Z~ + Z~]VI Note that this relationship gives the wrong slope of IOg(: ) versus VI, and if ZB 0, it does not reduce to k ko.
0 62 CHAPTER 2 Rate Constants of Elementary Reactions It is expected that the transition state for a unimolecular reaction may have a struc-ture similar to that of the reactant except for bond elongation prior to breakage. If this is the case, then List == 0 and (2.3.19) It has been experimentally verified that numerous unimolecular reactions have rate constants with pre-exponential factors on the order of 1013 S-I. However, the pre-exponential factor can be either larger or smaller than 1013 s-I depending on the details of the transition state.
Although it is clear that transition-state theory provides a molecular perspec-tive on the reaction and how to calculate the rate, it is difficult to apply since List, LiH(;, and YTS are usually not known a priori. Therefore, it is not surprising that the Arrhenius rate equation has been used to systemize the vast majority of exper-imental data.
Exercises for Chapter 2 1.
For a series of similar reactions, there is often a trend in the equilibrium constants that can be predicted from the rate constants. For example, the reaction coordi-nate diagrams of two similar reactions are given below. If the difference in free energy of formation of the transition state is proportional to the difference in the free energy change upon reaction, that is, LiGi LiGf = a(LiG1 -LiG1), derive the relationship between the rate constants k] and k1 and the equilibrium constants (K] and K1).
Reaction coordinate Schematic diagram of two similar reactions.
CHAPTER 2 Rate Constants of Elementary Reactions 63 2.
The decomposition of gaseous 2-propanol over a mixed oxide catalyst of magnesia and alumina produces both acetone and propene according to the fol-lowing equations: OH 0 ~ ===> ~ +Hz 2-propanol acetone OH ~ ===> ~ +HzO 2-propanol propene From the data presented below, calculate the activation energy for each reaction (A. L. McKenzie, M.S. Thesis, Univ. of Virginia, 1992). Assume the concentra-tion of 2-propanol is constant for each experiment. Selectivity to acetone is defined with respect to the products acetone and propene.
Temperature (K) 573 583 594 603 612 Rate of acetone 4.1 X 10 7 7.0 X 10-7 1.4 X 10-6 2.2 X 10-6 3.6 X 10-6 formation (mol gcat- I 5- 1) Selectivity to 92 86 81 81 81 acetone (%) 3,4-Dimethyl-~3 tetrahydro-benzaldehyde Acrolein + 3.
Use the data in Exercise 7 at the end of Chapter 1 to determine the activation energy and pre-exponential factor of the rate constant for the Diels-Alder reaction of 2,3-dimethyl-l,3-butadiene (DMB) and acrolein to produce 3,4-dimethyl-Li 3-tetrahydro-benzaldehyde.
o 0 ~H ===> H3C~H H3C~ 4.
Explain how the pre-exponential factor of a unimolecular reaction can be greater than 1013 S-I.
5.
Discuss the strengths and weaknesses of transition-state theory.
6.
Irradiation of water solutions with gamma rays can produce a very active intermediate known as a hydrated electron. This species can react with many different neutral and ionic species in solution. Devise an experiment to check the electrical charge of the hydrated electron. (Problem adapted from M.
Boudart, Kinetics of Chemical Processes, Prentice Hall, Englewood Cliffs, NJ, 1968, p. 55.) ~c~~3_ Reactors for Measuring Reaction Rates 3.1 I Ideal Reactors The confines in which chemical reactions occur are called reactors. A reactor can be a chemical reactor in the traditional sense or other entities, for example, a chem-ical vapor deposition apparatus for making computer chips, an organ of the human body, and the atmosphere of a large city. In this chapter, the discussion of reactors is limited to topics germane to the determination of reaction rates. Later in this text, strategies for attacking the problems of mathematically describing and predicting behavior of reactors in general are presented.
In practice, conditions in a reactor are usually quite different than the ideal re-quirements used in the definition of reaction rates. Normally, a reactor is not a closed system with uniform temperature, pressure, and composition. These ideal conditions can rarely if ever be met even in experimental reactors designed for the measure-ment of reaction rates. In fact, reaction rates cannot be measured directly in a closed system. In a closed system, the composition of the system varies with time and the rate is then inferred or calculated from these measurements.
There are several questions that can be put forth about the operation of reac-tors and they can be used to form the basis of classifying and defining ideal condi-tions that are desirable for the proper measurements of reaction rates.
The first question is whether the system exchanges mass with its surroundings.
If it does not, then the system is called a batch reactor. If it does, then the system is classified as a flow reactor.
The second question involves the exchange of heat between the reactor and its surroundings. If there is no heat exchange, the reactor is then adiabatic. At the other extreme, if the reactor makes very good thermal contact with the surroundings it can be held at a constant temperature (in both time and position within the reactor) and is thus isothermal.
64 CHAPTER 3 Reactors for Measllring Reaction Rates Table 3.1.1 I Limiting conditions of reactor operation" 1 65 Exchange of mass Exchange of heat Mechanical variables Residence time Space-time behavior Batch Isothermal Constant volume Unique Transient Flow Adiabatic Constant pressure Exponential distribution Stationary IFrom M. Boudart, Kinetics of Chemical Processes, Butterworth & Heinemann, 1991, p. 13.
The third question concerns the mechanical variables: pressure and volume. Is the reactor at constant pressure or constant volume? The fourth question is whether the time spent in the reactor by each volume element of fluid is the same. If it is not the same, there may exist a distribution of residence times and the opposite ex-treme of a unique residence time is an exponential distribution.
The fifth question focuses on a particular fixed volume element in the reactor and whether it changes as a function of time. If it does not, then the reactor is said to operate at a stationary state. If there are time variations, then the reactor is oper-ating under transient conditions. A nontrivial example of the transient situation is designed on purpose to observe how a chemically reactive system at equilibrium re-laxes back to the equilibrium state after a small perturbation. This type of relaxation experiment can often yield informative kinetic behavior.
The ten possibilities outlined above are collected in Table 3.1.1. Next, ideal re-actors will be illustrated in the contexts of the limiting conditions of their operation.
3.2 I Batch and Semibatch Reactors Consider the ideal batch reactor illustrated in Figure 3.2.1. If it is assumed that the contents of the reactor are perfectly mixed, a material balance on the reactor can be written for a species i as: or accumulation o input o output amount produced by reaction dn l -= vrV with n," = nO,@t=O dt ' (3.2.1) The material balance can also be written in terms of the fractional conversion and it is: °dft _ ( ) O( ) n -- vr V I + sF.
I dt ' ill where lSiI > 0 for nonconstant volume.
with/; O@t = 0 (3.2.2) 66 CHAPTER 3 Reactors for MeaslJring Reaction Rates v(t) Batch Semibatch EXAMPLE 3.2.1 I Figure 3.2.1 I Ideal batch and semibatch reactors. vet) is a volumetric flow rate that can vary with time.
An important class of carbon-carbon bond coupling reactions is the Diels-Alder reactions.
An example of a Diels-Alder reaction is shown below: o cyclopentadiene (A) + 0 ° CH/CO'CH ....
C"'CH I I=> II I CH2 II CH CHICH .........
/" ........
,./ CO CH ° benzoquinone tricycle [6.2.1.02,7J-undec-(B) 4,9-diene-3,6-dione If this reaction is performed in a well-mixed isothermal batch reactor, determine the time nec-essary to achieve 95 percent conversion of the limiting reactant (from C. Hill, An Introduc-tion to Chemical Engineering Kinetics and Reactor Design, Wiley, 1977, p. 259).
Data: k = 9.92 X 10- 6 m3/mol/s C~ = 100 mol/m3 C~ = 80 mol/m3 CHAPTER 3 Reactors for MeaslJring Reaction Rates 67 • Answer From the initial concentrations, benzoquinone is the limiting reactant. Additionally, since the reaction is conducted in a dilute liquid-phase, density changes can be neglected. The reac-tion rate is second-order from the units provided for the reaction rate constant. Thus, The material balance on the isothermal batch reactor is: with IB = 0 at t = O. Integration of this first-order initial-value problem yields: (fB dy t = J o kd(l - y)(M y) where y is the integration variable. The integration yields: Using the data provided, t = 7.9 X 103 s or 2.2 h to reach 95 percent conversion of the benzoquinone. This example illustrates the general procedure used for solving isother-mal problems. First, write down the reaction rate expression. Second, formulate the ma-terial balance. Third, substitute the reaction rate expression into the material balance and solve.
Consider the semibatch reactor schematically illustrated in Figure 3.2.1.
This type of reactor is useful for accomplishing several classes of reactions.
Fermentations are often conducted in semibatch reactors. For example, the concentration of glucose in a fermentation can be controlled by varying its concentration and flow rate into the reactor in order to have the appropriate times for: (1) the initial growth phase of the biological catalysts, and (2) the period of metabolite production. Additionally, many bioreactors are semibatch even if liquid-phase reactants are not fed to the reactor because oxygen must be continuously supplied to maintain the living catalyst systems (Le., bacteria, yeast, etc.).
Alternatively, semibatch reactors of the type shown in Figure 3.2.1 are useful for reactions that have the stoichiometry: A + B = products where B is already in the reactor and A is slowly fed. This may be necessary to: (1) con-trol heat release from exothermic reaction, for example, hydrogenations, (2) provide 68 CHAPTER 3 Reactors for MeaslJring Reaction Rates gas-phase reactants, for example, with halogenations, hydrogenations, or (3) alter reaction selectivities. In the network: A + B ~ desired product A ~ undesired product maintaining a constant and high concentration of B would certainly aid in altering the selectivity to the desired product.
In addition to feeding of components into the reactor, if the sign on vet) is neg-ative, products are continuously removed, for example, reactive distillation. This is done for reactions: (1) that reveal product inhibition, that is, the product slows the reaction rate, (2) that have a low equilibrium constant (removal of product does not allow equilibrium to be reached), or (3) where the product alters the reaction net-work that is proceeding in the reactor. A common class of reactions where product removal is necessary is ester formation where water is removed, o 0 0 2-~OH + HOCH2 CH2CH2 CH20H = -~OCH2CH2CH2CH20~-o + 2H20t A very large-scale reaction that utilizes reactive distillation of desired liquid prod-ucts is the hydroformylation of propene to give butyraldehyde: o II C -H ==> heavier products A schematic illustration of the hydroformylation reactor is provided in Figure 3.2.2.
The material balance on a semibatch reactor can be written for a species i as: or accumulation v(t)c9(t) input output amount produced by reaction (3.2.2) where C?(t) is the concentration of species i entering from the input stream of vol-umetric flow rate vet).
CHAPTER 3 Reactors for Meas!!ring Reaction Rates Propene, H2, CO 69 Unreacted propene, CO, Hz for recycle + product C4 Liquid-phase (solvent + catalyst + products) EXAMPLE 3.2.2 I Syn-gas (Hz: CO) Figure 3.2.2 I Schematic illustration of a propene hydrofonnylation reactor.
To a well-stirred tank containing 40 mol of triphenylmethylchloride in dry benzene (initial volume is 378 L) a stream of methanol in benzene at 0.054 mol/L is added at 3.78 L/min. A reaction proceeds as follows: CH30H + (C6Hs)3CCl =} (C6Hs)3COCH3 + HCl (A) (B) The reaction is essentially irreversible since pyridine is placed in the benzene to neutralize the fonned HC1, and the reaction rate is: r = 0.263 C~Cs (moUL/min) Detennine the concentration of the product ether as a function of time (problem adapted from N. H. Chen, Process Reactor Design, Allyn and Bacon, Inc., 1983, pp. 176-177).
• Answer The material balance equations for nA and ns are: dt (0.054 mol/L)(3.78 L/min) 0.263 C~CsV dns -= -0.263 C2 4CSV dt ' 70 CHAPTER 3 Beartors Inc MeaslJfjng Boaction Batos The volume in the reactor is changing due to the input of methanol in benzene. Thus, the volume in the reactor at any time is: V 3.78(100 + t) Therefore, 0.204 - 0.263 n~nsl[3.78(100 + where From Equation (1.2.6), dna dt -0.263 fI~flB/[3.78(100+ t)F n~ = O@ t 0 ng = 40@ t 0 nether = fig -fiB since no ether is initially present. The numerical solution to the rate equations gives flA(t) and fls(t) from which flether(t) can be calculated. The results are plotted below.
40 30 "0 g 20 i:-10 o o 2000 4000 Time (min) 6000 8000 For other worked examples of semibatch reactors, see H. S. Fogler, Elements of Chemical Reaction Engineering, 3rd ed., Prentice-Hall, 1992, pp. 190-200, and N. H. Chen, Process Reactor Design, Allyn and Bacon, Inc., 1983, Chap. 6.
3.3 I Stirred-Flow Reactors The ideal reactor for the direct measurement of reaction rates is an isothermal, con-stant pressure, flow reactor operating at steady-state with complete mixing such that the composition is uniform throughout the reactor. This ideal reactor is frequently CHAPTER 3 Reaclors for MeaslJring Reaction Rates 71 called a stirred-tank reactor, a continuous flow stirred-tank reactor (CSTR), or a mixedjlow reactor (MFR). In this type of reactor, the composition in the reactor is assumed to be that of the effluent stream and therefore all the reaction occurs at this constant composition (Figure 3.3.1).
Since the reactor is at steady-state, the difference in F! (input) and Fi (output) must be due to the reaction. (In this text, the superscript 0 on flow rates denotes the input to the reactor.) The material balance on a CSTR is written as: (3.3.1) + Fi output of i input of i o accumulation amount produced by reaction (Note that V is the volume of the reacting system and VR is the volume of the reactor; both are not necessarily equal.) Therefore, the rate can be measured directly as: (3.3.2) Equation (3.3.2) can be written for the limiting reactant to give: F[ F?
V (3.3.3) F?---..., Figure 3.3.1 I Stirred-flow reactor. The composition of the reacting volume, V, at temperature, T, is the same everywhere and at all times. F? is the molar flow rate of species i into the reactor while Fi is the molar flow rate of species i out of the reactor.
72 CHAPTER 3 Reactors for MeaslJ(jng Reaction Rates Recalling the definition of the fractional conversion: n?
nl F? - FI it = --0- = F O nl I Substitution of Equation (3.3.4) into Equation (3.3.3) yields: (3.3.4) (3.3.5) EXAMPLE 3.3.1 I If VI = -1, then the reaction rate is equal to the number of moles of the limiting reactant fed to the reactor per unit time and per unit volume of the reacting fluid times the fractional conversion. For any product p not present in the feed stream, a material balance on p is easily obtained from Equation (3.3.1) with F? = 0 to give: vpr = FiV (3.3.6) The quantity (FiV) is called the space-time yield.
The equations provided above describe the operation of stirred-flow reactors whether the reaction occurs at constant volume or not. In these types of reactors, the fluid is generally a liquid. If a large amount of solvent is used, that is, dilute solu-tions of reactants/products, then changes in volume can be neglected. However, if the solution is concentrated or pure reactants are used (sometimes the case for poly-merization reactions), then the volume will change with the extent of reaction.
Write the material balance equation on comonomer A for the steady-state CSTR shown be-low with the two cases specified for the reaction of comonomer A(CMA) and comonomer B(CMB) to give a polymer (PM).
CHAPTER 3 Reactors for MeaslJring Reaction Rates a, f3, (I) CMA + CMB =? polymer - I; r = kjCCMACCMB C Can Cf3n (II) 2CMA + MB =? polymer - II; r = kn CMA CMB • Answer The material balance equation is: input = output - removal by reaction + accumulation Since the reactor is at steady-state, there is no accumulation and: For case (I) the material balance is: F~MA = FCMA -(-kIC~:WAC~kB)V while case (II) gives: 73 If changes in the volume due to reaction can be neglected, then the CSTR ma-terial balance can be written in terms of concentrations to give (v = volumetric flow rate; that is, volume per unit time): o= vC?
vC, + (v,r)V C? - C, (-v,)r = (v/v) (3.3.7) (3.3.8) (3.3.9) The ratio (V/ v) is the volume of mixture in the reactor divided by the volume of mixture fed to the reactor per unit time and is called the space time, T. The in-verse of the space time is called the space velocity. In each case, the conditions for the volume of the feed must be specified: temperature, pressure (in the case of a gas), and state of aggregation (liquid or gas). Space velocity and space time should be used in preference to "contact time" or "holding time" since there is no unique residence time in the CSTR (see below). Why develop this terminology?
Consider a batch reactor. The material balance on a batch reactor can be written [from Equation (3.2.1)]; nflnaJ ffina!
I dn; of d/; t = -= c· n?
v;rV '0 (-v;)r(l +S;/;) Equation (3.3.9) shows that the time required to reach a given fractional conversion does not depend upon the reactor volume or total amount of reagents. That is to say, for a given fractional conversion, as long as C? is the same, 1,2, or 100 mol of i can be converted in the same time. With flow reactors, for a given C?, the fractional con-version from different sized reactors is the same provided T is the same. Table 3.3.1 compares the appropriate variables from flow and nonflow reactors.
74 CHAPTER 3 Reactors for MeaslJring Reaction Rates Table 3.3.1 I Comparison of appropriate variables for flow and nonflow reactors.
t (time) V = V°(1 + e;/;) (volume) ni = n?(l Ii) (mol) T (time) v vO(1 + e;/;) (volume/time) Fi = F?(1 - Ii) (mol/time) In order to show that there is not a unique residence time in a CSTR, consider the following experiment. A CSTR is at steady state and a tracer species (does not react) is flowing into and out of the reactor at a concentration Co. At t = 0, the feed is changed to pure solvent at the same volumetric flow. The material balance for this situation is: or o (accumulation) (input) C;v (output) (3.3.10) dC; -v- = Cv with C/· = CO att = O.
dt / Integration of this equation gives: C = CO exp[ -t/T] (3.3.11) This exponential decay is typical of first-order processes as shown previously. Thus, there is an exponential distribution of residence times; some molecules will spend lit-tle time in the reactor while others will stay very long. The mean residence time is: foOOtC(t)dt (t) = f(;C(t)dt (3.3.12) and can be calculated by substituting Equation (3.3.11) into Equation (3.3.12) to give: Since f(;texp[ -t/T]dt (t) = -"---..........,.--foOOexp[ -t/T]dt fOOxexp[-x]dx = 1 ° 1'2 (t) = = l' -1' exp[ -t/T] I~ (3.3.13) (3.3.14) Thus, the mean residence time for a CSTR is the space time. The fact that (t) = T holds for reactors of any geometry and is discussed in more detail in Chapter 8.
EXAMPLE 3.3.2 I CHAPTER 3 Reactors for Measllrjng Reaction Rates 75 The rate of the following reaction has been found to be first-order with respect to hydroxyl ions and ethyl acetate: In a stirred-flow reactor of volume V = 0.602 L, the following data have been obtained at 298 K [Denbigh et aI., Disc. Faraday Soc., 2 (1977) 263]: flow rate of barium hydroxide solution: flow rate of ethyl acetate solution: inlet concentration of OH-: inlet concentration of ethyl acetate: outlet concentration of OH-: 1.16 L/h 1.20 L/h 0.00587 mol/L 0.0389 mol/L 0.001094 mol/L Calculate the rate constant. Changes in volume accompanying the reaction are negligible.
(Problem taken from M. Boudart, Kinetics of Chemical Processes, Butterworth-Heinemann, Boston, 1991, pp. 23-24.) • Answer vH=1.16 CH =0.00587 VE =1.20 CE = 0.0389 ---i'" _+ CH = 0.001094 v = 2.36 SinceYH == 'IE' the limiting reactant is the hydroxyl ion. Thus, a material balance on OH- gives: 76 C HAP T E R 3 Reactors tQLMeascUlILU'inJjg,J--UR.CLe=ac"'-'t,,-iQ,,-"nJRJ<all'tec<>s,,--~~~~~~~~~~~~ where v = '1/1 + VE 2.36 Llh C~ C/lVH/v 0.00289 mol/L (-u/I)r = kC/lCE Since the outlet value of CH is known, the fractional conversion and CE can be calculated as: and 0.00289 - 0.001094 0.00289 0.62 Thus, C~ CEVE -- -C~fH = 0.0180 mol/L v and (-UH)r = c7/ - CH = 0.00289 0.00;094 T [0.602/2.36J 0.0070(mOI) L·h 0.0070 (0.001094)(0.0180) ( L ) 360 --mol·h 3.4 I Ideal Tubular Reactors Another type of ideal reactor is the tubular flow reactor operating isothermally at constant pressure and at steady state with a unique residence time. This type of re-actor normally consists of a cylindrical pipe of constant cross-section with flow such that the fluid mixture completely fills the tube and the mixture moves as if it were a plug traveling down the length of the tube. Hence the name plug flow reactor (PFR). In a PFR, the fluid properties are uniform over any cross-section normal to the direction of the flow; variations only exist along the length of the reactor. Ad-ditionally, it is assumed that no mixing occurs between adjacent fluid volume ele-ments either radially (normal to flow) or axially (direction of flow). That is to say each volume element entering the reactor has the same residence time since it does not exchange mass with its neighbors. Thus, the CSTR and the PFR are the two ideal limits of mixing in that they are completely mixed and not mixed at all, re-spectively. All real flow reactors will lie somewhere between these two limits.
Since the fluid properties vary over the volume of the reactor, consider a ma-terial balance on a section of a steady-state isothermal PFR, dL (see Figure 3.4.1): o F; (accumulation) = (input) (F; + dF;) + v;rAcdL (output) + (amount produced by reaction) (3.4.1) _--'C...,HLL-"'A"'P'-'Y u E ouR"-'3"--.LJ R&<A"""ac1illsJ0LMa""as=I.;LlrlL io<yQJRJ.<A""a"'ccu tio'-"OLLlR<J.at"'A"'S J.7-L7 Inlet, F? ---+ dL Figure 3.4.1 I Tubular reactor.
--.... Outlet where Ac is the cross-sectional area of the tube. Also, AcdL = dVR, so Equation (3.4.1) can be written as: f outlet _ VR _ of i dfi or T --C i ° ( ) v fi vir dP; -- = VT dVR I or Integration of Equation (3.4.3) gives: f outlet VR f i dfi pO = f O (vir) I I (3.4.2) (3.4.3) (3.4.4) (3.4.5) If changes in volume due to reaction are negligible, then [Pi = C;v; moles of i/time = (moles of i/volume) (volume/time)]: dC; dC; = = VT dT d(VR/v) i Note the analogy to batch reactors that have a unique residence time t and where dC; = VT dt i (3.4.6) Clearly, the space-time, T, in the ideal tubular reactor is the same as the residence time in the batch reactor only if volume changes are neglectable. This is easy to see from Equation (3.4.2) by substituting C;v for P; and recalling that for yolume changes v = y0(l + 8J} d (Cv) dVR i dC; dv v-+C dVR i dVp (3.4.7) 78 EXAMPLE 3.4.1 I CHAPTER 3 Reactors for Measllriog Reactioo Rates Thus, if dv/ dVR = 0, then there is an analogy between T and t in a PFR and batch reactor, respectively [Equations (3.4.5) and (3.4.6)], and if dv/ dVR =1= 0, then com-parison of Equations (3.4.7) and (3.4.6) shows that there is none.
A PFR operating isothennally at 773 K is used to conduct the following reaction: methylacetoxypropionate acetic acid methyl acrylate If a feed of pure methylacetoxypropionate enters at 5 atm and at a flow rate of 0.193 ft3/s, what length of pipe with a cross-sectional area of 0.0388 ft2 is necessary for the reaction to achieve 90 percent conversion (from C. G. Hill, An Introduction to Chemical Engineering Kinetics & Reactor Design. Wiley, 1977, pp. 266-267)?
Data: k = 7.8 X 109 exp[ -19,200/T] S-1 • Answer From Equation (3.4.2) and FA = F~ (1 - fA) CAY: or For this gas-phase reaction there is mole change with reaction and 2 -1 eA = 1-11 Therefore, Combination of the material balance and reaction rate expressions yields: CHAPTER 3 Reactors for Measllring Reaction Rates Integration of this equation yields: and at 773 K, k = 0.124 S-1 to give at fA = 0.9: T = 29.9 s 79 EXAMPLE 3.4.2 I Now, if mole change with reaction is ignored (i.e., SA = 0), T = 18.6 s. Notice the large dif-ference in the value of T when mole change with reaction is properly accounted for in the calculations. Since the gas is expanding with increasing extent of reaction, its velocity through the tube increases. Therefore, T must be likewise increased to allow for the specified con-version. The reactor volume and length of tube can be calculated as: VR = T VO = (29.9)(0.193) = 5.78 ft3 and A first-order reaction occurs in an isothermal CSTR (MFR) and PFR of equal volume. If the space-time is the same in both reactors, which reactor gives the largest space-time yield and why?
• Answer Assume that any changes in volume due to reaction can be neglected.
1 (CO) - In ---.i k C1 P Rearranging this equation gives (Cf and Ci are C1in the PFR and MFR, respectively): C? _ [c?-Ci] p - exp In Cl Cl 80 CHAPTER 3 Reactors for MeaslJring Reactioo Rates If the approximation: x2 exp[x] == 1 + x + + ...
2!
is used then or C?
Cf C?
Cf + C? - C7' + l(C?
C/')2 + ...
C/' 2 Cr C?
I (C?
C7')2 +--1+-+ ...
Cr 2 Cr Thus, Cf = cr[l +1(C:_ cm? + ...j 2C?Cr I I Since the term in the bracket will always be less than one, Cf < Cr, indicating that the fractional conversion and hence the space-time yield of the PFR is always higher than the CSTR. The rea-son for this is that for reaction-rate expressions that follow Rule I (r decreases with time or ex-tent of reaction) the rate is maintained at a higher average value in the PFR than the CSTR. This is easy to rationalize since the concentration of the feed (highest value giving the largest rate) in-stantaneously drops to the outlet value (lowest value giving the lowest rate) in a CSTR [schematic (b) below] while in the PFR there is a steady progression of declining rate as the fluid element traverses the reactor [schematic (a) below]. In fact, the PFR and the CSTR are the ideal maxi-mum and minimum space-time yield reactor configurations, respectively. This can be demon-strated by plotting the material balance equations for the PFR and CSTR as shown in the schematic (c) below. For the same conversion (i.e., the same outlet/A), the CSTR always requires a larger reactor volume. In other words, the area in the graph is larger for the CSTR than the PFR.
C~--+ PFR ~CA C~----.
CSTR ~CA CSTR (rectangular CO C~ A c c ,.
PFR 0 .S .~ '" ....
(area E C A t:J CA under <:: <l) <l) (;) (;) curve) c <:: 0 0 U u fJ fA Position Position Fractional conversion (a) (b) (c) EXAMPLE 3.4.3 I CHAPTER 3 Reactors for Measming Reaction Rates 81 Show that a large number of CSTRs in series approaches the behavior of a PFR.
• Answer Consider the following series of CSTRs accomplishing a first-order reaction (reactors of equal size).
The material balance for reactor i is: e i - 1 or --. = 1 + kT e' The ratio eO/eN can be written as: because T; = T (reactors of equal size). Also, eN/eo = 1 - / giving: 0/ 1 e eN = [1 + kT]N = 1 - / If TT = NT, then eO/eN can be written as: e% [kTT]N (kTT ) N(N - 1)(kTT )2 kT = 1+-=1+N-+ -+"'=e T eN N N 2!
N Therefore, as N gets large the series better approximates the exponential and eOI _ kTT leN - e or which is the material balance for a PFR. This result can be visualized graphically as follows.
In Example 3.4.2, r- 1 was plotted against/A' Note that if the area under this curve is inte-grated as illustrated: 82 CHAPTER 3 Reactors for MeaslJring Reaction Rates ,.
....
,.
....
17 V / 1// pP r-,....r-PP EXAMPLE 3.4.4 I the larger number of rectangles that are used to approximate the area under the curve, the less error. Thus, if each rectangle (see Example 3.4.2) represents a CSTR, it is clear that the larger the number of CSTRs in series, the better their overall behavior simulates that of aPFR.
Calculate the outlet conversion for a series of CSTRs accomplishing a first-order reaction with kTT = 5 and compare the results to that obtained from a PFR.
• Answer Using the equations developed in Example 3.4.3 gives with kTT = 5 the following results: 1 5 100 1000 PFR 0.1667 0.0310 0.0076 0.0068 0.0067 3.5 I Measurement of Reaction Rates What are the types of problems that need to be addressed by measuring reaction rates? The answers to this question are very diverse. For example, in the testing of catalysts, a new catalyst may be evaluated for replacement of another catalyst in an existing process or for the development of a new process. Accurate, reliable lab-oratory reaction rate data are necessary for the design of an industrial reactor CHAPTER 3 Reactors for Measllring Reaction Rates Table 3.5.1 I Rate of reactions in ideal isothermal reactors.
83 Batch 1 dn{ Equation (3.2.1) dC{ V---;Jt = v, r dt = vir Stirred-flow F?
= (-v{)r Equation (3.3.5) = (-v{)r Equation (3.3.8) V (V/v) Tubular o dft _ Equation (3.4.3) dC{ Equation (3.4.5) F,- - (-v,)r ---=v{r dVR d(VRjv) Nomenclature CI Molar concentration of limiting reactant, mol/volume C? Initial value of C1 It Fractional conversion of limiting reactant, dimensionless nl Number of moles of limiting reactant, mol F? Initial value of the molar flow rate of limiting reactant, mol/time r Reaction rate, mol/volume/time VI Stoichiometric coefficient of limiting reactant, dimensionless V Volume of reacting system, volume VR Volume of reactor, volume v Volumetric flow rate, volume/time whether the process is new or old. Another example of why reaction rate data are needed is to make predictions about how large-scale systems behave (e.g., the appearance of ozone holes and the formation of smog). The key issue in all of these circumstances is the acquisition of high-quality reaction rate data. In order to do this, a laboratory-scale reactor must be used. Although deviations from ideal behavior still exist in laboratory reactors, deliberate efforts can be made to approximate ideal conditions as closely as possible. Table 3.5.1 sum-marizes the material balance equations for the ideal reactors described above.
Examples of how these types of reactors are used to measure reaction rates are presented below.
When choosing a laboratory reactor for the measurement of reaction rate data, numerous issues must be resolved. The choice of the reactor is based on the characteristics of the reaction and for all practical matters by the availability of resources (i.e., reactors, analytical equipment, money, etc.). A good example of the issues involved in selecting a laboratory reactor and how they influence the ultimate choice is presented by Weekman [AIChE J., 20 (1974) 833]. Methods for obtaining reaction rate data from laboratory reactors that approximate the ideal reactors listed in Table 3.5.1 are now discussed.
84 CHAPTER 3 Reactors for Measllring Reaction Rates 3.5.1 Batch Reactors A batch reactor by its nature is a transient closed system. While a laboratory batch reactor can be a simple well-stirred flask in a constant temperature bath or a com-mercial laboratory-scale batch reactor, the direct measurement of reaction rates is not possible from these reactors. The observables are the concentrations of species from which the rate can be inferred. For example, in a typical batch experiment, the concentrations of reactants and products are measured as a function of time. From these data, initial reaction rates (rates at the zero conversion limit) can be obtained by calculating the initial slope (Figure 3.5.1b). Also, the complete data set can be numerically fit to a curve and the tangent to the curve calculated for any time (Fig-ure 3.5.la). The set of tangents can then be plotted versus the concentration at which the tangent was obtained (Figure 3.5.1c).
If, for example, the reaction rate function is first-order, then a plot of the tan-gents (dC/dt) versus concentration should be linear with a slope equal to the re-action rate constant and an intercept of zero. It is clear that the accuracy of the §I u Time (a) Time (b) Concentration (c) Figure 3.5.1 I (a) Plot of concentration data versus time, a curve for a numerical fit of the data and lines at tangents to the curve at various times, (b) plot of the initial slope, (c) plot of the slopes of the tangent lines in (a) versus concentrations.
CHAPTER 3 Reactors for MeastJring Reaction Rates 85 EXAMPLE 3.5.1 I data (size of the error bars) is crucial to this method of determining good reac-tion rates. The accuracy will normally be fixed by the analytical technique used.
Additionally, the greater the number of data points, the better the calculation of the rate. A typical way to measure concentrations is to sample the batch reactor and use chromatography for separation and determination of the amount of each component. In the best cases, this type of procedure has a time-scale of minutes.
If the reaction is sufficiently slow, then this methodology can be used. Note, how-ever, that only one datum point is obtained at each extent of reaction (i.e., at each time). If the reaction is fast relative to the time scale for sampling, then often it is not possible to follow the course of the reaction in a batch reactor.
P. Butler (Honors Thesis, Virginia Polytechnic Institute and State University, Blacksburg, VA, 1984) investigated the kinetics of the following reaction using rhodium catalysts: o II H -C -CH2CH2CH2CH2CH2CH3 CH2= CHCH2CH2CH2CH3 + CO + H2 ( (n-heptanal) (l-hexene) CH3-CH -CH2CH2CH2CH3 I C-H II o (2-methylhexanal) This homogeneous hydroformylation reaction was conducted in a batch reactor, and because of the nature of the catalyst, isomerization reactions of I-hexene to 2- and 3-hexenes and hy-drogenation reactions of hexenes to hexanes and aldehydes to alcohols were minimized. The following data were obtained at 323 K with an initial concentration of I-hexene at 1 mol/L in toluene and Pea = PH2 = PN2 (inert) = 0.33 atm. Calculate the initial rates of formation of the linear, rN, and branched, rs, aldehydes from these data.
0.17 0.67 1.08 1.90 2.58 0.0067 0.0266 0.0461 0.1075 0.1244 0.0000 0.0058 0.0109 0.0184 0.0279 • Answer Plot the concentration data as follows: 86 CHAPTER 3 Reactors for MeaslJring Reaction Rates 0.14 0.12 0.10 ~ ~ "0 5 0.08 Q oI0.06 u 0.04 0.02 0.00 o Time (h) 2 3 EXAMPLE 3.5.2 I The slopes of the lines shown in the plot give rN = 0.0515 mollL/h and rB = 0.0109 mollL/h (from linear regression).
Butler obtained the initial rate data given in the following table in a manner analogous to that illustrated in Example 3.5.1. Show that a reaction rate expression that follows Rules III and IV can be used to describe these data.
0.50 0.50 1.00 323 0.0280 0.0074 0.33 0.33 1.00 323 0.0430 0.0115 0.66 0.33 1.00 323 0.0154 0.0040 0.33 0.33 1.00 313 0.0156 0.0040 0.33 0.33 1.00 303 0.0044 0.0016 0.33 0.33 0.45 323 0.0312 0.0069 0.33 0.33 1.00 323 0.0410 0.0100 • Answer An empirical expression of the rate takes the form: CHAPTER 3 Reactors for MeaslJring Reaction Rates 87 and was used to correlate the data to give (Rg in cal): 13 [ ,/()J -1.5 0.45 0.40 rN = 2.0 X 10 exp -22,2001 RgT Pea PH2 CI-hexene rB = 4.9 X 1010 exp[ -19,200Y(RgT)JP~6·5p~:5C~·~~exene Ideally, much more data are required in order to obtain a higher degree of confidence in the reaction rate expressions. However, it is clear from Examples 3.5.1 and 3.5.2 how much ex-perimental work is required to do so. Also, note that these rates are initial rates and cannot be used for integral conversions.
3.5.2 Flow Reactors As pointed out previously, the use of flow reactors allows for the direct measure-ment of reaction rates. At steady state (unlike the batch reactor), the time scales of the analytical technique used and the reaction are decoupled. Additionally, since nu-merous samples can be acquired at the same conditions, the accuracy of the data dramatically increases.
Consider the following problem. In the petrochemical industry, many reac-tions are oxidations and hydrogenations that are very exothermic. Thus, to con-trol the temperature in an industrial reactor the configuration is typically a bun-dle of tubes (between 1 and 2 inches in diameter and thousands in number) that are bathed in a heat exchange fluid. The high heat exchange surface area per re-actor volume allows the large heat release to be effectively removed. Suppose that a new catalyst is to be prepared for ultimate use in a reactor of this type to con-duct a gas-phase reaction. How are appropriate reaction rate data obtained for this situation?
Consider first the tubular reactor. From the material balance (Table 3.5.1), it is clear that in order to solve the mass balance the functional form of the rate expres-sion must be provided because the reactor outlet is the integral result of reaction over the volume of the reactor. However, if only initial reaction rate data were re-quired, then a tubular reactor could be used by noticing that if the differentials are replaced by deltas, then: (3.5.1) Thus, a small tubular reactor that gives differential conversion (i.e., typically below 5 percent) can yield a point value for the reaction rate. In this case, the reaction rate is evaluated at Cr. Actually, the rate could be better calculated with the arithmetic mean of the inlet and outlet concentrations: CO + C exit C -1 1 1-2 However, since C? == Cfxit the inlet concentration is often used.
88 EXAMPLE 3.5.3 I CHAPTER 3 Reactors for Measuring Reaction Rates For the generic reaction A =} B the following three reaction rate expressions were proposed to correlate the initial rate data obtained. Describe how a differential tubular reactor could be used to discriminate among these models: klCA rl = 1 + kZCB k3CA rz = --".-C-'- I + k4CB + kSCA r3 = --"-'::"""-I + k7CA • Answer If initial rate data are obtained, and if there is no B in the feed stream, then the concentra-tion of B at low conversion is small. Thus, at these conditions the rate expressions are: r 1 = kjCA k3CA rz =----I + kSCA k6CA Clearly rj can be distinguished from rz and r3 by varying CA such that a plot of r versus CA can be obtained. Now suppose that rj does not describe the data. In a second set of experi-ments, B can be added to the feed in varying amounts. If r3 is the correct rate expression, then the measured rates will not change as CB is varied. If there is a dependence of the ob-served rate on the concentration of feed B, then r3 cannot describe the data.
Returning to the problem of obtaining reaction rate data from a new catalyst for a gas-phase reaction, if reaction rates are desired over the complete range of the extent of the reaction, the differential fixed bed is not an appropriate laboratory re-actor for this purpose. However, the ideal stirred-flow reactor can accomplish this objective. By varying 7, r can be directly obtained (see Table 3.5.1) at any extent of reaction. The problem with a gas phase reaction is the mixing. If the reaction oc-curs in the liquid phase, thorough mixing can be achieved with high agitation in many cases. With gases the situation is more difficult. To overcome this, several re-actor types have been developed and are commercially available on laboratory scale.
Referring to Figure 3.5.2, the Carberry reactor contains paddles in which the catalyst is mounted and the paddles are rapidly rotated via connection to a control shaft in order to obtain good mixing between the gas phase and the catalyst. A Berty reactor consists of a stationary bed of catalyst that is contacted via circulation of the gas phase by impeller blades. The quality of mixing in this type of configuration CHAPTER 3 Reactors for Measllring Reaction Rates 89 (a) (b) + II II t (c) Figure 3.5.2 I Stirred contained solids reactors. [Reproduced from V. W. Weekman, Jr., AIChE J., 20 (1974) p. 835, with pennission of the American Institute of Chemical Engineers.
Copyright © 1974 AIChE. All rights reserved.] (a) Carberry reactor, (b) Berty reactor (internal recycle reactor), (c) external recycle reactor.
depends on the density of the gas. For low densities (i.e., low pressures), the mix-ing is poor. Thus, Berty-type internal recycle reactors are most frequently used for reactions well above atmospheric pressure. For low-pressure gas circulation, exter-nal recycle can be employed via the use of a pump. At high recycle, these reactors approximate the behavior of a CSTR. This statement is proven below. Thus, these types of laboratory reactors have become the workhorses of the petrochemical in-dustry for measuring accurate reaction rate data.
Consider a generic recycle reactor schematically illustrated in Figure 3.5.3. First, denote R as the recycle ratio. The recycle ratio is defined as the volume of fluid re-turning to the reaction chamber entrance divided by the volume of fluid leaving the system. Simple material balances around the mixing point prior to the entrance of the reaction volume give: (3.5.2) and (3.5.3) If the density of the fluid is constant then ve = Vo and vr = Rvo. Using these rela-tionships with Equation (3.5.3) gives: (3.5.4) 90 CHAPTER 3 Reactors for Measuring Reaction Rates F~,Vi co FO ~,~ A' A --~r------yO, fA =0 yr=Ry' FJ.,C~ Figure 3.5.3 I Schematic diagram of a general recycle reactor. Superscripts i, e, and r refer to inlet, exit, and recycle. R is the recycle ratio.
or CO Re' el = A + __ A_ I +R 1+R (3.5.5) Notice that if R ~ 00, el ~ e: or the result obtained from an ideal CSTR. Also, if R ~ 0, el ~ e~, the inlet to the reaction volume. Thus, by fixing the value of R, the recycle reactor can behave like the two ideal limiting reactors (i.e., the CSTR and PFR, or anywhere between these limits). To see this further, a complete material balance on the reactor can be derived from Equation (3.4.2) and is: (3.5.6) However, since then FA = vOe~(R + 1)(1 - fA) = F~(R + 1)(1 - fA) -dFA = F~(R + l)dfA Substitution of Equation (3.5.7) into Equation (3.5.6) gives: VR If::' -F2 = (R + 1) fl -( --v-A-)r (3.5.7) (3.5.8) CHAPTER 3 Reactors for Measllring Reaction Rates 91 F2 + RveC~ vO + Rve Now, 11 must be related to inlet and/or outlet variables for ease of evaluation from readily measurable parameters. To do so, notice that Fi FO Fr ° rCe i A A + A FA + v A C A = -= ---:..;c:-----",,--vi vO + vr vO + vr In terms of lA' c1 is then i F2 + RF2(l - I~) F2[ I + R(l - I~) ] CA = vO + Rv°(l + BAI/.) = vO I + R(l + BA/~) Thus, C~ 1 + R -R/~ c2 I + R + RBA/~ Solving this equation for 11 gives: i RBA/~ + R/~ f ------:..::~-::....:.:.--A -I + R + BA + BAR R/~ I + R (3.5.9) (3.5.10) Substitution of Equation (3.5.9) into Equation (3.5.8) yields: VR Ilf dl A 0= (R + 1) 'Y1.-( ) FA I+R VA r Clearly, if R ---+ 0, then Equation (3.5.10) reduces to the material balance for a PFR.
However, it is not straightforward to recognize that Equation (3.5.10) reduces to the material balance for a CSTR as R ---+ 00. To do so, notice that the bottom limit on the integral goes to If.. as R ---+ 00. To obtain the value for the integral as R ---+ 00, Leibnitz's Rule must be used, and it is: Taking the limit of Equation (3.5.10) as R ---+ 00 gives (L'Hopital's Rule): rlf AI'.
() JRlf, (:~:)r lim _V R = lim 1+R--,- R-+oo F2 R-+oo I R + I To evaluate the numerator, use Leibnitz's Rule: d Jlf dlA Jlf a [ d l A ] I I d/~ I I d ( Rf: ) dR Rtf (-vA)r = RI; aR (-vA)r + (-VA)r Ie dR - (-VA)r l!lL dR I + R I+R I+R A I+R 92 CHAPTER 3 Reactors for Meas!J(ing Reaction Rates The fIrst and second terms on the right-hand side of this equation are zero. Therefore, 1 [~] V R -(-vA)r Rf; (1+R)2 lim -= w.l-,-+~R _ R->coFJ (1+lR)2 If.
EXAMPLE 3.5.4 I Note that this equation is the material balance for a CSTR (see Table 3.5.1). Thus, when using any recycle reactor for the measurement of reaction rate data, the effect of stirring speed (that fIxes recirculation rates) on extent of reaction must be inves-tigated. If the outlet conditions do not vary with recirculation rates, then the recy-cle reactor can be evaluated as if it were a CSTR.
AI-Saleh et al. [Chern. Eng. J., 37 (1988) 35] performed a kinetic study of ethylene oxida-tion over a silver supported on alumina catalyst in a Berty reactor. At temperatures between 513-553 K and a pressure of 21.5 atm, the observed reaction rates (calculated using the CSTR material balance) were independent of the impeller rotation speed in the range 350-1000 rpm (revolutions per minute). A summary of the data is: 553 51.0 0.340 3.145 2.229 553 106.0 0.272 5.093 2.676 553 275.0 0.205 9.336 3.564 533 9.3 0.349 0.602 0.692 533 51.0 0.251 2.661 1.379 533 106.0 0.218 4.590 1.582 533 275.0 0.162 8.032 2.215 513 9.3 0.287 0.644 0.505 513 51.0 0.172 1.980 0.763 513 106.0 0.146 3.262 0.902 513 275.0 0.074 3.664 0.989 Derive the equations necessary to determine rEO (rate of production of ethylene oxide) and rco2 (rate of production of CO2). Assume for this example that the volumetric flow rate is unaffected by the reactions.
• Answer From Equation (3.3.6): CHAPTER 3 Reactors for MeaSllring Reaction Rates 93 VIGNETTE 3.5.1 Since the reaction rates are reported on a per mass of catalyst basis rather than per volume, V is replaced with W (the mass of catalyst in the reactor). The reaction network is: ° /~ CH2 =CH2 + 1/202 ==> CH2--CH2 ° /~ CH2--CH2 + 5/202 ==> 2C02 + 2H20 CH2 = CH2 + 302 ==> 2C02 + 2H20 For the molar flow rates, ( coutlet] o EO FEo = FE --0-CE ( coutlet] o co, Feo, = FE C~ where F~ is the inlet molar flow rate of ethylene and C~ is the inlet concentration of ethyl-ene. Thus, the material balance equations are: _ F~ (C~~let] rEO -W C~ F~[c~~~et] reo, = W -co E Notice that F~, W, and C~ are all fixed for a particular experiment. Thus, measurement of the concentrations in the outlet stream directly yield values for the observed rates.
94 CHAPTER 3 Reactors for MeaslJring Reaction Rates EXAMPLE 3.5.5 I Using the data in Example 3.5.4 calculate the selectivity defined as the ratio of the moles of EO produced per mole of ethylene consumed times 100 percent, and plot the selectivity versus conversion.
Answer The selectivity can be calculated as: SEO = [ rEO ] X 100% rEO + rco2 From the plot shown below, the selectivity declines as the conversion is increased because of combustion reactions that produce carbon dioxide.
80 70 .~ .E; 60 u ~ en 50 40 0.0 0.1 0.2 Total conversion 0.3 0.4 CHAPTER 3 Reactors for MeaslJring Reaction Rates 95 In addition to the laboratory-scale reactors described here, there are numerous more specialized reactors in use. However, as mentioned previously, the perform-ance of these reactors must lie somewhere between the mixing limits of the PFR and the CSTR. Additionally, when using small laboratory reactors, it is often diffi-cult to maintain ideal mixing conditions, and the state of mixing should always be verified (see Chapter 8 for more details) prior to use. A common problem is that flow rates sufficiently large to achieve PFR behavior cannot be obtained in a small laboratory system, and the flow is laminar rather than turbulent (necessary for PFR behavior). If such is the case, the velocity profile across the reactor diameter is par-abolic rather than constant.
Exercises for Chapter 3 1.
The space time necessary to achieve 70 percent conversion in a CSTR is 3 h.
Determine the reactor volume required to process 4 ft3 min-I. What is the space velocity for this system?
2.
The following parallel reactions take place in a CSTR: A + B ~ Desired Product k l = 2.0 L (mol mintl B ~ Undesired Product k2 = 1.0 min-I If a liquid stream of A (4 mol L-I, 50 L min-I) and a liquid stream of B (2 mol L-I, 50 L min-I) are co-fed to a 100 L reactor, what are the steady-state effluent concentrations of A and B?
3.
The irreversible reaction 2A --+ B takes place in the gas phase in a constant temperature plug flow reactor. Reactant A and diluent gas are fed in equimolar ratio, and the conversion of A is 85 percent. If the molar feed rate of A is dou-bled, what is the conversion of A assuming the feed rate of diluent is unchanged?
4.
Consider the reversible first-order reaction of A = B in a CSTR of volume V = 2 L with forward and reverse rate constants of k l = 2.0 min-I and k_ 1 = 1.0 min-I. At time t = 0, the concentrations of A and B in the tank are both zero. The incoming stream ofA has a volumetric flow rate of 3 L min-I at con-centration C~ = 2 mol L-1. Find the concentrations of A and B as functions of time. You do not need a computer to solve this problem.
5.
Consider the liquid phase reaction: A => products with rate r = 0.20 cl (mol L-I min-I) that takes place in a PFR of volume 30 L.
(a) What is the concentration of A (C;) exiting the PFR?
10 Lmin-l c~= 1.2 mol L-I ce A 96 CHAPTER 3 Reactors for Measilring Reaction Rates (b) What is CX in a PFR with recycle shown below?
5 Lmin-1 10 Lmin-1 c;]= 1.2molL-l c e A (c) Now, add a separator to the system. Find C1, c1, C~ and c1. For the sepa-rator, assume cX = 5CX.
5 Lmin-1 10 Lmin-1 C~=1.2molL-l Separator .
c e A (Problem provided by Prof. J. L. Hudson, Univ. of Virginia.) 6.
(Adapted from H. S. Fogler, Elements ofChemical Reaction Engineering, 3rd ed., Prentice Hall, Upper Saddle River, NJ, 1999.) Pure butanol is fed into a semibatch reactor containing pure ethyl acetate to produce butyl acetate and ethanol in the reversible reaction: The reaction rate can be expressed in the Guldberg-Waage form. The reaction is carried out isothermally at 300 K. At this temperature, the equilibrium con-stant is 1.08 and the forward rate constant is 9.0 X 10-5 L (mol S)-I. Initially, there are 200 L of ethyl acetate in the reactor and butanol is fed at a rate of 0.050 L S-I. The feed and initial concentrations of butanol and ethyl acetate are 10.93 mol L-1 and 7.72 mol L-1, respectively.
(a) Plot the equilibrium conversion of ethyl acetate as a function of time.
(b) Plot the conversion of ethylacetate, the rate of reaction, and the concen-tration of butanol as a function of time.
7.
Dinitrogen pentoxide decomposes at 35°C with a first-order rate constant of 8 X 10-3 min- I [F. Daniels and E. H. Johnston, J. Am. Chem. Soc., 43 (1921) 53] according to: CHAPTER 3 Reactors for MeaslJring Reaction Rates However, the product N02 rapidly dimerizes to N20 4 : 97 If the decomposition reaction is carried out in a constant volume at 35°C, plot the pressure rise in the reactor as a function of time, for an initial charge of pure N20 S at 0.4 atm. Assume that the dimerization reaction equilibrates immedi-ately. The equilibrium constant of the N02 dimerization reaction at 35°C is 3.68. Assume ideal behavior.
8.
Ethanol can decompose in a parallel pathway: y ethylene + water ethanol ~ acetaldehyde + dihydrogen Assume that the reaction rates are both first-order in ethanol and that no prod-ucts are initially present. After 100 s in a constant volume system, there is 30 percent of the ethanol remaining and the mixture contains 13.7 percent ethyl-ene and 27.4 percent acetaldehyde. Calculate the rate constants k] and k2• 9.
Compound A is converted to B in a CSTR. The reaction rate is first-order with a reaction rate constant of 20 min.-] Compound A enters the reactor at a flow rate of 12 m3/min (concentration of 2.0 kInol/m3). The value of the product B is $1.50 per kmol and the cost of reactor operation is $2.50 per minute per cubic meter. It is not economical to separate unconverted A to recycle it back to the feed. Find the maximum profit.
10.
A very simplified model for the dynamical processes occurring when an ani-mal is given a drug can be represented by the single compartmental model shown below: Absorption rate Elimination rate Upon administration of the drug, there is absorption into the body.
Subsequently, the drug is converted to metabolites and/or is physically elimi-nated. As a result, the amount of drug in the body at any time is the net tran-sient response to these input and output processes.
Find the maximum "body" alcohol level in grams and the time when it occurs for a 70 kg human who quickly consumes one can of beer. Assume the absorption and elimination rates are first-order.
Data: The mass of one can of beer is 400 g and contains 5 wt. % alcohol.
kabsorption/kelimination = 5 kelimination = 0.008 min - I 98 CHAPTER 3 Reactors for Measllring Reaction Rates 11.
Titanium dioxide particles are used to brighten paints. They are produced by gas-phase oxidation of TiCl4 vapor in a hydrocarbon flame. The dominant reaction is hydrolysis, The reaction rate is first-order in TiC14 and zero-order in H20. The rate con-stant for the reaction is: k = 8.0 X 104 exp[ 88000J/mol]s 1 RgT The reaction takes place at 1200 K in a constant pressure flow reactor at 1 atm pressure (1.01 X 105 Pa). The gas composition at the entrance to the reactor is: COz 8% HzO 8% Oz 5% TiC14 3% Nz remainder (a) What space time is required to achieve 99 percent conversion of the TiC14 to Ti02?
(b) The reactor is 0.2 m diameter and 1.5 m long. Assuming that the reactor operates 80 percent of the time, how many kilograms of TiOz can be produced per year? (The molecular weight of TiOz is 80 g/mol.) Rg = 8.3144 J/mol/K (Problem provided by Richard Flagan, Caltech.) 12.
The autocatalytic reaction of A to form Q is one that accelerates with conver-sion. An example of this is shown below: A+Q~Q+Q However, the rate decreases at high conversion due to the depletion of reactant A.
The liquid feed to the reactor contains 1 mol L-] of A and 0.1 mol L-] of Q.
(a) To reach 50 percent conversion of A in the smallest reactor volume, would you use a PFR or a CSTR? Support your answer with appropriate calculations.
(b) To reach 95 percent conversion of A in the smallest reactor volume, would you use a PFR or a CSTR? Support your answer with appropriate calculations.
(c) What is the space time needed to convert 95 percent of A in a CSTR if k] = 1 L (mol S)-I?
13.
The irreversible, first-order, gas-phase reaction A=-- 2B + C CHAPTER 3 Reactors for Meas!!ring Reaction Rates 99 takes place in a constant volume batch reactor that has a safety disk designed to rupture when the pressure exceeds 1000 psi. If the rate constant is 0.01 s-1, how long will it take to rupture the safety disk if pure A is charged into the reactor at 500 psi?
14.
If you have a CSTR and a PFR (both of the same volume) available to carry out an irreversible, first-order, liquid-phase reaction, how would you connect them in series (in what order) to maximize the conversion?
15.
Find the minimum number of CSTRs connected in series to give an outlet con-version within 5 percent of that achieved in a PFR of equal total volume for: (a) first-order irreversible reaction of A to form B, kTPFR = 1 (b) second-order irreversible reaction of A to form B, kC~TPFR = 1.
16.
Davis studied the hydrogenation of ethylene to ethane in a catalytic recycle reactor operated at atmospheric pressure (R. J. Davis, Ph.D. Thesis, Stanford University, 1989.) The recycle ratio was large enough so that the reactor approached CSTR behavior. Helium was used as a diluent to adjust the par-tial pressures of the gases. From the data presented, estimate the orders of the reaction rate with respect to ethylene and dihydrogen and the activation ener-gy of the reaction.
Catalyst: 5%Pd/Alumina Hydrogenation of ethylene over 50 mg of Pd/alumina catalyst.
193 1.0 20 80 25.1 193 1.0 10 90 16.2 193 1.0 40 60 35.4 193 2.5 20 78.5 8.55 193 5.0 20 76 4.17 175 1.0 20 80 3.14 ~_4 The Steady-State Approximation: Catalysis 4.1 I Single Reactions One-step reactions between stable molecules are rare since a stable molecule is by definition a quite unreactive entity. Rather, complicated rearrangements of chemical bonds are usually required to go from reactants to products. This implies that most reactions do not proceed in a single elementary step as illustrated below for NO formation from Nz and Oz: N2 + 0 ( ) NO + N N····O NO III II==;>+ N····O NO N + O2 ( ) NO + 0 Normally, a sequence of elementary steps is necessary to proceed from reactants to products through the formation and destruction of reactive intermediates (see Section 1.1).
Reactive intermediates may be of numerous different chemical types (e.g., free radicals, free ions, solvated ions, complexes at solid surfaces, complexes in a homo-geneous phase, complexes in enzymes). Although many reactive intermediates may be involved in a given reaction (see Scheme 1.1.1), the advancement of the reaction can still be described by a single parameter-the extent of reaction (see Section 1.2). If this is the case, the reaction is said to be single. Why an apparently complex reaction re-mains stoichiometrically simple or single, and how the kinetic treatment of such reac-tions can be enumerated are the two questions addressed in this chapter.
There are two types of sequences leading from reactants to products through reactive intermediates. The first type of sequence is one where a reactive intermediate 100 CHAPTER 4 The Steady-State Approximatioo' Catalysis 101 is not reproduced in any other step of the sequence. This type of sequence is de-noted as an open sequence. The second type of sequence is one in which a reactive intermediate is reproduced so that a cyclic reaction pattern repeats itself and a large number of product molecules can be made from only one reactive intermediate. This type of sequence is closed and is denoted a catalytic or chain reaction cycle. This type of sequence is the best definition of catalysis.
A few simple examples of sequences are listed in Table 4.1.1. The reactive in-termediates are printed in boldface and the stoichiometrically simple or single reaction is in each case obtained by summation of the elementary steps of the sequence. While all reactions that are closed sequences may be said to be catalytic, there is a distinct difference between those where the reactive intermediates are provided by a separate entity called the catalyst that has a very long lifetime and those where the reactive intermediates are generated within the system and may survive only during a limited number of cycles. The first category encompasses truly catalytic reactions (catalytic reaction cycle) in the narrow sense of the word, while the second involves chain re-actions (chain reaction cycle). Both types exhibit slightly different kinetic features.
However, the two types are so closely related that it is conceptually straightforward to consider them together. In particular, both categories can be analyzed by means of the steady-state approximation that will be presented in the next section.
Chain and catalytic reaction cycles provide energetically favorable pathways for reactant molecules to proceed to product molecules. This point is illustrated below for both types of cycles. Consider the reaction between dihydrogen and dichlorine to produce HCl that can be brought about in the gas phase by irradiating the reactants with light. It is known that over 106 molecules of HCl can be formed per absorbed photon. The reaction proceeds as follows: light Clz ----+ 2CI CI + Hz ----+ HCl + H H + Clz ----+ HCl + Cl 2Cl ----+ Clz (initiation) (propagation) (propagation) (termination) Once chlorine atoms are produced (initiation), the propagation steps provide a closed cycle that can be repeated numerous times (e.g., 106) prior to the recombination of the chlorine atoms (termination).
The reason the chain reaction cycle dominates over a direct reaction between dihydrogen and dichlorine is easy to understand. The direct reaction between Hz and Clz has an activation energy of over 200 kJ/mol, while the activation energies of the two propagation steps are both less than 30 kJ/mol (see Figure 4.1.1). Thus, the difficult step is initiation and in this case is overcome by injection of photons.
Now consider what happens in a catalytic reaction cycle. For illustrative pur-poses, the decomposition of ozone is described. In the presence of oxygen atoms, ozone decomposes via the elementary reaction: Table 4.1.1 I Sequences and reactive intermediates.' 0 3 ~ O2 +° 0+ 0 3 ~ O2 + O2 203 ==} 302 open oxygen atoms in the gas phase open solvated (in liquid S02) benzhydryl ions + F (YCH'O (rCHD I I +F-~ I I ~ ~ , , Cl F (rCHY'fi (rCHY'fi ,I V+ F- ===>, I V+ Cl-° +N2 ~ NO +N N+02~NO+O N2 + O2 ==} 2NO 803 + O2 ~ 805 805 + SO~- ~ SO~- + 803 SO~- + SO~- ~ 2S0~-2S0~- + O2 ==} 2S0~- + H20 ~ H2 + 0 0 + CO ~ CO2 + Hp + CO ==} H2 + CO2 .
o O+o'~C chain chain catalytic chain oxygen and nitrogen atoms in the gas phase free radical ions 803 and 805 sites on a catalyst surface and the surface complex 0 cumyl and cumylperoxy free radicals in a solution of cumene IAdapted from M. Boudart, Kinetics of Chemical Processes, Butterworth-Heinemann, 1991, pp. 61-62.
CHAPTER 4 The Steady-State Approximatioo' Catalysis (a) Reaction coordinate 2HCI 103 Figure 4.1.1 I Energy versus reaction coordinate for H2 + Cl2 ~ 2HCl.
(a) direct reaction, (b) propagation reactions for photon assisted pathway.
The rate of the direct reaction can be written as: (4.1.1) where rd is in units of molecule/cm3/s, and are the number densities (mol-ecule/cm3) of° and 03, respectively, and k is in units of cm3/s/molecule. In these units, k is known and is: k = 1.9 X 10- 11 exp[-2300/T] where T is in Kelvin. Obviously, the decomposition of ozone at atmospheric con-ditions (temperatures in the low 200s in Kelvin) is quite slow.
The decomposition of ozone dramatically changes in the presence of chlorine atoms (catalyst): Cl + 0 3 ~ O2 + ClO ClO + ° ~ O2 + Cl 0+ 0 3 = 202 where: kj = 5 X 10- 11 exp(-140/T) k2 = l.l X 10- 10 exp(-220/T) cm3 /s/molecule cm3/s/molecule At steady state (using the steady-state approximation--developed in the next section), the rate of the catalyzed reaction rc is: 104 C HAP T E B 4 The Steady-State Approximation' Catalysis rc = k2[OJ[[CIJ + [CIOJ] and rc k2[[CIJ + [CIOJ] rd k[03J If [CIJ + [CIOJ ~ -3 [03J = 10 (a value typical of certain conditions in the atmosphere), then: (4.1.2) (4.1.3) (4.1.4) (4.1.5) rc k2 3 3,/ - = -X 10- = 5.79 X 10- exp(2080/ T) rd k At T = 200 K, rcyrd = 190. The enhancement of the rate is the result of the cata-lyst (Cl). As illustrated in the energy diagram shown in Figure 4.1.2, the presence of the catalyst lowers the activation barrier. The Cl catalyst first reacts with ° to (a) Reaction coordinate 2°2 Figure 4.1.2 I Energy versus reaction coordinate for ozone decomposition.
(a) direct reaction, (b) Cl catalyzed reaction.
__-"CuH:uAoo.r:P...LT-"'EUlJR'-4~'TJJhl:<.e -,-,Sw:te=ad~StateApproximation' Catalysis 105 give the reaction intermediate CIO, which then reacts with 0 3 to give O2 and regenerate Cl. Thus, the catalyst can perform many reaction cycles.
4.2 I The Steady-State Approximation Consider a closed system comprised of two, first-order, irreversible (one-way) ele-mentary reactions with rate constants k l and k2 : k j k2 A~B~C 106 CHAPTER 4 The Steady-State Approximation· Catalysis If C~ denotes the concentration of A at time t = 0 and C~ = C~ = 0, the material balance equations for this system are: dx -= -kx dt ] (4.2.1) (4.2.2) EXAMPLE 4.2.1 I where x = CA/C~, y = CB/C~, and w = CclC1. Integration of Equation (4.2.1) with x = 1, y = 0, w = 0 at t = 0 gives: x = exp(-kIt) k] y = k 2 -k1[exp(-k]t) - exp(-k2t)] k2 k] w = 1 -k exp(-k]t) + k exp(-k2t) 2 -k] k2 -I Show how the expression for y(t) in Equation (4.2.2) is obtained (see Section 1.5).
• Answer Placing the functional form of x(t) into the equation for ~ gives: y == 0 att == 0 This first-order initial-value problem can easily be solved by the use of the integration fac-tor method. That is, or after integration: k t k] exp[(k2 -k])t] ye 2 == + Y kz - k] Since y == 0 at t == 0: Substitution of the expression for y into the equation for y(t) gives: k t k] c 1 ye 2 == --lexp[(kz - k])t]-lJ kz k1 or y CHAPTER 4 The Steady-State Approximation" Catalysis 107 Time (a) or Time (b) Figure 4.2.1 I Two first-order reactions in series.
A B~C (a) k1 = 0.1 kJ, (b) k1 = kJ, (c) k1 = 10 k1• It is obvious from the conservation of mass that: x+y+w=l dx dy dw -+-+-=0 dt dt dt Time (c) (4.2.3) (4.2.4) The concentration ofA decreases monotonically while that ofB goes through a max-imum (see, for example, Figure 4.2.1b). The maximum in CB is reached at (Example 1.5.6): (4.2.5) (4.2.6) (4.2.7) and is: cy;ax = C~G~)[k2] At tmax, the curve of Cc versus t shows an inflection point, that is, (d1w)/(dt1) = O.
Suppose tlJ.at B is not a..Tl intermediate but a reactive intermediate (see Section 1.1).
Kinetically, this implies that k2 » k1• If such is the case, what happens to the solution of Equation (4.2.1), that is, Equation (4.2.2)? As kjk1 ~ 0, Equation (4.2.2) reduces to: x = exp(-kit) } y = k 1 exp(-kit) w = 1 - exp(-kit) Additionally, tmax ~ 0 as does Ymax (see Figure 4.2.1 and compare as ki/k1~ 0).
Thus, the time required for CB to reach its maximum concentration is also very (4.2.8) 108 CHAPTER 4 The Steady-State Approximation' Catalysis small. Additionally, the inflection point in the curve of Cc versus time is translated back to the origin.
Equation (4.2.7) is the solution to Equation (4.2.8): dx -= -kx dt 1 o = kjx k2y dw -= k2y dt Note that Equation (4.2.8) involves two differential and one algebraic equations. The algebraic equation specifies that: dy -=0 dt (4.2.9) (4.2.10) (1.2.4) This is the analytical expression of the steady-state approximation: the time deriv-atives of the concentrations of reactive intermediates are equal to zero. Equation (4.2.9) must not be integrated since the result that y = constant is false [see Equa-tion (4.2.7)]. What is important is that B varies with time implicitly through A and thus with the changes in A (a stable reactant). Another way to state the steady-state approximation is [Equation (4.2.4) with dy/dt = 0]: dx dw dt dt Thus, in a sequence of steps proceeding through reactive intermediates, the rates of reaction of the steps in the sequence are equal. It follows from Equation (4.2.10) that the reaction, however complex, can be described by a single parameter, the ex-tent of reaction (see Section 1.2): e})(t) = nj(t) -n?
Vj so that: de}) 1 dn j 1 dnj -=--== ... =--dt (4.2.11) For the simple reaction A =} C, Equation (4.2.11) simplifies to Equation (4.2.10).
The steady-state approximation can be stated in three different ways: 1.
The derivatives with respect to time of the concentrations of the reactive intermediates are equal to zero [Equation (4.2.9)].
2.
The steady-state concentrations of the reaction intermediates are small since as k/k2 « 1, tmax ~ 0 and Cllax ~ O.
3.
The rates of all steps involving reactants, products, and intermediates are equal [Equation (4.2.10)].
CHAPTER 4 The Steady-State AppIDxill-imLUa=tll..iowo~· vC"",atl.Qall.lly",swis~~~~~~~~---,1",O ....
9 0 (a) (b) (e) (d) '00 c ] <?H3 CHzD CHDz CD3 I I I I C C C C 100 300 Time (s) 500 Figure 4.2.2 I Display of the data of Creighton et al. (Left) reprinted from SUlface Science, vol. 138, no. ], J. R. Creighton, K. M. Ogle, and J. M. White, "Direct observation of hydrogen-deuterium exchange in ethylidyne adsorbed on Pt(l]l)," pp. Ll37-Ll41, copyright 1984, with permission from Elsevier Science. Schematic of species observed (right).
These conditions must be satisfied in order to correctly apply the steady-state approximation to a reaction sequence. Consider the H-D exchange with ethylidyne (CCH3 from the chemisorption of ethylene) on a platinum surface. If the reaction proceeds in an excess of deuterium the backward reactions can be ignored. The con-centrations of the adsorbed ethylidyne species have been monitored by a technique called secondary ion mass spectroscopy (SIMS). The concentrations of the various species are determined through mass spectroscopy since each of the species on the surface are different by one mass unit. Creighton et al. [Surf Sci., 138 (1984) L137] monitored the concentration of the reactive intermediates for the first 300 s, and the data are consistent with what are expected from three consecutive reactions. The re-sults are shown in Figure 4.2.2.
Thus, the reaction sequence can be written as: Dz ( ) 2D Pt=C~CH3 + D-+Pt=C~CHzD + H Pt C ~ CHzD + D -,> Pt = C ~ CHDz + H Pt~C-CHDz + D-+Pt-C-CD3 + H 2H ( ) Hz If the rate of Dz- Hz exchange were to be investigated for this reaction sequence, dCD, dCH, dt dt and the time derivative of all the surface ethylidyne species could be set equal to zero via the steady-state approximation.
110 EXAMPLE 4.2.2 I CHAPTER 4 The Steady-State Approximatioo' Catalysis Show how Equation (4.1.2) is obtained by using the steady-state approximation.
• Answer CI + 0 3 ~ O2 + CIO CIO + °~ O2 + CI 0+ 0 3 => 202 The reaction rate expressions for this cycle are: d[CIO] -- = k1[CI] - k2[CIO] dt drOol ----;z:- = k1[CI] + k2[CIO] Using the steady-state approximation for the reactive intermediate [ClO] specifies that: d[CIO] --=0 dt and gives: Also, notice that the total amount of the chlorine in any form must be constant. Therefore, [CI]o = [CI] + [CIO] By combining the mass balance on chlorine with the mass balance at steady-state for [CIO], the following expression is obtained: After rearrangement: Recall that: I d1> r = C V dt 1 d[A;] Vi dt 1 d 2 dt Thus, by use of the rate expressions for ozone and dioxygen: EXAMPLE 4.2.3 I CHAPTER 4 The Steady-State Approximation Gatalysis or The rate of ozone decomposition can be written as: or Substitution of the expression for [CIO] in this equation gives: or Equation (4.1.2).
111 Explain why the initial rate of polymerization of styrene in the presence of Zr(C6Hs)4 in toluene at 303 K is linearly dependent on the concentration of Zr(C6Hs)4 [experimentally observed by D. G. H. Ballard, Adv. Catat. 23 (1988) 285].
• Answer Polymerization reactions proceed via initiation, propagation, and termination steps as illus-trated in Section 4.1. A simplified network to describe the styrene polymerization is: where: (C6HshZrCHCH2C6Hs .
I C6Hs kp (C6HshZr(polymer)n + styrene --.. (C6HshZr(polymer)n+1 I I C6Hs C6Hs k (C6HshZr(polymer)n ~ (C6HshZr + C6Hs(polymer)n I C6Hs (i) (ii) (iii) is the species shown on the right-hand side of Equation (i). Equations (i-iii) are the initia-tion, propagation, and termination reactions (,a-hydrogen transfer terminates the growth of the polymer chains) respectively. The reaction rate equations can be written as: (initiation) 112 CHAPTER 4 The Steady-State Approximation· Catalysis (propagation) (termination) where Cz is the concentration of Zr(C6Hs)4, Cs is the styrene concentration, and C is the concentration of the Zr species containing (polymer)n=i' For simplicity, the rate constants for propagation and termination are assumed to be independent of polymer chain length (i.e., in-dependent of the value of n).
If the steady-state approximation is invoked, then ri = r, or Solving for the sum of the reactive intermediates gives: Substitution of this expression into that for r p yields: The rate of polymerization of the monomer is the combined rates of initiation and propaga-tion. The long chain approximation is applicable when the rate of propagation is much faster than the rate of initiation. If the long chain approximation is used here, then the polymer-ization rate is equal to rp • Note that the rp is linearly dependent on the concentration ofZr(ben-zyl)4' and that the polystyrene obtained (polystyrene is used to form styrofoam that can be made into cups, etc.) will have a distribution of molecular weights (chain lengths). That is, C6Hs(polymer)n in Equation (iii) has many values of n. The degree of polymerization is the average number of structural units per chain, and control of the molecular weight and its dis-tribution is normally important to the industrial production of polymers.
The steady-state approximation applies only after a time tn the relaxation time.
The relaxation time is the time required for the steady-state concentration of the reactive intermediates to be approached. Past the relaxation time, the steady-state approximation remains an approximation, but it is normally satisfactory. Below, a more quantitative description of the relaxation time is described.
k k Assume the actual concentration of species B in the sequence A --4- B ~ C, CB , is different from its steady-state approximation C; by an amount e: The expression: CB= C~(1 + B) (4.2.12) (4.2.13) CHAPTER 4 The Steady-State Approximation' Catalysis 113 still applies as does the equation [from taking the time derivative ofEquation (4.2.12)]: dCB C' de ( ) dC~ -= -+ l+e--dt B dt dt According to the steady-state approximation [see Equation (4.2.8)]: (4.2.14) (4.2.15) (4.2.16) dC~ ki -=--C dt k2 A Equating the right-hand sides of Equations (4.2.13) and (4.2.14) and substitut-ing the values for C; and dC;/dt from Equations (4.2.15) and (4.2.16), respec-tively, gives: de ( ) -+ kz - k 1 e - k1 = 0 dt (4.2.17) Integration of this initial-value problem with e= -1 (CB = 0) at t = 0 yields: e = 1 {K - exp[(K - 1)k2t]I (4.2.18) (K -1) where: Since B is a reactive intermediate, K must be smaller than one. With this qualifica-tion, and at "sufficiently large values" of time: e=K (4.2.19) What is implied by "sufficiently large values" of time is easily seen from Equation (4.2.18) when K « 1. For this case, Equation (4.2.18) reduces to: (4.2.20) The relaxation time is the time required for a quantity to decay to a fraction Ije of its original value. For the present case, tr = 1/k2• Intuitively, it would appear that the relaxation time (sometimes called the induction time) should be on the same order of magnitude as the turnover time, that is, the reciprocal of the turnover frequency (see Section 1.3). As mentioned previously, turnover frequencies around 1 s-1 are common. For this case, the induction time is short.
However, if the turnover frequency is 10- 3 S-1, then the induction time could be very long. Thus, one should not assume a priori that the induction time is brief.
114 VIGNETTE 4.2.1 CHAPTER 4 The Steady-State Approximation· Catalysis Now, let's consider an important class of catalysts-namely, enzymes. Enzymes are nature's catalysts and are made of proteins. The primary structure is the sequence of amino acids that are joined through peptide bonds (illustrated below) to create the protein polymer chain: 0 H /H ~ H /H ~ H 0 H II " " " /H II //C-OH N C-OH N C-OH N C C "C / "C / -H2O "C / " /" + ----+ N Rz / " / " / "H I H Rl H Rz R1 H amino acid amino acid The primary structure gives rise to higher order levels of structure (secondary, tertiary, quaternary) and all enzymes have a three-dimensional "folded" structure of the polymer chain (or chains). This tertiary structure forms certain arrangements ofamino acid groups that can behave as centers for catalytic reactions to occur (denoted as active sites). How an active site in an enzyme performs the chemical reaction is described in Vignette 4.2.1.
_.JC~HIiAAcJP~T[JEE1lRL4~=TllhlEec.;S:>Jt:eeaadffity::S1ataApprQxim· .
ll2D.:-eataL)iFS~ls'" .:1l1155 116 CHAPTER 4 The Steady-State Approximation· Catalysis VIGNETTE 4.2.2 CHAPTER 4 The Steady-State Approximation Catal}Ltilsili.s -:l1.:l1L7 In order to describe the kinetics of an enzyme catalyzed reaction, consider the following sequence: k1 k2 k3 Ez + S ( ) EzS EzP -----+ Ez + P L 1 where Ez is the enzyme, S is the substrate (reactant), EzS and EzP are enzyme bound complexes, and P is the product. An energy diagram for this sequence is shown in Figure 4.2.4. The rate equations used to describe this sequence are: dCEzP ~ = k2CEzS k 2CEzP k3CEzP dCp ----;Jt = k3CEzP (4.2.21) Using the steady-state approximation for the reactive intermediates CEzS and CEzP and the fact that: (4.2.22) 118 CHAPTER 4 The Steady-State Approximation' Catalysis where C~z is the concentration of the enzyme in the absence of substrate gives: (4.2.23) If the product dissociates rapidly from the enzyme (i.e., k3 is large compared to kz and k-z), then a simplified sequence is obtained and is the one most commonly employed to describe the kinetics of enzyme catalyzed reactions. For this case, k1 k Ez + S ( ) EzS --4 Es + P k_1 with dCEzS ~ = k,CSCEz - L,CEzS - k3CEzS dCp -:it = k3CEzS where Using the steady-state approximation for CEzS gives: with: (4.2.24) (4.2.25) (4.2.26) (4.2.27) Km is called the Michaelis constant and is a measure of the binding affinity of the substrate for the enzyme. If k_, >> k3 then Km = L jk, or the dissociation constant for the enzyme. The use of Equation (4.2.25) with Equation (4.2.26) yields an expression for CEzS in terms of C~z and Cs, that is, two measurable quantities: or CHAPTER 4 The Steady-State Approximation' Catalysis C = C~z Cs EzS K + C m S Substitution of Equation (4.2.28) into the expression for dCp/dt gives: dCp k3C~zCS dt K m + Cs If rmax = k3C~z, then Equation (4.2.29) can be written as: 119 (4.2.28) (4.2.29) dCp = dt dCs dt (4.2.30) EXAMPLE 4.2.4 I This fonn ofthe rate expression is called the Michaelis-Menton fonn and is used widely in describing enzyme catalyzed reactions. The following example illustrates the use of linear regression in order to obtain rmax and Km from experimental kinetic data.
Para and Baratti [Biocatalysis, 2 (1988) 39] employed whole cells from E. herbicola immo-bilized in a polymer gel to catalyze the reaction of catechol to form L-dopa: NH2 ° H0-o =;> HO-Q-CH2-{ -~OH HO HO H Catechol L-dopa (levodopa) Do the following data conform to the Michaelis-Menton kinetic model?
• Data The initial concentration of catechol was 0.0270 M and the data are: 0.00 0.00 0.25 11.10 0.50 22.20 0.75 33.30 1.00 44.40 1.25 53.70 1.50 62.60 2.00 78.90 2.50 88.10 3.00 94.80 3.50 97.80 4.00 99.10 4.50 99.60 5.00 99.85 120 CHAPTER 4 The Steady-State Approximatioo' Catalysis • Answer Notice that: [ dCs]-1 Km 1 --:it = rmaxCs + rmax [ dC ]-1 Thus, if - d/ is plotted as a function of Ijcs, the data should conform to a straight line with slope = Km/rmax and intercept = l/rmax• This type of plot is called a Lineweaver-Burk plot.
First, plot the data for Cs (catechol) versus time from the following data [note that Cs = C~ (1 - is)]: 0.00 0.25 0.50 0.75 1.00 1.25 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 0.027000 0.024003 0.021006 0.018009 0.015012 0.012501 0.010098 0.005697 0.003213 0.001404 0.000594 0.000243 0.000108 0.000041 0.03 100 0.025 80 0.02 ~ t ~ 0: 60 0.015 0 "§ u· " > 40 om 0: 0 u 0.005 20 2 3 4 5 0 0 2 3 4 5 Time (h) Time (h) From the data of Cs versus time, dCs/dt can be calculated and plotted as shown below. Ad-ditionally, the Lineweaver-Burk plot can be constructed and is illustrated next.
__-"C. Vent Differential pump (a) Figure 4.3.1 I (a) Schematic of apparatus used for isotopic transient kinetic analysis and (b) transients during ethane hydrogenolysis on Ru/Si02 at 180°C. (Figure from "Isotopic Transient Kinetic Analysis of Ethane Hydrogenolysis on Cu modified Ru/Si02" by B. Chen and J. G. Goodwin in Journal of Catalysis, vol. 158:228, copyright © 1996 by Academic Press, reproduced by permission of the publisher and the author.) F(t) in (b) represents the concentration of a species divided by its maximum concentration at any time.
1.0 0.8 0.6 I~ 0.4 0.2 0.0 0 10 20 30 Time (s) (b) 40 126 CHAPTER 4 The Steady-State Appmximatioo' Catalysis kinetic parameters for this system. This method is typically called "isotopic tran-sient kinetic analysis." Figure 4.3.la shows a schematic of an apparatus to perform the steady-state, isotopic transient kinetic analysis for the hydrogenolysis of ethane over a Ru/SiOz catalysis: A sampling of the type of data obtained from this experiment is given in Figure 4.3.1b. Kinetic constants can be calculated from these data using analyses like those presented above for the simple reversible, first-order system [Equation (4.3.1)].
Exercises for Chapter 4 1.
NzOs decomposes as follows: Experimentally, the rate of reaction was found to be: Show that the following sequence can lead to a reaction rate expression that would be consistent with the experimental observations: k1 NzOs ( ) NOz + N03 k_1 k2 N02 + N03 ~ N02 + O2 + NO NO + N03 ~ 2N02 2.
The decomposition of acetaldehyde (CH3CHO) to methane and carbon monoxide is an example of a free radical chain reaction. The overall reaction is believed to occur through the following sequence of steps: k CH3CHO --1...+ CH3" + CHO" k CH3CHO + CH3" ~ CH4 + CH3CO" k CH3CO" ---4 CO + CH3" CH3" + CH3" ~ C2H6 Free radical species are indicated by the" symbol. In this case, the free radical CHO" that is formed in the first reaction is kinetically insignificant. Derive a valid rate expression for the decomposition of acetaldehyde. State all of the assumptions that you use in your solution.
CHAPTER 4 The Steady-State Approximation Catalysis 127 3.
Chemical reactions that proceed through free radical intermediates can explode if there is a branching step in the reaction sequence. Consider the overall reaction of A =} B with initiator I and first-order termination of free radical R: I~R R+A~B+R R+A~B+R+R R ~ side product initiation propagation branching termination Notice that two free radicals are created in the branching step for everyone that is consumed.
(a) Find the concentration of A that leads to an explosion.
(b) Derive a rate expression for the overall reaction when it proceeds below the explosion limit.
4.
Molecules present in the feed inhibit some reactions catalyzed by enzymes. In this problem, the kinetics of inhibition are investigated (from M. L. Shuler and F. Kargi, Bioprocess Engineering, Basic Concepts. Prentice Hall, Englewood Cliffs, NJ, 1992).
(a) Competitive inhibitors are often similar to the substrate and thus compete for the enzyme active site. Assuming that the binding of substrate Sand inhibitor I are equilibrated, the following equations summarize the relevant reactions: k, Ez + S ~ EzS -----=-c. Ez + P K; Ez+I ~ EzI Show how the rate of product formation can be expressed as: (b) Uncompetitive inhibitors do not bind to the free enzyme itself, but instead they react with the enzyme-substrate complex. Consider the reaction scheme for uncompetitive inhibition: K m Ez+ S ~ EzS K; EzS+ I ~ EzSI Ez + P 128 C H A PT E R 4 The Steady-State Approximation Catalysis Show how the rate of product fonnation can be expressed as: (c) Noncompetitive inhibitors bind on sites other than the active site and reduce the enzyme affinity for the substrate. Noncompetitive enzyme inhibition can be described by the following reactions: Ki Ez + I +§:t EzI Ki EzS + I +§:t EzSI K m EzI + S +§:t EzSI The Michaelis constant, Km , is assumed to be unaffected by the presence of inhibitor I. Likewise, K; is assumed to be unaffected by substrate S.
Show that the rate of product fonnations is: r=( Cr)rn(ax K) I + ~ I + C: (d) A high concentration of substrate can also inhibit some enzymatic reactions. The reaction scheme for substrate inhibition is given below: 17 am kz Ez + S +§:t EzS ~ Ez + P KSi EzS + S +§:t EzSz Show that the rate of product fonnation is: CHAPTER 4 The Steady-State Approximation Catalysis 129 S.
Enzymes are more commonly involved in the reaction of two substrates to form products. In this problem, analyze the specific case of the "ping pong bi bi" mechanism [w. W. Cleland, Biochim. Biophys. Acta, 67 (1963) 104] for the irreversible enzymatic conversion of A+B=>P+W that takes place through the following simplified reaction sequence: KmA ko Ez+A ~ EzA ~ Ez'+P KmE k4 Ez'+B ~Ez'B ~Ez+W where Ez' is simply an enzyme that still contains a molecular fragment of A that was left behind after release of the product P. The free enzyme is regenerated after product W is released. Use the steady-state approximation to show that the rate of product formation can be written as: where K a , Kf3, and rmax are collections of appropriate constants.
Hints: C~z = CEz + CEzA + CEz' + CEz's and dCpjdt = dCVv/dt 6.
Mensah et al. studied the esterification of propionic acid (P) and isoamyl alcohol (A) to isoamyl propionate and water in the presence of the lipase enzyme [Po Mensah, J. L. Gainer, and G. Carta, Biotechnol. Bioeng., 60 (1998) 434.] The product ester has a pleasant fruity aroma and is used in perfumery and cosmetics. This enzyme-catalyzed reaction is shown below: Propionic acid Isoamyl alcohol CH,\ o H2 1-H,C JL C CH + H20 -~/~/~/~ C 0 C CH1 H2 H2-Isoamyl propionate 130 CHAPTER 4 The Steady-State Approximation' Catalysis This reaction appears to proceed through a "ping pong bi bi" mechanism with substrate inhibition. The rate expression for the forward rate of reaction is given by: Use nonlinear regression with the following initial rate data to find values of rmax, Kj, K2, and KPi. Make sure to use several different starting values of the parameters in your analysis. Show appropriate plots that compare the model to the experimental data.
Initial rate data for esterification of propionic acid and isoamyl alcohol in hexane with Lipozyme-IM (immobilized lipase) at 24°C, 0.15 0.15 0.15 0.15 0.15 0.15 0.33 0.33 0.33 0.33 0.33 0.33 0.33 0.60 0.60 0.60 0.60 0.60 0.60 0.60 0.72 0.72 0.72 0.72 0.72 0.72 0.72 0.93 0.93 0.93 0.93 0.93 0.10 0.20 0.41 0.60 0.82 1.04 0.10 0.11 0.20 0.41 0.60 0.81 1.01 0.13 0.13 0.20 0.42 0.62 0.83 1.04 0.14 0.20 0.41 0.61 0.82 0.85 1.06 0.21 0.42 0.65 0.93 1.13 1.19 1.74 1.92 1.97 2.06 2.09 0.90 1.00 1.29 1.63 1.88 1.94 1.97 0.80 0.79 1.03 1.45 1.61 1.74 1.89 0.73 0.90 1.27 1.51 1.56 1.69 1.75 0.70 1.16 1.37 1.51 1.70 From P. Mensah, J. L. Gainer and G. Carta, Biotechnol. Bioeng" 60 (1998) 434, CHAPTER 4 The Steady-State Approximation' Catalysis 131 7.
Combustion systems are major sources of atmospheric pollutants. The oxidation of a hydrocarbon fuel proceeds rapidly (within a few milliseconds) and adiabatically to establish equilibrium among the H/C/O species (COz, HzO, Oz, CO, H, OH, 0, etc.) at temperatures that often exceed 2000 K. At such high temperatures, the highly endothermic oxidation of Nz: 6.Hr = 180.8 (kJ mol-I) becomes important. The reaction mechanism by which this oxidation occurs begins with the attack on Nz by 0: with rate constants of: k i = 1.8 X 108e-384oo/T (m3 mol-1s- l ) L I = 3.8 X 107e-4Z5/T (m3 mol-1s- l ) The oxygen radical is present in its equilibrium concentration with the major combustion products, that is, 1/20z =€7= 0 The very reactive N reacts with Oz: kz N + Oz ( ) NO + 0 k_z kz = 1.8 X 104 Te-468o/T L z = 3.8 X 103 Te-zo8oo/T (m3 mol-1s- 1) (m3 mol-1s- 1) HBr+H (a) Derive a rate expression for the production ofNO in the combustion products of fuel lean (excess air) combustion of a hydrocarbon fuel. Express this rate in terms of the concentrations of NO and major species (Oz and Nz).
(b) How much NO would be formed if the gas were maintained at the high temperature for a long time? How does that concentration relate to the equilibrium concentration?
(c) How would you estimate the time required to reach that asymptotic concentration?
(Problem provided by Richard Flagan, Caltech.) 8.
The reaction of Hz and Brz to form HBr occurs through the following sequence of elementary steps involving free radicals: k i ( ) k_ I ko H + Br2 -::...-.. HBr + Br 132 CHAPTER 4 The Steady-State Approximation' Catalysis Use the fact that bromine radicals are in equilibrium with Br2 to derive a rate expression of the form: r= 9.
As a continuation of Exercise 8, calculate the concentration of H radicals present during the HBr reaction at atmospheric pressure, 600 K and 50 percent conversion.
The rate constants and equilibrium constant for the elementary steps at 600 K are given below.
kl = 1.79 X 107 cm3 mol- l S-l k_ l = 8.32 X 1012 cm3 mol- l S-l k2 = 9.48 X 1013 cm3 mol- l S-l K3 = 8.41 X 10- 17 mol cm- 3 10.
Consider the series reaction in which B is a reactive intermediate: As discussed in the text, the steady-state approximation applies only after a relaxation time associated with the reactive intermediates. Plot the time dependence of e (the deviation intermediate concentration from the steady-state value) for several values of krlk2 (0.5, 0.1, 0.05) and k2 = 0.1 S-l. What happens when: (a) k l = k2 and (b) k l > k2?
~---=---5 Heterogeneous Catalysis 5.1 I Introduction Catalysis is a term coined by Baron J. J. Berzelius in 1835 to describe the property of substances that facilitate chemical reactions without being consumed in them. A broad definition of catalysis also allows for materials that slow the rate of a reac-tion. Whereas catalysts can greatly affect the rate of a reaction, the equilibrium com-position of reactants and products is still determined solely by thermodynamics.
Heterogeneous catalysts are distinguished from homogeneous catalysts by the dif-ferent phases present during reaction. Homogeneous catalysts are present in the same phase as reactants and products, usually liquid, while heterogeneous catalysts are present in a different phase, usually solid. The main advantage of using a hetero-geneous catalyst is the relative ease of catalyst separation from the product stream that aids in the creation of continuous chemical processes. Additionally, heteroge-neous catalysts are typically more tolerant of extreme operating conditions than their homogeneous analogues.
A heterogeneous catalytic reaction involves adsorption of reactants from a fluid phase onto a solid surface, surface reaction of adsorbed species, and desorption of products into the fluid phase. Clearly, the presence of a catalyst provides an alter-native sequence of elementary steps to accomplish the desired chemical reaction from that in its absence. If the energy barriers of the catalytic path are much lower than the barrier(s) of the noncatalytic path, significant enhancements in the reaction rate can be realized by use of a catalyst. This concept has already been introduced in the previous chapter with regard to the CI catalyzed decomposition of ozone (Figure 4.1.2) and enzyme-catalyzed conversion of substrate (Figure 4.2.4). A sim-ilar reaction profile can be constructed with a heterogeneous catalytic reaction.
For example, G. Ertl (Catalysis: Science and Technology, 1. R. Anderson and M. Boudart, Eds., vol. 4, Springer-Verlag, Berlin, 1983, p. 245) proposed the ther-mochemical kinetic profile depicted in Figure 5.1.1 for the platinum-catalyzed oxidation of carbon monoxide according to the overall reaction CO + ±O2 =} CO2, The first step in the profile represents the adsorption of carbon monoxide and 134 CHAPTER 5 Heterogeneo!Js Catalysis -21 COZads 105 259 CO(g) + O(g) ++----r-..;,--...
~ .
• + • + • • • • • • • • / 249 : • • .
.
.
.
• • .
.
1" : CO(g) + "20z(g).
• ------...,~., ...-.'......!:.~".
---.-----.----..---,,---.,"-.__.'..--"".,,"',",-...---.-----: .------.--....,~-..,-· · · · · • · · • ~H = 283 • • • • + ++.+. COz(g) Figure 5.1.1 I Schematic energy diagram for the oxidation of CO and a Pt catalyst. (From data presented by G. Ertl in Catalysis: Science and Technology, J. R. Anderson and M. Boudart, Eds., vol. 4, Springer-Verlag, Berlin, 1983, p. 245.) All energies are given in kJ mol-I. For comparison, the heavy dashed lines show a noncatalytic route.
dioxygen onto the catalyst. In this case, adsorption of dioxygen involves dissocia-tion into individual oxygen atoms on the Pt surface. The product is formed by ad-dition of an adsorbed oxygen atom (Oads) to an adsorbed carbon monoxide mole-cule (COads)' The final step in the catalytic reaction is desorption of adsorbed carbon dioxide (C02ads) into the gas phase. The Pt catalyst facilitates the reaction by pro-viding a low energy path to dissociate dioxygen and form the product. The noncat-alytic route depicted in Figure 5.1.1 is extremely slow at normal temperatures due to the stability of dioxygen molecules.
VIGNETTE 5.1.1 CHAPTER 5 HeterogeneolJs Catalysis 135 136 CHAPTER 5 HeterogeneolJs Catalysis Since the number of atoms on the surface of a bulk metal or metal oxide is ex-tremely small compared to the number of atoms in the interior, bulk materials are often too costly to use in a catalytic process. One way to increase the effective sur-face area of a valuable catalytic material like a transition metal is to disperse it on a support. Figure 5.1.5 illustrates how Rh metal appears when it is supported as nanometer size crystallites on a silica carrier. High-resolution transmission electron microscopy reveals that metal crystallites, even as small as 10 nm, often expose the common low-index faces commonly associated with single crystals. However, the surface to volume ratio of the supported particles is many orders of magnitude higher than an equivalent amount of bulk metal. In fact, it is not uncommon to use cata-lysts with I nm sized metal particles where nearly every atom can be exposed to the reaction environment.
Estimation of the number of exposed metal atoms is rather straightforward in the case of a single crystal of metal since the geometric surface area can be mea-sured and the number density of surface atoms can be found from the crystal structure.
CHAPTER 5 HeterogeneolJs Catalysis (a) (b) Figure 5.1.5 I (a) Rhodium metal particles supported on silica carrier. (b) High-resolution electron micrograph shows how small supported Rh crystallites expose low-index faces. (Top photo courtesy of A. K. Datye. Bottom photo from "Modeling of heterogeneous catalysts using simple geometry supports" by A. K. Datye in Topics in Catalysis, vol. 13:131, copyright © 2000 by Kluwer Academic, reproduced by permission of the publisher and the author.) 137 138 CHAPTER 5 HetemgeneolJs Catalysis For supported metal catalysts, no simple calculation is possible. A direct mea-surement of the metal crystallite size or a titration of surface metal atoms is required (see Example 1.3.1). Two common methods to estimate the size of supported crys-tallites are transmission electron microscopy and X-ray diffraction line broadening analysis. Transmission electron microscopy is excellent for imaging the crystallites, as illustrated in Figure 5.1.5. However, depending on the contrast difference with the support, very small crystallites may not be detected. X-ray diffraction is usually ineffective for estimating the size of very small particles, smaller than about 2 nm.
Perhaps the most common method for measuring the number density of exposed metal atoms is selective chemisorption of a probe molecule like Hz, CO, or Oz.
Selective chemisorption uses a probe molecule that does not interact significantly with the support material but forms a strong chemical bond to the surface metal atoms of the supported crystallites. Chemisorption will be discussed in more detail in Sec-tion 5.2. Dihydrogen is perhaps the most common probe molecule to measure the frac-tion of exposed metal atoms. An example of Hz chemisorption on Pt is shown below: Pt atom H 2Ptsurface + H2 ==> 2PtsurfaceH The exact stoichiometry of the Pt-H surface complex is still a matter of debate since it depends on the size of the metal particle. For many supported Pt catalysts, an as-sumption of 1 H atom adsorbing for every 1 Pt surface atom is often a good one.
Results from chemisorption can be used to calculate the dispersion of Pt, or the frac-tion of exposed metal atoms, according to: 2 X (Hz molecules chemisorbed) Fraction exposed = Dispersion = f P Total number 0 t atoms If a shape of the metal particle is assumed, its size can be estimated from chemisorp-tion data. For example, a good rule of thumb for spherical particles is to invert the dispersion to get the particle diameter in nanometers: 1 Particle diameter (nm) =.
.
DIspersIOn Table 5.1.1 compares the average diameter of Pt particles supported on alumina determined by chemisorption of two different probe molecules, X-ray diffraction, and electron microscopy. The excellent agreement among the techniques used to CHAPTER 5 Heterogeneolls Catalysis Table 5.1.1 I Determination of metal particle size on Pt/AI20 3 catalysts by chemisorption of H2 and CO, X-ray diffraction, and transmission electron microscopy.
139 0.6a 1.2 1.3 1.3 2.0a 1.6 1.8 2.2 3.7" 2.7 2.9 2.7 3.7b 3.9 4.6 1.6 1.8 2.4 5.3 VIGNETTE 5.1.2 apretreatment temperature of 500°C.
bpretreatment temperature of 800°C.
Source: Renouprez et aI., 1. Calal., 34 (1974) 411.
characterize these model Pt catalysts shows the validity in using chemisorption to estimate particle size.
The reason for measuring the number of exposed metal atoms in a catalyst is that it allows reaction rates to be normalized to the amount of active component. As de-fined in Chapter I, the rate expressed per active site is known as the turnoverfrequency, rt• Since the turnover frequency is based on the number of active sites, it should not depend on how much metal is loaded into a reactor or how much metal is loaded onto a support. Indeed, the use of turnover frequency has made possible the comparison of rates measured on different catalysts in different laboratories throughout the world.
140 CHAPTER 5 Heterogeneolls Catalysis 5.2 I Kinetics of Elementary Steps: Adsorption, Desorption, and Surface Reaction The necessary first step in a heterogeneous catalytic reaction involves activation of a reactant molecule by adsorption onto a catalyst surface. The activation step implies that a fairly strong chemical bond is formed with the catalyst surface. This mode of adsorption is called chemisorption, and it is characterized by an enthalpy change typ-ically greater than 80 kJ mol- J and sometimes greater than 400 kJ mol- J. Since a chemical bond is formed with the catalyst surface, chemisorption is specific in nature, meaning only certain adsorbate-adsorbent combinations are possible. Chemisorption implies that only a single layer, or monolayer, of adsorbed molecules is possible since every adsorbed atom or molecule forms a strong bond with the surface. Once the avail-able surface sites are occupied, no additional molecules can be chemisorbed.
Every molecule is capable of weakly interacting with any solid surface through van der Waals forces. The enthalpy change associated with this weak adsorption mode, called physisorption, is typically 40 kJ mol- J or less, which is far lower than the enthalpy of chemical bond formation. Even though physisorbed molecules are not activated for catalysis, they may serve as precursors to chemisorbed molecules.
More than one layer of molecules can physisorb on a surface since only van der Waals interactions are involved. The number of physisorbed molecules that occupy CHAPTER 5 HeterogeneolJs Catalysis 141 a monolayer on the surface of any material can be used to calculate its geometric surface area if the cross-sectional area of the adsorbate molecule is known.
142 CHAPTER 5 HeterogeneolJs Catalysis CHAPTER 5 Heterogeneolls Catalysis 143 The potential energy diagram for the chemisorption of hydrogen atoms on nickel is schematically depicted in Figure 5.2.3. As molecular hydrogen approaches the surface, it is trapped in a shallow potential energy well associated with the physi-sorbed state having an enthalpy ofphysisorption flHp • The deeper well found closer to the surface with enthalpy flHc is associated with the hydrogen atoms chemisorbed on nickel. There can be an activation barrier to chemisorption, E, which must be overcome to reach a chemisorbed state from the physisorbed molecule. Since mo-lecular hydrogen (dihydrogen) is dissociated to form chemisorbed hydrogen atoms, this phenomenon is known as dissociative chemisorption.
Because rates of heterogeneous catalytic reactions depend on the amounts of chemisorbed molecules, it is necessary to relate the fluid phase concentrations of Energy (not to scale) I I -!'>Hc i I I I I I I 0.5 2Ni+H+H Dissociation energy 2Ni + Hz Distance from metal surface (nm) Figure 5.2.3 I Potential energy diagram for the chemisorption of hydrogen on nickel.
144 CHAPTER 5 HetemgeneOlls Catalysis reactants to their respective coverages on a solid surface. For simplicity, the following discussion assumes that the reactants are present in a single fluid phase (i.e., either liquid or gas), all surface sites have the same energetics for adsorption, and the adsorbed molecules do not interact with each other. The active site for chemisorption on the solid catalyst will be denoted by , with a surface concentra-tion (]. The adsorption of species A can then be expressed as: kads A + ( ) A kdes where A corresponds to A chemisorbed on a surface site and kads and kdes refer to the rate constants for adsorption and desorption, respectively. Thus, the net rate of adsorption is given by: (5.2.3) (5.2.4) Since the net rate of adsorption is zero at equilibrium, the equilibrium relationship is: kads [A] K -----ads -kdes -[A][] The fractional coverage 0A is defined as the fraction of total surface adsorption sites that are occupied by A. If (]0 represents the number density of all adsorption sites (vacant and occupied) on a catalyst surface, then: and (]0 = (] + (A] (5.2.5) (5.2.6) [A] °A = []0 Combining Equations (5.2.4-5.2.6) gives expressions for (l. (A], and 0A in terms of the measurable quantities (A] and Kads: [] = []0 1 + Kads[A] _ _ KnJA1110 lAJ = L.
--.-I + Kads[A] Kads[A] o A = ---=--=--I + Kads[A] (5.2.7) (5.2.8) (5.2.9) Equation (5.2.9) is known as the Langmuir Adsorption Isotherm, and it is depicted in Figure 5.2.4. At low concentrations of A, the fractional surface coverage is pro-portional to (A], with a proportionality constant of Kads, whereas, at high concen-trations of A, the surface coverage is unity and independent of (A] (i.e., saturation of the available sites).
CHAPTER 5 Heterogeneolls Catalysis Concentration of A Figure 5.2.4 I Langmuir Adsorption Isothenn: Fractional coverage 8A versus fluid phase concentration of A.
145 When more than one type of molecule can adsorb on the catalyst surface, the competition for unoccupied sites must be considered. For the adsorption of molecule A in the presence of adsorbing molecule B, the site balance becomes: []0 = [] + [A] + [B] and (JA, (JB can be expressed as: (5.2.10) (5.2.11) where the equilibrium constants are now associated with adsorption/desorption of either A or B.
As mentioned above, some molecules like H2 and O2 can adsorb dissociatively to occupy two surface sites upon adsorption. The reverse reaction is known as as-sociative desorption. In these cases, the rate of adsorption now depends on the num-ber of pairs of available surface sites according to: where refers to a pair of adjacent free sites and AA refers to a pair of adjacent occupied sites. M. Boudart and G. Djega-Mariadassou (Kinetics ofHeterogeneous Catalytic Reactions, Princeton University Press, Princeton, 1984) present expres-sions for [] and [A A], based on the statistics of a partially occupied square 146 CHAPTER 5 Heterogeneous Catalysis Isolated A cannot desorb as A2 Isolated vacan site cannot dissociate A2 A A A 1 1 A A A A A A A t A A A A -A A A A A A A if ~ A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A Figure 5.2.5 I Schematic depiction of a square lattice populated by A atoms. Empty squares represent vacant surface sites.
lattice, shown in Figure 5.2.5. The surface concentrations of adjacent pairs of un-occupied and occupied sites are: 2"J2 [J = l [Jo 2[AJ2 AAJ =--[ Jo Thus, the net rate of dissociative adsorption becomes: (5.2.12) (5.2.13) (5.2.14) and upon substitution for [] and [AA] with Equations (5.2.12) and (5.2.13), re-spectively, gives: = 2kadSr A~1r12 _ 2kdeSr AJ2 r [Jo l"'Ljl j [Jo l" The value of two before each rate constant in Equation (5.2.14) is a statistical factor that contains the number of nearest neighbors to an adsorption site on a square lattice and accounts for the fact that indistinguishable neighboring sites should not be dou-ble counted. For simplicity, this statistical factor will be lumped into the bimolecular rate constant to yield the following rate expression for dissociative adsorption: k k r = ~[A FJ2 -~rAJ2 [Jo L [Jo L (5.2.15) CHAPTER 5 Heterogeneolls I;StSIYSIS l"1l The total number of adsorption sites on the surface now appears explicitly in the rate expression for this elementary adsorption step. Since the site balance is the same as before (Equation 5.2.5), the equilibrium adsorption isotherm can be calculated in the manner described above: kads Kads = k = des 1 + (Kads [A2])1/2 (K 'lA ])1/2[] [A]= ads 2 L ~ 1 + (Kads[A 2 ])1;2 (Kads [A2])1/2 e A = -----=---=-"----1 + (Kads[A 2 ])1/2 (5.2.16) (5.2.17) (5.2.18) (5.2.19) At low concentrations of A2 , the fractional surface coverage is proportional to [A2 ] 1/2, which is quite different than the adsorption isotherm derived above for the case without dissociation. In terms of fractional surface coverage, the net rate of dissociative adsorption is expressed by: (5.2.20) As expected, the rate is proportional to the total number of adsorption sites []0' The next step in a catalytic cycle after adsorption of the reactant molecules is a surface reaction, the simplest of which is the unimolecular conversion of an ad-sorbed species into a product molecule. For example, the following two-step se-quence represents the conversion of A into products through the irreversible surface reaction of A: (1) (2) A + ( ) A ko A --"-+ products + A ==> products (reversible adsorption) (surface reaction) (overall reaction) Since the surface reaction is irreversible (one-way), the mechanistic details involved in product formation and desorption are not needed to derive the rate equation. The rate of the overall reaction is: r=k2[A] (5.2.21) If the adsorption of A is nearly equilibrated. the Langmuir adsorption isotherm can be used to find e A , and the final rate expression simplifies to: r (5.2.22) 19a EXAMPLE 5.2.1 I cnAPTl!fJ:t HeTerogeneous L8T81YSIS Use the steady-state approximation to derive the rate expression given in Equation (5.2.22).
kads (1) A + ( ) A (reversible adsorption) kdes k, (2) A ---"-+ products + (surface reaction) • Answer The rate of the reaction is: A ===;> products r = k2[A] (overall reaction) (5.2.23) To solve the problem, [A] and [] must be evaluated by utilizing the site balance: []0 = [A] + and the steady-state approximation: d[A] ---;Jt = 0 = kads[A][] kdes[A] k2[A] (5.2.24) (5.2.25) Solving Equations (5.2.24) and (5.2.25) simultaneously yields the following expression for the concentration of reactive intermediate: Gads)rA ][]0 [A] = des 1 + (kads) [A] + kdes kdes (5.2.26) The ratio kads/kdes is simply the adsorption equilibrium constant Kads' Thus, the rate expres-sion in Equation (5.2.23) can be rewritten as: k2Kads[A][]0 r = ------=-...:..::.--=--1 + Kads[A] + kdes (5.2.27) by using Equation (5.2.26) together with ll(ads, and is the same as Equation (5.2.22) except for the last term in the denominator. The ratio k 2 /kdes represents the likelihood of A reacting to form product compared to simply desorbing from the surface. Equation (5.2.22) is based on the assumption that adsorption of A is nearly equilibrated, which means kdes is much greater than k2• Thus, for this case, k2/kdes can be ignored in the denominator of Equation (5.2.27).
The previous discussion involved a two-step sequence for which the adsorp-tion of reactant is nearly equilibrated (quasi-equilibrated). The free energy change associated with the quasi-equilibrated adsorption step is negligible compared to CHAPTER 5 HeterogeneolJs Catalysis 149 that of the surface reaction step. The surface reaction is thus called the rate-determining step (RDS) since nearly all of the free energy change for the overall reaction is associated with that step. In general, if a rate determining step exists, all other steps in the catalytic sequence are assumed to be quasi-equilibrated. The two-step sequence discussed previously is written in the notation outlined in Chapter las: (1) (2) A + ~ A A ~ products + A ===> products (quasi-equilibrated) (rate-determining step) (overall reaction) The intrinsic rate at which a catalytic cycle turns over on an active site is called the turnover frequency, rt, of a catalytic reaction and is defined as in Chapter 1 [Equation (1.3.9)] as: 1 dn r=-t S dt (5.2.28) where S is the number of active sites in the experiment. As mentioned earlier, quantifying the number of active sites on a surface is problematic. With metals and some metal oxides, often the best one can do is to count the total number of ex-posed surface atoms per unit surface area as an approximation of []0' Thus, the turnover frequency for the reaction rate expressed by Equation (5.2.22) is calculated by simply dividing the rate by []0: (5.2.29) VIGNETTE 5.2.2 Surface atoms on real catalysts reside in a variety of coordination environments depending on the exposed crystal plane (see Figure 5.1.3) and may exhibit differ-ent catalytic activities in a given reaction. Thus, a turnover frequency based on []0 will be an average value of the catalytic activity. In fact, the calculated turnover fre-quency is a lower bound to the true activity because only a fraction of the total num-ber of surface atoms may contribute to the reaction rate. Nevertheless, the concept of a turnover frequency on a uniform surface has proven to be very useful in relat-ing reaction rates determined on metal single crystals, metal foils, and supported metal particles.
150 CHAPTER 5 HeterogeneQIJs Catalysis CHAPTER 5 HeterogeneQlls CatalYSIS 151 of ethane hydrogenolysis and cyclohexane dehydrogenation As an extension to concepts discussed earlier, a rate expression for the reaction of A and B to form products can be developed by assuming an irreversible, rate-determining, bimolecular surface reaction: (1) (2) (3) A + ~ A B + ~ B k, A + B + products + 2 A + B ===> products (quasi-equilibrated adsorption of A) (quasi-equilibrated adsorption of B) (surface reaction, RDS) (overall reaction) 152 CHAPTER 5 HeterogeneolJs Catalysis Table 5.2.2 I Pre-exponential factors for selected unimolecular surface reactions.
COads -+COg COads-+COg COads-+COg Au ads -+Au g CUads -+CUg Clads -+Clg NOads-+NOg COads -+ Cu(OOl) Ru(OOI) Pt(lll) W(llO) W(llO) Ag(llO) Pt(lll) Ni(lll) 1014 1019.5 1014 1011 1014.5 1020 1016 1015 Adapted from R. Masel, Principles ofAdsorption and Reaction on Solid Surfaces, Wiley, New York, 1996, p. 607.
Before A and B can react, they must both adsorb on the catalyst surface. The next event is an elementary step that proceeds through a reaction of adsorbed intermediates and is often referred to as a Langmuir-Hinshelwood step. The rate expression for the bimolecular reaction depends on the number density of adsorbed A molecules that are adjacent to adsorbed B molecules on the catalyst surface. This case is similar to the one developed previously for the recombinative desorption of diatomic gases [reverse reaction step in Equation (5.2.20)] except that two different atomic species are present on the surface. A simplified rate expression for the bimolecular reaction is: (5.2.30) where 0A and Os each can be expressed in the form of the Langmuir isotherm for competitive adsorption of A and B that are presented in Equations (5.2.10) and (5.2.11) and k3 is the rate constant for step 3. Thus, the overall rate of reaction of A and B can be expressed as: (5.2.31) Notice that the denominator is squared for a bimolecular surface reaction. In gen-eral, the exponent on the denominator is equal to the number of sites participating in a rate-determining surface-catalyzed reaction. Since trimolecular surface events are uncommon, the exponent of the denominator rarely exceeds 2.
It is instructive to compare the values or pre-exponential factors for elementary step rate constants of simple surface reactions to those anticipat~d by transition state theory. Recall from Chapter 2 that the pre-exponential factor A is on the order ofkT/ h = 1013 S-1 when the entropy change to form the transition state is negligible. Some pre-exponential factors for simple unimolecular desorption reac-tions are presented in Table 5.2.2. For the most part, the entries in the table are within a few orders of magnitude of 1013 s- 1. The very high values of the pre-exponential factor are likely attributed to large increases in the entropy upon formation of the transition state. Bimolecular surface reactions can be treated in the same way. However, one must explicitly account for the total number of surface CHAPTER 5 HeterogeneQlls Catalysis Table 5.2.3 I Pre-exponential factors for 2Hads -> Hz g on transition metal surfaces.
153 Pd(lll) Ni(100) Ru(OOl) Pd(llO) Mo(llO) Pt(lll) 1012 1013.5 1014.5 1013.5 1013 1012 aThe values have been normalized by []0' Adapted from R. Masel, Principles ofAdsorption and Reaction on Solid Surfaces, Wiley, New York, 1996,p. 607.
sites in the rate expression. As discussed above, the rate of associative desorption of H2 from a square lattice can be written as: (5.2.32) where the pre-exponential factor of the rate constant is now multiplied by 2/[]0 to properly account for the statistics of a reaction occurring on adjacent sites. For de-sorption of H2 at 550 K from a W(lOO) surface, which is a square lattice with []0 = 5 X 1014 cm-2, the pre-exponential factor is anticipated to be 4.6 X 10-2 cm2 S-2. The reported experimental value of 4.2 X 10-2 cm2 S-1 is very close to that predicted by transition state theory (M. Boudart and G. Djega-Mariadassou, Kinetics ofHeterogeneous Catalytic Reactions, Princeton University Press, Princeton, 1984, p. 71). The measured pre-exponential factors for associative desorption of dihydrogen from other transition metal surfaces (normalized by the surface site density) are summarized in Table 5.2.3. Clearly, the values in Table 5.2.3 are con-sistent with transition state theory.
A fairly rare elementary reaction between A and B, often called a Rideal-Eley step, occurs by direct reaction of gaseous B with adsorbed A according to the following sequence: A + ( ) A A + B ---)- products A + B ==> products (reversible adsorption of A) (Rideal-Eley step) (overall reaction) Theoretically, if reactions are able to proceed through either a Rideal-Eley step or a Langmuir-Hinshelwood step, the Langmuir-Hinshelwood route is much more preferred due to the extremely short time scale (picosecond) of a gas-surface collision. The kinetics of a Rideal-Eley step, however, can become important at extreme conditions.
For example, the reactions involved during plasma processing of electronic materials ,....
EXAMPLE 5.2.2 I I.inAr I en:l nH1Hrouefje01JS \ i8IalYSlS are believed to occur through Rideal-Eley steps. Apparently, the conditions typical of semiconductor growth favor Rideal-Eley elementary steps whereas conditions normally encountered with catalytic reactions favor Langmuir-Hinshelwood steps. This point is thoroughly discussed by R. Masel (Principles of Adsorption and Reaction on SoUd SUlfaces, Wiley, New York, 1996, pp. 444-448).
A reaction that is catalyzed by a Bronsted acid site, or H+, can often be accelerated by ad-dition of a solid acid. Materials like ion-exchange resins, zeolites, and mixed metal oxides function as solid analogues of corrosive liquid acids (e.g., H2S04 and HF) and can be used as acidic catalysts. For example, isobutylene (IE) reacts with itself to form dimers on cross-linked poly(styrene-sulfonic acid), a strongly acidic solid polymer catalyst: -0.2 -0.4 §' 'S: -0.6 Of) ..s -0.8 -1.0 The kinetics of IE dimerization are presented in Figure 5.2.7 for two different initial concentrations of IE in hexane solvent. The reaction appears to be first order at high 40 §' 30 :s: 20 100 200 Time (min) (a) 300 400 o 100 200 300 Time (min) (b) 400 500 600 Figure 5.2.7 I Kinetics of liquid phase oligomerization of isobutylene catalyzed by poly(styrene-sulfonic acid) at 20°C in hexane solvent. (a) Corresponds to high initial concentration of isobutylene. (b) Corresponds to low initial concentration of isobutylene.
[Figures from W. O. Haag, Chern. Eng. Prog. Syrnp. Ser., 63 (1967) 140. Reproduced with permission of the American Institute of Chemical Engineers. Copyright © 1967 AIChE.
All rights reserved.] CHAPTER 5 HeterogeneolJs Catalysis 155 concentrations of IB since a logarithmic plot of [IB] with respect to reaction time is linear.
At low concentrations, however, the reaction shifts to second order since a plot of l/[IB] as a function of time becomes linear. A rate expression consistent with these results is: aj[IB]Z r= I + £xz[IB] (5.2.33) where aj and az are constants. Show how a rate expression of this form can be derived for this reaction.
• Answer The following two-step catalytic cycle for dimerization of IB is proposed.
(1) (2) IB + ~ IB kz IB + IB ~ (IBh + 2 IB ==> (IBh (adsorption of IB on solid acid) (Rideal-Eley reaction) (overall dimerization reaction) In this case, represents a surface acid site and IB designates an adsorbed t-butyl cation.
Notice that step 2 involves the combination of an adsorbed reactant with one present in the liquid phase. The rate equation takes the form: r = kz[IB][IB] (5.2.34) The concentration of adsorbed IB is found by combining the equilibrium relationship for step I and the overall site balance to give: K [] [IB] [IB] = ads 0 " I + KadslIB] (5.2.35) where []0 represents the total number of acid sites on the catalyst surface, and Kads is the adsorption equilibrium constant of IB. Substitution of Equation (5.2.35) into (5.2.34) yields a final form of the rate expression: (5.2.36) that has the same functional form as Equation (5.2.33) with £X] = kzKads[]o and az = Kads and is consistent with the experimental data shown in Figure 5.2.7.
The rate of product desorption can also influence the kinetics of a surface-catalyzed reaction. Consider the following simple catalytic cycle: A + ( ) A A B B +==± B + A ==>B 156 CHAPTER 5 Heterogeneolls Catalysis If desorption of B from the surface is rate-determining, then all elementary steps prior to desorption are assumed to be quasi-equilibrated: K] (1) A + ~ A K2 (2) A ~B k3 (3) B~B+ A ==::;:. B The overall rate in the forward direction only is given by: (5.2.37) The rate expression is simplified by eliminating surface concentrations of species through the use of appropriate equilibrium relationships. According to step (2): (5.2.38) and thus: To find [A], the equilibrium adsorption relationship is used: (5.2.39) [A] K ---J -[A][] which gives: (5.2.40) (5.2.41) The concentration of vacant sites on the surface is derived from the total site balance: []0 = [] + [A] + [B] []0 = [] + [A] + K2[A] []0 = [] + KJ[A][] + K2KJ [A][] [] = []0 1 + (K1 + K2K1)[A] (5.2.42) Substitution of Equation (5.2.42) into Equation (5.2.41) gives the final rate expres-sion as: (5.2.43) Notice the rate expression for this case does not depend on the product concentration.
CHAPTER 5 HeterogeneolJs l;atalYSIS 5.3 I Kinetics of Overall Reactions Hit Consider the entire sequence of elementary steps comprising a surface-catalyzed reac-tion: adsorption ofreactant(s), surface reaction(s), and finally desorption ofproduct(s).
If the surface is considered uniform (i.e., all surface sites are identical kinetically and thermodynamically), and there are negligible interactions between adsorbed species, then derivation of overall reaction rate equations is rather straightforward.
For example, the reaction of dinitrogen and dihydrogen to form ammonia is postulated to proceed on some catalysts according to the following sequence of elementary steps: Step ai (1) Nz + 2 ( ) 2N 1 (2) N + H ( ) NH + 2 (3) NH + H ( ) NHz + 2 (4) NHz + H ( ) NH 3 + 2 2 (5) Hz + 2 ( ) 2H 3 N z + 3Hz = 2NH 3 where (ii is the stoichiometric number of elementary step i and defines the number of times an elementary step must occur in order to complete one catalytic cycle accord-ing to the overall reaction. In the sequence shown above, a z = 2 means that step 2 must occur twice for every time that a dinitrogen molecule dissociately adsorbs in step 1. The net rate of an overall reaction can now be written in terms of the rate of anyone of the elementary steps, weighted by the appropriate stoichiometric number: r= ri -Li (5.3.1) The final form of a reaction rate equation from Equation (5.3.1) is derived by re-peated application of the steady-state approximation to eliminate the concentrations of reactive intermediates.
In many cases, however, the sequence of kinetically relevant elementary steps can be reduced to two steps (M. Boudart and G. Djega-Mariadassou, Kinetics of Heterogeneous Cataiytic Reactions, Princeton University Press, Princeton, 1984, p. 90).
For example, the sequence given above for ammonia synthesis can be greatly sim-plified by assuming step I is rate-determining and all other steps are nearly equili-brated. The two relevant steps are now: ai (1) 2N (rate-determining step) (2) K2 N + 3/2 Hz =e= NH3 + 2 (quasi-equilibrated reaction) lilU \;"AI"U:.R;) HererogeoeOIJS \ ;atalYSIS It must be emphasized that step 2 is not an elementary step, but a sum of all of the quasi-equilibrated steps that must occur after dinitrogen adsorption. According to this abbreviated sequence, the only species on the surface of the catalyst of any kinetic rel-evance is N. Even though the other species (H, NH, etc.) may also be present, ac-cording to the assumptions in this example only N contributes to the site balance: []0 = [NJ + [] (5.3.2) In a case such as this one where only one species is present in appreciable concen-tration on the surface, that species is often referred to as the most abundant reac-tion intermediate, or mario The overall rate of reaction can be expressed as the rate of dissociative adsorption of N2: (5.3.3) (5.3.4) (5.3.5) (5.3.6) where [] and [N] are determined by the equilibrium relationship for step 2: K _ [NJ[Hz]3/Z 2 -[NH 3][J and the site balance. The constant K2 is written in such a way that it is large when [N] is also large. Solving Equations (5.3.2) and (5.3.4) for [] and [N], respectively, yields: ['J~ ?[~J) 1 + K __ 3_.
Z [HzF Z ( [NH 3]) _ Kz ~ []0 [N J -1 + K ([NH 3,J) z [HzP/z Substitution of Equations (5.3.5) and (5.3.6) into (5.3.3) gives the rate equation for the reaction as: r= (5.3.7) At very low conversion (far from equilibrium), the reverse reaction can be neglected thus simplifying the rate expression to: r= [1 +K([NH3 ])JZ Z [HzY/z (5.3.8) CHAPTER 5 Heterogeneolls Catalysis 159 EXAMPLE 5.3.1 I Ruthenium has been investigated by many laboratories as a possible catalyst for ammonia synthesis. Recently, Becue et al. [T. Becue, R. J. Davis, and 1. M. Garces, J. Catat., 179 (1998) 129] reported that the forward rate (far from equilibrium) of ammonia synthesis at 20 bar total pressure and 623 K over base-promoted ruthenium metal is first order in dinitrogen and inverse first order in dihydrogen. The rate is very weakly inhibited by ammonia. Pro-pose a plausible sequence of steps for the catalytic reaction and derive a rate equation con-sistent with experimental observation.
(quasi-equilibrated) (rate-determining step) (1) (2) • Answer In the derivation of Equation (5.3.8), dissociative adsorption of dinitrogen was assumed to be the rate-determining step and this assumption resulted in a first-order dependence of the rate on dinitrogen. The same step is assumed to be rate-determining here. However, the rate expression in Equation (5.3.8) has an overall positive order in dihydrogen. Therefore, some of the assumptions used in the derivation of Equation (5.3.8) will have to be modified. The observed negative order in dihydrogen for ammonia synthesis on ruthenium suggests that H atoms occupy a significant fraction of the surface. If H is assumed to be the most abun-dant reaction intermediate on the surface, an elementary step that accounts for adsorption of dihydrogen must be included explicitly. Consider the following steps: k1 Nz + 2 ~ 2N Kz Hz + 2 ~ 2H that are followed by many surface reactions to give an overall equilibrated reaction repre-sented by: (3) N + 3H ::§= NH3 + 4 (quasi-equilibrated) As before, the forward rate of the reaction (far from equilibrium) can be expressed in terms of the rate-determining step: (5.3.9) To eliminate [] from Equation (5.3.9), use the equilibrium relationship for step 2 combined with the site balance. Hence, the following equations: [Hf Kz = -r ",.=-.
l-=-Z[---'H'--l l j Zj (5.3.10) []0 = [H] + (5.3.11) are solved simultaneously to give: (5.3.12) .ou CHAPTER 5 HeterogeneolJs Catalysis Substitution of Equation (5.3.12) into Equation (5.3.9) provides the final rate equation: (5.3.13) If the surface is nearly covered with adsorbed H atoms, then: (5.3.14) and the rate equation simplifies to: (5.3.15) This expression is consistent with the experimental observations. For this example, the reaction equilibrium represented by step 3 is never used to solve the problem since the most abundant reaction intermediate is assumed to be H (accounted for in the equilibrated step 2.) Thus, a complex set of elementary steps is reduced to two kinetically significant reactions.
The concept that a multistep reaction can often be reduced to two kinetically significant steps is illustrated again by considering the dehydrogenation of methyl-cyclohexane to toluene on a PtjAh03 reforming catalyst [J. H. Sinfelt, H. Hurwitz and R. A. Shulman,1. Phys. Chern., 64 (1960)1559]: e5c5 M A + 3Hz The observed rate expression for the reaction operated far from equilibrium can be written in the form: (5.3.16) The reaction occurs through a complex sequence of elementary steps that includes adsorption of M, surface reactions of M, and partially dehydrogenated M, and fi-nally desorption of both Hz and toluene. It may be possible, however, to simplify this multistep sequence. Consider the following two step sequence that involves M as the rnari: (1) (2) Kj M+~M k, M~ .,.
(quasi-equilibrated adsorption) (rate-determining step) CHAPTER 5 Heterogeneolls Catalysis 161 The site balance and the Langmuir adsorption isotherm can be used to derive the forward rate expression: koK [] [M] r = r = k [M] = ~] 0 2 2 1 + K][M] (5.3.17) that has the same form as the observed rate expression, Equation (5.3.16). A word of caution is warranted at this point. The fact that a proposed sequence is consis-tent with observed kinetics does not prove that a reaction actually occurs by that pathway. Indeed, an alternative sequence of steps can be proposed for the above reaction: (1) (2) k] M+--'-+··· A ~ +A (irreversible adsorption) (surface reactions) (irreversible desorption) The application of the quasi-steady-state approximation and the site balance (as-suming A is the mari) gives the following expression for the reaction rate: (5.3.18) The functional form of the rate equation in Equation (5.3.18) is identical to that of Equation (5.3.17), illustrating that two completely different sets of assump-tions can give rate equations consistent with experimental observation. Clearly, more information is needed to discriminate between the two cases. Additional experiments have shown that benzene added to the methylcyclohexane feed inhibits the rate only slightly. In the first case, benzene is expected to compete with methylcyclohexane for available surface sites since M is equilibrated with the surface. In the second case, M is not equilibrated with the surface and the ir-reversibility of toluene desorption implies that the surface coverage of toluene is far above its equilibrium value. Benzene added to the feed will not effectively displace toluene from the surface since benzene will cover the surface only to the extent of its equilibrium amount. The additional information provided by the inclusion of benzene in the feed suggests that the second case is the preferred path.
It is possible to generalize the treatment of single-path reactions when a most abundant reaction intermediate (mari) can be assumed. According to M. Boudart and G. Djega-Mariadassou (Kinetics ofHeterogeneous Catalytic Reactions, Prince-ton University Press, Princeton, 1984, p. 104) three rules can be formulated: 1.
If in a sequence, the rate-determining step produces or destroys the mari, the sequence can be reduced to two steps, an adsorption equilibrium and the rate-determining step, with all other steps having no kinetic significance.
2.
If all steps are practically irreversible and there exists a mari, only two steps need to be taken into account: the adsorption step and the reaction (or desorption) step 162 EXAMPLE 5.3.2 I CHAPTER 5 Heterogeneolls Catalysis of the mario All other steps have no kinetic significance. In fact, they may be reversible, in part or in whole.
3.
All equilibrated steps following a rate-determining step that produces the mari may be summed up in an overall equilibrium reaction. Similarly, all equilibrated steps that precede a rate-determining step that consumes the mari may be represented by a single overall equilibrium reaction.
The derivation of a rate equation from two-step sequences can also be gener-alized. First, if the rate-determining step consumes the mari, the concentration of the latter is obtained from the equilibrium relationship that is available. Second, if the steps of the two-step sequence are practically irreversible, the steady-state approximation leads to the solution.
These simplifying assumptions must be adapted to some extent to explain the nature of some reactions on catalyst surfaces. The case of ammonia synthesis on supported ruthenium described in Example 5.3.1 presents a situation that is similar to rule I, except the rate-determining step does not involve the mario Nevertheless, the solution of the problem was possible. Example 5.3.2 involves a similar scenario.
If a mari cannot be assumed, then a rate expression can be derived through repeated use of the steady-state approximation to eliminate the concentrations of reactive intermediates.
The oxidation of carbon monoxide on palladium single crystals at low pressures (between 10-8 and 10-6 torr) and temperatures ranging from about 450 to 550 K follows a rate law that is first order in O2 and inverse first order in CO. An appropriate sequence of elementary steps is: (1) (2) (3) (4) CO + ~ CO O2 + ----+ 00 00 + ----+ 20 CO + 0 ----+ CO2 + 2 (quasi-equilibrated adsorption) (irreversible adsorption) (surface reaction) (surface reaction/desorption) If CO is assumed to be the mari, derive the rate expression.
• Answer The rate of reaction is the rate of any single step in the sequence weighted by its appropri-ate stoichiometric number. Thus, for the reaction: the rate can be written: (5.3.19) CHAPTER 5 Heterogeneolls Catalysis 163 (quasi-equilibrated adsorption) (irreversible adsorption) O2 + 00 (1) (2) The simplest solution involves the two-step sequence: Kl CO + +@:t CO where: (5.3.20) Application of the site balance, assuming CO is the mari gives: []0 = [] + [CO] (5.3.21) and the Langmuir isotherm for adsorption of CO can be written as: [COJ K l = -'-[ C-O-J [-=--J (5.3.22) Substitution of Equations (5.3.21) and (5.3.22) into Equation (5.3.20) yields: (5.3.23) At high values of surface coverage, Kl[CO] » 1, the rate equation simplifies to: r 2k2 [Jo[0 2J Kt[COJ (5.3.24) which is consistent with the observed rate law.
The rate constant for the reaction is composed of two terms, 2k2 and Ki l . Thus, the ap-parent activation energy contains contributions from the rate constant for O2 adsorption and the equilibrium constant for CO adsorption according to: Eapp = E2 -I1HadsCO (5.3.25) Since adsorption of O2 is essentially nonactivated, the apparent activation energy for CO ox-idation is simply the negative of the enthalpy of CO adsorption on Pd. This result has been experimentally observed [M. Boudart, J. Mol. Catal. A: Chern., 120 (1997) 271].
Example 5.3.2 demonstrates how the heat of adsorption of reactant molecules can profoundly affect the kinetics of a surface catalyzed chemical reaction. The ex-perimentally determined, apparent rate constant (2k2/ K1) shows typical Arrhenius-type behavior since it increases exponentially with temperature. The apparent activation energy of the reaction is simply Eapp = E2 -t::..HadsCO ~ -t::..HadsCO (see Example 5.3.2), which is a positive number. A situation can also arise in which a negative overall activation energy is observed, that is, the observed reaction rate lti4 CHAPTER 5 Heterogeneous Catalysis decreases with increasing temperature. This seemingly odd phenomenon can be understood in terms of the multistep mechanism of surface catalyzed reactions. Con-sider the rate of conversion of A occurring through a rate-determining surface reaction as described earlier: r= k2Kads[Jo[A] I + Kads[A] (5.3.26) Experimental conditions can arise so that I» Kads[AJ, and the reaction rate expression reduces to: (5.3.27) with an apparent rate constant k2Kads' The apparent activation energy is now: (5.3.28) 10-5 10-6 ~ ~ tij u OIl C/O t: B 10-7 ;:::;: 0 E 3 "" lO-s 10-9 1.20 1.30 1.40 1.50 1.60 1000/T (K-I) 1.70 Figure 5.3.1 I Temperature dependence of the cracking of n-alkanes. [Reprinted from J. Wei, "Adsorption and Cracking of N-Alkanes over ZSM-5: Negative Activation Energy of Reaction," Chern. Eng. Sci., 51 (1996) 2995, with permission from Elsevier Science.] Open sguare-Cs, inverted triangle-Cw, triangle-Cll. pIUS-C I4, filled sguare-C I6, diamond--C 1s, circle-Czo.
CHAPTER 5 Heterogeneous Catalysis 160 140 120 ~ 100 "0 .§.
..., 80 e 1 60 40 20 2 4 6 8 10 12 14 16 Carbon number Figure 5.3.2 I Heats of adsorption of n-alkanes on ZSM-5.
[Reprinted from J. Wei, "Adsorption and Cracking of N-Alkanes over ZSM-5: Negative Activation Energy of Reaction," Chem. Eng. Sci., 51 (1996) 2995, with permission from Elsevier Science.] 165 Since the enthalpy of adsorption is almost always negative, the apparent activation en-ergy can be either positive or negative, depending on the magnitudes of £z and tiHads• The cracking of n-alkanes over H-ZSM-5, an acidic zeolite, provides a clear illustration of how the apparent activation energy can be profoundly affected by the enthalpy of adsorption. An Arrhenius plot of the pseudo-first-order rate constant, k = kzKads, for n-alkanes having between 8 and 20 carbon atoms is shown in Figure 5.3.1. The reaction of smaller alkanes (CS-C14) has a positive apparent acti-vation energy that declines with chain length, whereas the reaction of larger alkanes (CIS' Czo) has a negative apparent activation energy that becomes more negative with chain length. Interestingly, the reaction of CI6 is almost invariant to tempera-ture. The linear relationship in Figure 5.3.2 demonstrates that the adsorption enthalpy is proportional to the number of carbons in the alkanes. Thus, if the activation energy of the surface reaction step, £z, is not sensitive to the chain length, then Equation (5.3.28) predicts that the apparent activation energy will decrease with increasing length of the alkane. The temperature invariance of the pseudo-first-order rate constant for cracking of CI6 apparently results from the cancellation of£z by tiHads' It is also worth mentioning that the magnitude of k, at a constant temperature, will be profoundly affected by carbon chain length through the equilibrium adsorption constant.
ltiti CHAPTER 5 HeterogeneQus Catalysis 10-4 10-5 "" u 10-6 0/) en t:: B ::s: 0 g 10-7 ""'" 10-8 10-9 5 10 15 20 25 30 VIGNETTE 5.3.1 I Carbon number Figure 5.3.3 I The apparent first-order rate constant of n-alkane cracking over ZSM-5 at 380°C.
[Reprinted from J. Wei, "Adsorption and Cracking of N-Alkanes over ZSM-5: Negative Activation Energy of Reaction," Chern. Eng. Sci., 51 (1996) 2995, with permission from Elsevier Science.] Indeed, Figure 5.3.3 illustrates the exponential dependence of k on the size of the alkane molecule. The high enthalpies of adsorption for long chain alkanes means that their surface coverages will far exceed those associated with short chain mol-ecules, which translates into much higher reaction rates.
CHAPTER 5 Heterogeneolls Catalysis 167 168 CHAPTER 5 HeterogeneolJs Catalysis 169 170 CHAPTER 5 HeterogeneolJs Catalysis The discussion to this point has emphasized kinetics of catalytic reactions on a uniform surface where only one type of active site participates in the reaction.
Bifunctional catalysts operate by utilizing two different types of catalytic sites on the same solid. For example, hydrocarbon reforming reactions that are used to up-grade motor fuels are catalyzed by platinum particles supported on acidified alu-mina. Extensive research revealed that the metallic function of Pt/Al20 3 catalyzes hydrogenation/dehydrogenation of hydrocarbons, whereas the acidic function of the support facilitates skeletal isomerization of alkenes. The isomerization of n-pentane (N) to isopentane (1) is used to illustrate the kinetic sequence associated with a bi-functional Pt/Al20 3 catalyst: CH3 Pt!AI20 3 I CH3-CHZ CHz CHz CH3 > CH3-CH-CHz-CH3 n-pentane (N) isopentane (I) CHAPTER 5 HeterogeneolJs Catalysis 171 The sequence involves dehydrogenation of n-pentane on Pt particles to fonn intennediate n-pentene (N=), which then migrates to the acidic alumina and reacts to i-pentene (1=) in a rate-detennining step. The i-pentene subsequently migrates back to the Pt particles where it is hydrogenated to the product i-pentane. The following sequence describes these processes (where represents an acid site on alumina): (1) (Pt-catalyzed) (3) (4) (5) K 2 (2) N= + ~ N= k3 N= -A+ 1= 1= ~ 1= + 1=+ Hz :::§:: I N ==> I (Pt-catalyzed) Interestingly, the rate is inhibited by dihydrogen even though it does not appear in the stoichiometric equation. The rate is simply: The equilibrium relationships for steps I and 2 give: [N=] = Kz[N=][] [N=] = K [N] 1 [Hz] (5.3.29) (5.3.30) (5.3.31) The site balance on alumina (assuming N= is the mari, far from equilibrium): (5.3.32) and its use along with Equations (5.3.30) and (5.5.31) allow the rate expression for the forward reaction to be written as: r== K1Kzk3[Jo[N] [Hz] + K1Kz[N] Thus, the inhibitory effect of Hz in pentane isomerization arises from the equili-brated dehydrogenation reaction in step I that occurs on the Pt particles.
5.4 I Evaluation of Kinetic Parameters Rate data can be used to postulate a kinetic sequence for a particular catalytic reac-tion. The general approach is to first propose a sequence ofelementary steps consistent with the stoichiometric reaction. A rate expression is derived using the steady-state 172 CHAPTER 5 HeterogeneQlls Catalysis approximation together with any other assumptions, like a rate-determining step, a most abundant reaction intermediate, etc. and compared to the rate data. If the func-tional dependence of the data is similar to the proposed rate expression, then the se-quence of elementary steps is considered plausible. Otherwise, the proposed sequence is discarded and an alternative one is proposed.
Consider the isomerization of molecule A far from equilibrium: A=-B that is postulated to occur through the following sequence of elementary steps: (I) A + ( ) A (2) A ( ) B (3) B ( ) B + Case 1.
If the rate of adsorption is rate determining, then the forward rate of re-action can be simplified to two steps: k) (1) A + ~ A K2a (2a) A =€F B+ where step 2a represents the overall equilibrium associated with surface reaction and desorption of product. A rate expression consistent with these assumptions is (derived according to methods described earlier): r = r1 = k][A][] [A] K ---2a -[B][] []0 = [] + [A] (5.4.1) (5.4.2) (5.4.3) (5.4.4) The equilibrium constant is written such that K2a is large when [A] is large. Com-bining Equations (5.4.1-5.4.3) results in the following expression for the forward rate: k][Jo[A] r= 1 + K2a[B] Case 2.
If the surface reaction is rate-determining, the following sequence for the forward rate is appropriate: K) (1) A + +Q:t A (2) A 4 B K3 (3) B +Q:t B + CHAPTER 5 Heterogeoeolls Catalysis 173 This particular sequence assumes both A and B are present on the surface in ki-netically significant amounts. The rate expression for this case is: r = rz = kz[A] [A] K =--I [A ][] [B] K =--3 [B][] []0 = [] + [A] + [B] kz[Jo[A] r = ------=--"---'---1 + KI[A] + K3[B] (5.4.5) (5.4.6) (5.4.7) (5.4.8) (5.4.9) Case 3.
In this last case, the desorption of product is assumed to be rate deter-mining. Similar to Case 1, two elementary steps are combined into an overall equil-ibrated reaction: Kia (la) A + =9= B k2 (3) B ~ B + The expression for the forward rate is derived accordingly: r = r3 = k 3[B] [B] Kia = [A][] []0 = [] + [B] k3 Kla[Jo[A] r= 1 + Kla[A] (5.4.10) (5.4.11) (5.4.12) (5.4.13) A common method used to distinguish among the three cases involves the meas-urement of the initial rate as a function of reactant concentration. Since B is not present at early reaction times, the initial rate will vary proportionally with the con-centration of A when adsorption of A is the rate-determining step (Figure 5.4.1).
Similarly, the initial rate behavior is plotted in Figure 5.4.2 for Cases 2 and 3, in which the rate-determining step is either surface reaction or desorption of prod-uct. Since the functional form of the rate expression is the same in Cases 2 and 3 when B is not present, additional experiments are required to distinguish between the two cases. Adding a large excess of product B to the feed allows for the difference between kinetic mechanisms to be observed. As shown in Figure 5.4.3, the presence of product in the feed does not inhibit the initial rate when desorption of B is the rate-determining step. If surface reaction of adsorbed A were the rate-determining 174 CHAPTER 5 Heterogeneous Catalysis [A] in feed Figure 5.4.1 I Results from Case I where adsorption is the rate-determining step.
[A] in feed Figure 5.4.2 I Results from Cases 2 and 3 where surface reaction or product desorption is the rate-determining step.
step, then extra B in the feed effectively competes for surface sites, displaces A from the catalyst, and lowers the overall rate.
Once a rate expression is found to be consistent with experimental observations, then rate constants and equilibrium constants are obtained from quantitative rate data. One way to arrive at numerical values for the different constants in a Langmuir-Hinshelwood rate expression is to first invert the rate expression. For Case 2, the rate expression: (5.4.14) becomes: CHAPTER 5 Heterogeoeo!!s Catalysis Desorption is RDS ----....
........................
...
Surface reaction ..........
isRDS ....................
....
""" [B] in feed Figure 5.4.3 I The influence of extra B on the initial rate (at constant [AJ) for cases with different rate-determining steps (RDS).
175 (5.4.15) Multiplying by [A] gives an equation in which the groupings of constants can be calculated by a linear least squares analysis: [A] 1 K] K 2 -r- = k2[]o + k2[]o [A] + k2[]o [B] The above expression is of the form: with (5.4.16) y = , r K2 a ----3 -k r] 2L 0 If an independent measure of []0 is available from chemisorption, the constants k2, Kj, and K 2 can be obtained from linear regression. It should be noted that many kineticists no longer use the linearized form of the rate equation to obtain rate con-stants. Inverting the rate expression places greater statistical emphasis on the low-est measured rates in a data set. Since the lowest rates are usually the least precise, a nonlinear least squares analysis of the entire data set using the normal rate expression is preferred.
176 EXAMPLE 5.4.1 I CHAPTER 5 Heterogeneolls Catalysis The reaction CO + CI2 => COCI2 has been studied over an activated carbon catalyst. A sur-face reaction appears to be the rate-determining step. Fashion a rate model consistent with the following data: 4.41 0.406 0.352 0.226 4.4 0.396 0.363 0.231 2.41 0.310 0.320 0.356 2.45 0.287 0.333 0.376 1.57 0.253 0.218 0.522 3.9 0.610 0.113 0.231 2.0 0.179 0.608 0.206 • Answer The strategy is to propose a reasonable sequence of steps, derive a rate expression, and then evaluate the kinetic parameters from a regression analysis of the data. As a first attempt at solution, assume both CI2 and CO adsorb (nondissociatively) on the catalyst and react to form adsorbed product in a Langmuir-Hinshelwood step. This will be called Case I. Another possible sequence involves adsorption of CI2 (nondissociatively) followed by reaction with CO to form an adsorbed product in a Rideal-Eley step. This scenario will be called Case 2.
Case 1 K1 CI2 + ~ C12 K2 CO + ~ CO k2 CO + C12 ~ COCI2 + K4 COCI2 ~ COCI2 + The rate expression derived from the equilibrium relations for steps 1, 2, and 4, assuming all three adsorbed species are present in significant quantities, is: The data fit well the above expression. However, some of the constants (not shown) have negative values and are thus unrealistic. Therefore, Case I is discarded.
Case 2 CHAPTER 5 HeterogeneolJs Catalysis K 1 Clz + Clz kz CO + Clz -A+ COClz K1 COClz ~ COClz + 177 Assuming only Clz and COClz are present on the surface, the following rate expression is derived: Fitting the data to the above equation results in the following rate model: 1.642[COJ[ClzJ ( mOl) r = 1 + 124.4[ClzJ + 58.1 [COClzJ gcat-h where concentrations are actually partial pressures expressed in atm. Even though all the kinetic parameters are positive and fit the data set reasonably well, this solution is not guaranteed to represent the actual kinetic sequence. Reaction kinetics can be consistent with a mechanism but they cannot prove it. Numerous other models need to be constructed and tested against one an-other (as illustrated previously in this section) in order to gain confidence in the kinetic model.
Rate constants and equilibrium constants should be checked for thermody-namic consistency if at all possible. For example, the heat of adsorption 6.Hads derived from the temperature dependence of Kads should be negative since adsorption reactions are almost always exothermic. Likewise, the entropy change 6.Sads for nondissociative adsorption must be negative since every gas phase molecule loses translational entropy upon adsorption. In fact, I6.Sads I< So (where ., Sg is the gas phase entropy) must also be satisfied because a molecule cannot lose more entropy than it originally possessed in the gas phase. A proposed kinetic sequence that produces adsorption rate constants and/or equilibrium constants that do not satisfy these basic principles should be either discarded or considered very suspiciously.
Exercises for Chapter 5 1.
(a) Calculate the BET surface area per gram of solid for Sample I using the full BET equation and the one-point BET equation. Are the values the same? What is the BET constant?
(b) Calculate the BET surface area per gram of solid for Sample 2 using the full BET equation and the one-point BET equation. Are the values the same? What is the BET constant and how does it compare to the value obtained in (a)?
178 CHAPTER 5 HeterogeoeolJs Catalysis Oinitrogen adsorption data 0.02 0.Q3 0.04 0.05 0.10 0.15 0.20 0.25 0.30 23.0 25.0 26.5 27.7 31.7 34.2 36.1 37.6 39.1 0.15 0.23 0.32 0.38 0.56 0.65 0.73 0.81 0.89 2.
A 0.5 wt. % Pt on silica catalyst gave the data listed below for the sorption of H2.
Upon completion of Run 1, the system was evacuated and then Run 2 was per-formed. Find the dispersion and average particle size ofthe Pt particles. Hint: Run 1 measures the total sorption of hydrogen (reversible + irreversible) while Run 2 gives only the reversible hydrogen uptake. Calculate the dispersion based on the chemisorbed (irreversible) hydrogen.
10.2 1.09 10.7 1.71 12.9 1.30 14.2 2.01 16.1 1.60 18.2 2.33 20.4 1.93 23.2 2.73 25.2 2.30 28.9 3.17 30.9 2.75 35.9 3.71 37.9 3.30 44.4 4.38 46.5 3.96 55.2 5.22 57.2 4.79 66.2 6.05 68.2 5.64 77.2 6.90 79.2 6.49 88.3 7.72 90.2 7.32 99.3 8.57 101 8.18 110 9.42 112 9.03 121 10.3 123 9.87 132 ILl 134 10.7 143 12.0 145 11.6 154 12.8 156 12.4 3.
A. Peloso et al. [Can. J. Chem. Eng., 57 (1979) 159] investigated the kinetics of the following reaction over a CuO/Cr203/Si02 catalyst at temperatures of 225-285°C: CH 3CH 20H (Et) CH 3CHO + H2 (Ad) (DH) CHAPTER 5 HeterogeneOlls Catalysis A possible rate expression to describe the data is: 179 Write a reaction sequence that would give this rate expression.
4.
1. Franckaerts and G. F. Froment [Chern. Eng. Sci., 19 (1964) 807] investigated the reaction listed in Exercise 3 over the same temperature range using a CuO/CoO/CrZ03 catalyst. The rate expression obtained in this work was of the form: Write a reaction sequence that gives a rate expression of this form. What is different from the sequence used in Exercise 3?
S.
The following oxidation occurs over a solid catalyst: OH 0 I " CH3CHCH3 + 112 02 = CH3CCH3 + H20 (IP) (DO) (At) (W) If acetone cannot adsorb and the rate of surface reaction between adsorbed isopropanol and adsorbed oxygen is the rate-determining step, a rate expression of the form: k[IP][DO]l can be obtained from what reaction sequence?
6.
For the reaction: CO + Clz = COClz the rate expression given below can be obtained: r = ------------::-[I + KIPCl, + KzPCoCl,?
What does the exponent on the denominator imply and what does the lack of a K3Pco term in the denominator suggest?
7.
For the reaction of A to form B over a solid catalyst, the reaction rate has the form: 180 CHAPTER 5 Hetemgeneolls Catalysis However, there is a large excess of inert in the reactant stream that is known to readily adsorb on the catalyst surface. How will this affect the reaction order with respect to A?
8.
G. Thodor and C. F. Stutzman [Ind. Eng. Chern., 50 (1958) 413] investigated the following reaction over a zirconium oxide-silica gel catalyst in the presence of methane: If the equilibrium constant for the reaction is 35 at reaction conditions, find a reaction rate expression that describes the following data: 2.66 7.005 0.300 0.370 0.149 2.61 7.090 0.416 0.215 0.102 2.41 7.001 0.343 0.289 0.181 2.54 9.889 0.511 0.489 0.334 2.64 10.169 0.420 0.460 0.175 2.15 8.001 0.350 0.250 0.150 2.04 9.210 0.375 0.275 0.163 2.36 7.850 0.400 0.300 0.208 2.38 10.010 0.470 0.400 0.256 2.80 8.503 0.500 0.425 0.272 9.
Ammonia synthesis is thought to take place on an iron catalyst according to the following sequence: Nz + 2 ( ) 2N Hz + 2 ( ) 2H N + H ( ) NH + NH + H ( ) NHz + NHz + H ( ) NH3 + 2 (1) (2) (3) (4) (5) Obviously, step 2 must occur trliee times for every occurrence of step 1, and steps 3-5 each occur twice, to give the overall reaction of Nz + 3Hz = 2NH3.
Experimental evidence suggests that step 1is the rate-determining step, meaning that all of the other steps can be represented by one pseudo-equilibrated overall reaction. Other independent evidence shows that nitrogen is the only surface species with any significant concentration on the surface (most abundant reaction intermediate). Thus, []0 = [] + [N]. Equation (5.3.7) gives the rate expression consistent with the above assumptions.
(a) Now, assume that the rate-determining step is actually: Nz + ~ Nz (associative adsorption of Nz) CHAPTER 5 Heterogeneolls Catalysis 181 with Nz being the most abundant reaction intermediate. Derive the rate expression for the reversible formation of ammonia.
(b) Can the rate expression derived for part (a) and the one given in Equation (5.3.7) be discriminated through experimentation?
10.
Nitrous oxide reacts with carbon monoxide in the presence of a ceria-promoted rhodium catalyst to form dinitrogen and carbon dioxide. One plausible sequence for the reaction is given below: (1) NzO + +§:t NzO (2) NzO ~ Nz + 0 (3) CO + +§:t CO (4) CO + 0 ~ COz + 2 NzO + CO ===;> Nz + COz (a) Assume that the surface coverage of oxygen atoms is very small to derive a rate expression of the following form: K 1PN20 r=-----"----1 + K2PN,o + K3PCO where the K's are collections of appropriate constants. Do not assume a rate-determining step.
(b) The rate expression in part (a) can be rearranged into a linear form with respect to the reactants. Use linear regression with the data in the following table to evaluate the kinetic parameters Kb Kz, and K3• Rates of N20 + CO reaction at 543 K 30.4 30.4 30.4 30.4 30.4 "7" I.V 15.2 45.6 76 7.6 15.2 30.4 45.6 76 '}r\ A JV.'"t 30.4 30.4 30.4 0.00503 0.00906 0.0184 0.0227 0.0361 0.0386 0.0239 0.Q117 0.00777 Source: J. H. Holies, M. A. Switzer, and R. J. Davis, J. Catal.. 190 (2000) 247.
(c) Use nonlinear regression with the data in part (b) to obtain the kinetic parameters. How do the answers compare?
(d) Which species, NzO or CO, is present on the surface in greater amount?
182 CHAPTER 5 Heterogeneolls Catalysis 11.
The reaction of carbon monoxide with steam to produce carbon dioxide and dihydrogen is called the water gas shift (WGS) reaction and is an important process in the production of dihydrogen, ammonia, and other bulk chemicals.
The overall reaction is shown below: Iron-based solids that operate in the temperature range of 360 to 530°C catalyze the WGS reaction.
(a) One possible sequence is of the Rideal-Eley type that involves oxidation and reduction of the catalyst surface. This can be represented by the following steps: HzO + ( ) Hz + 0 CO + 0 ( ) COz + Derive a rate expression.
(b) Another possible sequence for the WGS reaction is of the Langmuir-Hinshelwood type that involves reaction of adsorbed surface species. For the following steps: CO + ~ CO (1) HzO + 3 =@= 2H + 0 (2) CO + 0 ~ COz + 2 (3) 2H ~ Hz + 2 (4) derive the rate expression. (Notice that step 2 is an overall equilibrated reaction.) Do not assume that one species is the most abundant reaction intermediate.
12.
For the hydrogenation of propionaldehyde (CH3CHzCHO) to propanol (CH3CHzCHzOH) over a supported nickel catalyst, assume that the rate-limiting step is the reversible chemisorption of propionaldehyde and that dihydrogen adsorbs dissociatively on the nickel surface.
(a) Provide a reasonable sequence of reaction steps that is consistent with the overall reaction.
(b) Derive a rate expression for the rate of consumption of propionaldehyde.
At this point, do not assume a single mario (c) Under what conditions would the rate expression reduce to the expe-rimentally observed function (where prop corresponds to propionaldehyde, and P represents partial pressure): kPprop r=--p o.s H, CHAPTER 5 Heterogeneolls Catalysis 183 13.
Some of the oxides of vanadium and molybdenum catalyze the selective oxidation of hydrocarbons to produce valuable chemical intermediates. In a reaction path proposed by Mars and van Krevelen (see Section 10.5), the hydrocarbon first reduces the surface of the metal oxide catalyst by reaction with lattice oxygen atoms. The resulting surface vacancies are subsequently re-oxidized by gaseous O2, The elementary steps of this process are shown below. Electrons are added to the sequence to illustrate the redox nature of this reaction.
hydrocarbon lattice oxygen product surface vacancy Derive a rate expression consistent with the above sequence. For this problem, assume that the vacancies do not migrate on the surface. Thus, 20- 2 and 20 can be considered as a single occupied and a single unoccupied surface site, respectively. Show that the expression can be transformed into the following form: where the K's are collections of appropriate constants.
~~6 Effects of Transport Limitations on Rates of Solid-Catalyzed Reactions 6.1 I Introduction To most effectively utilize a catalyst in a commercial operation, the reaction rate is often adjusted to be approximately the same order of magnitude as the rates of transport phenomena. If a catalyst particle in an industrial reactor were oper-ating with an extremely low turnover frequency, diffusive transport of chemicals to and from the catalyst surface would have no effect on the measured rates.
While this "reaction-limited" situation is ideal for the determination of intrinsic reaction kinetics, it is clearly an inefficient way to run a process. Likewise, if a catalyst particle were operating under conditions that normally give an extremely high turnover frequency, the overall observed reaction rate is lowered by the inadequate transport of reactants to the catalyst surface. A balance between reaction rate and transport phenomena is frequently considered the most effec-tive means of operating a catalytic reaction. For typical process variables in in-dustrial reactors, this balance is achieved by adjusting reaction conditions to give a rate on the order of 1 p,mol!(cm3-s) [Po B. Weisz, CHEMTECH, (July 1982) 424]. This reaction rate translates into a turnover frequency of about 1 s-I for many catalysts (R. L. Burwell, Jr. and M. Boudart, in "Investigations of Rates and Mechanisms of Reactions," Part 1, Ch. 12, E. S. Lewis, Ed., John Wiley, New York, 1974).
Figure 6.1.1 depicts the concentration profile of a reactant in the vicinity of a catalyst particle. In region 1, the reactant diffuses through the stagnant bound-ary layer surrounding the particle. Since the transport phenomena in this region occur outside the catalyst particle, they are commonly referred to as external, or CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 185 Region 1 j Boundary layer Region 2 Distance Porous catalyst pellet Figure 6.1.1 I Concentration profile of a reacting species in the vicinity of a porous catalyst particle. Distances are not to scale.
interphase, transport effects. In region 2, the reactant diffuses into the pores of the particle, and transport phenomena in this region are called internal, or intraphase, transport effects. Both external and internal transport effects may be important in a catalytic reaction and are discussed separately in the following sections. In addition to mass transfer effects, heat transfer throughout the catalyst particle and the stag-nant boundary layer can dramatically affect observed reaction rates.
6.2 I External Transport Effects For a solid-catalyzed reaction to take place, a reactant in the fluid phase must first diffuse through the stagnant boundary layer surrounding the catalyst particle. This mode of transport is described (in one spatial dimension) by the Stefan-Maxwell equations (see Appendix C for details): Uy = , ,("-1 (6.2.1) where Xi is the mole fraction of component i, C is the total concentration, Ni is the flux of component i, and Dij is the diffusivity of component i in j. The following relationship for diffusion of A in a two component mixture at constant pressure (constant total concentration) can be obtained from simplifying the Stefan-Maxwell equations: (6.2.2) 186 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Since there are only two components in the mixture, Xs = 1 - XA and the above expression reduces to: (6.2.3) Equimolar counterdiffusion (NA = -Ns ) can often be assumed and further simpli-fication is thus possible to give: (6.2.4) (6.2.5) The same equation can also be derived by assuming that concentrations are so di-lute that XA(NA + Ns ) can be neglected. Equation (6.2.3) is known as Fick's First Law and can be written as: (equimolar counterdiffusion and/or dilute concentration of A) The diffusivities of gases and liquids typically have magnitudes that are 10- 1 and 10- 5 cm2 s-1, respectively. The diffusivity of gases is proportional to TI.5 and inversely proportional to P, whereas, the diffusivity of liquids is proportional to T and inversely proportional to viscosity Ii (may strongly depend on T).
To obtain the flux of reactant A through the stagnant boundary layer surround-ing a catalyst particle, one solves Equation (6.2.~ with the appropriate boundary conditions. If the thickness of the boundary layer 8 is small compared to the radius of curvature of the catalyst particle, then the problem can be solved in one dimen-sion as depicted in Figure 6.2.1. In this case, Fick's Law reduces to: (6.2.6) Catalyst surface Boundary layer x=O x=b Bulk fluid Figure 6.2.1 I Concentration profile of reactant A in the vicinity of a catalyst particle.
C H A PT E R 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 187 (6.2.7) Since the flux of A must be constant through the stagnant film (conservation of mass), the derivative of the flux with respect to distance in the film must vanish: dNA, =0 dx Differentiating Equation (6.2.6) (assuming constant diffusivity) and combining with Equation (6.2.7), yields the following differential equation that describes diffusion through a stagnant film: with boundary conditions: (6.2.8) atx = 0 atx = 8 (6.2.9) (6.2.10) The solution of Equation (6.2.8) results in a linear concentration profile through the boundary layer: (6.2.11) and the molar flux of A through the film is simply: (6.2.12) Although diffusion of reacting species can be wr:!tten in terms of the diffusiv-ity and boundary layer thickness, the magnitude of 8 is unknown. Therefore, the mass-transfer coefficient is normally used. That is, the average molar flux from the bulk fluid to the solid surface is (-x direction in Figure 6.2.1) (6.2.13) where kc is the mass transfer coefficient over the surface area of the particle. The mass transfer coefficient is obtained from correlations and is a function of the fluid velocity past the particle. If the fluid is assumed to be well mixed, the concentra-tion of A at the edge of the stagnant boundary layer is equivalent to that in the bulk fluid, CAB, and Equation (6.2.13) can therefore be written as: (6.2.14) At steady-state, the flux of A equals the rate of reaction thus preventing accumula-tion or depletion. For a simple first-order reaction, the kinetics depend on the sur-face rate constant, ks, and the concentration of A at the surface: (6.2.15) 188 C H A PT E R 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Solving for CAS yields: (6.2.16) Substitution of the above expression for CAS into Equation (6.2.15) gives a rate ex-pression in terms of the measurable quantity, CAB, the reactant concentration in the bulk fluid: kskc r = --- C = --'='--k+f AB 11 s c _ + =-ks kc An overall, observed rate constant can be defined in terms of ks and kc as: (6.2.17) (6.2.18) so that the rate expressed in terms of observable quantities can be written as: robs = kobsCAB (6.2.19) For rate laws that are noninteger or complex functions of the concentration, CAS is found by trial and error solution of the flux expression equated to the reaction rate.
The influence of diffusional resistance on the observed reaction rate is especially apparent for a very fast surface reaction. For that case, the surface concentration of reactant is very small compared to its concentration in the bulk fluid. The observed rate is then written according to Equation (6.2.15), but ignoring CAS: (6.2.20) (6.2.21) The observed rate will appear to be first-order with respect to the bulk reactant con-centration, regardless of the intrinsic rate expression applicable to the surface reac-tion. This is a clear example of how external diffusion can mask the intrinsic ki-netics of a catalytic reaction. In a catalytic reactor operating under mass transfer limitations, the conversion at the reactor outlet can be calculated by incorporating Equation (6.2.20) into the appropriate reactor model.
Solution of a reactor problem in the mass transfer limit requires an estimation of the appropriate mass transfer coefficient. Fortunately, mass transfer correlations have been developed to aid the determination of mass transfer coefficients. For ex-ample, the Sherwood number, Sh, relates the mass transfer coefficient of a species A to its diffusivity and the radius of a catalyst particle, Rp : kc Sh = ----'-DAB For flow around spherical particles, the Sherwood number is correlated to the Schmidt number, Sc, and the Reynolds number, Re: Sh = 2 + 0.6Re 1/ 2 SC1/ 3 (6.2.22) CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Sc = pDAS up(2Rp) Re = ----'------'---t-t 189 (6.2.23) (6.2.24) where t-t is the viscosity (kg m-] s-]), p is the fluid density (kg m-3), and u is the linear fluid velocity (m s-]). However, most mass-transfer results are correlated in terms of Colburn J factors: Sh J=---SC I/ 3Re (6.2.25) that are plotted as a function of the Reynolds number. These J factor plots are avail-able in most textbooks on mass transfer. If one can estimate the fluid density, ve-locity, viscosity, diffusivity, and catalyst particle size, then a reasonable approxi-mation of the mass-transfer coefficient can be found.
It is instructive to examine the effects of easily adjustable process variables on the mass-transfer coefficient. Combining Equations (6.2.21-6.2.24) gives the func-tional dependence of the mass-transfer coefficient: ( ) 2/3 ]/2 ]/6 kG. ex DAsSh ex DAS Re]/2 SCI /3 ex DAS (RpUP)I/2(---,!:-)1/3 ex DAS U . p' Rp Rp Rp Ii pDAS (Rpt 2(1it 6 or (6.2.26) Equation (6.2.26) shows that decreasing the catalyst particle size and increasing the fluid velocity can significantly increase the mass-transfer coefficient. These simple variables may be used as process "handles" to decrease the influence of external mass-transfer limitations on the observed reaction rate.
To quickly estimate the importance of external mass-transfer limitations, the magnitude of the change in concentration across the boundary layer can be calcu-lated from the observed rate and the mass-transfer coefficient: (6.2.27) If ~CA « CAS' then external mass-transfer limitations are not significantly affect-ing the observed rate.
The effects of heat transfer are completely analogous to those of mass trans-fer. The heat flux, q, across the stagnant boundary layer shown in Figure 6.2.1 is related to the difference in temperature and the heat-transfer coefficient, h" according to: (6.2.28) ..L1...
9""O'----"C'-'H ....
A"'-"'P-LT-"E ..
R"-"6'---'F""'ff~e"'cd.>ts'"QLTransport I imitations on Rates Qf SQlid Catalyzed Reactions Steady state requires that the heat flux is equivalent to the heat generated (or con-sumed) by reaction: (6.2.29) (6.2.30) where ~Hr is the heat of reaction per mole of A converted. To estimate the influ-ence of heat-transfer limitations on the observed rate, the change in temperature across the film is found by evaluating the observed rate of heat generated (or con-sumed) and the heat-transfer coefficient (obtained from J factor correlations, simi-lar to the case of mass-transfer coefficients): robs(~Hr) = ~T hi If I~TI « TB, then the effect of external heat-transfer limitations on the observed rate can be ignored. Equation (6.2.30) can also be used to find the maximum tem-perature change across the film. Using Equation (6.2.15) to eliminate the observed rate, the resulting equation relates the concentration change across the film to the temperature change: (6.2.31) (6.2.32) The maximum temperature change across the film will occur when CAS ap-proaches zero, which corresponds to the maximum observable rate. Solving Equa-tion (6.2.31) for ~Tmax with CAS = 0 gives the following expression: kc~Hr ~Tmax = --- CAS hi that can always be calculated for a reaction, independent of an experiment. If both external heat and mass transfer are expected to affect the observed reaction rate, the balances must be solved simultaneously.
6.3 I Internal Transport Effects Many solid catalysts contain pores in order to increase the specific surface area available for adsorption and reaction, sometimes up to 103 m2 g ~ 1. Since nearly all of the catalytically active sites in highly porous solids are located in the pore net-work, diffusion of molecules in confined spaces obviously plays a critical role in the observed rate of reaction.
The preceding section assumed that the mass-transport mechanism in a fluid medium is dominated by molecule-molecule collisions. However, the mean free path of gases often exceeds the dimensions of small pores typical of solid catalysts. In this situation, called Knudsen diffusion, molecules collide more often with the pore walls than with other molecules. According to Equation (6.3.1), the Knudsen diffu-sivity of component A, DKA , is proportional to T 1/2, but is independent of both pressure and the presence of other species: CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 191 (6.3.1) where Rpore is the pore radius in em, T is the absolute temperature in Kelvins, and MA is the molecular weight of A. Recall that the diffusivity DAB for molecular dif-fusion depends on the pressure and the other species present but is independent of the pore radius. In cases where both molecule-molecule and molecule-wall colli-sions are important, neither molecular diffusivity nor Knudsen diffusivity alone can adequately describe the transport phenomena. Under the conditions of equimolar counterdiffusion of a binary mixture, a transition diffusivity of component A, DTA , can be approximated by the Bosanquet equation (see Appendix C for derivation): III -=-+-D TA DAB D KA (6.3.2) VIGNETTE 6.3.1 192 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Consider the idealized cylindrical pore in a solid catalyst slab, as depicted in Figure 6.3.3. For an isothermal, isobaric, first-order reaction of A to form B that oc-curs on the pore walls, the mole balance on a slice of the pore with thickness Ax can be written as: (Rate of input A) -(Rate of output A) + (Rate of generation A) = 0 1TR~oreNAlx 1TR~oreNAlxHx - ksCA (21TRpore)(Ax) = 0 (6.3.3) (6.3.4) where NA is the flux of A evaluated at both sides of the slice, ks is the first-order rate constant expressed per surface area of the catalyst (volume/ {surface area}/time), and 21TRpore(AX) is the area of the pore wall in the catalyst slice. Rearranging CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 193 t'Rpore o x L Figure 6.3.3 I Schematic representation of component A diffusing and reacting in an idealized cylindrical pore.
Equation (6.3.4) and taking the limit as Ilx approaches zero gives the following differential equation for the mole balance: (6.3.5) Recall that the Stefan-Maxwell equation relates the molar flux of A to its concen-tration gradient according to: For diffusion in one dimension in the absence of bulk flow: ( 1 1 )-1 DAB + D KA which is Pick's First Law that can be written as: dCA N --D -A -TA dx (6.3.6) (6.3.7) (6.3.8) where D TA is defined by Equation (6.3.2). Substitution of Fick's Law into the mole balance, Equation (6.3.5), yields the following second-order differential equation: (6.3.9) 194 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Assuming D TA is constant: (6.3.10) The surface rate constant can be rewritten on a volume basis by using the surface to volume ratio of a cylindrical pore: Area Volume and To simplify the mole balance, let: 27TRporeL 2 7TR~oreL R pore 2ks k=-R pore (6.3.11) (6.3.12) x x== L -j;f 1J =L -D TA and their substitution into Equation (6.3.10) gives: d 2C __ A _ A,2C = 0 dX 2 'f' A (6.3.13) (6.3.14) (6.3.15) Boundary conditions at each end of the pore are needed to solve the mole balance.
At x = 0 (the pore mouth), the concentration of A is equal to CAS' At the other end of the pore, the gradient in concentration is equal to zero. That is, there is no flux at the end of the pore. These conditions can be written as: CA = CAS dCA -=0 dX atx = 0 at X = 1 (6.3.16) (6.3.17) The solution of Equation (6.3.15) using boundary conditions given in Equations (6.3.16) and (6.3.17) is: (6.3.18) The term 1J, also known as the Thiele modulus, is a dimensionless number com-posed of the square root of the characteristic reaction rate (kCAS) divided by the characteristic diffusion rate (DI~AS)' The Thiele modulus indicates which process CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 0.8 0.6 C/o ~~ cJ 0.4 0.2 x/[ Figure 6.3.4 I Effect of Thiele modulus on the normalized concentration profiles in a catalyst pore with first-order surface reaction.
195 is rate-limiting. Figure 6.3.4 illustrates the concentration profile of reactant A along the pore for various values of the Thiele modulus. When cP is small, the diffusional resistance is insufficient to limit the rate of reaction and the concentration can be maintained near CAS within the catalyst particle. However, when cP is large, a sig-nificant diffusional resistance prevents a constant concentration profile of A within the catalyst particle and thus lowers the observed rate.
Now consider a catalyst pellet with a random network of "zig-zag" pores.
The surface of the pellet is composed of both solid material and pores. The flux equation derived earlier must be modified to account for the fact that the flux, NA , is based only on the area of a pore. A parameter called the porosity of the pellet, or Bp , is defined as the ratio of void volume within the pellet to the to-tal pellet volume (void + solid). The flux can be expressed in moles of A diffusing per unit pellet surface area (containing both solids and pores) by using 8p as fo11o\vs: Since the porosity of many solid catalysts falls between 0.3 and 0.7, a reasonable es-timate of Bp in the absence of experimental data is 0.5. The second parameter needed to modify the flux is the tortuosity, T, which accounts for the deviations in the path length of the pores. Since the concentration gradient is based on the pellet geometry, the flux equation must be corrected to reflect the actual distance molecules travel in the pores. The tortuosity is the ratio of the "zig-zag length" to the "straight length" of the pore system. Obviously, T must be greater than or equal to one. For example, an 196 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions ideal, cylindrical pore has T equal to 1, and a network of randomly oriented cylindri-cal pores has T equal to approximately 3. The flux equation can be written to take into account the "true" diffusion path length as: Since the tortuosity of many solid catalysts falls between 2 and 7, a reasonable es-timate of T in the absence of experimental data is 4. The diffusivity in a unimodal pore system can now be defined by the flux of reactant into the pellet according to: (6.3.19) where the superscript e refers to the efiective diffusivity. Likewise, the following effective diffusivities can be written: DAB = - DAB T (6.3.20) (6.3.21) Now consider several ideal geometries of porous catalyst pellets shown in Figure 6.3.5. The first pellet is an infinite slab with thickness 2xp- However, since I Xp I (a) (b) (c) Figure 6.3.5 I Schematic representations of ideal catalyst pellet geometries. (a) Infinite slab. (b) Infinite right cylinder. (c) Sphere.
CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Spherical shell Catalyst sphere Figure 6.3.6 I Schematic of the shell balance on a spherical catalyst pellet.
197 pores can have openings on both faces of the slab because of symmetry, the char-acteristic length associated with the slab is half of the thickness, or xp- The sec-ond pellet is an infinite right cylinder with radius Rp , and the third pellet is a sphere with radius Rp • End effects (or edge effects) are ignored in the cases of the slab and the cylinder.
As an example, simultaneous diffusion and reaction in a spherical catalyst pel-let is described in detail below. The results are then generalized to other pellet shapes.
The reaction is assumed to be isothermal, since the high thermal conductivity of most solid catalysts ensures a fairly constant temperature within a single pellet. In addition, the reaction is assumed to be isobaric, which implies negligible mole change upon reaction. For reactions with a significant mole change with conversion, the presence of a large excess of inert material can reduce the impact of reacting species on the total pressure. Isobaric conditions can therefore be achieved in a va-riety of catalytic reactions, regardless of reaction stoichiometry.
A diagram of the shell balance (material balance) for simultaneous diffusion and first order reaction of component A in a sphere is shown in Figure 6.3.6. The material balance in the spherical shell is given by: (Rate of input A) (Rate of output A) + (Rate of generation A) = 0 41Tr 2NA lr 41Tr 2NAlrHr - kCA 41Tr 2Ar = 0 (6.3.22) The third term in Equation (6.3.22) is the rate of consumption of A in the differen-tial volume defined between rand r + Ar. Simplifying Equation (6.3.22) and tak-ing the limit as Ar approaches zero yields the following differential equation: (6.3.23) The flux of A can be expressed in terms of concentration for binary systems ac-cording to Fick's Law (in spherical coordinates): (equimolar counterdiffusion) (6.3.24) 198 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Substitution of Equation (6.3.24) into (6.3.23) gives: dr De [zdZCA + T dCA ] TA r dr Z r dr DTA[dZ~: + ~ dd~A] kCA = 0 dr r r (6.3.25) The above equation can be made dimensionless by the following substitutions: (6.3.26) (6.3.27) r w=-Rp where CAS is the concentration of A on the external surface and Rp is the radius of the spherical particle. Rewriting Equation (6.3.25) in terms of dimensionless con-centration and radius gives: The Thiele modulus, ¢, for a sphere is defined as: ¢ = Rp ) D~A so that Equation (6.3.28) becomes: dZljJ 2 dljJ Z dwz + -; dw - ¢ ljJ = 0 with boundary conditions: (6.3.28) (6.3.29) (6.3.30) ljJ= dljJ = 0 dw at w = 1 (surface of sphere) at w = 0 (center of sphere) (6.3.31) (6.3.32) The zero-flux condition at the center results from the symmetry associated with the spherical geometry. The solution of the above differential equation with the stated boundary conditions is: sinh(¢w) wsinh(¢) (6.3.33) CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 0.8 vo 0.6 s1 cJ 0.4 0.2 0 0 0.2 0.4 0.6 0.8 Figure 6.3.7 I Effect of Thiele modulus on the nonnalized concentration profiles in a spherical catalyst particle with first-order reaction. The external surface of the particle is located at r/Rp = 1.
199 Figure 6.3.7 illustrates the effect of the Thiele modulus on the concentration profile within a spherical catalyst pellet.
An effectiveness factor, 1], can be defined as the ratio of the observed rate (robs) to the rate that would be observed in the absence of internal diffusional limitations (rmax); fPr (C A )41Tr 2dr o SpfPr(cA)(ItYdr Vpr(CAS) (6.3.34) where Vp is the volume of the catalyst pellet, Sp is the external surface area of the pellet, and r(CA) is the reaction rate determined '"'lith concentration CA' The de-nominator in Equation (6.3.34) is simply the rate of reaction in the catalyst pellet assuming the reactant concentration is equal to that on the external surface, CAS' For a first-order reaction in a spherical particle, the rate observed in the absence of diffusional limitations is; (6.3.35) At the steady state, the flux of A entering the pellet must be equivalent to the net rate of consumption of A in the pellet. Thus, the flux entering the sphere can be 200 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions used to determine the observed rate of reaction in the presence of diffusional limitations.
(6.3.36) (6.3.37) 3 robs TJ=-= f max Substituting Equations (6.3.35) and (6.3.36) into (6.3.34) gives: o dCA I 47T(Rp)-D'TA~ r=R p 4 3 "37T (Rp) kCAS The above equation is made nondimensional by the substitutions defined in Equa-tions (6.3.26) and (6.3.27) and is: TJ = ;2 ~~ Iw=I (6.3.38) The derivative is evaluated from the concentration profile, (6.3.33), to give: (6.3.39) 3 dl{! I 3 d [Sinh (4>w)] I TJ = 4>2 dw w = I = 4>2 dw w sinh (4>) w = I TJ = 32[ . \ ) (~COSh(4>W) ~ sinh(4)w))] I 4> smh 4> w w w= I 3[ 4> ] 3 1 1] TJ = 4>2 tanh (4» - 1 = 4> tanh (4» - -;;; Figure 6.3.8 illustrates the relationship between the effectiveness factor and the Thiele modulus for a spherical catalyst pellet.
0.1 0.1 10 100 Figure 6.3.8 I Effectiveness factor for a first-order reaction in a sphere as a function of the Thiele modulus.
CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Table 6.3.1 I Influence of catalyst particle geometry on concentration profile and effectiveness factor for a first-order, isothermal, isobaric reaction.
201 r-;;-Ik Ik c/J x 1-Rp'V Dh Rp-V Dh P'V Dh l/J = CA/CAS cosh (c/Jw) Io(c/Jw) sinh(c/Jw) cosh(c/J) Io(c/J) wsinh(c/J) tanh(c/J) 2I1(c/J) 3 1 1 ] 1] c/J c/JIo(c/J) c/J tanh(c/J) c/J (aJIi is a modified Bessel function of order i.
Table 6.3.2 I Characteristic length parameters of common pellet shapes.
Slab Cylinder Sphere A low value of the Thiele modulus results from a small diffusional resistance.
For this case, the effectiveness factor is approximately 1 (values of 4> typically less than 1). Large values of the Thiele modulus are characteristic of a diffusion-limited reaction with an effectiveness factor less than 1. For 4> » 1, the value of the ef-fectiveness factor in a sphere approaches 3/4>, as illustrated in Figure 6.3.8.
The concentration profile and the effectiveness factor are clearly dependent on the geometry of a catalyst particle. Table 6.3.1 summarizes the results for catalyst particles with three common geometries.
Aris was the first to point out that the results for the effectiveness factor in dif-ferent pellet geometries can be approximated by a single function of the Thiele mod-ulus if the length parameter in 4> is the ratio of the pellet volume, Vp , to the pellet external surface area, Sp [R. Aris, Chern. Eng. Sci., 6 (1957) 262]. Thus, the length parameter, Lp , is defined by: (6.3.40) and the Thiele modulus is defined by: .~-1 k L j--1'\; De V TA (6.3.41) where Table 6.3.2 summarizes the characteristic length parameter for common geometries.
202 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 0.1 0.1 1 0 10 Figure 6.3.9 I Effectiveness factor [T/ = tanh(4>0)/4>0] for a first-order reaction in a catalyst as a function of the Thiele modulus with generalized length parameter.
According to the above definitions, the effectiveness factor for any of the above shapes can adequately describe simultaneous reaction and diffusion in a catalyst par-ticle. The equation for the effectiveness factor in a slab is the simplest in Table 6.3.1 and will be used for all pellet shapes with the appropriate Thiele modulus: 77= (6.3.42) EXAMPLE 6.3.1 I This relationship is plotted in Figure 6.3.9. The effectiveness factor for a severely diffusion-limited reaction in a catalyst particle is approximated by the inverse of the Thiele modulus.
The double bond isomerization of I-hexene to form 2-hexene was studied in a laboratory reactor containing rhodium particles supported on alumina at 150°C and atmospheric pressure: The reaction was found to be first order in I-hexene with a rate constant of 0.14 S-I. Find the largest pellet size that can be used in an industrial reactor to achieve 70 percent of the maximum rate. The pore radius of the alumina is 10 nm, and DAB is 0.050 cm2 s-1 .
• Answer It is desired to find the particle size that gives an internal effectiveness factor equal to 0.70.
For any geometry, the Thiele modulus is determined from: CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions tanh (0 4 I 6 I I 8 10 Figure 6.3.10 I Effectiveness factors for sphere, infinite cylinder, and finite cylinder pellet geometries where the Thiele modulus is based on equal Vp/Sp. Individual points correspond to the numerical solutions of the material balance on a finite cylinder.
(Cumene) 206 EXAMPLE 6.3.2 I CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions length in the Thiele modulus by (VI' / Sp) is an excellent approximation to the true solution.
The rate constant for the first-order cracking of cumene on a silica-alumina catalyst was mea-sured to be 0.80 cm3j(s'gcat) in a laboratory reactor: o =a=tm=>0 + CH3CHCH2 (Benzene) (Propylene) Is the observed rate constant the true rate constant or is there influence of pore diffusion?
Additional data: Rp = 0.25 cm p 1.2gcatcm-3 D TA = 1.0 X 10-3 cm2 S-1 • Answer Recall that the effectiveness factor can be approximated by: tanh(cPo) 71= cPo when the Thiele modulus is defined in terms of the characteristic length of a pellet: cPo =L p& For the spherical particles in this problem: / ( -) ( ) 'cm" gcat ((0.25)(cm)\ Ik ~ . 1.2 ~ _. r: \ )V'~ e -'/ '~•• / = L,9Vk \ 3 (cm") 1.0 X 10-3 -s-Since the observed rate is first-order: and, by definition, the effectiveness factor is: observed rate 71 = rate in absence of diffusion kobsC,s kCAS CHAPTER 6 Effects of Transport I imitations on Rates of SoW!lidU::\..C LC allita;U'IYYJ.7:.t;ecud-rR::u:8iCalLcciltiownJ.;:sL----'2,..OLL 7 Substitute the Thiele modulus into the expression for the effectiveness factor to solve for k, the true rate constant, by trial and error: tanh (1>0) tanh (2.9vk) k obs 0.80 TJ 1>0 2.9vk k k cm 3 k 5.4 TJ = 0.15 s' gcat Since TJ is small, there is a great influence of pore diffusion on the observed rate.
The material balance for simultaneous reaction and diffusion in a catalyst pel-let can be extended to include more complex reactions. For example, the general-ized Thiele modulus for an irreversible reaction of order n is: n + 1 kC~Sl 2 Dh n> (6.3.51) The generalized modulus defined in Equation (6.3.51) has been normalized so that the effectiveness factor is approximately 1/4>0 at large values of 1Jo, as illustrated in Figure 6.3.9.
The implications of severe diffusional resistance on observed reaction kinetics can be determined by simple analysis of this more general Thiele modulus. The ob-served rate of reaction can be written in terms of the intrinsic rate expression and the effectiveness factor as: robs = YJkC~s (6.3.52) As discussed earlier, the effectiveness factor is simply the inverse of the Thiele mod-ulus for the case of severe diffusional limitations (Figure 6.3.9.) Thus, the observed rate under strong diffusional limitations can be written as: 1 robs = 1Jo kC~s (6.3.53) (6.3.54) Substitution of the generalized Thiele modulus, Equation (6.3.51), into (6.3.53) gives the following expression for the observed rate: s (2 )~ -P De k c!n+ robs -V p n + 1 TA AS The order of reaction observed under conditions of severe diffusional limitations, nabs' becomes (n + 1)/2 instead of the true reaction order n. The temperature de-pendence of the rate is also affected by diffusional limitations. Since the ob-served rate constant, kobs, is proportional to (D~Ak)2, the observed activation energy is (ED E)/2, where ED is the activation energy for diffusion and E is the activation en-ergy for reaction. Diffusional processes are weakly activated compared to chemical 208 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions J .s EXAMPLE 6.3.3 I If[ Figure 6.3.11 I Temperature dependence of the observed rate constant of a reaction occurring in a porous catalyst pellet.
reactions, and the value of ED can often be neglected compared to E. Thus, the ob-served activation energy for a severely diffusion-limited reaction is approximately one half the true value. An Arrhenius plot of the observed rate constant, shown in Figure 6.3.11, illustrates the effect of diffusional resistances on the observed activation en-ergy. At low temperatures, the reaction rate is not limited by diffusional resistances, and the observed activation energy is the true value. At high temperatures, the reac-tion rate is inhibited by diffusional resistances, and the activation energy is half the true value.
Develop expressions for the Thiele modulus and the concentration profile of A for the fol-lowing reversible first-order reaction that takes place in a flat plate catalyst pellet: A = B, K = L j Catalyst slab .""'--x=o CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions • Answer The material balance for diffusion/reaction in one dimension is given by: 209 (6.3.55) With C representing the total concentration, CA + CB, the rate expression can be rewritten in terms of C and CA , thus eliminating the explicit dependence on Cs: ( 1) k l (K+l)( C) C)=k 1+- C --C=k --C ---A I K A K I K A K+l Assuming D~A is constant, the following equation can be solved for the concentration profile: with boundary conditions: K~ J (6.3.56) dCA -=0 d.x at x = xp (external surface of the slab) at x = 0 (center line, point of symmetry) (6.3.57) (6.3.58) The following change ofvariables facilitates solution ofthe problem. Let ~ = CA -C/ (K + 1) and X = x/xp so that the material balance can be written as: (6.3.59) By expressing the material balance in this form, the Thiele modulus appears as the dimen-sionless constant c/J: (6.3.60) The general solution of Equation (6.3.59), \vith arbitra..ry constants al a..Tld a2' is: (6.3.61) The boundary conditions expressed in terms of the variables that are used to evaluate the con-stants (Xl and 0'2 are: C (6.3.62) l/J = CAS K + 1 at X = 1 dl/J at X = 0 (6.3.63) -=0 dX 210 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions The constant al vanishes due to the second boundary condition [Equation (6.3.63)]: dl/J = alcPcosh (O) + azcPsinh(O) = 0 dX and cosh(O) = 1 so The first boundary condition is used to evaluate az and thus completes the solution of the problem as shown below: EXAMPLE 6.3.4 I (6.3.64) (6.3.65) (6.3.60) Set up the equations necessary to calculate the effectiveness factor for a flat-plate catalyst pellet in which the following isothermal reaction takes place: Benzene Cyclohexane CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 211 • Answer To solve this problem, the Stefan-Maxwell relations for molecular diffusion in a multicom-ponent gas mixture (see Appendix C for details) should be used: For the case of diffusion in one dimension within a porous medium, these equations yield the following expression, which is derived in Appendix C: (6.3.66) The above equation reduces to a familiar form for two components if equimolar counterdif-fusion of A and B (NA -NB) at constant temperature is assumed: 1 dPA -----R~T dx (6.3.67) Recall this equation is similar to Equation (6.3.7) for the flux of A in one dimension. To solve the multicomponent diffusion/reaction problem of benzene hydrogenation in one dimension, Equation (6.3.66) must instead be used. First, let: Benzene = component 1 Dihydrogen component 2 Cyclohexane = component 3 The following diffusivities (Knudsen and binary) need to be determined from tabulated data, handbooks, correlations, theoretical equations, etc.: Benzene: DKb D12, DB The porosity (ep) and tortuosity (1') of the flat plate catalyst pellet are then used to calculate the effective diffusivities associated with each component according to: 8p Dk, = =DKi T De IJ (6.3.68) (6.3.69) 212 C H APT E B 6 Effects of Transport I Imitations on Bates of Solid-Catalyzed Reactions From the stoichiometry of the hydrogenation reaction, the ratios of the fluxes of the compo-nents are: 3 and Substitution of these relations into Equation (6.3.66) gives the appropriate flux equations: (6.3.70) (6.3.71) Finally, the material balance on a slice of the catalyst pellet that is needed to completely spec-ify the reaction/diffusion problem is: (6.3.72) Rate (Xl> X2) is the rate expression for benzene hydrogenation that depends on XI and X2.
For example, the following rate equation could be used if the constants Cl'A and Cl'B were known at the reaction temperature: (6.3.73) (6.3.74) The three equations representing the material balance and the flux relations can be solved si-multaneously to determine the dependent variables (Xl> X2• and N I ) as a function of the inde-pendent variable x. (Recall that X3 can be expressed in terms ofXl and X2: I = Xl + X2 + X3.) The boundary conditions for these equations are: at x Xs (external surface of pellet) at x = 0 (center line of the slab) To calculate the effectiveness factor, the actual reaction rate throughout the catalyst is divided by the rate determined at the conditions of the external surface. that is, Rate(X IS. X2S)' The overall reaction rate throughout the particle is equivalent to the flux N I evaluated at the ex-ternal surface of the catalyst. Thus, the final solution is: N) Cp) 71 = Rate (XIS. X2S)' VI' The temperature profile in the catalyst pellet can be easily incorporated into this solution by including the energy balance in the system of equations.
The previous discussion focused on simultaneous diffusion and reaction in isothermal catalyst pellets. Since !1H, is significant for many industrially relevant reactions, it is necessary to address how heat transfer might affect solid-catalyzed CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 213 reactions. For isothennal catalyst pellets, the effectiveness factor is less than or equal to unity, as illustrated in Figure 6.3.9. This is rationalized by examining the tenns that comprise the effectiveness factor for the reaction of A in a flat plate catalyst pellet: (6.3.75) For an isothennal pellet: (6.3.76) If, as stated in Chapter 1, the rate is separable into two parts, one dependent on the temperature and the other dependent on the concentrations of reacting species, that is, (6.3.77) then the concentration of A inside the pellet is less than that on the external surface, (6.3.78) for reactions with nonnegative reaction orders. Thus, (6.3.79) and explains the upper limit of unity for the isothennal effectiveness factor in a cat-alyst pellet. The situation can be very different for a nonisothennal pellet. For ex-ample, the temperature dependence of the reaction rate constant, k(T), is generally expressed in an Arrhenius fonn k(T) = Aexp( - E) RgT (6.3.80) For an endothennic reaction in the presence of significant heat-transfer resistance, the temperature at the surface of the pellet can exceed the temperature of the inte-rior, which according to Equation (6.3.80), gives: (6.3.81) Since F(CA) :5 F(CAS) as discussed above, the effectiveness factor is always less than or equal to unity for an endothermic reaction. For an exothennic reaction, the opposite situation can occur. The temperature of the interior of the particle can ex-ceed the surface temperature, T > Ts, which leads to: k(T) 2: k(Ts) (6.3.82) 214 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions Recall that k(T) is a strong function of temperature. The effectiveness factor for an exothermic reaction can be less than, equal to, or greater than unity, depending on how k(T) increases relative to P(CA ) within the particle. Thus, there are cases where the increase in k(T) can be much larger than the decrease in P(CA)' for example, k(T) . P(CA ) > k(Ts)' P(CAS) To evaluate the effectiveness factor for a first-order, isobaric, nonisothermal, flat plate catalyst pellet, the material and energy balances must be solved simulta-neously. As shown previously, the mole balance in a slab is given by: dNA -= -k(T) 'CA dx where the rate constant is of the Arrhenius form: k(T) = Aexp( - E) RgT The flux of A can be written in terms of Fick's Law: and substituted into the mole balance to give: (6.3.83) (6.3.84) (6.3.85) (6.3.86) The energy balance is written in the same manner as the mole balance to give: (6.3.87) where the flux is expressed in terms of the effective thermal conductivity of the fluid-solid system, A e, and the gradient in temperature: dT a = -Ae -, .. dx (6.3.88) (6.3.89) The heat of reaction, Li H,., is defined to be negative for exothermic reactions and pos-itive for endothermic reactions. Substitution of Equation (6.3.88) into (6.3.87) results in: d 2T A e dx 2 = (-LiHr) 'k(T) 'CA To render the material and energy balances dimensionless, let: (6.3.90) CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 215 (6.3.91) (6.3.92) (6.3.94) r=~ Ts The rate constant k(T) is expressed in terms of Ts by first forming the ratio: :&!J ~ exp[ - : G -;J] ~ exp[-'(f-I)] (6.3.93) where: E y=-RgTs The dimensionless group y is known as the Arrhenius number. Substitution of the dimensionless variables into the material and energy balances gives: ~~ = [(Xp)~~(Ts)] . exp[-Y(f- 1)] . if/ (6.3.95) ~~ = _ [(Xp )2. k(Ts)~e~~ LlHJ CAS] . exp[-Y(f- 1)] . if/ (6.3.96) Both equations can be expressed in terms of the Thiele modulus, 4>, according to: (6.3.97) (6.3.98) A new dimensionless grouping called the Prater number, f3, appears in the energy balance: (- LlHr ) • D TA . CAS f3 = AeTs Thus, the energy balance is rewritten: (6.3.99) (6.3.100) The material and energy balances are then solved simultaneously with the follow-ing boundary conditions: dif/ dr -=-=0 dX dX if/=r=l atx = 0 at X = I (6.3.101) (6.3.102) 216 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 1000.0 500.0 100.0 50.0 10.0 5.0 "'" 1.0 0.5 0.1 0.05 0.01 0.005 0.001 -t--r-->--r"f"'CTT 0.01 0.05 0.1 0.5 1.0 2'J! exp[ -Y(¥ - 1)] ~:~ = ~4>2'J! exp[ -Y(¥ - 1)] (6.4.14) (6.4.15) 4>2 = (xp?· k(Ts) D'fA Notice that aU of the parameters are based on bulk fluid values of the concentration and temperature. The boundary conditions are: d'J!
df -=-=0 atX = 0 (6.4.16) dX dX d'J!
dX = Bim(l - 'J!) atx = 1 (6.4.17) dr -= Bih(l f) at X = 1 (6.4.18) dx In general, solution of these equations requires a numerical approach.
EXAMPLE 6.4.1 I Find an expression for the overall effectiveness factor of a first-order isothermal reaction in a flat plate catalyst pellet.
• Answer Since the reaction is isothennal, the energy balance can be ignored and the mass balance re-duces to: (6.4.19) 222 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions with boundary conditions: d'IJI -=0 dX d'IJI dX = Bim(1 -'IJI) at X = 0 at X = 1 (6.4.20) (6.4.21) As discussed previously, the general solution of the differential equation is: (6.4.22) The constants are evaluated by using the appropriate boundary conditions. At the center of the pellet: (6.4.23) (6.4.24) Substitution of Equation (6.4.24) into (6.4.22) eliminates one of the integration constants: 'IJI = ajexp(c/>x) + ajexp(-c/>X) _ exp(c/>x) + exp(-c/>X) 'IJI = 2aj 2 (6.4.25) (6.4.26) (6.4.27) (6.4.28) The boundary condition at the external surface provides another relation for d'IJI/dX: (6.4.29) Equating Equations (6.4.28) and (6.4.29), at X = 1, enables the determination of aj: and therefore, Bim aj = -------------2[c/>sinh(c/» + Bimcosh(c/»J (6.4.30) (6.4.31) (6.4.32) Since the concentration profile is determined by Equation (6.4.32), evaluation of the over-all effectiveness factor, TJO' is straightforward. By definition, TJo is the observed rate di-vided by the rate that would be observed at conditions found in the bulk fluid. Recall C H A PT E R 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 223 that the observed rate must equal the flux of A at the surface of the pellet at the steady state: '10 (6.4.33) tanh (4:» '10 = 4:>[1 + _4:>_tan_h-,--( 4:>-,--)] Bim (6.4.34) The overall effectiveness factor is actually comprised of the individual effec-tiveness factors for intraphase and interphase transport: 1]0 = 1]intraphase • 1]interphase = 1] . 1] (6.4.35) For example, an isothermal, first-order reaction in a flat plate catalyst pellet has in-dividual effectiveness factors that are: tanh(1)) 1]= 1> 17 = [I + 1> tanh(1>)] -1 Bim (6.4.36) (6.4.37) VIGNETTE 6.4.2 Common ranges of diffusivities, thermal conductivities, mass transfer coeffi-cients, heat transfer coefficients, and catalyst pore sizes can be used to estimate the relative magnitude of artifacts in kinetic data obtained in industrial reactors. For gas-solid heterogeneous systems, the high thermal conductivity of solids compared to gases suggests that the temperature gradient in the film surrounding the catalyst par-ticle is likely to be greater than the temperature gradient in the particle. Since the Knudsen diffusivity of gaseous molecules in a small pore of a catalyst particle is much lower than the molecular diffusivity in the stagnant film, intraphase gradients in mass are likely to be much greater than interphase gradients. For liquid-solid heterogeneous systems, internal temperature gradients are often encountered. A typ-ical range of Bim/Bih is from 10 to 104 for gas-solid systems and from 10-4 to 10- 1 for liquid solid systems.
224 CHAPTER 6 Effects of Transport I imitations on Bates of Solid CatalyzedBeactions Reactor Flow ===i> CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions at 0 225 (6.4.38) with the following atx Lc is the half of the macrocavity) 226 CHAPTER 6 Effects of Transport I imitatiolliUlliBates of Solid-Catalyzed Reactions CHAPTER 6 Effects of Transport I imitations on Rates oLSuoilllidJ.:-"-C,",a1LtaClJIL¥YLze,",d-.u::Rll:e,,,al.l..cclltiuoilns,,-~~~~~~~---, 2 ...
2 oc 7 L ious temperatures. Solid curves were obtained diffusion and reaction. Watanabe and 573-723 K on the bottom of 1.2 /km on simultaneous rate of the film at trenches was less than that on the surface for 228 CHAPTER 6 Effects of Transport I imitations on Rates of Solid:Ca1alyzed Reaction~ (6.5.1) (6.5.2) 6.5 I Analysis of Rate Data To arrive at a rate expression that describes intrinsic reaction kinetics and is suit-able for engineering design calculations, one must be assured that the kinetic data are free from artifacts that mask intrinsic rates. A variety of criteria have been pro-posed to guide kinetic analysis and these are thoroughly discussed by Mears [D. E.
Mears, Ind. Eng. Chern. Process Des. Develop., 10 (1971) 541].
A lack ofsignificant intraphase diffusion effects (i.e., YJ 2': 0.95) on an irreversible, isothermal, first-order reaction in a spherical catalyst pellet can be assessed by the Weisz-Prater criterion [P. B. Weisz and C. D. Prater, Adv. Catal., 6 (1954) 143]: robs (Rp)2 ----'-- < 1 DhCAs where robs is the observed reaction rate per unit volume and Rp is the radius of a catalyst particle. An important aspect of this criterion is that it uses the observed rate and the reactant concentration at the external surface. The intrinsic rate and the concentration profile inside the pellet are not needed. For power law kinetics where n is the reaction order (other than 0), the following expression can be used: robs (Rp)2 1 ----'-- < -DTACAS n The influence of mass transfer through the film surrounding a spherical cata-lyst particle can also be examined with a similar expression. Satisfaction of the following inequality demonstrates that interphase mass transfer is not significantly affecting the measured rate: (6.5.3) where k c is the mass transfer coefficient and the reactant concentration is determined in bulk fluid. The above relationship is analogous to the modified Weisz-Prater cri-terion with ke replacing Dh/Rp.
Criteria have also been developed for evaluating the importance of intraphase and interphase heat transfer on a catalytic reaction. The Anderson criterion for es-timating the significance of intraphase temperature gradients is [J. B. Anderson, Chern. Eng. Sci., 18 (1963) 147]: (6.5.4) C H A PT E R 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 229 where ,\e is the effective thermal conductivity of the particle and E is the true acti-vation energy. Satisfying the above criterion guarantees that robs does not differ from the rate at constant temperature by more than 5 percent. Equation (6.5.4) is valid whether or not diffusional limitations exist in the catalyst particle. An analogous cri-terion for the lack of interphase temperature gradients has been proposed by Mears [D. E. Mears, 1. Catal., 20 (1971) 127]: (6.5.5) where ht is the heat transfer coefficient and TB refers to the bulk fluid temperature.
The Mears criterion is similar to the Anderson criterion with ht replacing ,\e/Rp- In addition, the Mears criterion is also valid in the presence of transport limitations in the catalyst particle.
While the above criteria are useful for diagnosing the effects of transport limitations on reaction rates of heterogeneous catalytic reactions, they require knowledge of many physical characteristics of the reacting system. Experimen-tal properties like effective diffusivity in catalyst pores, heat and mass transfer coefficients at the fluid-particle interface, and the thermal conductivity of the cat-alyst are needed to utilize Equations (6.5.1) through (6.5.5). However, it is dif-ficult to obtain accurate values of those critical parameters. For example, the diffusional characteristics of a catalyst may vary throughout a pellet because of the compression procedures used to form the final catalyst pellets. The accuracy of the heat transfer coefficient obtained from known correlations is also ques-tionable because of the low flow rates and small particle sizes typically used in laboratory packed bed reactors.
Madon and Boudart propose a simple experimental criterion for the absence of artifacts in the measurement of rates of heterogeneous catalytic reactions [R. J.
Madon and M. Boudart, Ind. Eng. Chern. Fundarn., 21 (1982) 438]. The experiment involves making rate measurements on catalysts in which the concentration of ac-tive material has been purposely changed. In the absence of artifacts from transport limitations, the reaction rate is directly proportional to the concentration of active material. In other words, the intrinsic turnover frequency should be independent of the concentration of active material in a catalyst. One way of varying the concen-tration of active material in a catalyst pellet is to mix inert particles together with active catalyst particles and then pelletize the mixture. Of course, the diffusional characteristics of the inert particles must be the same as the catalyst particles, and the initial particles in the mixture must be much smaller than the final pellet size.
If the diluted catalyst pellets contain 50 percent inert powder, then the observed reaction rate should be 50 percent of the rate observed over the undiluted pellets.
An intriguing aspect of this experiment is that measurement of the number of ac-tive catalytic sites is not involved with this test. However, care should be exercised when the dilution method is used with catalysts having a bimodal pore size distri-bution. Internal diffusion in the micropores may be important for both the diluted and undiluted catalysts.
230 CHAPTER 6 Effects of Transport I imitations on Bates of Solid-Catalyzed Beaction~ Another way to change concentration of active material is to modify the cata-lyst loading on an inert support. For example, the number of supported transition metal particles on a microporous support like alumina or silica can easily be varied during catalyst preparation. As discussed in the previous chapter, selective chemisorption of small molecules like dihydrogen, dioxygen, or carbon monoxide can be used to measure the fraction of exposed metal atoms, or dispersion. If the turnover frequency is independent of metal loading on catalysts with identical metal dispersion, then the observed rate is free of artifacts from transport limitations. The metal particles on the support need to be the same size on the different catalysts to ensure that any observed differences in rate are attributable to transport phenomena instead of structure sensitivity of the reaction.
A minor complication arises when dealing with exothermic reactions, since the effectiveness factor for a catalyst pellet experiencing transport limitations can still equal one. To eliminate any ambiguity associated with this rare condition, the Madon-Boudart criterion for an exothermic reaction should be repeated at a differ-ent temperature.
The simplicity and general utility of the Madon-Boudart criterion make it one of the most important experimental tests to confirm that kinetic data are free from artifacts. It can be used for heterogeneous catalytic reactions carried out in batch, continuous stirred tank, and tubular plug flow reactors.
Development of rate expressions and evaluation of kinetic parameters require rate measurements free from artifacts attributable to transport phenomena. Assum-ing that experimental conditions are adjusted to meet the above-mentioned criteria for the lack of transport influences on reaction rates, rate data can be used to pos-tulate a kinetic mechanism for a particular catalytic reaction.
CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 3.5 3 ::::--'::: 2.5 '8 N I6 2 "0 6 '0 1.5 x ~ B '" 0:: 0.5 500 1000 1500 2000 2500 3000 3500 4000 Stirring speed (rpm) Figure 6.5.1 I Effect of agitation on the rate of 2-propanol dehydrogenation to acetone at 355 Kover Ni catalysts.
[Rates are calculated at constant conversion level from the data in D. E. Mears and M. Boudart, AIChE 1., 12 (1966) 313.] In this case, increasing the stirring speed increased the rate of acetone diffusion away from the catalyst pellet and decreased product inhibition.
231 If mass and heat transfer problems are encountered in a catalytic reaction, various strategies are employed to minimize their effects on observed rates. For example, the mass transfer coefficient for diffusion through the stagnant film sur-rounding a catalyst pellet is directly related to the fluid velocity and the diameter of the pellet according to Equation (6.2.26). When reactions are not mass trans-fer limited, the observed rate will be independent of process variables that affect the fluid velocity around the catalyst pellets. Conversely, interphase transport lim-itations are indicated if the observed rate is a function of fluid flow. Consider the results illustrated in Figure 6.5.1 for the dehydrogenation of 2-propanol to ace-tone over powdered nickel catalyst in a stirred reactor. The dependence of the rate on stirring speed indicates that mass transfer limitations are important for stirring speeds less than 3600 rpm. Additional experiments with different surface area cat-alysts confirmed that rates measured at the highest stirring speed were essentially free of mass transfer limitations [D. E. Mears and M. Boudart, AIChE J., 12 (1966) 313].
Both interphase and intraphase mass transfer limitations are minimized by decreasing the pellet size of the catalyst. Since a packed bed of very small catalyst particles can cause an unacceptably large pressure drop in a reactor, a compro-mise between pressure drop and transport limitations is often required in commercial reactors. Fortunately, laboratory reactors that are used to obtain 232 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 10 8 ~ ~ '2 ::> 6 Q IE ~ 4 ~ '" r::>:: 2 Pellet diameter (mm) Figure 6.5.2 I Schematic illustration of the influence of catalyst pellet size on the observed reaction rate.
intrinsic reaction kinetics require relatively small amounts of catalyst and can be loaded with very small particles. For the case presented in Figure 6.5.2, observed rates measured on catalyst pellets larger than I mm are affected by transport limitations.
Exercises for Chapter 6 1.
The isothermal, first-order reaction of gaseous A occurs within the pores of a spherical catalyst pellet. The reactant concentration halfway between the external surface and the center of the pellet is equal to one-fourth the concentration at the external surface.
(a) What is the relative concentration of A near the center of the pellet?
(b) By what fraction should the pellet diameter be reduced to give an effectiveness factor of 0.77 2.
The isothermal, reversible, first-order reaction A = B occurs in a flat plate catalyst pellet. Plot the dimensionless concentration of A (CA/ CAS) as a function of distance into the pellet for various values of the Thiele modulus and the equilibrium constant. To simplify the solution, let CAS = 0.9(CA + CB) for all cases.
3.
A second-order. irreversible reaction with rate constant k = 1.8 L mol- 1 s- J takes place in a catalyst particle that can be considered to be a one-dimensional slab of half width I em.
CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 233 lern The concentration ofreactant in the gas phase is 0.1 mol L- 1 and at the surface is 0.098 mol L- I. The gas-phase mass-transfer coefficient is 2 cm s-1. Determine the intraphase effectiveness factor. (Contributed by Prof. J. L. Hudson, Univ.
of Virginia.) 4.
Consider the combustion of a coal particle occurring in a controlled burner.
Assume the rate expression of the combustion reaction at the surface of the particle is given by: E in kJ mol- 1 where the rate is in units of moles O2 reacted min- 1 (m2 external surface)-l, Ts is the surface temperature of the coal particle in Kelvins, and Cs is the surface concentration of O2 . The estimated heat and mass-transfer coefficients from the gas phase to the particle are ht = 0.5 kJ min- I K- I m-2 and kc = 0.5 m min- I, and the heat of reaction is -ISO kJ (mol O2) -1. Consider the bulk temperature of the gas to be TB = 500°C and the bulk concentration of O2 to be 2 mol m- 3.
(a) What is the maximum temperature difference between the bulk gas and the particle? What is the observed reaction rate under that condition?
(b) Determine the actual surface temperature and concentration, and therefore the actual reaction rate.
5.
The irreversible, first-order reaction of gaseous A to B occurs in spherical catalyst pellets with a radius of 2 mm. For this problem, the molecular diffusivity of A is 1.2 X 10- 1 cm2 s- 1 and the Knudsen diffusivity is 9 X 10- 3 cm2 s-1. The intrinsic first-order rate constant determined from detailed laboratory measurements was found to be 5.0 s-1. The concentration of A in the surrounding gas is 0.0 I mol L-1. Assume the porosity and the tortuosity of the pellets are 0.5 and 4, respectively.
(a) Determine the Thiele modulus for the catalyst pellets.
(b) Find a value for the internal effectiveness factor.
234 C H A PT E R 6 Effects oj Transport I imitations on Rates oj Solid-Catalyzed Reactions (c) For an external mass-transfer coefficient of 32 s-I (based on the external area of the pellets), determine the concentration of A at the surface of the catalyst pellets.
(d) Find a value for the overall effectiveness factor.
6.
J. M. Smith (J. M. Smith, Chemical Engineering Kinetics, 2nd ed., McGraw-Hill, New York, 1970, p. 395) presents the following observed data for Pt-catalyzed oxidation of SOz at 480°C obtained in a differential fixed-bed reactor at atmospheric pressure and bulk density of 0.8 g/cm3.
251 171 119 72 0.06 0.06 0.06 0.06 0.0067 0.0067 0.0067 0.0067 0.2 0.2 0.2 0.2 0.1346 0.1278 0.1215 0.0956 The catalyst pellets were 3.2 by 3.2 mm cylinders, and the Pt was superficially deposited upon the external surface. Compute both external mass and temperature gradients and plot LlCsoz and LlT versus the mass velocity. Can you draw any qualitative conclusions from this plot? If the reaction activation energy is 30 kcal/mol, what error in rate measurement attends neglect of an external LlT? What error prevails if, assuming linear kinetics in SOz, external concentration gradients are ignored?
Hints: J = kc Sc Z/3 = 0.817 Re- 1/2 u catalyst pellet is nonporous reaction carried out with excess air /iair = 1.339 g/h/cm @ 480°C DSO,-air = 2.44 ft2/h @ 480°C BB = void fraction of bed = 0.4 Sc = 1.28 /L Cp Prandtl number = -A- = 0.686 Cp 7.514 cal/mol/K 18.75 mm- I surface area volume LlHr -30 kcal/mol a CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 235 7.
The importance of diffusion in catalyst pellets can often be determined by measuring the effect ofpellet size on the observed reaction rate. In this exercise, consider an irreversible first-order reaction occurring in catalyst pellets where the surface concentration of reactant A is CAS = 0.15 M.
Data: Diameter of sphere (em) 0.2 0.06 0.02 0.006 rob, (mol/h /em3) 0.25 0.80 1.8 2.5 (a) Calculate the intrinsic rate constant and the effective diffusivity.
(b) Estimate the effectiveness factor and the anticipated rate of reaction (robs) for a finite cylindrical catalyst pellet of dimensions 0.6 cm X 0.6 cm (diameter = length).
8.
Isobutylene (A) reacts with water on an acidic catalyst to form t-butanol (B).
(CH3)2C=CH2 + H2 0 = (CH3 )3COH When the water concentration greatly exceeds that of isobutylene and t-butanol, the reversible hydration reaction is effectively first order in both the forward and reverse directions.
V. P. Gupta and W. J. M. Douglas [AIChE J., 13 (1967) 883] carried out the isobutylene hydration reaction with excess water in a stirred tank reactor utilizing a cationic exchange resin as the catalyst. Use the following data to determine the effectiveness factor for the ion exchange resin at 85°C and 3.9 percent conversion.
Data: Equilibrium constant @ 85°C = 16.6 = [B]/[A] D~A = 2.0 X 10-5 cm2 S-l Radius of spherical catalyst particle = 0.213 mm Density of catalyst = 1.0 g cm- 3 Rate of reaction at 3.9 percent conversion = 1.11 X 10-5 mol s-1 gcaC I CAS = 1.65 X 10-2 M (evaluated at 3.9 percent conversion) C~ = 1.72 X 10-2 M (reactor inlet concentration) (Problem adapted from C. G. Hill, Jr., An Introduction to Chemical Engineering Kinetics and Reactor Design, Wiley, NY, 1977.) 9.
Ercan et al. studied the alkylation of ethylbenzene, EB, with light olefins (ethylene and propylene) over a commercial zeolite Y catalyst in a fixed-bed reactor with recycle [Co Ercan, F. M. Dautzenberg, C. Y. Yeh, and H. E. Barner, Ind. Eng. Chem. Res., 37 (1998) 1724]. The solid-catalyzed liquid-phase reaction was carried out in excess ethylbenzene at 25 bar and 190°C. Assume 236 CHAPTER 6 Effects of Transport I imitations on Rates of Solid-Catalyzed ReactiorlS-- the reaction is pseudo-first-order with respect to olefin. The porosity of the catalyst was 0.5, the tortuosity was 5.0, and the density was 1000 kg m-3. The observed rate (robsJ and rate constant (kobs) were measured for two different catalyst pellet sizes. Relevant results are given below: 0.63 0.17 5.69 X 10-4 1.07 X 10-3 4.62 17.13 8.64 X 10-6 11.7 X 10-6 0.33 X 10-3 1.06 X 10-3 (a) Determine whether or not external and internal mass transfer limitations are significant for each case. Assume the diffusivity of olefins in ethylbenzene is DAB = 1.9 X 10-4 cm2 S-l.
(b) Calculate the Thiele modulus, cP, and the internal effectiveness factor, y), for each case.
(c) Determine the overall effectiveness factor for each case.
10.
Reaction rate expressions of the form: reveal zero-order kinetics when KCA » 1. Solve the material balance (isothermal) equation for a slab catalyst particle using zero-order kinetics. Plot CA(X)/CAS for a Thiele modulus of 0.1, 1.0, and 10.0. If the zero-order kinetics were to be used as an approximation for the rate form shown above when KCAS » I, would this approximation hold with the slab catalyst particle for the Thiele moduli investigated?
11.
Kehoe and Butt [1. P. Kehoe and J. B. Butt, AIChE 1., 18 (1972) 347] have reported the kinetics of benzene hydrogenation of a supported, partially reduced Ni/kieselguhr catalyst. In the presence of a large excess of hydrogen (90 percent) the reaction is pseudo-fIrst-order at temperatures below 200°C with the rate given by: where PB = benzene partial pressure, torr P~ dihydrogen partial pressure, torr KO 4.22 X 10- 11 torr- I k?
4.22 mol!gcat/s/torr E -2.7 kcal/mol C H A PT E R 6 Effects of Transport I imitations on Rates of Solid-Catalyzed Reactions 237 For the case of P~ = 685 torr, Pe 75 torr, and T l50aC, estimate the effectiveness factor for this reaction carried out in a spherical catalyst particle of density 1.88 gcat/cm3, DYB 0.052 cm2/s, and Rp = 0.3 cm.
12.
A first-order irreversible reaction is carried out on a catalyst of characteristic dimension 0.2 cm and effective diffusivity of 0.015 cm2/s. At 100aC the intrinsic rate constant has been measured to be 0.93 s-I with an activation energy of 20 kcallmoI.
(a) For a surface concentration of 3.25 X 10-2 mollL, what is the observed rate of reaction at looac?
(b) For the same reactant concentration, what is the observed rate of reaction at l50aC? Assume that DTA is independent of temperature.
(c) What value of the activation energy would be observed?
(d) Compare values of the Thiele modulus at 100aC and l50aC.
13.
The catalytic dehydrogenation of cyclohexane to benzene was accomplished in an isothermal, differential, continuous flow reactor containing a supported platinum catalyst [L. G. Barnett et aI., AIChE J., 7 (1961) 211].
O===>O+3HZ C6H12 =0.002 mol s-1 Hz =0.008 mol 8-1 1=15.5 % Dihydrogen was fed to the process to minimize deposition ofcarbonaceous residues on the catalyst. Assuming the reaction is first-order in cyclohexane and the diffusivity is primarily of the Knudsen type, estimate the tortuosity 7 of the catalyst pellets.
Additional Data: Diameter of catalyst pellet = 3.2 mm Pore volume of the catalyst = 0.48 cm3 g-I Surface area = 240 m2 g-I Pellet density Pp = 1.332 g em- 3 Pellet Porosity op = 0.59 Effectiveness factor TJ = 0.42 14.
For a slab with first-order kinetics: TJo tanhcP cP[ 1 + cP tanh(cP)/Bi",J (1) How important is the mass Biot number in Equation (1) with respect to its influence upon TJo for (a) cP 0.1, (b) cP 1.0. (c) cP 5.0, (d) cP 10.07 238 CHAPTER 6 Effects of Transport I imitations on Rates of Solid Catalyzed Reactions Consider the effect of the Biot number significant if it changes 1]0 by more than I percent. Can you draw any qualitative conclusions from the behavior observed in parts (a)-(d)?
15.
The liquid-phase hydrogenation of cyclohexene to cyclohexane (in an inert solvent) is conducted over solid catalyst in a semibatch reactor (dihydrogen addition to keep the total pressure constant).
P=constant Draw a schematic representation of what is occurring at the microscopic level. Provide an interpretation for each of the following figures.
Time Figure 1 I Linear relationship between cyclohexene concentration in the reactor and reaction time. Results apply for all conditions given in the figures provided below.
Weight of catalyst Figure 2 I Effect of catalyst weight on reaction rate.
CHAPTER 6 Effects of Transport I imitations on Rates of Soiid-Catalyzed Reactions (lIT) Figure 3 I Evaluation of temperature effects on the reaction rate constant.
239 __~_7 Microkinetic Analysis of Catalytic Reactions 7.1 I Introduction A catalytic reaction consists of many elementary steps that comprise an overall mechanistic path of a chemical transformation. Although rigorous reaction mecha-nisms are known for simple chemical reactions, many catalytic reactions have been adequately described by including only the kinetically significant elementary steps.
This approach was used in Chapter 5 to simplify complex heterogeneous catalytic reaction sequences to kinetically relevant surface reactions. The next step in fur-thering our understanding of catalytic reactions is to consolidate the available ex-perimental data and theoretical principles that pertain to elementary steps in order to arrive at quantitative models. This is the role of microkinetic analysis, which is defined as an examination of catalytic reactions in terms of elementary steps and their relation with each other during a catalytic cycle (1. A. Dumesic, D. F. Rudd, L. M. Aparicio, J. E. Rekoske, and A. A. Trevino, The Microkinetics of Heteroge-neous Catalysis, American Chemical Society, Washington, D.C., 1993, p. 1). In this chapter, three catalytic reactions will be examined in detail to illustrate the concept of microkinetic analysis and its relevance to chemical reaction engineering. The first example is asymmetric hydrogenation of an olefin catalyzed by a soluble organometallic catalyst. The second and third examples are ammonia synthesis and olefin hydrogenation, respectively, on heterogeneous transition metal catalysts.
7.2 I Asymmetric Hydrogenation of Prochiral Olefins The use of soluble rhodium catalysts containing chiral ligands to obtain high stere-oselectivity in the asymmetric hydrogenation of prochiral olefins represents one of the most important achievements in catalytic selectivity, rivaling the stereoselectiv-ity of enzyme catalysts [1. Halpern, Science, 217 (1982) 401]. Many chiralligands 240 C H A PT E R 7 Microkinetic Analysis of Catalytic Reactions 241 Methyl-(Z)-a-acetamidocinnamate (MAC) H '" ~C V o Ii /,C-O-CH3 c'" + H2 N-C-CH3 H II o H 0 I If II H 3C ~ ~ ~ N ----'--Y): C. ~ 0 ~.. CH 3 o ~ I ~ N-acetYI-(R)-PhenYlalanine-::thYI ester (R) (Rh catalyst) N-acetyl-(S)-phenylalanine methyl ester (8) Figure 7.2.1 I Hydrogenation of MAC catalyzed by a homogeneous Rh catalyst to give Rand S enantiomers of N-acetylphenylalanine methyl ester.
have been used to create this class of new asymmetric hydrogenation catalysts. In addition, stereoselectivity has been observed with a variety of olefins during hydro-genation reactions in the presence of these chiral catalysts, demonstrating the general utility of the materials. As an example of microkinetic analysis, the asymmetric hy-drogenation of methyl-(Z)-a-acetamidocinnamate, or MAC, to give the enantiomers (R and S) of N-acetylphenylalanine methyl ester will be discussed in detail. The overall reaction is illustrated in Figure 7.2.1.
C. R. Landis and J. Halpern found that cationic rhodium, Rh(I), with the chiral ligand R,R-l,2-bis[(phenyl-o-anisol)phosphino]ethane, or DIPAMP (see Figure 7.2.2), was very selective as a catalyst for the production of the S enantiomer of OpQ HoCOe OCH 3 6 :~ Prj 'f ~ R.R-DIPAMP R,R-I ,2-bis[(phenyl-o-anisol)phosphino]ethane Figure 7.2.2 I Chiral ligand for enantioselective rhodium catalyst 242 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions N-acetylphenylalanine methyl ester during MAC hydrogenation in methanol sol-vent [CO R. Landis and J. Halpern, 1. Am. Chern. Soc., 109 (1987) 1746]. The DI-PAMP coordinates to the rhodium through the phosphorus atoms and leaves plenty of space around the Rh cation for other molecules to bind and react. The unique feature of this system is that the chirality of the DIPAMP ligand induces the high stereoselectivity of the product molecules. An analogous nonchiral ligand on the Rh cation produces a catalyst that is not enantioselective.
To understand the origin of enantioselectivity in this system, the kinetics of the relevant elementary steps occurring during hydrogenation had to be determined. The catalytic cycle can be summarized as: (a) reversible binding of the olefin (MAC) to the Rh catalyst, (b) irreversible addition of dihydrogen to the Rh-olefin complex, (c) reaction of hydrogen with the bound olefin, and (d) elimination of the product into the solution to regenerate the original catalyst. Landis and Halpern have mea-sured the kinetics of step (a) as well as the overall reactivity of the adsorbed olefin with dihydrogen [CO R. Landis and J. Halpern, 1. Am. Chern. Soc., 109 (1987) 1746].
Figure 7.2.3 shows two coupled catalytic cycles that occur in parallel for the pro-duction of Rand S enantiomers from MAC in the presence of a common rhodium catalyst, denoted as . Since the catalyst has a chiral ligand, MAC can add to the catalyst in two different forms, one that leads to the R product, called MACR, and one that leads to the S product, called MACs. The superscripts refer to the cat-alytic cycles that produce the two enantiomers.
MAC Major \1AC· R intermediate k R -I • k S -I Minor intermediate R Minor product S Major product Figure 7.2.3 I Scheme of the coupled catalytic cycles for the asymmetric hydrogenation of MAC.
CHAPTER 7 Microkinetic Analys;s of Catalytic Reactions 243 Expressions for relative concentrations of the intermediates and products are developed from concepts discussed in earlier chapters, namely, the use of a total catalytic site balance and the application of the steady-state approximation. The to-tal amount of rhodium catalyst in the reactor is considered constant, []0, so that the site balance becomes: (7.2.1) The steady-state approximation indicates that the concentrations of reactive inter-mediates remain constant with time. In other words, the net rate of MAC binding to the catalyst to form an intermediate must be the same as the rate of hydrogena-tion (or disappearance) of the intermediate. The steady-state approximation for the coupled catalytic cycles is expressed mathematically as: drMAcR] , - 0 - rR - rR - rR dt --1 -I 2 d[MACS] ----= 0 = rS r~1 -r~ dt 1 or in terms of the rate expressions as: o = kf[MAC][]-k~I[MACR]-k~[MACR][H2] o = knMAC][]-k~I[MAC S]-k~ [MACS][H2] (7.2.2) (7.2.3) (7.2.4) (7.2.5) Rearranging Equations (7.2.4 and 7.2.5) gives the concentrations of reactive intermediates: kR rMAC][] [MAC R] = _I_L _ k~1 + k~[H2] kS [MAC][] [MAC S] = _I _ k~1 + kHH2] that can be used to derive an equation for their relative concentration: (7.2.6) (7.2.7) (7.2.8) The ratio of concentrations of the products R and S can be derived in a similar fashion: rR rf -r~1 r~ (7.2.9) rs = rf r~1 r~ (7.2.10) rs r~ k~ [MAC S][H2] C~) ,[MAC S] (7.2.11) [R] rR rf k~[MAC k~ ~MAC R] 244 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions Table 7.2.1 I Kinetic parameters for the asymmetric hydrogenation of MAC catalyzed by Rh(R,R-DIPAMP) at 298 K, k1 (L mmol- I S-l) L 1 (S-l) KI (L mmol- I ) k2 (L mmol- 1 S-I) 5.3 0.15 35 1.1 x 10-3 11 3.2 3.3 0.63 (7.2.12) Source: C. R. Landis and J. Halpern, 1. Am, Chem, Soc., 109 (1987) 1746.
The relative concentration of reactive intermediates given in Equation (7.2.8) is then substituted into Equation (7.2.11) to give the ratio of final products: [S] = (k~ k~) k~l + k~ [H2 ] [R] k~ k~ k~l + kHH2 ] Notice that the relative concentrations of the intermediates and products depend only on temperature (through the individual rate constants) and the pressure of dihydrogen.
At sufficiently low dihydrogen pressures, the reversible binding of MAC to the catalyst is essentially equilibrated because: (7.2.13) and (7.2.14) In other words, the rate-determining step for the reaction at low dihydrogen pres-sures is the hydrogenation of bound olefin. Simplification of Equations (7.2.8) and (7.2.12) by assuming low dihydrogen pressure gives the following expressions for the relative concentrations of intermediates and products: [MAC S] = (k~) k~l = K~ [MAC R] k~ k~l K~ [S] (k~ k~) k~l _ K~ k~ [R] \k~ k~ k S I -K~ k~ (7.2.15) (7.2.16) Landis and Halpern have measured the relevant rate constants for this reaction sys-tem and they are summarized in Table 7.2.1 [C. R. Landis and 1. Halpern, J. Am.
Chern. Soc., 109 (1987) 1746].
The values of the measured constants indicate that the S enantiomer is greatly favored over the R enantiorner at 298 K: [S] [R] K~ k~ K~ k~ 3.3 35 __ 0_.6_3_ = 54 1.1 X 10-3 (7.2.17) CHAPTER 7 Microkinetic Analysis Qf Catalytic RAactiQos 245 This result is rather surprising since the reactive intennediate that leads to the S enan-tiomer is in the minority. Equation (7.2.15) illustrates the magnitude of this difference: K~ R = 0.094 K 1 (7.2.18) The reason that the minor reactive intennediate leads to the major product is due to the large rate constant for hydrogenation (k2) associated with the S cycle compared to the R cycle. Clearly, the conventional "lock and key" analogy for the origin of enantioselectivity does not apply for this case since the selectivity is determined by kinetics of hydrogenation instead of thennodynamics of olefin binding.
A microkinetic analysis of this system also adequately explains the dependence of enantioselectivity on dihydrogen pressure. At sufficiently high pressures of dihydrogen, (7.2.19) and (7.2.20) Since the binding of olefin is not quasi-equilibrated at high pressure, the subsequent hydrogenation step cannot be considered as rate-detennining. The relative concen-trations of reactive intennediates and products are now given as: and [MAC s] = (k~) k~ = (~) l.l X 10-3 = 3.6 X 10-3 [MAC R] kf k~ 5.3 0.63 ~ = (k~) = (~) = [R] R 2.1 k] 5.3 (7.2.21) (7.2.22) The effect of dihydrogen is to lower the enantioselectivity to the S product from [S]/[R] = 54 at low pressures to [S]/[R] = 2.1 at high pressures. The reason for the drop in selectivity is again a kinetic one. The kinetic coupling of the olefin bind-ing and hydrogenation steps becomes important at high dihydrogen pressure [M.
Boudart and G. Djega-Mariadassou, Caral. Lett., 29 (1994) 7]. In other words, the rapid hydrogenation of the reactive intennediate prevents quasi-equilibration of the olefin binding step. The relative concentration of the minor intennediate MACs (that leads to the major product S) decreases significantly with increasing dihydro-gen pressure since k~ » k~. In this example, the kinetic coupling on the S cycle is much stronger than that on the R cycle, and that leads to an overall reduction in selectivity with increasing dihydrogen pressure.
It should be pointed out that selectivity does not depend on dihydrogen in the limit of very high or very low pressure. For intennediate dihydrogen pressures, Equa-tions (7.2.8) and (7.2.12)-that depend on dihydrogen-should be used to calculate selectivity. The microkinetic methodology satisfactorily explains the surprising 246 CHAPTER 7 MicrQkinetic Analysis Qf Catalytic ReactiQns inverse relationship between intermediate and product selectivity and the unusual effect of dihydrogen pressure on product selectivity observed during asymmetric hydrogenation of prochiral olefins with a chiral catalyst.
7.3 I Ammonia Synthesis on Transition Metal Catalysts The ammonia synthesis reaction is one of the most widely studied reactions and it has been discussed previously in this text. (See Example 1.1.1, Table 5.2.1, and Section 5.3.) In this section, results from the microkinetic analysis of ammonia synthesis over transition metal catalysts containing either iron or ruthenium will be presented.
A conventional ammonia synthesis catalyst, based on iron promoted with Al20 3 and K20, operates at high temperatures and pressures in the range of 673-973 K and 150-300 bars to achieve acceptable production rates. Metallic iron is the active catalytic component while A120 3 and K20 act as structural and chemical promot-ers, respectively. The goal of microkinetic analysis in this case is to study the con-ditions under which data collected on single crystals in ultrahigh vacuum or on model powder catalysts at ambient pressures can be extrapolated to describe the per-formance of a working catalyst under industrial conditions of temperature and pres-sure. Unfortunately, the kinetic parameters of all possible elementary steps are not known for most heterogeneous catalytic reactions. Thus, a strategy adopted by Dumesic et al. involves construction of a serviceable reaction path that captures the essential surface chemistry, and estimation of relevant parameters for kinetically sig-nificant elementary steps (J. A. Dumesic, D. F. Rudd, L. M. Aparicio, J. E. Rekoske, and A. A. Trevino, The Microkinetics ofHeterogeneous Catalysis, American Chem-ical Society, Washington, D.C., 1993, p. 145).
In the case of ammonia synthesis on transition metal catalysts, a variety of reasonable paths with various levels of complexity can be proposed. For the sake of clarity, only one such path will be presented here. Stolze and Norskov suc-cessfully interpreted the high-pressure kinetics of ammonia synthesis based on a microscopic model established from fundamental surface science studies [Po Stolze and J. K. Norskov, Phys. Rev. Lett., 55 (1985) 2502; P. Stolze and J. K. Norskov, J. Catal., 110 (1988) 1]. Dumesic et al. re-analyzed that sequence of elementary steps, which is presented in Table 7.3.1, for ammonia synthesis over iron catalysts [1. A. Dumesic and A. A. Trevino, 1. Catal.. 116 (1989) 119]. The rate constants for adsorption and desorption steps were estimated from results obtained on iron sin-gle crystal surfaces at ultrahigh vacuum conditions. The surface hydrogenation rates were estimated from the relative stabilities of surface NHx species. In the follow-ing microkinetic analysis, an industrial plug flow reactor was modeled as a series of 10,000 mixing cells (see Example 3.4.3) in which the steady-state rate equations, the catalyst site balance, and the material balances for gaseous species were solved simultaneously. Thus, no assumptions with respect to a possible rate-determining step and a most abundant reaction intermediate were needed to complete the model.
The conditions chosen for the microkinetic analysis correspond to an industrial reactor with 2.5 cm3 of catalyst operating at 107 bar with a stoichiometric feed of __-"CLlH:LAOU"P'-JTLE....,RLL7--.JMlYillicdr.lQ"'kilJn"'elltic~filla4'1:>s'"is'-'Q.l1f-'C..ua:lJt.aa""ly'lJtiJ..,cJB:lle:«allc-.itLLiQlJJn"'s ~~2"'4~7 Table 7.3.1 I A proposed mechanism of ammonia synthesis over iron catalysts~ a 2 3 4 5 6 7 2 X 1O IPN/I, 4 X 109 e -29/(R,T)()N 2 (), 2 X 10ge- 81/(R,T)()H()N 1 X 1013e- 36/(R,T) ()NH ()H 4 X 1013e- 39/(R,T)()NH 2 ()H 4 X 1012e-39/(R,T)()NH 3 7X N2 + ( ) 2N H + N ( ) NH + NH + H ( ) NH2 + NH2 + H ( ) NH3 + NH3 ( ) NH3 + H2 + 2 ( ) 2H X 109e- 155/(RJ)(()N?
X 107e- 23/(R,T)()NH () X 1012 ()NH, (), 2 X 1013 ()NH, () 2 X 103 PNH , () 3 X 1013e-94/(R,T) (()H?
a Rates are in units of molecules per second per site, pressures (P) are in pascals, and activation energies are in kilojoules per mole.
Concentrations of surface species are represented by fractional surface coverages as discussed in Chapter 5. Kinetic parameters are adapted from the data of Stolze and Norskov [Po Stolze and J. K. Norskov, Phys. Rev. Lett., 55 (1985) 2502; P. Stolze and J. K. Norskov, Surf Sci. Lett., 197 (1988) L230; P. Stolze and J. K. Norskov, J. Catal., 110 (1988) 1].
20 '" 15 'a 0 8 8 10 '" E '" e '" 0-S o 0.00 0.20 0.40 0.60 0.80 1.00 Dimensionless distance Figure 7.3.1 I Ammonia concentration calculated from microkinetic model versus longitudinal distance from reactor inlet. (Figure adapted from "Kinetic Simulation of Ammonia Synthesis Catalysis" by J. A. Dumesic and A. A. Trevino, in Journal of Catalysis, Volume I 16: 119, copyright © 1989 by Academic Press, reproduced by permission of the publisher and the authors.) dinitrogen and dihydrogen. The space velocity was 16,000 h-1, the catalyst bed den-sity was 2.5 g cm -3, and the catalyst site density was 6 X 10-5 mol g-1. The am-monia concentration in the effluent of the reactor operating at 723 K was measured experimentally to be 13.2 percent. Figure 7.3,1 illustrates the ammonia concentra-tion calculated from the microkinetic model as a function of axial distance along 248 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions 1.0 0.8 (£ :::J t:: " c:. 0.6 '" "0 § 'j:; 0.4 ~ ~ U.I 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Dimensionless distance Figure 7.3.2 I Departure from equilibrium for the two slowest elementary steps in ammonia synthesis. Squares are for step 2 and circles are for step 3. (Figure adapted from "Kinetic Simulation of Ammonia Synthesis Catalysis" by J. A. Dumesic and A. A. Trevino, in Journal of Catalysis, Volume 116:119, copyright © 1989 by Academic Press, reproduced by permission of the publisher and the authors.) the reactor. The effluent concentration approached the experimental value of 13.2 percent, which illustrates the consistency of the model with observation. The util-ity of a microkinetic model, however, is that it allows for a detailed understanding of the reaction kinetics. For example, a rate-determining step can be found, if one exists, by examining the departure of the elementary steps from equilibrium under simulated reaction conditions. Figure 7.3.2 presents the departure from equilibrium, (forward rate reverse rate)/forward rate, for the slowest steps as the reaction proceeds through the reactor. Step 2, the dissociation of dinitrogen, is clearly the rate-determining step in the mechanism throughout the reactor. All of the other steps in Table 7.3.1 are kinetically insignificant. Thus, the turnover frequency of ammo-nia synthesis depends critically on the kinetics of dissociative adsorption of dini-trogen (steps 1 and 2), as discussed earlier in Section 5.3. As long as the individual quasi-equilibrated elementary steps can be summed to one overall quasi-equilibrated reaction with a thermodynamically consistent equilibrium constant, small errors in the individual pre-exponential factors and activation energies are irrelevant. A mi-crokinetic model also allows for the determination of relative coverages of various intermediates on the catalyst surface. Figure 7.3.3 illustrates the change in fractional surface reaction of N and H through the reactor. Adsorbed nitrogen is the most abundant reaction intermediate throughout the iron catalyst bed except at the entrance.
__..
Ccl:HI.JA.u=PJ.T-"E"-'B"-'7'--lYMJjjic..dr'-'OLkiLUnJ;;;9JJ'tic~,Analysis of Catalytic Reactions 249 1.0 0.0 1.0 0.8 0.6 0.4 0.2 0.001 0.002 0.003 0.004 0.005 0.01 1.0 0.0 0.8 0.8 " OJ) " ...
" > 0.6 0.6 0 u " u iJ ::> en ., <:: 0.4 0.4 0 ';:l al .t 0.2 0.2 Dimensionless distance Figure 7.3.3 I Fractional surface coverages of predominant adsorbed species versus dimensionless distance from the reactor inlet. (Figure adapted from "Kinetic Simulation of Ammonia Synthesis Catalysis" by J. A. Dumesic and A. A. Trevino, in Journal of Catalysis, Volume 116:119, copyright © 1989 by Academic Press, reproduced by permission of the publisher and the authors.) Since the formation of N is the rate-determining step, its high coverage during steady-state reaction on iron results from the equilibrium of H2 and NH3. Thus, the high coverage of H instead of N at the entrance to the reactor (Figure 7.3.3) results from the vanishingly low pressure of ammonia at that point.
Another test of validity is to check the performance of the model against ex-perimental rate data obtained far from equilibrium. The microkinetic model pre-sented in Table 7.3.1 predicts within a factor of 5 the turnover frequency of ammo-nia synthesis on magnesia-supported iron particles at 678 K and an ammonia concentration equal to 20 percent of the equilibrium value. This level of agreement is reasonable considering that the catalyst did not contain promoters and that the site density may have been overestimated. The model in Table 7.3.1 also predicts within a factor of 5 the rate of ammonia synthesis over an Fe( Ill) single crystal at 20 bar and 748 K at ammonia concentrations less than 1.5 percent of the equilibrium value.
It should be emphasized that since the rate of ammonia synthesis on iron de-pends critically on the dissociative adsorption of dinitrogen (steps I and 2), any 250 CHAPTER 7 Microkine>tic Analysis of Catalytic Reactions Table 7.3.2 I A proposed mechanism of ammonia synthesis over ruthenium catalysts, a 1 2 3 4 5 6 6 X 1O-Ze-33;(R,JiPN2(8?
6 X 1013 e-86/(R/J8H8N 5 X 1013e-60/(R/i8NH8H 3 X 1013e-17/(RJ!8NH28H 6 X lOI3 e-84/(R/)8NH, 6 X (8f Nz + 2 ( ) 2N H + N ( ) NH + NH + H ( ) NH2 + NHz + H ( ) NH3 + NH3 ( ) NH3 + Hz + 2 ( ) 2H 2 X lOJOe-137/(RJ!(8N?
3 X 1014e-41/(R/7J8NH8 2 X 1OI3e-9/(R,7)8NH28 9 X 10IZ e-65/(R/7) 8NH,8 2 X 103PNHj 8 2 X 1013e-89/(R,7) a Rate8 are in unit8 of molecules per second per site, pressures (P) are in pascals, and activation energies are in kilojoules per mole, Concentrations of surface species are represented by fractional surface coverages as discussed in Chapter 5, Kinetics parameters are adapted from the data of Hinrichsen et aL [0, Hinrichsen, E Rosowski, M, Muhler, and G, Ertl, Chern, Eng, Sci" 51 (1996) 1683,] microkinetic model that properly accounts for those steps and that is thermody-namically consistent with the overall reaction will effectively describe the kinetics of ammonia synthesis. This feature explains the success of the microkinetic model used to describe an industrial reactor even though the kinetic parameters for dini-trogen adsorption/desorption match those obtained from studies on iron single crystals in ultrahigh vacuum.
Ammonia synthesis on supported ruthenium catalysts has also been the sub-ject of microkinetic analysis. Table 7.3.2 presents the kinetic parameters for one model of ammonia synthesis catalyzed by cesium-promoted ruthenium particles supported on MgO [0. Hinrichsen, F. Rosowski, M. Muhler, and G. Ertl, Chern.
Eng. Sci., 51 (1996) 1683]. In contrast to the model in Table 7.3.1, the dissocia-tive adsorption of dinitrogen is represented by a single step. Many of the rate con-stants in Table 7.3.2 were determined from independent steady-state and transient experiments on the same catalyst. The experiments included temperature-programmed adsorption and desorption of dinitrogen, isotopic exchange of labeled and unlabeled dinitrogen, and temperature-programmed reaction of adsorbed nitrogen atoms with gaseous dihydrogen. The unknown rate constants were esti-mated by regression analysis of the experimental data and checked to ensure that they were within physically reasonable limits. As in the previous example with the iron catalyst, the steady-state reactor containing the supported Ru catalyst was modeled as a series of mixing cells in which the steady-state rate equations, the catalyst site balance and the material balances for gaseous species were solved simultaneously.
Similar to the results found with iron catalysts, the rate-determining step dur-ing ammonia synthesis on ruthenium catalysts is the dissociative adsorption of dinitrogen. However, the overall reaction rate is strongly inhibited by dihydrogen, indicating that adsorbed hydrogen is the most abundant reaction intermediate that covers most of the surface [B. C. McClaine, T. Becue, C. Lock, and R. J. Davis, CHAPTER 7 Microkinetic Analysis of Gataly.uticI.-LBu:e"'a.L-, u '" 0 '" '-~ 0.0 100 Dihydrogen pressure (torr) 1000 Figure 7.4.1 I Comparison of results from microkinetic model and experimental observation. [Adapted with permision from J. E. Rekoske, R. D. Cortright, S. A. Goddard, S. B.
Sharma, and J. A. Dumesic, J. Phys. Chern., 96 (1992) 1880. Copyright 1992 American Chemical Society.] 254 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions of dihydrogen pressure and a 100 K span in temperature. The observed order of reaction with respect to dihydrogen increases from 0.47 at 223 K to 1.10 at 336 K.
Clearly, the kinetic model reproduces the observed rates at all of the conditions tested. The model also predicted the order of reaction with respect to ethylene (not shown) quite well. This microkinetic analysis suggests that ethylene hydrogenates mostly through the noncompetitive route at low temperatures whereas the compet-itive route dominates at high temperatures.
The rapid increase in computing power and the advent of new quantum chem-ical methods over the last decade allow kinetic parameters for surface reactions to be estimated from ab initio quantum chemical calculations and kinetic simulation schemes. Indeed, the adsorption enthalpies of gas-phase species can be calculated, the activation barriers to form transition states from surface bound species can be pre-dicted, and the evolution of surface species with time can be simulated. Figure 7.4.2 shows, for example, the transition state for the hydrogenation of adsorbed ethylene (step 3) on a model palladium surface consisting of 7 Pd atoms arranged in a plane.
The activation energy associated with ethyl formation on a Pd(lll) metal surface was calculated from first principles quantum mechanics to be about 72 kJ mol-I [M. Neurock and R. A. van Santen, J. Phys. Chern. B, 104 (2000) 11127]. The con-figuration of the adsorbed ethyl species, the product of step 3, is illustrated for two different Pd surfaces in Figure 7.4.3. The energies for chemisorption of this reactive intermediate on a Pd l9 cluster and a semi-infinite Pd(lll) slab are -130 and -140 kJ mol-I, respectively, and are essentially the same within the error of the calculation.
Results involving adsorbed ethylene, adsorbed hydrogen and adsorbed ethyl indicate Figure 7.4.2 I The isolated transition state structure for the fonnation of ethyl from ethylene and hydrogen on a model Pd7 cluster.
The vectors correspond to the motion along the reaction coordinate. [Adapted with pennission from M. Neurock and R. A. van Santen, J. Phys. Chern. B, 104 (2000) 11127. Copyright 2000 American Chemical Society.] CHAPTER 7 Microkinetic Analysis of Catalytic Reac'"'t.n.ioLLnws~ ....
2"",5"",5 LiE =-130 kllmol (a) LiE=-140kllmol (b) Figure 7.4.3 I Structures and energies for the chemisorption of ethyl on (a) a Pd 19 cluster model and (b) a model Pd(lll) surface. [Reproduced with permission from M. Neurock and R. A. van Santen, J. Phys. Chem. B, 104 (2000) 11127. Copyright 1992 American Chemical Society.] 256 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions that the overall heat of reaction for step 3 is about 3 kJ mol- J endothermic on a Pd surface. In principle, these types of calculations can be performed on all of the species in the Horiuti-Polanyi mechanism, and the results can be used in a subse-quent kinetic simulation to ultimately give a reaction rate. Rates obtained in this fashion are truly from first principles since regression of experimental data is not required to obtain kinetic parameters.
Hansen and Neurock have simulated the hydrogenation of ethylene on a Pd cat-alyst using a Monte Carlo algorithm to step through the many reactions that occur on a model surface [E. W. Hansen and M. Neurock, 1. Catal., 196 (2000) 241]. In essence, the reaction is simulated by initializing a grid of Pd atoms and allowing all possible reactions, like adsorption of reactants, elementary surface reactions, and desorption of products, to take place. The simulation moves forward in time, event by event, updating the surface composition as the reaction proceeds. The details of the method are too complicated to describe here but can be found in E. W. Hansen and M. Neurock, 1. Catal., 196 (2000) 241. The input to the simulation involves a complete set of adsorption and reaction energies. An important feature of this sim-ulation is that it also accounts for interactions among the species on the surface.
Figure 7.4.4 I Snapshot of a Pd(100) surface during a simulation of ethylene hydrogenation at 298 K, 25 torr of ethylene and 100 torr of dihydrogen. [Figure from "First-Principles-Based Monte Carlo Simulation of Ethylene Hydrogenation Kinetics on Pd," by E. W. Hansen and M. Neurock, in Journal of Catalysis. Volume 196:241, copyright © 2000 by Academic Press, reproduced by permission of the publisher and authors.) ___________-'C>LI:IHLliAuP~T .......
EnRL7L~~l'f.lyliJtic..cDRJ;;1eC.'acvJtlJ..iollJn"'"s_~ ~2...
5'.L7 Although many kinetic models assume that the catalyst is an ideal Langmuir sur-face (all sites have identical thermodynamic properties and there are no interactions among surface species), modern surface science has proven that ideality is often not the case.
Results from the simulation reproduced the experimentally observed kinetics (activation energy, turnover frequency, and orders of reaction) for ethylene hydro-genation on a Pd catalyst. A snapshot of the Pd surface during a simulation of eth-ylene hydrogenation is given in Figure 7.4.4. At this point in the simulation, highly mobile hydrogen atoms moved rapidly around the surface and reacted with fairly stationary hydrocarbons. In this model, no distinction was made between the com-petitive and noncompetitive adsorption steps in the mechanism. The results suggest that lateral interactions among species can explain some of the experimental obser-vations not easily accounted for in the Horiuti-Polanyi mechanism.
7.5 I Concluding Remarks This chapter illustrated the concepts involved in microkinetic analysis of catalytic reactions. The first example involving asymmetric hydrogenation ofprochiral olefins with a chiral homogeneous catalyst illustrated how precise rate measurements of critical elementary steps in a catalytic cycle can yield a kinetic model that describes the overall performance of a reaction in terms of enantioselectivity and response to reaction conditions. The second example regarding ammonia synthesis showed how kinetic parameters of elementary steps measured on single crystals under ultrahigh vacuum conditions can be incorporated into a kinetic model that applies to reactors operating under industrial conditions of high pressure. The last example involving ethylene hydrogenation revealed how quantum chemical calculations provide esti-mates of the energies associated with elementary steps on catalytic surfaces that can be subsequently used in reaction simulations. Finally, microkinetic analysis is clearly moving toward the routine use of quantum chemical calculations as inputs to ki-netic models, with the ultimate goal of describing industrial chemical reactors.
In Section 7.2, the reaction paths of the Rh-catalyzed asymmetric hydrogenation of MAC were described in detail. The ratio of the S product to the R product, [S]/[R], was expressed in the limit of low [Equation (7.2.16)] and high [Equation (7.2.22)] pressure (or concentration) of dihydrogen. Use the rate constants in Table 7.2.1 to plot [S]/[R] as a function of H2 concentration. At what concentrations are Equation (7.2.16) and Equation (7.2.22) valid?
Landis and Halpern studied the temperature dependence of the various rate constants in Table 7.2.1 [CO R. Landis and J. Halpern, 1. Am. Chem. Soc., 109 (1987) 1746]. The values of the individual activation energies and pre-exponential factors are: Exercises for Chapter 7 1.
2.
258 CHAPTER 7 Microkinetic Analysis of Catalytic Reactions Kinetic parameters for the asymmetric hydrogenation of MAC catalyzed by Rh(R,R-DIPAMP).
k l (L mmo!-I 5- 1 ) k- I (5- 1) k2 (L mmo!-I 5- 1 ) 4.9 13.3 10.7 2.32 X 104 8.53 X 108 7.04 X 104 6.9 13.0 7.5 1.21 X 106 9.94 X 109 1.84 X 105 a Activation energy, kcal mol-I.
b Pre-exponential factor in the units of the rate constant.
Source: C. R. Landis and J. Halpern, 1. Am. Chern. Soc., 109 (1987) 1746.
Detennine the ratio of the S product to the R product, [S]/[R], as a function of temperature, over the range of 0 to 37°C. Keep the H2 concentration at a constant value between the limits of high and low pressure as discussed in Exercise 1.
3.
A microkinetic analysis of ammonia synthesis over transition metals is presented in Section 7.3. Use the results of that analysis to explain how adsorbed nitrogen atoms (N) can be the most abundant reaction intennediate on iron catalysts even though dissociative chemisorption of N2 is considered the rate-detennining step.
4.
Describe the main differences in the kinetics of ammonia synthesis over iron catalysts compared to ruthenium catalysts.
5.
The Horiuti-Polanyi mechanism for olefin hydrogenation as discussed in Section 7.4 involves 4 steps: Step 1 Step 2 Step 3 Step 4 Overall H2 + 2 ( ) 2H C2H4 + 2 ( ) C2H4 C2H4 + H ( ) C2Hs + 2 C2Hs + H ( ) C2H6 + 2 Derive a rate expression for the hydrogenation of ethylene on Pt assuming steps 1, 2, and 3 are quasi-equilibrated, step 4 is virtually irreversible, and C2Hs is the most abundant reaction intennediate covering almost the entire surface ([]0 - [C 2Hs]). Discuss why the rate expression cannot properly account for the experimentally observed half order dependence in H 2 and zero-order dependence in ethylene. Could the observed reaction orders be explained if adsorbed ethylene (C2H4) were the most abundant reaction intennediate?
Explain your answer.
6.
Read the paper entitled "Microkinetics Modeling of the Hydroisomerization of n-Hexane," by A. van de Runstraat, J. van Grondelle, and R. A. van Santen, Ind. Eng. Chem. Res., 36 (1997) 3116. The isomerization of n-hexane is a classic CHAPTER 7 Microkinetic Analysis of Catalytic Reactions 259 example of a reaction involving a bifunctional catalyst, where a transition metal component facilitates hydrogenation/dehydrogenation and an acidic component catalyzes structural rearrangement.
Section 5.3 illustrates the important reactions involved in hydrocarbon isomerization over bifunctional catalysts.
Summarize the key findings of the microkinetic analysis by van de Runstraat et al.
~~8 Nonideal Flow in Reactors 8.1 I Introduction In Chapter 3, steady-state, isothermal ideal reactors were described in the context of their use to acquire kinetic data. In practice, conditions in a reactor can be quite different than the ideal requirements used for defining reaction rates. For example, a real reactor may have nonuniform flow patterns that do not conform to the ideal PFR or CSTR mixing patterns because of comers, baffles, nonuniform catalyst pack-ings, etc. Additionally, few real reactors are operated at isothermal conditions; rather they may be adiabatic or nonisothermal. In this chapter, techniques to handle non-ideal mixing patterns are outlined. Although most of the discussion will center around common reactor types found in the petrochemicals industries, the analyses presented can be employed to reacting systems in general (e.g., atmospheric chem-istry, metabolic processes in living organisms, and chemical vapor deposition for microelectronics fabrication). The following example illustrates how the flow pat-tern within the same reaction vessel can influence the reaction behavior.
EXAMPLE 8.1.1 I In order to approach ideal PFR behavior, the flow must be turbulent. For example, with an open tube, the Reynolds number must be greater than 2100 for turbulence to occur. This flow regime is attainable in many practical situations. However, for laboratory reactors conduct-ing liquid-phase reactions, high flow rates may not be achievable. In this case, laminar flow will occur. Calculate the mean outlet concentration of a species A undergoing a first-order reaction in a tubular reactor with laminar flow and compare the value to that obtained in a PFR when (kL)/u = I (u average linear flow velocity).
• Answer The material balance on a PFR reactor accomplishing a first-order reaction at constant density is: 260 CHAPTER 8 Nonideal Flow in Beacto,.wrs~ --,,2 ..
6 .......
1 li(i',) = 0 li(O) = 2u ~ -: \ I V \ V \ I I ,,~ v V L-..
i -Parabolic velocity distribution li(i') Figure 8.1.1 I Schematic representation of laminar velocity profile in a circular tube.
uAc dCA ----Ac dz dCA u--dz Integration of this equation with CA = C~ at the entrance of the reactor (z 0) gives: kz U For laminar flow: u(1') = 2U[ I Ci1'}] where 1', is the radius of the tubular reactor (see Figure 8.U).
The material balance on a laminar-flow reactor with negligible mass diffusion (discussed later in this chapter) is: __ C!C4 u(r) -' -kCA C!Z Since u(1') is not a function of z, this equation can be solved to give: C (1')' = CO expr A A I L k:: To obtain the mean concentration, follows: must be integrated over the radial dimension as 262 CHAPTER 8 Nonideal Flow in Reactors (I CACr)u(r)2r.rdr ·0 u(r)2r.rdr Thus, the mean outlet concentration of A, C~, can be obtained by evaluating C A at z L. For (kL)/u I the outlet value of CA from the PFR, C~, is 0.368 C~ while for the laminar-flow re-actor C~ 0.443 C~. Thus, the deviation from PFR behavior can be observed in the outlet con-version of A: 63.2 percent for the PFR versus 55.7 percent for the laminar-flow reactor.
8.2 I Residence Time Distribution (RTD) In Chapter 3, it was stated that the ideal PFR and CSTR are the theoretical limits of fluid mixing in that they have no mixing and complete mixing, respectively. Al-though these two flow behaviors can be easily described, flow fields that deviate from these limits are extremely complex and become impractical to completely model. However, it is often not necessary to know the details of the entire flow field but rather only how long fluid elements reside in the reactor (i.e., the distribution of residence times). This information can be used as a diagnostic tool to ascertain flow characteristics of a particular reactor.
The "age" of a fluid element is defined as the time it has resided within the reac-tor. The concept of a fluid element being a small volume relative to the size of the re-actor yet sufficiently large to exhibit continuous properties such as density and con-centration was first put forth by Danckwerts in 1953. Consider the following experiment: a tracer (could be a particular chemical or radioactive species) is injected into a reac-tor, and the outlet stream is monitored as a function of time. The results of these ex-periments for an ideal PFR and CSTR are illustrated in Figure 8.2.1. If an impulse is injected into a PFR, an impulse will appear in the outlet because there is no fluid mix-ing. The pulse will appear at a time t1 = to + T, where T is the space time (T = Vjv).
However, with the CSTR, the pulse emerges as an exponential decay in tracer con-centration, since there is an exponential distribution in residence times [see Equation (3.3.11)]. For all nonideal reactors, the results must lie between these two limiting cases.
In order to analyze the residence time distribution of the fluid in a reactor the following relationships have been developed. Fluid elements may require differing lengths of time to travel through the reactor. The distribution of the exit times, de-fined as the E(t) curve, is the residence time distribution (RTD) of the fluid. The exit concentration of a tracer species C(t) can be used to define E(t). That is: such that: E(t) = coC(t) f C(t)df o (8.2.1 ) (8.2.2) ___~C~H~A~P~T.LOOEoJRI>8~JJNQDidealFlow in Beac1oJjr"'s'--"'2..
6t>;t3 " .g S " OJ U " o U " .g S § " o U Inlet Time Inlet " .g " !:J " OJ g o U (a) " .g E " OJ U § U Outlet Outlet Time (b) Time Inlet Outlet Time (c) ~I JI~ '----r'1'------to Time Figure 8.2.1 I Concentrations of tracer species using an impulse input.
(a) PFR (tl = to + r). (b) CSTR. (c) Nonideal reactor.
264 CHAPTER 8 Nonideal Flow in Reactors With this definition, the fraction of the exit stream that has residence time (age) between t and t + dt is: EXAMPLE 8.2.1 I E(t)dt while the fraction of fluid in the exit stream with age less than t] is: I Ii E(t)dt o (8.2.3) (8.2.4) Calculate the RTD of a perfectly mixed reactor using an impulse of n moles of a tracer.
• Answer The impulse can be described by the Dirac delta function, that is: o(t such that: The unsteady-state mass balance for a CSTR is: dC V =no(t)-vC dt accumulation input output where to in the Dirac delta is set to zero and: fOO o(t)dt = 1 -00 Integration of this differential equation with C(0) 0 gives: (a) (b) (c) dC (n) r + C = -o(t) dt v I,--j -(n)Ilo(t) -odlCexp(tjr)J = ~ 0 -r-exp(t/r)dt I' (n)Ilo(t) C exp(t/r) = -exp(t/r)dt 10 v 0 r CHAPTER 8 Nonideal Flow in Reactors (d) C(t) exp(tI7) - C(O) exp(O) = (~) r5~) exp(1/7)d1 (e) C(t) = (~) exp(-tI7)fo~) exp(l/7)d1 (f) another property of the Dirac delta function is: f') o(t to)f(t)dt f(to) -00 265 (g) C(t) ( n) exp(-tl7) -exp(O/7) v 7 (h) (i) (j) ( n) exp(-t17) C(t) = -V 7 E(t) = C(t) (~)looexp(;1/7)d1 ~rJ-t/T) 7 E(t)=-----exp( -tI7) I;;" (k) E(t) exp(-tI7) 7 VIGNETTE 8.2.1 Thus, for a perfectly mixed reactor (or often called completely backmixed), the RTD is an exponential curve.
(8.2.5) 266 EXAMPLE 8.2.2 I CHAPTER 8 Nonideal Flow in Reactors Two types of tracer experiments are commonly employed and they are the in-put of a pulse or a step function. Figure 8.2.1 illustrates the exit concentration curves and thus the shape of the E(t)-curves (same shape as exit concentration curve) for an impulse input. Figure 8.2.2 shows the exit concentration for a step input of tracer. The E(t)-curve for this case is related to the time derivative of the exit concentration.
By knowing the E(t)-curve, the mean residence time can be obtained and is: fXltE(l)dt o (t) = --:-:-::----fXlE(t)dt o Calculate the mean residence time for a CSTR.
• Answer The exit concentration profile from a step decrease in the inlet concentration is provided in Equation (3.3.11) and using this function to calculate the E(t)-curve gives: E(t) exp(-tiT) T Therefore application of Equation (8.2.5) to this E(t)-curve yields the following expression: (t) =.! (OOtexp(-tIT)dt do Since: 1 00 x exp( - x)dx = 1 o (t) = ~ r oo T2 (tjT)exp(-tIT)d(tjT) = T . Jo As was shown in Chapter 3, the mean exit time of any reactor is the space time, T.
The RTD curve can be used as a diagnostic tool for ascertaining features of flow patterns in reactors. These include the possibilities of bypassing and/or re-gions of stagnant fluid (i.e., dead space). Since these maldistributions can cause unpredictable conversions in reactors, they are usually detrimental to reactor op-eration. Thus, experiments to determine RTD curves can often point to problems and suggest solutions, for example, adding or deleting baffles and repacking of catalyst particles.
CHAPTER 8 hJonideal Flow in Reactors 267 Inlet c: o .'" E c: " u § U Time Inlet .8 g c: " u c: o U (a) Outlet Outlet Time (b) Time Inlet Outlet J r--I-'ILL u 8 Time (c) J .g E c: " u § U Time Figure 8.2.2 I Concentrations of tracer species using a step input.
(a) PFR (tl to + 7). (b) CSTR. (c) Nonideal reactor.
268 EXAMPLE 8.2.3 I CHAPTER 8 Nonideal Flow in Reactors In Section 3.5 recycle reactors and particularly a Berty reactor were described. At high im-peller rotation speed, a Berty reactor should behave as a CSTR. Below are plotted the di-mensionless exit concentrations, that is, c(t)1Co, of cis-2-butene from a Berty reactor con-taining alumina catalyst pellets that is operated at 4 atm pressure and 2000 rpm impeller rotation speed at temperatures of 298 K and 427 K. At these temperatures, the cis-2-butene is not isomerized over the catalyst pellets. At t = 0, the feed stream containing 2 vol % cis-2-butene in helium is switched to a stream of pure helium at the same total flow rate. Reac-tion rates for the isomerization of cis-2-butene into I-butene and trans-2-butene are to be measured at higher temperatures in this reactor configuration. Can the CSTR material bal-ance be used to ascertain the rate data?
• Answer The exit concentrations from an ideal CSTR that has experienced a step decrease in feed concentration are [from Equation (3.3.11)]: clc O = exp[ If the RTD is that of an ideal CSTR (i.e., perfect mixing), then the decline in the exit con-centration should be in the form of an exponential decay. Therefore, a plot of In(CI Co) versus time should be linear with a slope of -7- 1. Using the data from the declining por-tions of the concentration profiles shown in Figure 8.2.3, excellent linear fits to the data are obtained (see Figure 8.2.4) at both temperatures indicating that the Berty reactor is behaving as a CSTR at 298 K :=; T :=; 427 K. Since the complete backmixing is achieved over such a large temperature range, it is most likely that the mixing behavior will also occur at slightly higher temperatures where the isomerization reaction will occur over the alumina catalyst.
c .8 ;; .... 0.8 E " u c 0 0.6 u z; ~ "2 0.4 0 .~ c " 0.2 E is 0 0 10 20 30 40 50 60 70 80 90 Time (min) (a) c .8 ;; 1::: 0.8 c " u C 0 0.6 u 'lJ Y) " "2 0.4 .~ c " 0.2 E is 0 0 5 10 15 20 25 30 35 40 45 50 Time (min) (b) Figure 8.2.3 I Dimensionless concentration of cis-2-butene in exit stream of Berty reactor as a function of time. See Example 8.2.3 for additional details.
__"'C'-'H...,AOOJ:p"'r"'E"'-lR"--'s"---"-'N'"-onideaLElow in Beacto~ 269 0.00 0.00 -0.50 -0.50 -1.00 ~ -1.00 =0 ...) ~ Q -1.50 .s -1.50 .s -2.00 -2.00 -2.50 -2.50 -3.00 10 20 30 40 50 60 70 80 90 10 Time (min) (a) 15 20 25 30 35 40 45 50 Time (min) (b) (8.3.1) Figure 8.2.4 I Logarithm of the dimensionless concentration of cis-2-butene in exit stream of Berty reactor as a function of time. See Example 8.2.3 for additional details.
8.3 I Application of RTD Functions to the Prediction of Reactor Conversion The application of the RTD to the prediction of reactor behavior is based on the as-sumption that each fluid element (assume constant density) behaves as a batch re-actor, and that the total reactor conversion is then the average conversion of all the fluid elements. That is to say: o [concentration of ] [fraction of exit stream] [ mean concentratIOn] 0 0 0 h .
f fl °d o 2: reactant remammg m t at consists 0 UI of reactant 111 = 0 • 0 reactor outlet a flUId element of age elements of age between t and t + elt between t and t elt where the summation is over all fluid elements in the reactor exit stream. This equa-tion can be written analytically as: (CAl = j''''CA(t)E(t)dt o (8.3.2) where CA(t) depends on the residence time of the element and is obtained from: with dCA dt (8.3.3) 270 eM APTER 8 NonideaL8o.wJnBea'-.Lcl'-'-'o"-'[s'-- For a first-order reaction: (8.3.4) or CA = C~ exp[ -kt] Insertion of Equation (8.3.5) into Equation (8.3.2) gives: \CA) = fXJC~ exp[ -kt]E(t)dl ° (8.3.5) (8.3.6) Take for example the ideal CSTR. If the E(t)-curve for the ideal CSTR is used in Equation (8.3.6) the result is: COfoo \CA) = -:-exp(-kl) exp(-1/7)dl ° or \~~) = ~1 00 exp[ -(k + ~)lJdl that gives after integration: \C A )= ~[_(_1)eXP[-(k + ~)tJIOO] =_1 7 k+ 1 7 ° kT+l T (8.3.7) Notice that the result shown in Equation (8.3.7) is precisely that obtained from the material balance for an ideal CSTR accomplishing a first-order reaction. That is: or c1 kT + 1 (8.3.8) Unfortunately, if the reaction rate is not first-order, the RTD cannot be used so di-rectly to obtain the conversion. To illustrate why this is so, consider the two reac-tor schemes shown in Figure 8.3.1.
Froment and Bischoff analyze this problem as follows (G. F. Froment & K. B. Bischoff, Chemical Reactor Analysis and Design, Wiley, 1979). Let the PFR and CSTR have space times of 71 and 72, respectively. The overall RTD for either system will be that of the CSTR but with a delay caused by the PFR. Thus, a tracer experiment cannot distinguish configuration (I) from (II) in Figure 8.3.1.
CHAPTER 8 Nonideal Flow in Reactors 271 (I) c~ (II) Figure 8.3.1 I PFR and CSTR in series. (I) PFR follows the CSTR, (II) CSTR follows the PFR.
A first-order reaction occurring in either reactor configuration will give for the two-reactor network: C~ e-krj C~ 1 + kT2 This is easy to see; for configuration (I); C~ CA = ------I + kT2 C ~ = exp[ -kT I ] CA or C~ e-krj C~ 1 + kT2 and for configuration (II); CA C~ = exp -kTd C~ 1 CA 1 + kT2 or C~ 1 + kT2 (CSTR) (PFR) (PFR) (CSTR) (8.3.9) Now with second-order reaction rates, configuration (1) gives: 1(1 + 4(kC~'T2)) 1 + \/ ~ 1 + (8.3.10) (8.3.11) 272 CHAPTER 8 Nonideal Flow in Reactors while configuration (II) yields: C~ -1 +V 1 + 4kC~T2 C~ 2kC~T2 + kC~T If kC~ = 1 and TdT] = 4, configurations (I) and (II) give outlet dimensionless concentrations (C~/C~) of 0.25 and 0.28, respectively. Thus, while first-order kinetics (linear) yield the same outlet concentrations from reactor configurations (1) and (II), the second-order kinetics (nonlinear) do not. The reasons for these dif-ferences are as follows. First-order processes depend on the length of time the mol-ecules reside in the reactors but not on exactly where they are located during their trajectory through the reactors. Nonlinear processes depend on the encounter of more than one set of molecules (fluid elements), so they depend both on residence time and also what they experience at each time. The RTD measures only the time that fluid elements reside in the reactor but provides no information on the details of the mixing. The terms macromixing and micromixing are used for the RTD and mixing details, respectively. For a given state of perfect macromixing, two extremes in micromixing can occur: complete segregation and perfect micromixing. These types of mixing schemes can be used to further refine the reactor analysis. These methods will not be described here because they lack the generality of the proce-dure discussed in the next section.
In addition to the problems of using the RTD to predict reactor conversions, the analysis provided above is only strictly applicable to isothermal, single-phase sys-tems. Extensions to more complicated behaviors are not straightforward. Therefore, other techniques are required for more general predictive and design purposes, and some of these are discussed in the following section.
8.4 I Dispersion Models for Nonideal Reactors There are numerous models that have been formulated to describe nonideal flow in vessels. Here, the axial dispersion or axially-dispersed plug flow model is described, since it is widely used. Consider the situation illustrated in Figure 8.4.1. (The steady-state PFR is described in Chapter 3 and the RTD for a PFR discussed in Section 8.2.) The transient material balance for flow in a PFR where no reaction is occur-ring can be written as: or ac Acaz a/ = uACCi -[uACCi + a(uAcCJ] (accumulation) (in) (out) (8.4.1 ) (8.4.2) CHAPTER 8 Nonideal Flow in Reactors 1 1 ----: 1 I Fi~ :~Fi+dFi ----1_-1-----1 I 1 dz I I I 273 "'AXiallY-diSpersed ~PFR de Fi = uACCi - ACDadf Figure 8.4.1 I Descriptions for the molar flow rate of species i in a PFR and an axially-dispersed PFR. Ac: cross-sectional diameter of tube, u: linear velocity, Da: axial dispersion coefficient.
where Ac is the cross-sectional area of the tube. If u == constant, then: ac aCi = -u-at az (8.4.3) Now if diffusion/dispersion processes that mix fluid elements are superimposed on the convective flow in the axial direction (z direction), then the total flow rate can be written as: dCi Fi = uACCi -AcDa -dz (convection) (dispersion) (8.4.4) Note that Da is called the axial-dispersion coefficient, and that the dispersion term of the molar flow rate is formulated by analogy to molecular diffusion. Fick's First Law states that the flux of species A (moles/area/time) can be formulated as: dCA N4 = -DAB -,- + uCA ..
az (8.4.5) for a binary mixture, where DAB is the molecular diffusion coefficient. Since axial dispersion processes will occur by molecular diffusion during laminar flow, at this condition the dispersion coefficient will be the molecular diffusion coefficient. How-ever, with turbulent flow, the processes are different and Da must be obtained from correlations. Since D a is the molecular diffusion coefficient during laminar flow, it is appropriate to write the form of the dispersion relationship as in Equation (8.4.4) and then obtain Da from correlations assuming this form of the molar flow rate ex-pression. Using Equation (8.4.4) to develop the transient material balance relation-ship for the axially-dispersed PFR gives: 274 CHAPTER 8 Nonideal FiQllLirlBeactors-aCi A az-e at (accumulation) or for constant u and Da : lin) ( ac'l + a uAeC; - AeDa -.-,) I , dz J (out) ac) AeDa -' az (out) (8.4.6) aCi ac +u-' at az (8.4.7) If 8 = t/(t) = (tu)/L (L: length of the reactor), Z z/L and Pea = (Lu)/Da (axial Peclet number), then Equation (8.4.7) can be written as: ac; + ac; = 1 a 2c; a8 az Pea az2 (8.4.8) The solution of Equation (8.4.8) when the input (i.e., Cat t = 0) is an impulse is: Thus, for the axially-dispersed PFR the RTD is: E (8) = (~Pc~?a)1 expr --,-(l_8-,-)_2 P~ea ] 47T8 _ 48 (8.4.9) (8.4.10) A plot of E(8) versus 8 is shown in Figure 8.4.2 for various amounts of dispersion.
Notice that as Pea -+ CXJ (no dispersion), the behavior is that of a PFR while as Pea -+ 0 (maximum dispersion), it is that of a CSTR. Thus, the axially-dispersed reactor can simulate all types of behaviors between the ideal limits of no back-mixing (PFR) and complete backmixing (CSTR).
The dimensionless group Pea is a ratio of convective to dispersive flow: convective flow ------ in the axial direction dispersive flow (8.4.11 ) The Peclet number is normally obtained via correlations, and Figure 8.4.3 illus-trates data from Wilhelm that are plotted as a function of the Reynolds number for packed beds (i.e.. tubes packed with catalyst particles). Notice that both the Pea and Re numbers use the particle diameter, dp , as the characteristic length: Pea dpu Re dpup Da ' M Impulse CHAPTER 8 Nonideal Flow in Reactors (a) I I o Jil Measuring point 275 2.0 1.5 § ~ 1.0 -< 0.5 o o 0.5 1.0 8=1/<1> (b) 1.5 2.0 Figure 8.4.2 I (a) Configuration illustrating pulse input to an axially-dispersed PFR. (b) Results observed at measuring point.
It is always prudent to check the variables uscd in each dimensionless group prior to their application. This is especially true with Peelet numbers, since they can have many different characteristic lengths.
Notice that in packed beds, Pea 2 for gases with turbulent flow Re (dpup)/Ji > 40, while for liquids Pea is below 1. Additionally, for unpacked tubes, .}IQ' 10 3 1\ 10-4 10-3 Figure 8.4.3 I Axial and radial Peclet numbers as a function of Reynolds number for packed-beds.
[Adapted from R. H. Wilhelm, Pure App. Chem., 5 (1962) 403, with pennission of the International Union of Pure and Applied Chemistry.] Pea (dtu/ Da) is about 10 with turbulent flow (Re = (dtup)/Ii greater than 2100) (not shown). Thus, all real reactors will have some effects of dispersion. The ques-tion is, how much? Consider again Equation (8.4.7) but now define Pea = deu/ Da where de is an effective diameter and could be either dp for a packed bed or dt for an open tube. Equation (8.4.7) can be then written as: ac aCt' 1+ a8 az (8.4.12) If the flow rate is sufficiently high to create turbulent flow, then Pea is a constant and the magnitude of the right-hand side of the equation is determined by the aspect ra-tio, L/de- By solving Equation, (8.4.12) and comparing the results to the solutions of the PFR [Equation (8.4.3)], it can be shown that for open tubes, L/dt > 20 is suf-ficient to produce PFR behavior. Likewise, for packed beds, L/dp > 50 (isothermal) and L/dp > 150 (nonisothermal) are typically sufficient to provide PFR character-istics. Thus, the effects of axial dispersion are minimized by turbulent flow in long reactors.
VIGNETTE 8.4.11 CHAPTER 8 Nonldeal Flow in Reactors 277 B. G. Anderson et aL [Ind.
Chern. Res.. 37 (1998) 815] obtained in situ of pulses of IlC-Iabeled alkanes that were passing through packed beds of zeolites us-ing positron emission tomography (PET). PET is a technique developed primarily for nuclear medicine that is able to create three-dimensional images of gamma-ray emitting species within various organs of the human body. By using PET, Anderson et aL could obtain profiles of IIC-Iabeled alkanes as a function of time in a pulse of the tracer alkane. Using analyses sim-ing data were obtained: y ,the value calculated with the information presented in Figure 8.4.3 is in good agreement with the findings.
8.5 I Prediction of Conversion with an Axially-Dispersed PFR Consider (so that an analytical solution can be obtained) an isothermal, axially-dispersed PFR accomplishing a first-order reaction. The material balance for this reactor can be written as: dCA u-kC4 = 0 dz (8.5.1) If y = CA/C~, Z = z/L, and Pea = uL IDa' then Equation (8.5.1) can be put into di-mensionless form as: d 2y dy Pea dZ2 dZ (kL) -y U • o (8.5.2) The proper boundary conditions used to solve Equation (8.5.2) have been exhaus-tively discussed in the literature. Consider the reactor schematically illustrated in Figure 8.5.1. The conditions for the so-called" open" configuration are: 278 CHAPTER 8 Nooideal Flow io Bear,tors 0) Pre-reaction zone 2=0 Reaction zone 2=1 Post-reaction zone Figure 8.5.1 I Schematic of hypothetical reactor.
Z = -00, y = 1 ) Z = +00, Y = is finite Z = 0, y(O) = y(O+) = y(O) Z = 1, y(L) = y(l+) Note that the use of these conditions specifies that the flux continuity: gives: dCA dCA D -=D-a, dz a2 dz (8.5.3) That is to say that if the dispersion coefficients in zones 1 and 2 are not the same, then there will be a discontinuity in the concentration gradient at z = 0. Alterna-tively, Danckwerts fonnulated conditions for the so-called "closed" configuration that do not allow for dispersion in zones land 3 and they are: uCAlo_ = [UCA - Da d~A Jlo+ dCAI =° dz IL_ (8.5.4) EXAMPLE 8.5.1 I The Danckwerts boundary conditions are used most often and force discontinuities in both concentration and its gradient at z = 0.
Consider an axially-dispersed PFR accomplishing a first-order reaction. Compute the di-mensionless concentration profiles for L/dp 5 and 50 and show that at isothermal con-ditions the values for = 50 are nearly those from a PFR. Assume Pe 2.
dp = 0.004 m and k/u = 25 m- I .
CHAPTER 8 Nonideal Flow in Reactors • Answer The material balance for the axially-dispersed PFR is: , dy (L 2kPea) Pe (Lid) -- -- y = 0 a P dZ udp or d 2y dy -- (500L) -- (12500L2)y = 0 dZ 2 dZ The solution to this equation using the Danckwerts boundary conditions of: 279 1 (dp) dy 1 = Y - Pea L dZ dy dZ = 0 atZ = 0 atZ = 1 gives the desired form of y as a function of Z and the result is: y = (Xl exp(524LZ) + (X2 exp(-23.9LZ) where (Xl and (X2 vary with L/dp • The material balance equation for the PFR is: or dy = _(Lk)y dZ u with y=l atZ=O The solution to the PFR material balance gives: YPF = exp[ -(kLZ)/u] exp[ -25LZ] Note that the second-term of y is nearly (but not exactly) that of the expression for YPF and that the first-term of y is a strong function of L. Therefore, it is clear that L will significantly affect the solution y to a much greater extent than YPF and that Y " YPF even for very long L. However, as shown in Figure 8.5.2, at L/dp = 50, Y = YPF for all practical matters. No-tice that Y " I at Z 0 because of the dispersion process. There is a forward movement of species A because of the concentration gradient within the reaction zone. The dispersion al-ways produces a lower conversion at the reactor outlet than that obtained with no mixing (PFR)-recall conversion comparisons between PFR and CSTR.
"'2 ...
8"'O'---"C ....
H1Aou:P"-'TuE<.JRlL<8'"--i'N'l.QlLOuuideaLEill'LinBeac.lauL_ . .
.
0.95 0.9 0.85 ~, 0.8 0.75 0.7 0.65 0.6 0.0 0.20 0.40 0.60 0.80 1.0 Z 0.8 0.6 0.4 0.2 o 0.0 0.20 0.40 z 0.60 0.80 1.0 VIGNETTE 8.5.1 Figure 8.5.2 I Dimensionless concentration profiles for axially-dispersed (y) and plug flow reactors.
CHAPTER 8 Nonideal Flow in Reactors 281 282 CHAPTER 8 Nonideal Flow in ReactQ[~.~ ~.
~ 8.6 I Radial Dispersion Like axial dispersion, radial dispersion can also occur. Radial-dispersion effects nor-mally arise from radial thermal gradients that can dramatically alter the reaction rate across the diameter of the reactor. Radial dispersion can be described in an analo-gous manner to axial dispersion. That is, there is a radial dispersion coefficient. A complete material balance for a transient tubular reactor could look like: ac ac (PC [a2c 1 ac] +u-=D +D -+--at az a az2 ,. ar2 r ar (8.6.1 ) Ife = t/(t) = (tu)/L , Z z/L, R r/ de (de is dp for packed beds, dt for unpacked tubes), Pea = (ude)/Da and Pe,. = (ude)/D,. (D,. is the radial-dispersion coefficient), then Equation (8.6.1) can be written as: (8.6.2) The dimensionless group Pe,. is a ratio of convective to dispersive flow in the radial direction: Pe,.
convective flow ~----~ in radial direction dispersive flow (8.6.3) Referring to Figure 8.4.3, for packed beds with turbulent flow, Pe,. = 10 if de = dp-For unpacked tubes, de = dt and Pe,. = 1000 with turbulent flow (not shown).
Solution of Equation (8.6.1) is beyond the level of this text.
8.7 I Dispersion Models for Nonideal Flow in Reactors As illustrated above, dispersion models can be used to described reactor behavior over the entire range of mixing from PFR to CSTR. Additionally, the models are not confined to single-phase, isothermal conditions or first-order, reaction-rate func-tions. Thus, these models are very general and, as expected, have found widespread use. What must be kept in mind is that as far as reactor performance is normally concerned, radial dispersion is to be maximized while axial dispersion is minimized.
The analysis presented in this chapter can be used to describe reaction con-tainers of any type~they need not be tubular reaetors. For example, consider the situation where blood is flowing in a vessel and antibodies are binding to cells on the vessel wall. The situation can be described by the following material balance: with ac u az (8.7.1) CHAPTER 8 Nonideal Flow in Reactors 283 ac uCo = uC - Da az ac -=0 az ac = ° ar ac -Dr _ = AB ar at z = 0, all r at z = L, all r at r = 0, all z at r = rn all z where C is the concentration of antibody, rt is the radius of the blood vessel and AB is the rate of antibody binding to the blood vessel cells. Thus, the use of the dis-persion model approach to describing flowing reaction systems is quite robust.
Exercises for Chapter 8 1.
Find the residence time distribution, that is, the effluent concentration of tracer A after an impulse input at t = 0, for the following system of equivolume CSTRs with a volumetric flow rate of liquid into the system equal to v: How does the RTD compare to that of a single CSTR with volume 2V?
2.
Sketch the RTD curves for the sequence of plug flow and continuous stirred tank reactors given in Figure 8.3.1.
3.
Consider three identical CSTRs connected in series according to the diagram below.
(a) Find the RTD for the system and plot the E curve as a function of time.
(b) How does the RTD compare to the result from an axially-dispersed PFR (Figure 8.4.2)? Discuss you answer in terms of the axial Peelet number.
(c) Use the RTD to calculate the exit concentration of a reactant undergoing first-order reaction in the series of reactors. Confirm that the RTD method 284 CHAPTER 8 Nonideal Flow in ReactillS>--- gives the same result as a material balance on the system (see Example 3.4.3).
4.
Calculate the mean concentration of A at the outlet (z L) of a laminar flow, tubular reactor (C ~) accomplishing a second-order reaction (kC1), and compare the result to that obtained from a PFR when [(C~kL)/ u] I. Referring to Example 8.1.1, is the deviation from PFR behavior a strong function of the reaction rate expression (i.e., compare results from first- and second-order rates)?
5.
Referring to Example 8.2.3, compute and plot the dimensionless exit concentration from the Berty reactor as a function of time for decreasing internal recycle ratio to the limit of PFR behavior.
6.
Consider the axially-dispersed PFR described in Example 8.5.1. How do the concentration profiles change from those illustrated in the Example if the second boundary condition is changed from: to y=o at Z = I at Z 00 7.
Write down in dimensionless form the material balance equation for a laminar flow tubular reactor accomplishing a first-order reaction and having both axial and radial diffusion. State the necessary conditions for solution.
8.
Falch and Gaden studied the flow characteristics of a continuous, multistage fermentor by injecting an impulse of dye to the reactor [E. A. Falch and E. L. Gaden, Jr., Biotech. Bioengr., 12 (1970) 465]. Given the following RTD data from the four-stage fermentor, calculate the Pea that best describes the data.
Impulse Response Curve from a Four-Stage Fermenter 1 == .S 0.8 :; .0 .~ 0.6 '8 <U .§ <U 0.4 u == 2 2.5 CHAPTER 8 Nonideal Flow in Reactors 0.000 0.000 0.950 0.745 0.050 0.001 0.990 0.710 0.090 0.040 1.035 0.675 0.120 0.080 1.080 0.630 0.170 0.140 1.110 0.600 0.210 0.220 1.180 0.550 0.245 0.330 1.200 0.525 0.295 0.420 1.240 0.485 0.340 0.530 1.290 0.460 0.370 0.590 1.365 0.380 0.420 0.670 1.410 0.360 0.460 0.730 1.450 0.325 0.500 0.810 1.490 0.310 0.540 0.830 1.580 0.250 0.590 0.860 1.670 0.220 0.630 0.870 1.760 0.180 0.670 0.875 1.820 0.150 0.710 0.850 1.910 0.140 0.750 0.855 2.000 0.125 0.790 0.845 2.120 0.100 0.825 0.840 2.250 0.080 0.870 0.810 2.320 0.070 0.920 0.780 285 9.
Using the value of the Pea determined in Exercise 8, compute the concentration profile in the reactor for the reaction of catechol to L-dopa catalyzed by whole cells according to the following rate of reaction: where Cs is the concentration of substrate catechol. This reaction is discussed in Example 4.2.4. The values of the various parameters are the same as those determined by the nonlinear regression analysis in Example 4.2.5, that is, cf = 0.027 mol L-1, rmax = 0.0168 mol L-I h- I , and Km = 0.00851 mol L- I. Assume that the mean residence time of the reactor is 2 h.
10.
Using the same rate expression and parameter values given in Exercise 9, compute the concentration profile assuming PFR behavior and compare to the results in Exercise 9.
________---""C"'--H APT E RL-- 9""'----Nonisothermal Reactors 9.1 I The Nature of the Problem In Chapter 3, the isothennal material balances for various ideal reactors were derived (see Table 3.5.1 for a summary). Although isothermal conditions are most useful for the measurement of kinetic data, real reactor operation is nonnally nonisothennal.
Within the limits of heat exchange, the reactor can operate isothennally (maximum heat exchange) or adiabatically (no heat exchange); recall the limits of reactor be-havior given in Table 3.1.1. Between these bounds of heat transfer lies the most com-mon fonn of reactor operation-the nonisothermal regime (some extent of heat ex-change). The three types of reactor operations yield different temperature profiles within the reactor and are illustrated in Figure 9.1.1 for an exothennic reaction.
If a reactor is operated at nonisothennal or adiabatic conditions then the ma-terial balance equation must be written with the temperature, T, as a variable. For example with the PFR, the material balance becomes: vir(Fi,T) (9.1.1) Since the reaction rate expression now contains the independent variable T, the material balance cannot be solved alone. The solution of the material balance equation is only possible by the simult\final) (Tinitial' cI>initial) \ is the extent of reaction (s~ Chapter 1), !::..Hr is the heat of reaction, MS is the total mass of the system, and (Cp) is an average heat capacity per unit mass for the system. Since enthalpy is a state variable, the solution of the integral in Equation (9.3.4) is path independent. Referring to Figure 9.3.2, a simple way to analyze the in-tegral [pathway (I)] is to allow the reaction to proceed isothermally [pathway (II)] and then evaluate the sensible heat changes using the product composition [pathway (III)].
That is, I T,;,,,, = !::..H \ .
-\ ..
+ MS IC dT Q rI Tinitial (fmal Illitlal) final \ Pfinaj) Tinitial (9.3.5) where a positive value for the heat of reaction denotes an endothermic reaction.
Since the reaction-rate expressions normally employ moles of species in their evaluation: I T,;"", MSfinal (Cp..)dT = .
tuwl Tinitial (9.3.6) where ni is the moles of species i, and CPi is the molar heat capacity of species i. Note also that the heat exchange is the integral over time of the heat transfer rate, that is, Q = J Qdt = J UAH(T - T)dt (9.3.7) where U is an overall heat transfer coefficient, AH is the heat transfer area, and T is the reference temperature. Combining Equations (9.3.7), (9.3.6), and (9.3.5) CHAPTER 9 Nooisothermal Reactors 291 and recalling the definition of the extent of reaction in terms of the fractional conversion [Equation (1.2.10)] gives: (9.3.8) or in differential form: -6.Hr l ro dIe:L . dT UAH(T - T) = n~- + (nC)-vee dt j I P, dt Notice that [see Equations (1.2.10) and (1.3.2)]: (9.3.9) n~ dIe -Ve dt d\ = rV dt (9.3.10) Using Equation (9.3.10) in Equation (9.3.9) yields: "dT UAH (T - T) = 6.Hr l rorV + LJ (nj Cp)--;tt I (9.3.11) EXAMPLE 9.3.1 I Equations (9.3.11), (9.3.9), and (9.3.8) each define the energy balance for a batch reactor.
Show that the general energy balance, Equation (9.3.9), can simplify to an appropriate form for either adiabatic or isothermal reactor operation.
• Answer For adiabatic operation there is no heat transfer to the surroundings (i.e., U = 0). For this case, Equation (9.3.9) can be written as: _ - ~HrITO 0 die L ( .) dT 0-ne- + nC-Ve dt j I p, dt or when integrated as: (9.3.12) If the reactor is operated at isothermal conditions, then no sensible heat effects occur and Equation (9.3.9) becomes: (9.3.13) The description of the nonisothermal batch reactor then involves Equation (9.3.1) and either Equation (9.3.9) or (9.3.11) for nonisothermal operation or Equation (9.3.12) 292 EXAMPLE 9.3.2 I CHAPTER 9 NQoisothe.unalBeacJ:Qr.s....
for adiabatic operation. Notice that for adiabatic conditions, Equation (9.3.12) provides a relationship between T and Insertion of Equation (9.3.12) into Equation (9.3.1) allows for direet solution of the mass balance.
The hydration of 1-hexene to 2-hexanol is accomplished in an adiabatic batch reactor: OH ~+H20===>~ (A) (8) (C) The reactor is charged with 1000 kg of a 10 wI. % H2S04 solution and 200 kg of 1-hexene at 300 K. Assuming that the heat capacities for the reactants and products do not vary with temperature, the heat of reaction does not vary with temperature, and the presence of H2S04 is ignored in the calculation of the heat capacity, determine the time required to aehieve 50 pereent conversion and the reactor temperature at that point.
Data: I-hexene H20 2-hexanol 43.8 16.8 54.0 -10.0 -68.0 -82.0 reaction rate constant: k 104exp -IO;(RJ)] (S-I) • Answer The material balanee on the batch reactor is: or dt k(l -since the reaction is first-order (units on k and large excess of water: I-hexene is the limit-ing reagent). The energy balance is: where Vi I. In order to calculate the heat of reaction, the standard heats of formation can be used as follows: IO = -4 kcal/mol CHAPTER 9 Nooisotherma l Reactors Thus, the energy balance equation can be written: where The values of n~ and n~ are: n~ = (2 X 105 g)(l mol/84 g) = 2381 mol n~ = (9 X 105 g)(l mol1l8 g) = 50000 mol so that: nVn~ = 21 By placing the values for n~, Ai, TO, and the Cp into the energy balance, the result is: 4000 fA T = 300 + 421.8 7.8 fA 293 EXAMPLE 9.3.3 I The material balance equation is then solved with k(D being first converted to k(j~) by sub-stitution of the energy balance for T: l -l~ -k = 10 4 exp ( 4000 fA Rg 300 + ) 421.8 - 7.8 fA The material balance equation must be solved numerically to give t = 1111 s or 18.52 min.
The reactor temperature at this point is obtained directly from the energy balance with fA = 0.5 to give T = 304.8 K.
Consider accomplishing the reaction A + B ~ C in a nonisothermal batch reactor. The re-action occurs in the liquid phase. Find the time necessary to reach 80 percent conversion if the coolant supply is sufficient to maintain the reactor wall at 300 K.
Data: 6.Hr -15 kJ/mol VA H 50 J/(s-K) Cpr = 150 J/(mol-K) k C~ = 0.5 mol/L C~ = 0.6 mol/L Cp , CPR 65 J/(mol-K) n~ 100 mol 1 ll20000 J/mol ( I 5 X 10-' exp ---Rg 300 1)l T J (Llmol/s) • Answer The material balance for the batch reactor is: d'j' ~=kCO(I-t·)·(1.2 j" dt A . A A) The energy balance can be written as: '(' )_A (Jdj~ lJAH T - T -I..>. H, nA di + The material and energy balance equations must be solved simultaneously. A convenient form for solution by numerical techniques is: dfA dt dT dt n~(1 with fA 0 and T 300 K at t O. The results are shown in Figure 9.3.3. From the data given in Figure 9.3.3, it takes 462 s to reach a fractional conversion of 0.8. Additionally, the final temperature is 332 K. Notice that the final temperature is not the maximum tempera-ture achieved during the reaction, in contrast to adiabatic operation.
0.8 c: .S '" 0.6 '-< c: 0 <l) c;J 0.4 c .S t) J: 100 I I I I I I I 200 300 Time (s) I !
I 400 I I 500 340 g 330 ~ '" '§ 320 <l) 0< E ~ 310 o 100 200 300 Time (s) 400 500 Figure 9.3.3\ Fractional conversion and temperature profiles for the reactor described in Example 9.3.3.
VIGNETTE 9.3.11 CHAPTER 9 Nonisothermal Re.ac..ul"-'ou;rs'---"2 ..
9 .....
5 M. K1adko [CHEMTECH, 1 (971) 141] presented an case study of per-isomerization reaction. In a 50 gal steam-heated 1;1 ratio of A to sol-with a reactant-to-out of con-ffects of the in this case is higher reaction more efficiently ..
2...
9,.;6L.---"CLlH1A ......
P'-JTLEI:JlRLOi (E) (C) If equimo1ar butadiene and ethylene at 450°C and 1 atm are fed to a PFR operating adiabat-ically, what is the space time necessary to reach a fractional conversion of 0.1 O?
k 1075 exp[ -27,500/(RgT)] L/mol/s /::,. H, - 30000 cal/mol Cpu 36.8 cal/mol/K CP1 20.2 cal/mol/K Cpr = 59.5 cal/mol/K • Answer Assume that each CPi is not a strong function of temperature over the temperature range obtained within the PFR (i.e., each is not a function of T). The material and energy balance equations are: and -kC~ [ Ii' ]2(TO )2 -k(ci\f B_ 1+ sldB T Thus, the material balance can be written as: -CO diB B dT df' r B _ kCo I dT -B II 1 - iB 0.5 iB T since £B 0.5 2 lI 1, I !J -0,5. Now for the energy balance, dT since 300 CHAPTER 9 Nooisothermal Reactors Thus, 2: FiCPir dT = FZ (1 - IB)(36.S)(T -TO) + F2 (1 fe)(20.2)(T TO) 1 "TO + F2!B(59.5)(T TO) or 2: FiCPifTdT = (57 + 2.5 fe) F2 (T - TO) l TO The energy balance then becomes: ° ° _ (-30,000) ° (57 + 2.5Ie)Fs (T - T) -(-1) Fsfs or T (30,000)Is 723 + -'------'--""'-57 + 2.51e VIGNETTE 9.4.1 The solution of the material balance equation: with T from the energy balance gives a value of 7 = 47.1 s. Additionally, the exit tempera-ture is 775 K.
The energy balance for the PFR can also be written as follows by combining Equations (9.4.1) and (9.4.2): C H A PT E R 9 NonisolhermaLR""e""'8c"'1"-'ow:ts'----"3 ....
0 LL 1 or U(T (9.4.6) U(T - T)i = '" FC dT _ !:.H,ITo FO die d ~ I Pi dV Vo e dV t I R , R (9.4.7) Using the fact that: Equation (9.4.7) can be written as: dT '" FC = -7' I Pi dVR 4 !:.H,ITo)r - U(T - T )-dt (9.4.8) Thus, the material and energy balances for the PFR can be written as (Fe = Cev = Ce7Td;u/4; dVR = 7Td;dz/4): 4U (T - T) } dt (9.4.9) with Ce = c2 and T = TO at z = 0, ~here u is the superficial linear velocity, p is the average density of the fluid, and Cp is the average heat capacity per unit mass.
Equation (9.4.9) is on~ applicable when the density of the fluid is not changing in the reactor and when Cp is not a function of temperature, since the following rela-tionship is used to obtain Equation (9.4.9): EXAMPLE 9.4.3 (9.4.10) A PFR of dimensions L = 2 m and d1 = 0.2 m is accomplishing a homogeneous reaction. The inlet concentration of the limiting reactant is c2 0.3 kmol/m3 and the inlet temperature is 700K. Other data are: - !J.Hr 104 kJ/kmol,C" 1kJ/(kg-K),E = lOOkJ/mol,p 1.2kg/m3, u = 3 mis, and A 5 s-I. Calculate the dimensionless concentration (y = Ce/c2) and tem-perature (8 profiles for adiabatic (U = 0) and nonisothennal (U 70 J/(m2-s-K)) operations. (Example adapted from J. Villadsen and M. L. Michelsen, Solution of Differential Equation Models by Polynomial Approximation, Prentice-Hall, Englewood Cliffs, 1978, p. 59.) • Answer From the units on A, it is clear that the reaction rate is first order. Using Equation (9.4.9) with a first-order reaction rate expression gives: 302 CHAPTER 9 Nonisotbermal Reactors -u dC, Aexp[ -E/(R~T)]C, dz -dT -upCp dz = (-I1H,)A exp[ -E/(RgT)]C, Let x = z/L, y = c,/c2 and (j = T/To. Using these dimensionless variables in the material and energy balance relationships yields: dy dx de -[( 1)] -dx = f3T(Da)y exp y 1 - e - Hw(e with y = 1 at x = 0 and where: 1) Notice that all the groupings Da, f3T' y, and H w are dimensionless. For adiabatic operation, Hw O. For this case, the mass balance equation can be multiplied by f3T and added to the energy balance to give: d -(e + f3TY) = 0 dx Integration of this equation with y e = 1 at x = 0 leads to: (j = 1 + f3T(1 - y) (9.4.11) Equation (9.4.11) is the adiabatic relationship between temperature and conversion in dimensionless form. The mass balance can then be written as: dy dx with y=latx=O The solution of this differential equation is straightforward and is shown in Figure 9.4.2. For the nonisothermal case, the material and energy balances must be solved simultaneously by _________________~C>LLHuA"_P"'_LT"EiWlR""9'____L'LN,illlisQtbBImaLBaaclo~ ~3 ...0 ...3 ...
1.2 1.35 1.0 1.30 1.25 0.8 1.20 ;>-, 0.6 1.15 "'I 0.4 1.10 1.05 0.2 1.00 0.0 0.95 0.0 0.2 0.4 0.6 0.8 1.0 x Figure 9.4.2 I Dimensionless concentration and temperature profiles for adiabatic and nonisothermal operati~. Ya and 8a are for adiabatic conditions while Yn; and 8n; are for nonisothermal operation.
numerical methods. The nonisothermal results are plotted also in Figure 9.4.2. Notice that there is a maximum in the value of 8,,; that occurs at x 0.57 and gives T = 860 K. The maximum in the temperature profile from nonisothermal operation is normally denoted as the hot spot in the reactor.
9.5 I Temperature Effects in a CSTR Although the assumption of perfect mixing in the CSTR implies that the reactor contents will be at uniform temperature (and thus the exit stream will be at this tem-perature), the reactor inlet may not be at the same temperature as the reactor. If this is the case and/or it is necessary to determine the heat transferred to or from the re-actor, then an energy balance is required.
The energy balance for a CSTR can be derived from Equation (9.2.7) by again carrying out the reaction isothermally at the inlet temperature and then evaluating sensible heat effects at reactor outlet conditions, that is, Q (9.5.1) where.the superscript f denotes the final or outlet conditions. For adiabatic opera-tion, Q = O.
EXAMPLE 9.5.1 I The nitration of aromatic compounds is a highly exothermic reaction that generally uses cat-alysts that tend to be corrosive (e.g., HN03IH2S04). A less corrosive reaction employs N20 s as the nitrating agent as illustrated below: NOz N,o,+ 2U= 2O+H'O (A) (8) (e) (D) If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of NzOs? The reaction rate is first order in A and second order inB.
Data: I::.Hr = -370.1 kllmol Cp, = 84.5 J/(mol-K) Cps = 137 J/(mol-K) Cpc = 170 ll(mol-K) CPD = 75 J/(mol-K) TO = 303 K F~ = 10 mol/min F2 = 30 mol/min v = 1000 L/min C~ = 0.01 mol/L k r(40kJ/mOl)( 1 1)] 0.090 expl - - -R~ 303 T (Llmol? (mint] EXAMPLE 9.5.2 I • Answer The reaction occurs in the liquid-phase and the concentrations are dilute so that mole change with reaction does not change the overall density of the reacting fluid. Thus, CA = C~ (1 fA)' FA = F~ (1 fA) Cs = C~ (3 2fA)' Fs = F~(3 2fA) Fe = 2FUA' FD = F~fA The material balance on the CSTR can be written as: The energy balance for the adiabatic CSTR is: 0= I::.HrFUA F~ (1 - fA) CPA (T TO) + F~ (3 2fA)CPB (T - TO) + 2FUACpc (T - TO) + F~.t~CPD (T - TO) For fA = 0.35, the energy balance yields T = 554 K. At 554 K, the value of the rate constant is k = 119.8 L2/(mo12-min). Using this value of k in the material balance gives V 8,500 L and thus T 8.5 min.
Consider the aromatic nitration reaction illustrated in Example 9.5.1. Calculate the reactor volume required to reach 35 percent conversion if the reactor is now cooled.
Data: VAil 9000 J/(min-K) T~.
323 K (and is constant) v 100 L/min c2 O. 10 mol/L All other data are from Example 9.5. I.
• Answer The material balance equation remains the same as in Example 9.5.1. The energy balance is now: VAIl (T~ - T) 6.HrF~j~ + F~ (I - fA)Cp, (T TO) + F~ (3 2j~)CpfJ (T -TO) + 2F~fA CfJc (T -TO) + FV~ CPD (T TO) whenj~ 0.35, the energy balance yields T 407 K. At 407 K, the reaction rate constant is k 5.20 L2/(mof-min). Using this value of k and c2 0.10 mol/L in the material bal-ance equation gives V 196 L.
9.6 I Stability and Sensitivity of Reactors Accomplishing Exothermic Reactions Consider the CSTR illustrated in Figure 9.6.1. The reactor is accomplishing an exothermic reaction and therefore must transfer heat to a cooling fluid in order to remain at temperature T. Assume that the heat transfer is sufficiently high to maintain the reactor wall temperature at Te . Therefore, Q = UAH(T - TJ heat removed from reactor vcCpc(Tc - T?) sensible heat change of coolant +-TO (9.6.1) Figure 9.6.1 I Schematic illustration of a CSTR that is maintained at temperature T by transferring heat to a coolant t1uid (Tc > (9.6.2) 306 CHAPTER 9 Nonisotbermal Reactors where Vc is the volumetric flow rate of the coolant and Cp, is the heat capacity of the coolant and is not a function of temperature. Solving for Tc and then substitut-ing the expression back into the heat transfer equation yields: .
UAHvcCp Q = ' (T T~) = AA (T -T~) UAH + veCpc The energy balance on the CSTR can be written as [from Equation (9.5.1) with a first-order reaction rate expression]: or since: as: -VerV = F~(je f~) (material balance) (first-order reaction rate) (9.6.3) where vP is the volumetric flow rate of the product stream and p and Cp are the av-erage density and heat capacity of the. outlet stream. Rea:ranging Equation (9.6.3) and substituting Equation (9.6.2) for Q gives (note that Q is heat removed): (9.6.4) Let: AA (l'Cl'l=--vPpCp -kT(t:.HrITo) pCp so that Equation (9.6.4) can be written as: (1 + Cl'Cl'l)T (TO + Cl'Cl'lT~) = Cl'Cl'2Ce Since: Ce = C~/(l + kT) (9.6.5) from the solution of the material balance equation, Equation (9.6.5) can be formu-lated as: ° ° _ Cl'Cl'2C~ (l + Cl'Cl'l)T -(T + Cl'Cl'lTcJ ---1 + kT Qr (heat removed) = Qg (heat generated) (9.6.6) CHAPTER 9 Nooisothermal Reactors 307 (1) :~O:O:I= 00, T= I I I I I I I I I I T TO+0:0:1T2 1+0:0:1 Figure 9.6.2 I Schematic illustration of Q,. and Qg as functions of T.
(II) ",--- Qo / I:'> / I I I I I I I I / " Q,.
-Q 3 g T (III) Q,.
,,'" --Qg / I I I I I I I I / " T T T Figure 9.6.3 I Steady-state solutions to Equation (9.6.3).
If Q,. and Qg are plotted versus the reaction temperature, T, the results are illustrated in Figure 9.6.2. A solution of Equation (9.6.6) occurs when Q,. equals Qg, and this can happen as shown in Figure 9.6.3. Notice that for cases (I) and (III) a single solution exists. However, for case (II) three steady-states are possible. An impor-tant question of concern with case (II) is whether all three steady-states are sta-ble. This is easy to rationalize as follows. At steady-state 1, if T is increased then Q,. > Qg so the reactor will return to point 1. Additionally, if T is decreased, Qg > Qr so the reactor will also return to point 1 in this case. Thus, steady-state 1 is stable since small perturbations from this position cause the reactor to return to the steady-state. Likewise steady-state 3 is a stable steady-state. However, for steady-state 2, if T is increased, Qg > Q,. and the reactor will move to position 3.
If T is decreased below that of point 2, Q,. > Qg and the reactor will move to point 1. Therefore, steady-state 2 is unstable. It is important to determine the stability of reactor operation since perturbations from steady-state always occur in a real system. Finally, what determines whether the reactor achieves steady-state 1 or 3 is the start-up of the reactor.
308 EXAMPLE 9.6.1 I CHAPTER 9 Nonisothermal Reactors Calculate the steady-states for the following reactor configuration. Is there an unstable steady-state?
Data: A + B =} 2C in the liquid phase, V = I L, k 33 X 109 exp [-20,000/(RgDJ L/(mol . min), - tJ.Hr = 20 kcal/mol, C~ 20 mol/L, C~ = 3 mol/L, v = 100 cm3/min, TO l7°C, T~ = 87°C, pCp = 650 cal/(L' °C), U = 0.1 cal/(cm2 • min' K), and AH = 250 cm2• • Answer The reaction rate expression is: giving the following material balance: Therefore.
Qr (1 + aal) T - (TO + aal T~) -k(C~)2 r(tJ.Hr)(1 - fs)(iIi - fs) Qg = pCp First, solve the material balance equation for fE (gives two values for fE-one will have phys-ical meaning) at a particular T and next calculate Qr and Qg- A sampling of results is pro-vided below (assume vcCPc » UAH for the calculation of Qr): 310 320 330 340 350 360 370 If these data are plotted, they yield: 0.049 0.124 0.264 0.461 0.658 0.807 0.898 0.77 14.62 28.46 42.31 56.20 70.00 83.80 4.55 11.45 24.40 42.50 60.77 74.78 82.85 314.2 340.0 T 368.8 CHAPTER 9 Nooisothermal ReactoCGrs"--.~~~~~~~-,,3 ...
0~9 Thus, there are three steady-states and they are: 1 2 3 314.2 340.0 368.8 41.2 67.0 95.8 0.079 0.460 0.888 Steady-state 2 is unstable. Notice the vast differences in i8 for the three steady-states. Thus, it is clear that attempts by a naive designer to operate at steady-state 2 would not succeed since the reactor would not settle into this steady-state at all.
In addition to knowing the existence of multiple steady-states and that some may be unstable, it is important to assess how stable the reactor operation is to variations in the processing parameters (i.e., the sensitivity). In the above example for the CSTR, it is expected that the stable steady-states have low sensitivity to variations in the processing parameters, while clearly the unstable steady-state would not.
To provide an example of how the sensitivity may be elucidated, consider a tu-bular reactor accomplishing an exothermic reaction and operating at nonisothermal conditions. As described in Example 9.4.3, hot spots in the reactor temperature pro-T~(4) I f f I, , , , I I I I / / / / / / !
......•......•.•......•..••..•...
, T~(3) .......
. ... TOe') ...•..• c .:..
Length Figure 9.6.4 I Temperature profiles in a tubular reactor operating nonisothermically and conducting an exothermic reaction. T? is the temperature of the coolant fluid.
310 EXAMPLE 9.6.2 I CHAPTER 9 Nooisothermal Reactors file can occur for this situation. Figure 9.6.4 illustrates what a typical reactor tem-perature profile would look like. Consider what could happen as the temperature of the coolant fluid, T~, increases. As T~ becomes warmer T~ (i + 1) > T~ (1), i = 1, 2, 3, then less heat is removed from the reactor and the temperature at the reactor hot spot increases in value. Eventually, heat is generated at a sufficiently high rate that it cannot be removed [illustrated for T~ (4)] such that the hot spot temperature exceeds some physical limit (e.g., phase change of fluid, explosions or fire, the cat-alyst melts, etc.) and this condition is called runaway. Thus, reactor operation close to a runaway point would not be prudent, and determining the sensitivity towards the tendency of runaway a critical factor in the reactor analysis. Several criteria have been developed to assess the sensitivity of reactors; each involves the use of critical as-sumptions [see, for example, G. F. Froment and K. B. Bischoff, Chemical Reaction Analysis and Design, Wiley, New York, 1977, Chapter 1]. Here, an example of how reactor stability can be assessed is illustrated and was provided by J. B. Cropley.
A tubular reactor packed with a heterogeneous catalyst is accomplishing an oxidation reac-tion of an alcohol to an aldehyde. The reactions are: alcohol + air => aldehyde + air => CO2 + H20 In this series reaction pathway, the desired species is the aldehyde. Since both reactions are exothermic (second reaction is highly exothermic), the reactor is operated nonisothermally.
The reactor is a shell-and-tube heat exchanger consisting of 2500 tubes of I inch diameter.
Should the heat exchanger be operated in a cocurrent or countercurrent fashion in order to provide a greater stabilization against thermal runaway?
• Answer Since this example comes from the simulation of a real reactor, the amount of data neces-sary to completely describe it is very high. Thus, only the trends observed will be illustrated in order to conserve the length of presentation.
Schematically, the reactor can be viewed as: (A) '-'IL----, Alcohol + -':;111111:11: Air Alcohol, aldehyde, NJ llllJJ!lI!1 IJ1!!1! HIH 'Ill Hill!!! lllIIIlJlll!ll11l!lllIIIHIIJ!1!111 lfrTfTI1TIT -+----02' CO2, H20 I ® The cooling fluid is fed at point A or at point B for cocurrent and countercurrent opera-tion, respectively. Next, the reactor temperature profiles from the two modes of operation CHAPTER 9 NnnlsDIDBrmal Reacto~ 311 are illustrated. The three profiles in eaeh graph are for different coolant feed temperatures, T;).
Notice that a hot spot occurs with countercurrent operation. In order to access the sensitivity, h Cocunen!
.......
.............................
.--..-----..- ..-------t h Countercurrent ......................................
"T?t Length Length Cropely suggests plotting the maximum temperature in the reactor as a function of the inlet coolant temperature. The results look like: Inlet coolant temperature The slope of the line would be an indication of the reactor stability to variations in the inlet coolant temperature. Clearly, cocurrent operation provides better sensitivity and this conclu-sion is a general one. The reason for this is that by operating in a cocurrent manner the great-est ability to remove heat [largest D.T Teoolan,)] can occur in the region of highest heat generation within the reactor.
Calculate the final temperature and time required to reach 50 percent conversion in the batch reactor described in Example 9.3.2 if the heat of reaction is now -40 kcallmol. Do you think that this time is achievable in a large reactor system?
Find the final temperature and time required to reach 90 percent conversion in the reactor system of Exercise I.
2.
Exercises for Chapter 9 1.
312 CHAPTER 9 Nonisothermal Reactors 3.
Plot the fractional conversion and temperature as a function of time for the batch reactor system described in Example 9.3.3 if the reactor is now adiabatic (U = 0). Compare your results to those for the nonisothermal situation given in Figure 9.3.3. How much energy is removed from the reactor when it is operated nonisothermally?
4.
Consider what happens in the batch reactor given in Example 9.3.3 if the wall temperature does not remain constant. For comparison to the constant wall temperature, calculate the fractional conversion and reactor temperature as a function of time when: T = 300 + O.1t where t is in seconds.
5.
Calculate the exit temperature and T for the PFR described in Example 9.4.2 when mole change with reaction is ignored (i.e., Bs = 0). How much error is introduced by making this change?
6.
Calculate the exit temperature and T for the PFR described in Example 9.4.2 when the temperature effects on the concentration are ignored. Is this a reasonable simplification or not?
7.
An adiabatic PFR can be described by the following set of equations: ;: = -4y exp [18(1 )J ~~ = O.2y exp [ 18(1 )J y = (J = 1 at x = 0 Solve these equations and plot y and (J as a function of x for o(entrance) :s x :s 1 (exit). What happens if the heat of reaction is doubled?
8.
Ascertain whether the following exothermic reaction: k j A+A( )P+P k2 (k1 and k1 at 80°C) could bc carried out in the reactor shown below: CSTR 1 4 =0.4 T=80°C CHAPTER 9 Nooisothermal Reactors 313 Calculate the volume and heat removed from the CSTR and the PFR. Do the magnitudes of the heat being removed appear feasible? Why or why not?
Data: C = 45 cal mol- l K- 1 PA C = 40 cal mol- 1 K- 1 Pp - liHr 10,000 cal mol- l C~ = 1.5 mol L- 1 F~ = 100 mol min- l 9.
The ester of an organic base is hydrolyzed in a CSTR. The rate of this irre-versible reaction is first-order in each reactant. The liquid volume in the vessel is 6500 L. A jacket with coolant at 18°C maintains the reactant mixture at 30°C. Additional data: Ester feed stream-l M, 30°C, 20 Lis Base feed stream-4 M, 30°C, lOLls Rate constant = 1014 exp(-ll,OOO/T) M- 1s- 1, Tin K Ii Hr = -45 kcal/mol ester The average heat capacity is approximately constant at 1.0 kcal L-1 °c- 1.
(a) What is the conversion of ester in the reactor?
(b) Calculate the rate at which energy must be removed to the jacket to maintain 30°C in the reactor. If the heat transfer coefficient is 15 kcal S-lm-2 K- 1, what is the necessary heat transfer area?
(c) If the coolant supply fails, what would be the maximum temperature the reactor could reach?
10.
A reaction is carried out in an adiabatic CSTR with a volume of 10,000 L.
The feed solution with reactant A, at a concentration of 5 M, is supplied at 10 L s-1. The reaction is first-order with rate constant: k = 1013 exp( -12500/T) S-1, where T is in K The density is 1000 kg m- 3 and Cp = 1.0 kcal kg-1K- 1. The heat of reaction is Ii H, = -70 kJ mol- 1• (a) Calculate the reactor temperature and exit concentration for feed temperatures of 280, 300, and 320 K.
(b) To maintain the reactor temperature below 373 K, a well-mixed cooling jacket at 290 K is used. Show that it is possible to get 90 percent conversion in this reactor with a feed temperature of 320 K. Do you anticipate any start-up problems?
11.
The reversible, first-order reaction shown below takes place in a CSTR.
A B 314 CHAPTER 9 Nonisotbermal Reactors The following data are known: k1 = 103 exp(-2500/T)s-l, TinK AHr = -lOkcalmol- 1 K = 8 at 300 K Cp = 1kcal kg- 1K- 1 P = 1 kg L- 1 (a) For a reactor space time of 10 min, what is the conversion for a 300 K operating temperature? What is the conversion at 500 K? (Remember: the equilibrium constant depends on temperature.) (b) If the feed temperature is 330 K and the feed concentration is 5 M, what is the necessary heat-removal rate per liter of reactor volume to maintain a 300 K operating temperature?
-----------~ 10 Reactors Accomplishing Heterogeneous Reactions 10.1 I Homogeneous versus Heterogeneous Reactions in Tubular Reactors In earlier chapters, tubular reactors of several forms have been described (e.g., laminar flow, plug flow, nonideal flow). One of the most widely used industrial reactors is a tubular reactor that is packed with a solid catalyst. This type of reac-tor is called a fixed-bed reactor since the solid catalyst comprises a bed that is in a fixed position. Later in this chapter, reactors that have moving, solid catalysts will be discussed.
A complete and perfect model for a fixed-bed reactor is not technically possi-ble. However, such a model is not necessary. Rather, what is needed is a reasonably good description that accounts for the major effects. In this chapter, the fixed-bed reactor is analyzed at various degrees of sophistication and the applicability of each level of description is discussed.
Consider the PFR illustrated in Figure 1O.1.Ia. A mass balance on the reactor volume located within dz can be written as: (10.1.1) Since Fi = uACCi and V Acz, Equation (10.1.1) gives: (10.1.2) provided u and Ac are not functions of z. The units of rare moles/(time . volume).
Now let the tube be packed with a solid catalyst and write the material balance again (situation depicted in Figure 10.1.1b): (10.1.3) 316 C H A PT E R 10 Reactors Accomplishing Heterogeneolls Reactions where the units on r (rate of reaction over the solid catalyst) are now mole/(mass of catalyst)/(time) and PB is the bed density (mass of catalyst)j(volume of bed).
Equation (10.1.3) can be written as: dCA -u- = YfoPB( -vJr dz (10.1.4) provided u and Ac are not functions of z. Note the differences between the PFR accomplishing a homogeneous [Equation (10.1.2)] and a solid catalyzed [Equa-tion (10.1.4)] reaction. First of all, the solid-catalyzed reaction rate is per unit mass of catalyst and must therefore be multiplied by the bed density to obtain a reaction rate per unit volume in the mass balance. This is because the amount of catalyst packed into a reactor can vary from packing to packing with even the same solid. Thus, the bed density must be elucidated after each packing of the reactor, while the reaction rate per mass of catalyst need not be. Second, as discussed in Chapter 6, the overall effectiveness factor can be used to relate the reaction rate , Flow ----..------I~~~I _ F, ~ dz f-+- F, +dF, (a) (b) Figure 10.1.1 I Illustrations of tubular reactors: (a) unpacked tube, (b) packed tube. F, = C,v = uAcC,.
1.0 ----I I I I //---..
/ / / Catalyst particle size -+ Figure 10.1.2 I Effects of catalyst particle size on TJo and limits of flow rates.
C HAP T E RiO Reactor~8cCDmplishinQ-He1erngBlliillU5Beactions 317 that occurs within the catalyst particle to the rate that would occur at the local, bulk fluid conditions.
To maximize the reaction rate in the fixed-bed reactor, 710 should be equal to one. In order to do this, smaller particles are necessary (see Chapter 6). As the catalyst particles get smaller, the pressure drop along the reactor increases. (For example, ponder the differences in how difficult it is to have water flow through a column of marbles versus a column of sand.) Thus, a major consideration in the design of a fixed-bed reactor is the trade-off between pressure drop and transport limitations of the rate (illustrated in Figure 10.1.2). A practical design typically in-volves 710 - 1. Methodologies used to describe fixed-bed reactors given this situa-tion are outlined in the next section.
10.2 lOne-Dimensional Models for Fixed-Bed Reactors Consider a PFR operating at nonisothermal conditions and accomplishing a solid-catalyzed reaction (Figure 10.1.1b). If the fluid properties do not vary over the cross-section of the tube, then only changes along the axial direction need to be consid-ered. If such is the case, then the mass, energy, and momentum balances for a fixed-bed reactor accomplishing a single reaction can be written as: de -u-d ' = TJoPB( -v;)r z (mass) -dT upCp dz 4U (T -T) d, (energy) (10.2.1) dP fjpu 2 dz gcdp Ci = C?, T = TO, P pO at z = 0 (momentum) where u and Ac are not functions of z,if is a friction factor, and gc is the gravitational constant. Mass and energy balances of this type were discussed in Chapter 9 for ho-mogeneous reactions [Equation (9.4.9)] and their solutions were illustrated (Example 9.4.3). Here, the only differences are the use of the heterogeneous reaction rate terms and the addition of the momentum balance.
For packed columns with single-phase flow, the Ergun equation can be used for the momentum balance and is: (10.2.2) where 813 is the void fraction (porosity) of the bed and the Reynolds number is based on the particle size, dp . The Ergun equation only considers frictional losses because of the packing. However, for d/dp < 50, frictional losses from the wall of the tube are also significant. D. Mehta and M. C. Hawley [Ind. Eng. Chem. Pmc.
318 C H A PT E R 10 Reactors Accomplishing Heterogeneolls Reactions Des. Dev., 8 (1969) 280] provided a correction factor to the Ergun equation to consider both the frictional losses from the wall and the packing, and their fric-tion factor expression is: [ 1 liB] [ It = (liB )3 1 + 150 (1 liB)l (10.2.3) Re J EXAMPLE 10.2.1\ Solution of Equation (10.2.1) provides the pressure, temperature, and con-centration profiles along the axial dimension of the reactor. The solution of Equation (10.2.1) requires the use of numerical techniques. If the linear velocity is not a function of z [as illustrated in Equation (10.2.1)], then the momentum bal-ance can be solved independently of the mass and energy balances. If such is not the case (e.g., large mole change with reaction), then all three balances must be solved simultaneously.
J. P. Kehoe and J. B. Butt [AIChE J., 18 (1972) 347] have studied the kinetics of benzene hydrogenation on a Ni/kieselguhr catalyst. In the presence of excess dihydrogen, the reaction rate is given by: (mollgcat/s) where ko = 4.22 mol/(gcat-s-torr) Ko = 1.11 X 10-3 cm3/(mol) PH, = in torr T. H. Price and J. B. Butt [Chern. Eng. Sci., 32 (1977) 393] investigated this reaction in an adiabatic, fixed-bed reactor. Using the following data: PH, = 685 torr Ps = 1.2 gcat/cm3 Llu = 0.045 s TO l50aC Cp = 1.22 X 105 J/ (kmol-0c) - f::,.Hr!r" 2.09 X 108 J/kmol calculate the dimensionless concentration of benzene and the dimensionless temperature along the axial reactor dimension assuming TJo = 1.
CHAPTER 10 Reactors Accomplishing Heterogeneolls Reactions 319 • Answer The mass and energy balances can be written as (assume u is constant to simplify the solu-tion of the material and energy balances): dCB -u- = PBrB dz -dT upCp -= (-D.HrITo)PBrB dz CB C~, T = TO at z = 0 Let y CB/C~,"8 T/To and Z = z/L, so that the mass and energy balances can be ex-pressed as: GJ:~ (PB/ c2)rB pcpCiJ:~ = (-LlHrITo{~)rB y=o latZ 0 Using the data given, the mass and energy balances reduce to (note that C~ = 0.1 Ctota] = O.lp): dy dZ dO dZ [ 3.21 J -0.174 exp T Y 3.21 0.070 exp y o y o=latZ=O Numerical solution of these equations gives the following results: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Thus, at the reactor exit CB 1.00 0.70 0.53 0.41 0.33 0.27 0.22 0.18 0.15 0.12 0.10 1.00 1.12 1.19 1.23 1.27 1.29 1.31 1.33 1.34 1.35 1.36 ..
3""'2...
0 -'C ....
Hr:LfO.A.."P.....
T-"E ....
R:LJ1...
0'--lR.=eacJ:ills Accomplishing Heterogeneolls Reactions Example 10.2.1 illustrates the simultaneous solution of the mass and energy balances for an adiabatic, fixed-bed reactor with no fluid density changes and no transport limitations of the rate, that is, 170 = I. Next, situations where these sim-plifications do not arise are described.
If there is significant mole change with reaction and/or large changes in fluid density because of other factors such as large temperature changes, then Equation (10.2.1) is not appropriate for use. If the fluid density varies along the axial reactor dimension, then the mass and energy balances can be written as: (10.2.4) (10.2.5) Recall that: v = Acu FA = F~(1 - fA) FA = vCA = v o (1 + eAfA)C~C l+-~~J(;o) Using these expressions in Equation (10.2.4) gives: dFA - dz = Ac 17oPs( -vA)r -dT 4U upCp IiHrIT o)17oPs(-vA)r - -(T T) dz d, dP ffpu 2 dz gcdp FA = F~, T = TO, P = pO at z = 0 assuming Cp is not a strong function of T. All three balances in Equation (10.2.5) must be solved simultaneously, since u = uo(1 + eAfA)(po/P)(T/TO).
Thus far, the overall effectiveness factor has been used in the mass and energy balances. Since 170 is a function of the local conditions, it must be computed along the length of the reactor. If there is an analytical expression for 17m for example for an isothermal, first-order reaction rate: (6.4.34) CHAPTER 10 Reactors Accomplishing HeterogeneollS Reactions 321 then the use of 710 is straightforward. As the mass and energy balances are being solved along the axial reactor dimension, 710 can be computed by calculating s~~~~~~~~ and CAB' TB = bulk fluid phase concentration of A and temperature Pp = density of the catalyst particle a y = external catalyst particle surface area per unit reactor volume Notice how the catalyst particle balances are coupled to the reactor mass and en-ergy balances. Thus, the catalyst particle balances [Equation (10.2.7)] must be solved at each position along the axial reactor dimension when computing the bulk mass and energy balances [Equation (10.2.6)]. Obviously, these solutions are lengthy.
Since most practical problems do not have analytical expressions for the ef-fectiveness factor, the use of Equations (10.2.6) and (10.2.7) are more generally applicable. If Equations (10.2.6) and (10.2.7) are to be solved numerous times (e.g., when performing a reactor design/optimization), then an alternative ap-proach may be applicable. The catalyst particle problem can first be solved for a variety of CAB' TB, 50 Lldp > 150 Additionally, L. C. Young and B. A. Finlayson [Ind. Eng. Chem. Fund.. 12 (1973) 412] showed that if: then axial dispersion can be neglected. However, for laboratory reactors that can have low flow rates and are typically of short axial length, axial dispersion may be-come important. If such is the case, then the effects of axial dispersion may be needed to accurately describe the reactor.
at Z = I at Z = 1 at Z = 0 324 EXAMPLE 10.2.21 C H A PT E R 10 Reactors Accomplishing Heterogeneolls Reactions An axially-dispersed, adiabatic tubular reactor can be described by the following mass and energy balances that are in dimensionless form (the reader should verify that these descrip-tions are correct): 1 d 2y dy _ -2 ---r(y, 0) = 0 Pe" dZ dZ I d 2e de --BOa dZ 2 -dZ - Hwr(y, 0) = 0 with 1 dy --= y -I at Z = 0 Pea dZ 1 dO --- = 0 -I BOa dZ dy -=0 dZ dO = 0 dZ If H w = -0.05, and r(y, e) 4y exp[ l8(1 ~)l calculate y and efor Pea = BOa = 10, 20, and 50 (BOa is a dimensionless group analogous to the axial Peelet number for the en-ergy balance).
• Answer The dimensionless system of equations shown above are numerically solved to give the fol-lowing results.
0.0 0.2 0.4 0.6 0.8 1.0 0.707 0.276 0.092 0.029 0.009 0.004 1.01 1.03 1.05 1.05 1.05 1.05 0.824 0.293 0.082 0.021 0.005 0.002 1.01 1.04 1.05 1.05 1.05 1.05 0.923 0.309 0.072 0.015 0.003 0.001 1.00 1.03 1.05 1.05 1.05 1.05 As Pea -+ 00 the reactor behavior approaches PFR. From the results illustrated, this trend is shown (for PFR, y = 0 = I at Z = 0).
In general, a one-dimensional description for fixed-bed reactors can be used to capture the reactor behavior. For nonisothermal reactors, the observation of hot spots (see Example 9.4.3) and reactor stability (see Example 9.6.2) can often be described.
~~~~~~~---,C~H~ALrP~Tufr..,J R<LJ1l'oO<--JRDJe""al..cilllo./L[s",-,=,A..<'ccCLCo./Lmlljp.J.il""iswhiiloO+lHeterrle",[Q40,",e",o""eQUJ!£:lsuR-,"e,,",aL.,cll!ljo.J.io.L:;s~~~~~~~~~325 However, there are some situations where the one-dimensional descriptions do not work well. For example, with highly exothermic reactions, a fixed-bed reactor may contain several thousand tubes packed with catalyst particles such that elt / elp ~ 5 in order to provide a high surface area per reaction volume for heat transfer. Since the heat capacities of gases are small, radial temperature gradients can still exist for highly exothermic gas-phase reactions, and these radial variations in temperature produce large changes in reaction rate across the radius of the tube. In this case, a two-dimensional reactor model is required.
10.3 I Two-Dimensional Models for Fixed-Bed Reactors Consider a tubular fixed-bed reactor accomplishing a highly exothermic gas-phase reaction. Assuming that axial dispersion can be neglected, the mass and energy bal-ances can be written as follows and allow for radial gradients: a 1 a (~ acA ) } -(uC) - ---= rD.-+ 17 P (-v )r = 0 az A rar 'ar 0 S A (10.3.1) a 1 a - aT -(upC/,T) -~ (rAr-=) + (-6.H,hops(-vA)r = 0 az rar ar where Dr is the radial dispersion coefficient (see Cha~er 8) and Ar is the effective thermal conductivity in the radial direction. If Dr and Ar are not functions ofrthen Equations (10.3.1) can be written as (u not a function of z): The conditions necessary to provide a solution to Equation (10.3.2) are: CA = C~, T = TO at z = 0 for 0::; r::; elt/~ (10.3.2) aCA ar aCA ar aT ar o at r = 0 for 0::; z (10.3.3) for 0::; z where hi is the heat transfer coefficient at the reactor wall and T" is the wall temperature. Solution of Equations (10.3.2) and (10.3.3) gives CA(r, z) and T(r, EXAMPLE 10.3.11 CHAPTER 10 Reactors Accomplishing l::ie1Brog~e:un",eol..lJl..L;ls:u::lR",ea",cwtJ.<.ioJ.LnE:>s~~~~~~~~~ G. F. Froment [Ind. Eng. Chem., 59 (1967) 18J developed a two-dimensional model for a fixed-bed reactor accomplishing the following highly exothermic gas phase reactions: k, ko o-xylene ==> phthalic anhydride ====> C02, CO, H20 (A) (B) (C) Ii k \1'1 COlo CO, H20 (C) The steady-state mass and energy balances are (in dimensionless form): with YI Yz = 0 () = 1 aYI aY2 ae -==-==-:==0 aR aR aR o where YI = CB/C~ Y2 () dimensionless temperature 2 dimensionless axial coordinate R dimensionless radial coordinate r l kl(l -YI [2 k3(1 YI-f3f3i dimensionless groups dimensionless wall temperature at 2 0 for O:s;R:S;l at 2=0 for O:s;R:S;l at R=O for 0:s;2:S;1 at R = 1 for 0:s;2:S;1 at R for 0:s;2:S;1 CHAPTER 10 Reactors Accomplishing HeterogeneolJS Reactions Using the data given by Froment, Per = 5.706, BOr 10.97, Hw 2.5 f3f31 5.106,f3f32 3.144,f3f33 11.16 Additionally, k1 exp[ 1.74 + 21.6( I ~)J k2 eXP [-4.24 + 25.1(1 ~)J k3 = exp [ -3.89 + 22.9( I ~)J If e w = 1, compute Yl(Z, R), Y2(Z, R) and e(Z, R).
327 • Answer This reactor description is a coupled set of parabolic partial differential equations that are solved numerically. The results shown here were obtained using the software package PDECOL [N. K. Madsen and R. F. Sincovec, ACM Toms, 5 (1979) 326].
Below are listed the radial profiles for two axial positions within the reactor. Notice that at Z = 0.2, there are significant radial gradients while by Z = 0.6 these gradients are all but eliminated. If the inlet and coolant temperatures are around 360°C, then the centerline and inner wall temperatures at Z = 0.2 are 417°C and 385°C, respectively, giving a radial tem-perature difference of 32°C.
0.0 0.458 0.782 1.09 0.2 0.456 0.777 1.09 0.4 0.451 0.763 1.08 0.6 0.445 0.747 1.07 0.8 0.440 0.734 1.05 1.0 0.438 0.728 1.04 0.0 0.613 0.124 1.02 0.2 0.613 0.124 1.02 0.4 0.613 0.124 1.01 0.6 0.612 0.124 1.01 0.8 0.612 0.124 1.01 1.0 0.612 0.124 1.01 As illustrated in Example 10.3.1, the radial temperature gradient can be significant for highly exothermic, gas-phase reactions. Therefore, one must make a decision for a particular situation under consideration as to whether a I-D or 2-D analysis is needed.
328 CHAPTER 10 Reactors Accomplishing Heterogen.CLeo'-"leuls:LL1Rs;;;.eacuc'"'t.LI.ioan""s _ 10.4 I Reactor Configurations Thus far, fixed-bed reactor descriptions have been presented. Schematic represen-tations of fixed-bed reactors are provided in Figure 10.4.1, and a photograph of a commercial reactor is provided in Figure 10.4.2.
Feed __ =-------, Product (a) Fuel Flue --- gas '---'-= Product (b) Figure 10.4.1 I Fixed-bed reactor schematics: (a) adiabatic, (b) nonadiabatic. [From "Reactor Technology" by B. L. Tarmy, Kirk-Othmer Encyclopedia of Chemical Technology, vol. 19, 3rd ed., Wiley (1982). Reprinted by permission of John Wiley and Sons, Inc., copyright © 1982.] Figure 10.4.2 I Photograph of a commercial reactor. (Image provided by T. F. Degnan of Exxon Mobil.) ~~~~~~~~~...
CUlH:L>1AJ<P~TLE....,RL--<1...
0'-----LliORHacJ:DJ:sAcrQmpiisbiogJ=Jeterogeneo' JS ne"ctions _ ...32.9 P~Tf:#~- Perforated plate ~~~~~~l- Distribution caps I3l Liquid (a) Liquid feed Gas (b) Continuous __ ~ ir=--iIO- Product phase .....ll.l L...
o C::>o oC> 0° 0 .. 0 .0 eO .. " Dispersed~ phase (c) Gas Liquid Deentrained vapor Product t--..,... Spent catalyst Fresh catalyst Feed (d) Liquid-I'----tofl gas bubbles ,~ Liquid recycle Liquid (e) Figure 10.4.3 I Examples of multiphase reactors: (a) trickle-bed reactor, (b) countercurrent packed-bed reactor, (c) bubble column, (d) slurry reactor, and (e) a gas-liquid fluidized bed. [From "Reactor Technology" by B. L. Tarmy, Kirk-Othmer Encyclopedia oj' Chemical '!e'io!lIlOlog:v, vol. 19, 3rd ed., Wiley (1982). Reprinted by permission of John Wiley and Sons, Inc., copyright © 1982.] The reactor models developed for fixed-bed reactors have been exploited for use in other situations, for example, CVD reactors (see Vignette 6.4.2) for micro-electronics fabrication. These models are applicable to reaction systems involving single fluid phases and nonmoving solids. There are numerous reaction systems that involve more than one fluid phase. Figure J0.4.3 illustrates various types of reactors Gas bubble 330 CHAPTER 10 Reactors Accomplishing Heterogeneolls Reactions -------------- Film Film //----------...X( 'Bulk Porous J catalyst ,/1 liquid particle , , , " , ~/ : ~/1 !
I \ ,,.," I I I \ I ~..../ !
l:,i ! \!
r 1'..., 1------';...
", !
_----Distance Figure 10.4.4 I Concentration profile of a gas-phase reactant in a porous solid-catalyzed reaction occurring in a three phase reactor.
that process multiple fluid phases. Figure lOo4.3a shows a schematic of a trickle-bed reactor. In this configuration, the solid catalyst remains fixed and liquid is sprayed down over the catalyst to create contact. Additionally, gas-phase compo-nents can enter either the top (cocurrent) or the bottom (countercurrent; Figure lOo4.3b) of the reactors to create three-phase systems. The three-phase trickle-bed reactor provides large contact between the catalyst and fluid phases. Typical reac-tions that exploit this type of contact are hydrogenations (Hz-gas, reactant to be hydrogenated-liquid) and halogenations. Bubble (Figure lOo4.3c), slurry (Figure lOo4.3d), and fluidized (Figure lOo4.3e) reactors can also accomplish multiphase re-actions (solid catalyst in liquid phase for both cases). As with all multiphase reac-tors, mass transfer between phases (see Figure 100404) can be very important to their overall performance.
The mathematical analyses of these reactor types will not be presented here.
Readers interested in these reactors should consult references specific to these re-actor types.
CHAPTER 10 Rear-tors Accomplishing Hoterogeneolls Rear-tions 331 VIGNETTE 10.5.1 at Engelhard Corporation, who have developed fixed-bed reactors for treating cooking fumes (u.s. Patent Exhaust plumes from fast food restaurants contain com-ponents that contribute to air pollution. Engelhard has implemented in some fast food restaurants a fixed-bed catalytic reactor to oxidize the exhaust components to CO2, A sec-ond example where fixed beds are used to clean up exhaust streams involves wood stoves.
In certain areas of the United States, the chimneys of wood stoves now contain fixed beds of catalysts to assist in the cleanup of wood stove fumes prior to their release into the atmosphere. These two examples show how the fixed-bed reactor is not limited to use in the petrochemical industries.
10.5 I Fluidized Beds with Recirculating Solids Fluidized beds with recirculating solids developed primarily as reactors for the cat-alytic cracking of oil to gasoline. Today their uses are much broader. Prior to the early 1940s, catalytic cracking of oil was performed in fixed-bed reactors. After that time, fluidized-bed reactors have been developed for this application. Today, flu-idized beds with recirculating zeolite-based catalysts are used for catalytic cracking of oiL Figure 10.5.1 shows a schematic of a fluidized bed with recirculating solids.
The oil feed enters at the bottom of the riser reactor and is combined with the solid catalyst. The vapor moves upward through the riser at a flow rate sufficient to flu-idize the solid catalyst and transport it to the top of the riser in a millisecond time-frame. The catalytic cracking reactions occur during the transportation of the solids to the top of the riser and the catalyst becomes covered with carbonaceous residue (coke) that deactivates it by that point. The solids are separated from the products and enter a fluidized-bed, regeneration reactor. In the regeneration reactor, the car-bonaceous residue is removed from the catalyst by combustion with the oxygen in the air. The "clean" catalyst is then returned to the bottom of the riser reactor. Thus, the solid catalyst recirculates between the two reactors. Typical fluidized beds with recirculating solids contain around 250 tons of catalyst of 20-80 microns in particle 332 C H A PT E R 1 0 ReactOL~Dillpljshing Heterogeneolls Reactions ::.
.
:': ~~r Transfer- :' line reactor ;; ..; Product Steam stripping ~,.----Flue gas regenerator Figure 10.5.1 I Schematic of fluidized-bed reactor with recirculating solids. [From "Reactor Technology" by B. L. Tarmy, Kirk-Othmer Encyclopedia of Chemical Technology, vol.
19, 3rd ed., Wiley (1982). Reprinted by permission of John Wiley and Sons, Inc., copyright © 1982.] size and have circulation rates of around 25 ton/min. As might be expected, the solids are continuously being degraded into smaller particles that exit the reactor.
Typical catalyst losses are around 2 ton/day for a catalyst reactor inventory of 250 ton. Thus, catalyst must be continually added.
The fluidized-bed reactor with recirculating solids is an effective reactor con-figuration for accomplishing reactions that have fast deactivation and regenera-tion.
In addition to catalytic cracking, this configuration is now being used for another type of reaction. The oxidation of hydrocarbons can be accomplished with oxide-based catalysts. In the mid-1930s, Mars and van Krevelen postulated that the oxidation of hydrocarbons by oxide-containing catalysts could occur in two steps: Overall: RH2 + O2 = RO + H 20 (hydrocarbon) (oxidized product) Step 1: Step 2: (lattice oxygen) O2 + 20 + 4e- =20- 2 (surface oxygen vacancy) In the first step the hydrocarbon is oxidized by lattice oxygen from the oxide-containing catalyst to create surface oxygen vacancies in the oxide. The second step involves filling the vacancies by oxidization of the reduced oxide surface with C H A PT E R 10 Rear,tors Accomplishing Heterogeneolls Reactions 333 dioxygen. This mechanism has been shown to hold for numerous oxide-based cat-alytic oxidations.
Dupont has recently commercialized the reaction of n-butane to maleic anhy-dride using a vanadium phosphorus oxide (VPO) catalyst in a recirculating solid re-actor system. The overall reaction is: A proposed reaction pathway over the VPO catalyst is as shown below: where 0L denotes lattice oxygen. In addition to these reactions, each of the organic compounds can be oxidized to CO2 and CO. In the absence of gas-phase O2 the se-lectivity to CO2 and CO is minimized. A solids recirculating reactor system like that schematically illustrated in Figure 10.5.1 was developed to accomplish the butane oxidation. In the riser section, oxidized VPO is contacted with butane to produce maleic anhydride. The reduced VPO solid is then re-oxidized in the regeneration re-actor. Thus, the VPO is continually being oxidized and reduced (Mars-van Krevelen mechanism). This novel reactor system opens the way to explore the performance of other gas-phase oxidations in solids recirculating reactors.
There are numerous other types of reactors (e.g., membrane reactors) that will not be discussed in this text. Readers interested in reactor configuration should look for references specific to the reactor configuration of interest.
Exercises for Chapter 10 1.
Consider the reactor system given in Example 10.2.1.
(a) Compare the dimensionless concentration profile from the adiabatic case given in the example to that obtained at isothennal conditions.
(b) What happens if the reactor is operated adiabatically and the heat of reaction is doubled?
(c) If the reactor in part (b) is now cooled with a constant wall temperature of ISO°C, what is the value of 4Ujdt that is necessary to bring the outlet temperature to approximately the value given in Example 1O.2.1?
2.
Plot the dimensionless concentration and temperature profiles for the reactor system given in Example 10.2.2 for Pea = BOa = 00 and compare them to those obtained when Pea = BOa = 15.
3.
Reproduce the results given in Example 10.3.1. What happens as Hw is decreased?
4.
Design a multitube fixed-bed reactor system to accomplish the reaction of naphthalene (N) with air to produce phthalic anhydride over a catalyst of vanadium pentoxide on silica gel at a temperature of about 610-673 K.
334 C H APT E R 10 Reactors Accomplishing Heterogeneolls Reactions The selective oxidation reaction is carried out in excess air so that it can be considered pseudo-first-order with respect to naphthalene. Analysis of available data indicates that the reaction rate per unit mass of catalyst is represented by: 1.14 X 1013 exp(-19,000jT) (cm3 S-I gcaC I) where CN = the concentration of naphthalene in mol cm- 3, and T = absolute temperature in K. (Data for this problem were adapted from C. G. Hill, Jr.
An Introduction to Chemical Engineering Kinetics and Reactor Design, Wiley, New York, 1977, p. 554.) Assume for this problem that the overall effectiveness factor is unity (although in reality this assumption is likely to be incorrect!). To keep the reactant mixture below the explosion limit of about 1 percent naphthalene in air, consider a feed composition of 0.8 mol % naphthalene vapor in 99.2 percent air at 1.7 atm total pressure. Although there will be some complete oxidation of the naphthalene to CO2 and H20 (see part c), initially assume that the only reaction of significance is the selective oxidation to phthalic anhydride as long as the temperature does not exceed 673 K anywhere in the reactor: The heat of reaction is -429 kcal mol- I of naphthalene reacted. The properties of the reaction mixture may be assumed to be equivalent to those of air (Cp = 0.255 cal g-] K-], Ii = 3.2 X 10-4 g cm-I s-]) in the temperature range of interest.
It is desired to determine how large the reactor tubes can be and still not exceed the maximum temperature of 673 K. The reactor will be designed to operate at 95 percent conversion of CIOHs and is expected to have a production rate of 10,000 Ib/day of phthalic anhydride. It will be a multitube type reactor with heat transfer salt circulated through its jacket at a temperature equivalent to the feed temperature.
The catalyst consists of 0.32 cm diameter spheres that have a bulk density of 0.84 g cm-3 and pack into a fixed bed with a void fraction of 0.4. The mass velocity of the gas through each tube will be 0.075 g cm-2 s-I, which corresponds to an overall heat transfer coefficient between the tube wall and the reacting fluid of about 10-3 cal cm-2 s-] K- 1.
(a) Show the chemical structures of the reactants and products. Give the relevant material, energy, and momentum balances for a plug flow reactor. Define all terms and symbols carefully.
(b) Determine the temperature, pressure, and naphthalene concentration in the reactor as a function of catalyst bed depth using a feed temperature of 610 K and tubes of various inside diameters, so that a maximum temperature of 673 K is never exceeded. Present results for several different tube diameters. What is the largest diameter tube that can be used without C H A PT E R 10 Reactors Accomplishing HeterogeneolJs Rear,tions exceeding 673 K? How many tubes will be required, and how long must they be to attain the desired conversion and production rates? For this optimal tube diameter, vary the feed temperature (615, 620 K) and naphthalene mole fraction (0.9 percent and I percent) to examine the effects of these parameters on reactor behavior.
(c) Combustion can often occur as an undesirable side reaction. Consider the subsequent oxidation of phthalic anhydride (PA) that also occurs in the reactor according to: CsH40 3 + 7.5 O2 =? 2 H20 + 8 CO2 with rate: r = k2CPA where k2 = 4.35 X 104 exp(-10,000IT) cm3 s~ 1 gcaC 1 and I1Hr = -760 kcal mol ~ 1. Redo part (b) and account for the subsequent oxidation of PA in the reactor. Include the selectivity to phthalic anhydride in your analysis.
S.
When exothermic reactions are carried out in fixed-bed reactors, hot spots can develop. Investigate the stability of the fixed-bed reactor described below for: (a) cocurrent coolant flow and (b) countercurrent coolant flow when the inlet coolant temperature (T~) is 350 K and higher. Calculate the sensitivity by plotting the maximum temperature in the reactor versus the inlet coolant temperature (the slope of this line is the sensitivity). Which mode of cooling minimizes the sensitivity? (This problem is adapted from material provided by Jean Cropley.) Data: Reaction system (solid-catalyzed): I Dammitol + 202 ~ Valualdehyde + H20, (DT) 5 Valualdehyde + 2 O 2 (VA) (VA) with tJ.Hf , -74.32 kBTU/lbmol, tJ.Hr, expressions for these reactions are: -474.57 kBTU/ibmol. The rate RT (~ 3~3)/1.987 Po, Xo,Pr PDT XDTPr PYA = XyAPr 336 CHAPTER 10 Reactors Accomplishing Heterogeneolls Reactions k1exp[ -k2RT]P&P~r r] = -1-+-k-P-k6-+-k-P-k8---'~'---k-P-k-1O 1bmoI/(lbeat-h) S 0, 7 DT 9 VA kl1exp[ -k12RT]P&JP0A k k" k lbmoI/(lbcat-h) 1 + ksPo~ + k7P DT + kgP..j'A where PT is the total pressure and: k] = 1.771 X 10-3 k2 = 23295.0 k3 = 0.5 k4 = 1.0 ks = 0.8184 k6 = 0.5 k7 = 0.2314 ks = 1.0 kg = 1.25 klO = 2.0 kl1 = 2.795 X 10-4 k12 = 33000.0 k13 = 0.5 k]4 = 2.0 The governing differential equations are (the reader should derive these equations): mass balances: dX; (NT)R;PBAc . MW Xi dMW .
-d = W + =-d-' I = 1(DT),2(VA),3(02),4(C02 ),5(H2 0) Z MW z energy balance on reacting fluid: dTR -(NT)Ac(r]D.Hr , + r2D.Hr,)PB - 1T(NT)dt U(TR - Tc) m W.~ energy balance on coolant: dTc (mode)1T(NT)dtU(TR - Tc) dz FcCpc mole change due to reactions-reactions in the presence of inert N2: dMW NC dX; NC dX; -= ~M;--28~-dz i=] dz ;=] dz momentum balance [alternative form from Equation 00.2.3) that accounts for the losses from both the wall and catalyst particles]: dPr -o?
1150I:Lu (1 -1.75MWu 2 (1 dz = (32.2)(144)l~ -'----::-~ + aRgTRdp -'---,:-:::":'" fJ~ CHAPTER 10 Reactors Accomplishing Heterogeneolls Reactlons where 33Z (273)(3600)(NT) 0 Ac 0 MW 0 PT (conversion of mass flow rate to superficial velocity) u oT X j = mole fraction of component i u = superficial velocity (ft/s) TR = reacting fluid temperature (K)(T2 = inlet value) Tc = coolant temperature (K) (T ~ = inlet value) MW = mean molecular weight of the reacting gas PT = total pressure (psia) (P~ = inlet value) W = total mass flow rate (lb/h) NT = number of reactor tubes Ac = cross-sectional area of a reactor tube dt = diameter of tube dp = catalyst pellet diameter P8 = bed density (lbcat/ft3) z = axial coordinate (ft) R j = net reaction rate of component i (lbmol/lbcat/h), for example, R] = -rj, Rz = r]- rz, R3 = -0.5r]- 2.5rz: assumes effectiveness factors are equal to one 88 = porosity of bed U = overall heat transfer coefficient (BTU/h/ftz;oC) mode 1) for cocurrent coolant flow, (- 1) for countercurrent coolant flow FcCpc coolant rate (BTU/h;oC) J-t = reacting fluid viscosity (lb/ft/h) Cp = gas heat capacity (BTU/lb/oC) R o = universal gas constant (psia-fr3/lbmol/K) ,':) NC = number of components = 5 I 2 3 4 5 Dammitol Valuadehyde O2 CO2 H20 46 44 32 44 18 Ib/(ft-h) BTU/(IVC) BTU/(h-ftZ_0C) 12 ft (ID) x 20 ft (length) Ib/h 33iL~ ----"CLlH"-8A-""P,-,T,-,E"-JR",-~ Reactors Ar'C'omplisbing.He.tea[.\0.l.\gjJ:e1Lou:;e.uo,",1 )s:LDRzceaclJc..ltl..iollJo""'-s ~ reactor 2500 tubes W 100,000 X) 0.1 X3 0.07 Xs 0.02 remainder is Nz (Oz comes from air) FcCpc 106 BTU/(h-°C) P~ 114.7 psia T~ 353 K dp 0.25 inches (spherical) Ps 100 Ib/ft3 Ss 0.5 JL 0.048 Cp 0.50 U 120.5 6.
Refer to the reactor system described in Exercise 5. Using the mode of cooling that minimizes the sensitivity, what does the effect of changing BS from 0.5 to 0.4 (a realistic change) have on the reactor performance?
7.
If the effectiveness factors are now not equal to one, write down the additional equations necessary to solve Exercise 5. Solve the reactor equations without the restriction of the effectiveness factors being unity.
Review of Chemical Equilibria A.1 I Basic Criteria for Chemical Equilibrium of Reacting Systems The basic criterion for equilibrium with a single reaction is: NCOMP D..G = :L VilLi = ° i~l (Al.l ) where D..G is the Gibbs function, NCOMP is the number of components in the sys-tem, Vi is the stoichiometric coefficient of species i, and lLi is the chemical poten-tial of species i. The chemical potential is: (A.I.2) where Rg is the universal gas constant, IL? is the standard chemical potential of species i in a reference state such that ai = 1, and ai is the activity of species i. The reference states are: (1) for gases (i.e., JO = 1) (ideal gas, P = 1 atm) where 1 is the fugacity, (2) for liquids, the pure liquid at T and one atmosphere, and (3) for solids, the pure solid at T and one atmosphere. If multiple reactions are occurring in a network, then Equation (AI. 1) can be extended to give: NCOMP D..Gj :L Vi,jlLi = 0, j 1, "', NRXN i~ I (Al.3) where NRXN is the number of independent reactions in the network.
In general it is not true that the change in the standard Gibbs function, D..Go, is zero. Thus, NCOMP D..Go :L VilL;) '° i~ I (A.I.4) 340 AppEND. X A Review of Chemical EqIJilibrLa...
_ Therefore, NCOMP I1C - I1Co 2: Vj(ILj IL?) i~l or by using Equation (A.I.2): Now consider the general reaction: CiA + bB + ... = wW + sS + ...
(Al.5) (Al.6) (A 1.7) Application of Equation (Al.6) to Equation (Al.7) and recalling that I1C = 0 at equilibrium gives: Thus, the equilibrium constant Ka is defined as: NCOMP Ka II aY; i=l Differentiation of Equation (AI.8) with respect to T yields: (AI.8) (A.I.9) (A 1.10) Note that I1Co = I1Ho - TI1So, where I1Ho and I1So are the standard enthalpy and entropy, respectively, and differentiation of this expression with respect to T gives: (A 1.1 I) Equating Equations (A l. 10) and (A1.1!) provides the functional form for the tem-perature dependence of the equilibrium constant: I1Ho RgT 2 or after integration (assume I1Ho is independent of T): (A.1.12) (Al.13) Notice that when the reaction is exothennic (I1Ho is negative), Ka increases with decreasing T. For endothermic reactions the opposite is true. From Equation (Al.8): _Jl2J>LN~JX A Revjnw of CJocBmicaLEqlJi'ibria (A.1.l4) and (A. l.l 5) Since LlCo is not a function of pressure, it is clear that pressure has no influence on Ka .
A.2 I Determination of Equilibrium Compositions Consider a gas-phase reaction. If the Lewis and Randall mixing rules are used (simplest form of mixing rules-more sophisticated relatiol1ships could be applied if deemed necessary) to give for the fugacity of species i, Ii: where - /-0//-0 0 a -. i.
i (A.2.1) and 1Ji fugacity coefficient of pure i at T and P of system for any mole fraction Xi' Substituting the above expression into Equation (A. 1.9) for the reaction given in Equation (A.I.7) yields (let all .17 I): (A.2.2) or where n· X.= I f J1inert + n .I Equation (A.2.2) can be written in terms of moles as: rn.
·.. lr P = K, 1 tv _ 1.
1 -----Ka (/) I b 11 ,""" L nB • .. J L f1inert + ~ J (A.2.3) Note that Ka is not a function of pressure and that KJ, is only weakly dependent on pressure. Thus, if: { then Pt,KJ --b ...
then Pt,K,+ tV S a 0 then no effect 342 APPENDIX A Review of Chemical Eqllilibria and the effect of inert species are: then add inert, K).
then add inert, Kxt then no effect Finally, just to state clearly once again, a catalyst has no effect on equilibrium yields.
______------'A""""--J~ B Regression Analysis B.1 I Method of Least Squares Below is illustrated the method of least squares to fit a straight line to a set of data points (Yh x;). Extensions to nonlinear least squares fits are discussed in Section BA.
Consider the problem of fitting a set of data (Y" x;) where y and x are the dependent and independent variables, respectively, to an equation of the form: (B.Ll) by determining the coefficients O:'J and a2 so that the differences between y, and y, = O:'t + 0:'2 x, are minimized. Given O:'J and 0:'2' the deviations ily, can be calcu-lated as: (B.1.2) For any value of x = x" the probability PP, for making the observed measurement y, with a Gaussian distribution and a standard deviation (J', for the observations about the actual value y(x;) is (P. R. Bevington, Data Reduction and Error Analysis for the Physical Sciences. McGraw-Hill, New York, 1969, p. 101): (8.13) The probability of making the observed data set of measurements of the N values of y, is the product of the individual PP, or: N PP(a l , 0:'2) TI PP, n( 1 )' exp[ 12: y, - --rII !
(J',y!2; 2 , (J', J ~ (8. 104) The best estimates for aJ and 0:'2 are the values that maximize PP(al> 0:'2) (method of maximum likelihood). Define: 344 A PPE N 0 I X B RAgres"'SlliiO.lLnLACliUna:ui?ys=is"-- N 2: (B.l.5) Note that in order to maximize PP(ajo (2), X2 is minimized. Thus, the method to find the optimum fit to the data is to minimize the sum of the squares of the devi-ations (i.e., least-squares fit).
As an example of how to calculate at and a2' consider here the case where U i = U constant. To minimize X 2 , the partial derivatives of X2 with respect to aI and a2 must be set equal to zero: ax 2 a -=0= aal aal ax 2 a -=0= aa2 aa2 :2 2: (Yi - al - (X2 X Y] 1 7 2: (Yi -(XI -(X2 X Y] U-These equations can be rearranged to give: 2:Yi = 2: a ]+ 2: (X2X j = (XIN + a22: Xi 2: X iYi 2: a I X i+ 2: (X2 X ?=al2: X i +a22:x?
The solutions to these equations yield a] and (X2 in the following manner: I2:Yi 2:xjI 2:xf 0'] 12:~ti 2:x j I 2:xf (B.l.6) 12:~~ 2:Yi I 2:XjYi N2:X iYi - 2:xj2:Yi Ct'2 = I2: NXi 2: x, I N2:xf (2:XJ2 2:xf The calculation is straightforward. First compute 2:x" 2:y" 2:xl, and 2:x,y,. Second use the summed values in Equation (B.l.6) to obtain (XI and a2' B.2 I Linear Correlation Coefficient Referring to Equations (B.l.l) and (B.1.6), if there is no correlation between x and y, then there are no trends for Y to either increase or decrease with increasing x.
Therefore, the least-squares fit must yield a2 = O. Now, consider the question of whether the data correspond to a straight line of the form: The solution for (X; is: N2:x,y, 2:Xi 2:Yi N2:l - (2:yY (B.2.! ) (B.2.2) APPENDIX B Regression Analysis 345 Again, if there is no correlation between x and y, then a; = O. At the other extreme, if there is complete correlation between x and y, then there is a relationship between al> al> af, and a; that is: Thus, a perfect correlation gives Cl'.za; = 1 and no correlation yields Cl'.zCl'.; = 0 since both Cl'.z and a; are zero for this condition. The linear correlation coefficient is there-fore defined as: (B.2.3) The values of Ree are -1 ::5 Ree ::5 1 with Rec = 1 and Rec = 0 defining perfect and no correlations, respectively. Although the linear correlation coefficient is commonly quoted as a measure of "goodness of fit," it is really not appropriate as a direct meas-ure of the degree of correlation. If the data can be represented in a manner such that the fit should result in a y-intercept equal to zero, then a simple method can be used to determine the "goodness of fit." B.3 I Correlation Probability with a Zero Y.lntercept Numerous kinetic expressions can be placed into a form that would yield a zero y-intercept when using the linear least-squares method. A survey of a few of these mod-els is provided in Table B.3.1. Given that the y-intercept is a known value (i.e., zero), if a perfect correlation could be achieved, the hypothesis that the true value of the param-eter, ai' is equal to the specified value, a~, could be tested by referring the quantity: (B.3.l) to the table of Student's [ values with N-2 degrees of freedom. The standard error, SE, can be calculated as follows. The standard deviation U o of the determination of a parameter 75 is via the chain rule: (B.3.2) where Ui is the standard deviation of each datum point i. If (J"i = (J" = constant, then u; is approximately equal to the sample variance, which is (P. R. Bevington, Data r.li I • I = kt ----,-,-TCC_I G~} [ kl(~+ 1)] = [~~ +fA~ t t -= kt In = (C~ -C~)kt C~ C4 kt A P k --A --'t P A P A+B ~P • product present at t = 0, c~ 4.
Second-order, irreversible (one-way), (constant volume) • C~ C~ • C~ ' C~ 2.
First-order, irreversible (one-way) 3.
First-order, reversible (two-way) • no product present at t 0 1.
Oth-order, irreversible (one-way), (constant volume) 5.
Second-order, reversible (two-way) C~ C~, cg. = C~ 0 k A+B( ')C+D k2 = 2kl[~ -l]C~t 6.
Third-order, irreversible (one-way), (constant volume) A+2B ~p A+B+C ~p (CO C) -A A (0)2 7] _ In 0/ 0 ., CA -~ kt (C B CA)C B (C::) [C2/C~ 1] (C~) [c~/c~ I] (C() Inc"';; + --(~ _ <1:) In c;, + (~ --c:r-) In (c~ c~{ c;: c~ c;~ (C~?
1] kt 7.
nth-order, irreversible (one-way), (constant volume) A ~P (CA)(i -11) -(C~)(i -11) (n 1)kt APPENDIX B Regression Analysis 347 Reduction and Error Analysis for the Physical Sciences, McGraw-Hill, New York, 1969, p. 114): ~2 == N _1 2 '" (y,.
~ ~ X)2 v .:::;.; '-<] -'-<2 i (B.3.3) for the linear equation (B.1.1). (The sample variance is the sum of squares of the residuals divided by the number of data points minus the number of parameters fitted.) Now Equation (B.3.2) written for the linear equation (B.1.1) gives: (B.3.4) Using Equation (B.1.6) to calculate the partial derivatives in Equation (B.3.4) yields: (B.3.5) (B.3.6) where Thus, SE(ii]) = (Tal and SE(ii2) = (Ta,' Now returning to Equation (B.3.1), t is the experimental deviation over the standard error and if this value is larger than the value in a Student's t-distribution table (see any text on statistics for this table) for a given degree of confidence, for example, 95 percent Ci:xp = expected deviation/standard er-ror), then the hypothesis is rejected, that is, the y-intercept is significantly different than zero. If t < t:xp then the hypothesis is accepted and 0'2 can be reported as: (B.3.7) B.4 I Nonlinear Regression There are numerous methods for performing nonlinear regression. Here, a simple analysis is presented in order to provide the reader the general concepts used in per-forming a nonlinear regression analysis.
To begin a nonlinear regression analysis, the model function must be known. Let: y = f(x, a) (B.4.1) where the function f is nonlinear in the dependent variable x and unknown param-eters designated by the set a = [a], a2, ..., all]. A least-squares fit of the observed 348 APPENDIX B Regression Analysis measurements y; to the function shown in Equation (B.4.1) can be perfonned as fol-lows. First, define X2 [for linear regression see Equation (B.l.5)] as: (B.4.2) As with linear least squares analysis, X2 is minimized as follows. The partial de-rivatives of X 2 with respect to the parameters of a are set equal to zero, for exam-ple, with al: ax2 a {I N } - 2N af(x;, a) 0=- = -2 ~ [Yi - f(x;, a)]2 = 2 ~ [Yi - f(x;, a)] (B.4.3) aa1 aa1 (J';= 1 (J' ;= 1 aal Thus, there will be n equations containing the n parameters of a. These equations involve the functionf(x;, a) and the partial derivatives of the function, that is, for j = I, ... , n The set of n equations of the type shown in Equation (B.4.3) needs to be solved.
This set of equations is nonlinear if f(x;, a) is nonlinear. Thus, the solution of this set of equations requires a nonlinear algebraic equation solver. These are readily available. For infonnation on the type of solution, consult any text on numerical analysis. Since the solution involves a set nonlinear algebraic equation, it is per-fonned by an iterative process. That is, initial guesses for the parameters a are re-quired. Often, the solution will tenninate at local minimum rather than the global minimum. Thus, numerous initial guesses should be used to assure that the final so-lution is independent of the initial guess.
The issue of "goodness-of-fit" with nonlinear regression is not straightforward.
Numerous methods can be used to explore the "goodness-of-fit" of the model to the data (e.g., residual analysis, variance analysis, and Chi-squared analysis). It is al-ways a good idea to inspect the plot of the predicted [y(x;)] versus observed Y; val-ues to watch for systematic deviations. Additionally, some analytical measure for "goodness-of-fit" should also be employed.
Transport in Porous Media C.1 I Derivation of Flux Relationships in One-Dimension Consider a tube filled with an isobaric binary gas mixture of components A and B. When component A moves it exerts a force on B. This frictional force, flAB, can be described as: ffAB = (proportionality constant) (concentration of A) (concentration of B) (relative velocity of A to B) or number of collisions momentum exchange per collision (C.l.l) where Vi is the molecular velocity of species i. The total frictional losses have to equal the driving force. Thus, for the section dz, <1: (C.l.2) Multiply Equation (C.l.2) by C (total concentration) and rearrange to give: .3.:19 350 APPENDIX C Transport in PorOllS Media If the proportionality factor is defined as: then Equation (C.l.3) can be written as follows: (C.l.3) (C.IA) (C.l.5) -CD dX _--"A=..B _A = C(VA -VB) XAXB dz Equation (C.1.5) is Fick's First Law. To see this, rearrange Equation (C.1.5) as shown below: dXA -CD -AB dz (C.l.6) Recall that Fick's First Law is: (C.l.7) where fA is the flux of A with respect to a coordinate system that is moving at Vlolal and: Further arrangements of Equation (C.l.6) can be made as shown below: dXA CA(-XAVA - XBVB + XAVA + XBVA) = -CDAB -dz dXA Vlolal) = -CDAB dz which is Equation (C.1.7).
APPENDIX C Transport in PorOllS Media 351 Now consider a multicomponent system. For a multicomponent mixture of NCOMP species: (Cl.8) by analogy to the frictional force for a binary mixture [Equation (C.l.l) with Equa-tion (CIA)]. Using Equation (C.l.8) gives a total force balance of: -dX.
NCOMP XX(V - V) 1 = 2: 1) 1 ) (Cl.9) dz j=l Dij ij that is the Stefan-Maxwell equations.
The Stefan-Maxwell equations are normally written in terms of fluxes, Ni. Since N i = CiVi, Equation (Cl.9) can be expressed in terms of fluxes as: -dX.
NCOMP [XN XN] __ 1= 2: )1 1) dz j=l CDij ij (CUO) C.2 I Flux Relationships in Porous Media Consider the movement of a binary gas mixture in a pore: dz Pore wall The loss of momentum in dz due to molecule-wall collisions is: where gc = gravitational constant A~ = cross-sectional area of the pore DKA = Knudsen diffusion coefficient of A (dPA)K = change in pressure from molecule-wall collisions (C2.I) 352 APPENDIX C Transport in PorOllS Media Fick's First Law is (momentum losses due to molecule-molecule collisions): (C.2.2) (diffusive flux) + (bulk flow) Let: FR=I+NBI INA so that Equation (C.2.2) can be written as: or (C.2.3) (C.2.3) (C.2A) p _ R~T[ -(dPA)mmgcAc - NA-- I DAB where (dPA)mm is the change in pressure from molecule-molecule collisions. Now the total momentum loss due to molecule-wall and molecule-molecule collisions is the sum of Equation (C.2.1) and Equation (C.2A) [(dPA) = (dPA)K + (dPA)mm]: or after rearrangement: -1 1 dPA --[1- (FR)XA + 1] RgT dz DAB D KA (C.2.S) If there is equimolar counterdiffusion (NA = -NB ) and/or if XA is small, Equation (C.2.S) reduces to: where I 1 dPA + 1 R~T dz DAB D KA -DTA dPA ---RgT dz (C.2.6) (C.2.7) Equation (C.2.7) is called the Bosanquet equation.
A momentum balance for multicomponent mixtures can be fonnulated in a man-ner analogous to that used to derive Equation (C.2A) using molecule-wall and molecule-molecule (Stefan-Maxwell) relationships to give: or APPENDIX C Transport in PorOIJS Media NR T NCOMP R T (d) P-Ig d P '" g( -P; gcAc - -- z gcAc + £.; -- Xp; D~ j=!
Dij ij 353 -1 dPi RgT dz NCOMP [XN - XN] N.
'" J I I J l £.; +-j=!
Dij DKi ij (C.2.8) lHDEX _ A Absorption of beer, modeling of 97 Aeetaldehyde ~ , decomposition of, 126 formation of, 178-179 Acetic acid, formation of, 78-79 Aeetone formation of, 47-48, 63, 179, 231 rate function properties, 20 (figure) Acetylene, vinyl chloride from, 47 N-Acetyl-phenylalanine methyl ester, formation of, 241-246 Acid catalysis, with zeolites, 169-170 Acrolein/DMB reaction, 48, 63 Activated carbon catalysts, 176-177 Activation energy apparent, 163-165 diffusional limitations and, 207-208 direct vs. catalyzed reactions, 10 I, 104 hcat of adsorption and, 163-165 in transition state theory, 59-60 Active sites in bifunctional catalysts, 170-171 calculating number of, 149 defined, 18 ensembles of surface atoms, 150 Koshland induced fit theory, 114-116 notation for, 144 Activity coeffIcients, 60 Adiabatic batch reactors, 291-293 Adiabatic continuous flow stirred tank reactors, 304 Adiabatic fixed-bed reactors, 318-319, 324 Adiabatic operation, 64 Adiabatic plug flow reactors butadiene/ethylene reaction in, 299-300 dimensionless concentrations in, 301-302 energy balance for, 297-298, 299 temperature profile, 287 (figure) Adiabatic temperature rise, 217-218 Adsorption BET, 141-143 chemisorption, 140, 143-147 dissociative, 143, 145-147 physisorption, 140-143 as rate-determining step, 172 Adsorption isothenTIs BET, 141-143 dissociative. 147 Langmuir, 144, 145 (tlgure) Agitation, effect of, 231 Air pollution from hydrocarbon fuels, 131 ozone loss and. 105 pollution standard index, 11-13 Alkanes cracking of, 164 (fIgure), 165-166 dehydrogenation of, 42 Alkylation, of ethylbenzene, 235-236 Alloys, in catalysts, 150-151 AI-Saleh, M., 92 Alumina supported catalysts in carbon monoxide oxidation, 149-150 in methylcyclohexane dehydrogenation, 160-161 pellet size calculation for, 202-203 in n-pentane isomerization, 170-171 Aluminosilicate frameworks, in zeolite catalysts, 166-167 Ammonia synthesis iron catalysts for, 2, 141-143, 150,246-250 microkinetic analysis of, 246-252 ruthenium catalysts for, 49-50, 159-160 stoichiometric relationship, 8 thermodynamic constraints of, 1-2 turnover frequency, 150 two-step assumption and, 157-158 Anderson, B. G., 277 Anderson criterion, 228-229 Anderson, J. B.. 228 Antarctic ozone loss, 105 Aparicio, L. M., 240, 246 Apparent activation energy, 21, 55 Areal rate of reaction. 17 Ads, R.. 201 Aromatic compounds, nitration of, 303-305 Arrhenius' Law rate constants and, 21, 54-56 vs. transition state theory, 60 Arrhenius number, 215 Arrhenius plots cracking of n-alkanes, 164 (figure) general~form, 22 (fIgure) ~ observed rate constants, 208 (figure) turnover frequency, 140 (figure) Associative desorption, 145 Automotive catalytic converters, 12-13 Axial dispersion coefficient, 273 Axial dispersion model described. 272-277 industrial vs. laboratory reactors, 323-324 PFR behavior and. 324 reactor conversion prediction and, 277-280 B Ballard, D. G. H., III Baratti, 1.. 119 355 356 Index Barner, H, K, 235 Barnett, L. G" 237 Bateh reactors adiabatic operation, 291-293 analogy to PFRs, 77 defined, 64, 65-67 laboratory scale, 84-87 material balance for, 65, 83 (table) nonisothelmal operation, 288-294 reactor conversion predietion assumptions and, 269 variables for, vs. flow reactors, 74 (table) Becue, T., 159,250 Bed density, 316-317 Beer, metabolism of, 97 Benzene hydrogenation of, 210-212, 236-237, 318-319 in methylcyclohexane dehydrogenation feed, 161 production' of, 47 Benzoquinone, 66-67 Berty reactors described, 88-89 ethylene oxidation kinetics, 92-93 residence time distributions, 268-269 Berzelius, J. J., 133 BET equation, 142 Bevington, P. R., 343, 345 Bifunctional catalysts, 170-171 Bimetallic catalysts, 150-151 Bimolecular reactions rate expressions for, 25 (table), 29-31 surface-catalyzed, 151-153 Binding, of substrates, 114-116 Bioreactors modeling of, 280, 281 (figure), 322-323 semibatch reactors as, 67 for vaccines, 300 Biot numbers, 220 Bischoff, K. B., 270, 310 Blood flow, dispersion model for, 282-283 Body-centered eubic crystal structure, 135 (figure) Bohlbro, H., 150 Bosanquet equation derivation of, 352 transition diffusivity approximation, 191 Boudart, M., 18, 20,22,23,59,63,65,75, 102, 145, 149, 150, 153, 157, 161,163,184,229,230,231,245 Boundary conditions axial-dispersion model, 277-278 cylindrical pores, ideal, 194 flat plate catalyst pellets, 209-210, 215, 221, 222 spherical catalyst pellets, 198 two-dimensional fixed-bed model, 325 Boundary layer, diffusion through, 185-187 Bromine. See Dibromine I-Bromobutane, reaction with DEA 51 Brunauer, Stephen, 141, 142 Bubble column reaetors, 329 (figure), 330 Bulk fluid phase, reactor balanees for, 321 Burwell, R. L., Jr., 184 Butadiene, reaction with ethylene, 299-300 n-Butane, maleic anhydride from, 333 n-Butanol, dehydration of, 169 Butene isomers, 28-29, 46, 268-269 Butler, P., 85, 86 Butt, J. B., 218, 219, 236, 318 C Carberry reactors, 88, 89 (figure) Carbon dioxide, formation of, 181 Carbon monoxide dichlorine, reaction with, 176-177 nitrous oxide, reaction with, 181 oxidation of, 10-11, 133-134, 149-150, 162-163 water gas shift reaetion, 182 Carta, G., 129 Carter, J. L., 150, 151 Catalysis. See also Enzyme catalysis; Heterogeneous catalysis; Steady-state approximation defined, 101 reaetion sequences in, 101-105, 102 (table) Catalysts. See also specific types, e.g., Platinum catalysts bifunetional, 170-171 bimetallic, 150-151 in eommercial reactors, 2 turnover frequency, 18 Catalytic converters (automobile), 12-13 Catechol, L-dopa synthesis from, 119-121,285 Chain reaction sequences, 101, 102 (table) Characteristic length parameters "equivalent" spheres, 205-206 finite cylindrical pellet eatalysts, 204 ideal pellet geometries, 201-202 Chemieal potential, 339 Chemical vapor deposition (CVD), 223-228 Chemisorption active site ensembles and, 150 defined, 140 dissociative, 143, 145-147 palladium supported catalysts, 18-19 platinum supported catalysts, 138-139 Chen, E., 125 Chen, N. H., 69, 70 Chiral reactions L-dopa synthesis, 123-124 olefin hydrogenation, 240-246 Chlorine, as catalyst for ozone decomposition, 103-105, 110-111. See also Dichlorine Chlorofluorocarbons, ozone loss and, 105 Cleland, W. w., 129 Clontz, C. R., 48 Closed reaction sequences, 101, 102 (table) Cocurrent vs. eountercurrent coolant flow, 310-311, 329 (figure), 330 Colburn J factors, 189 Combined internal/external transport chemical vapor deposition, 223-228 flat plate eatalyst pellets, 219-223 thermal conductivity effect, 218-219 Combustion, of hydrocarbon fuels, 131 Commercial scale reactions. See Large scale reactions Comonomers, material balance for, 72-73 Competition, among adsorbing molecules, 145 Competitive inhibitors, 127 Completely backmixed reactor, 265, 281 (figure).
See also Continuous flow stirred tank reactors (CSTRs) Computer simulation, of ethylene hydrogenation, 256-257 Concentration profiles axial dispersion model, 278-279, 280 (figure) multiphase reactors, 330 (figure) stagnant film diffusion, 187 Concentrations. See also Surface concentrations defined,17 notation for, 26 for transition states, 58 Configurational diffusion, 191 Confonnational changes, in enzyme catalysis, 114-115 Constraints. See Thennodynamic constraints Continuous flow stirred tank reactors (CSTRs) axial dispersion model and, 274 Berty reactor behavior vs., 268-269 defined,70-74 multiple steady-states in, 305-309 nonisothermal operation, 303-305 rate constants for, 75-76 as recycle reactor limit, 91-92 residence time distributions, 262-267, 263 (figure), 267 (figure), 270-272 in series, vs. PFRs, 81-82, 322-323 space-time yield, vs. PFRs, 79-80 Cooking fumes, exhausting of, 331 Coolant temperature, effect of, 309-311 Correlation coefficients, in linear regression, 32, 344-345 Cortright, R. D., 139, 140,253 Countercurrent vs. cocurrent coolant flow, 310-311, 329 (figure),330 Coupled catalytic cycles, 242-245 Cracking reactions of alkanes, 164 (figure), 165-166, 169-170 of cumene, 206-207 fluidized-bed reactors for, 331-332 Creighton, J. R., 109 Cronstedt, A. E, 166 Cropley, J. B., 310, 311, 335 Crystal structures metal catalysts, 134-136 zeolite catalysts, 166-169 Crystallites, size of, 138-139 CSTRs. See Continuous flow stirred tank reactors (CSTRs) Cumene, cracking of, 206-207 CVD (chemical vapor deposition), 223-228 Cyclohexane dehydrogenation of, 150-151,237 fonnation of, 18-19,210-212,230,238-239 Cyclohexene fonnation of, 299-300 hydrogenation of, 18-19,230,238-239 Cyclopentadiene, 66-67 Cylindrical catalyst pellets diffusion/reaction in, 203-206 Index infinite, 196-197,201 (table) thennal conductivity effect, 218-219 Cylindrical pores, diffusion through, 192-195 D Damkohler number, 219 Danckwerts, P., 262, 278 Daniels, E, 96 Data analysis. See Experimentation; Statistical analysis Datye, A. K., 137 Dautzenberg, EM., 235 Davis, M. E., 167, 168, 169,280,281,322,323 Davis, R. J., 49, 99, 159, 181,250 DEA (diethylamine), 51 DEBA (diethylbutylamine), 51 Debye-Htickel theory, 61 Decoloring, of effluent streams, 48-49 Decomposition of acetaldehyde, 126 of dinitrogen pentoxide, 15-16,27-28,54-55, 126 of ethanol, 97 of ozone, 101, 103-105, 110-111 photocatalytic, 49 of 2-propanol, 59-60, 63 rate expressions for, 26 (table) Degnan, T. E, 328 Degree of polymerization, 112 Dehydration, of alcohols, 169 Dehydrogenation alkanes, 42 cyclohexane, 150-151,237 methylcyclohexane, 160-161 n-pentane, 171 propane, 45 2-propanol,231 Denbigh, K., 2, 75 Density of catalyst beds, 316-317 Desorption associative, 145 as rate-detennining step, 155-156, 173 Detergents, 37 Deuterium-hydrogen exchange, 109 Dibromine dihydrogen, reaction with, 131-132 hydrogenation of, 4 Dichlorine carbon monoxide, reaction with, 176-177 dihydrogen, reaction with, 101 Diels-Alder reactions DMB/acrolein reaction, 48, 63 in isothennal batch reactor, 66-67 Diethylamine (DEA), 51 Diethylbutylamine (DEBA), 51 Diffusion catalyst boundary layer, 185-188 configurational, 191 Knudsen, 190-191 observed kinetics and, 202, 207-208 357 Diffusivity effective, 196 transition, 191 typical values, 186 Dihydrogen, See also Hydrogen in ammonia synthesis, 1-3, 159-160,250-251 associative desorption of, 153 chemisorption, on platinum, 138-139 dibromine, reaction with, 131-132 dichlorine, reaction with, 101 as inhibitor, 171, 250-251 olefin hydrogenation, eflect on, 244-246, 252-253 Dimensionless material and energy balances t1at plate catalysts, 221 nonisotherma1 PFRs, 302-303 Dimerization, of isobuty1ene, 154-155 2,3-Dimethyl-1 ,3-butadiene (DMB), 48, 63 Dimethyl-p-to1udine (PT), 34-35 Dinitrogen in ammonia synthesis, 1-3,49-50, 141-143 chemisorption of, 150 dissociative adsorption of, 157-158,248-249,250 formation of, 181 oxidation of, 131 Dinitrogen pentoxide decomposition of, 15-16,27-28,54-55, 126 as nitrating agent, 303-304 DIPAMP, 241-245, 257-258 Dirac delta function, 265 Direct vs. catalyzed reactions, 101, 103, 104 (figure), 116 (figure), 134 (figure) Dispersion models axial-dispersion model, 272-277 radial-dispersion model, 282 reactor conversion prediction and, 277-280, 282-283 Dispersion of supported metal catalysts, 136-138 Dissociative chemisorption defined, 143 rate expressions for, 145-147 Distillation, reactive, 68 Distribution of residence times, See Residence time distribution function Djega-Mariadassou, G., 145, 153, 157, 161,245 I-DodecanoL 37 Domain closures, in enzyme catalysis, 114-116 L-Dopa synthesis, 119-121, 122-124,285 Dopamine, 123 Douglas, W. J, M" 235 Drug metabolism, tracer experiments for, 265 Dumesic, J. A" 139,140,150,240,246,247,248,249,253 Dupont, 333 Dyes, in effluent streams, 48-49 E E.
119 Effective diffusivitv, 196 Effectiveness facto~rs, See also Overall effectiveness factors "equivalent" sphere characteristie length and, 205-206 nat plate catalyst pellets, 212 interphase, 220 nonisothermal reactions, 212-217 spherical catalyst pellets, 199-202 Effluent streams, dyes in, 48-49 Elementary steps adsorption, in heterogeneous catalysis, 140-147 Arrhenius' Law and, 54-56 defined, 4 desorption, in heterogeneous catalysis, 155-156 microkinetic analysis and, 240, 248 notation for, xvi (table), 4 (table) rate expressions for, 23-24, 25 (table) Rideal-Eley vs. Langmuir-Hinshelwood, 153-154 vs, stoichiometric reactions, 8, 53-54 surface reactions, in heterogeneous catalysis, 147-155 transition state theory for,S, 56-62 two-step assumption for, 157-162 Elimination of beer, modeling of, 97 Emmet, Paul, 141, 142 Enantioselectivity, in olefin hydrogenation, 240-246 Endothermic reactions effectiveness factors and, 213-214 potential energy profile, 57 (figure) temperature dependence, 56 Energy balance expressions batch reactors, 291 coolant, in fixed-bed reactor, 335 CSTRs, 303, 304 dimensionless, 221 fixed-bed reactors, 317, 320, 321, 325, 335 t1at plate catalyst pellets, 214, 220 general form, nonisothermal reactors, 286-288 PFRs,297-298,299,300-301 Energy diagrams carbon monoxide oxidation, 134 (figure) enzyme catalysis, 116 (figure) hydrochloric acid formation, 103 (figure) ozone decomposition, 104 (figure) Engelhard Corporation, 331 Ensembles, in active sites, 150 Enthalpy, in nonisothermal operation, 288-290, 290 (figure) Enthalpy of adsorption 162-166 Enthalpy of reactions. See Heat of reactions Entropy, of activation, 59-60 Environmental cleanup, with fixed-bed reactors, 330-331 Enzyme catalysis kinetics of. 117-119 Koshland induced fit theory, 114-116 Lineweaver-Burk analysis, 119-121 nonlinear regression analysis, 121-122 Epoxidation, of olefins, 43-45 Equilibrium criteria for, 339-341 notation for, xvi (table), 4 (table) Equilibrium composition ammonia synthesis example, 2-3 determination of. 341-342 Equilibrium constants from experimentation, 174-175 in nonideal systems, 60 rate constant and, 24 temperature dependence of, 340-341 for transition states, 58 Equilibrium extent of reactions. See Extent of reactions "Equivalent" sphere characteristic lengths, 205-206 Ercan, C., 235 Ergun equation, 317-318 Ertl, G., 133, 134,250,251 Ester formation isoamyl propionate, 129-130 product removal in, 68 Ethane, hydrogenolysis of, 125-126, ISO-lSI Ethanol acetaldehyde from, 178-179 decomposition of, 97 oxidation of, 38 Ether, production of, 69-70 Ethyl acetate, reaction with hydroxyl ions, 75-76 Ethyl species, formation of, 254-256 Ethylbenzene, alkylation of, 235-236 Ethylene butadiene, reaction with, 299-300 hydrogenation of, 99, 139-140,252-257 oxidation of, 92-94 Ethylidyne, 109 Exhaust streams, cleanup of, 331 Exothermic reactions effectiveness factors and, 213-214 heat release for, 87 Madon-Boudart criterion and, 230 potential energy profile, 57 (figure) temperature dependence, 56 temperature profile, in PFR, 287 (figure) Experimentation. See also Microkinetic analysis; Statistical analysis criteria for kinetic analysis, 228-230 and kinetic sequences, postulation of, 171-177 minimizing transport effects, 230-232 Exponential residence times in CSTRs, 74 defined, 65 Exposed metal atoms, in catalysts, 138-139 Extent of reactions defined, 9-10 equilibrium composition, 10-11 fractional conversion, 13-14 molar expansion factor, 14-15 vs. rate function, 20 (figure) steady-state approximation and, 108 External recycle reactors, 89 External transport effects defined, 184-185 heat transfer effects, 189-190 mass transfer coefficients and, 187-189 molecular diffusion, 185-188 F Face-centered cubic crystal structure, 135 (figure) Falch, E. A., 284 Fast food restaurants, exhausting of, 331 Faujasite frameworks, 167 Index Femtochemistry, 5, 57-58 Fermentation reactors residence time distribution data for, 284-285 semibatch reactors as, 67 Fertilizers, 2 Fick's First Law axial dispersion model and, 273 cylindrical pores, ideal, 193 derivation of, 350, 352 external transport effects and, 186 Finlayson, B. A., 323 First order reactions rate expressions for, 24, 26 (table), 26-29 residence time distributions, 270-272 vs. second order, 31-33, 32 (figure) in series, 105-108 Fishel, C. T., 49 Fixed-bed reactors described, 315-317, 328 (figure) environmental cleanup with, 330-331 one-dimensional model, 317-325 Peclet number for, 274-275, 276 (figure), 282 two-dimensional model, 325-327 Flagan, R., 98, 131 Flat plate catalyst pellets combined internal/external transport in, 219-223 diffusion/reaction in, 208-212 nonisothermal effectiveness factors, 213-215 Flow reactors. See also specific types, e.g., Continuous flow stirred tank reactors (CSTRs) defined, 64 energy balances for, 286-288 laboratory scale, 87-95 relaxation times and, 124-126 variables for, 74 (table) Fluidized-bed reactors, 329 (figure), 331-333 Flux expressions. See Heat flux expressions; Molar flux expressions Fogler, H. S., 70, 96 Fractional conversion, 13, 14 Fractional surface coverage in ammonia synthesis, 248-249 defined, 144 for dissociative adsorption, 147 Franckaerts, J., 179 Free energy change. See Gibbs free energy change Free radical chain reactions acetaldehyde decomposition, 126-127 hydrochloric acid formation, 131-132 Frequency, in transition state theory, 59 Friction factors, for fixed-bed reactors, 318 Frictional forces, in gas mixtures, 349 Friend, C. M., 134 Froment, G. F., 179,270,310,326,327 Fructose, 116-117 Fugacity, 3, 339 G Gaden, E. L., Jr., 284 Gainer, J. L., 49, 129 359 360 Index Garces, J, M., 49, 159 Gases diffusivity of, 186 ideal gas law, 15 Lewis and Randall mixing rules for, 3, 341 mole changes and, 78-79 quality of mixing, 88-89 Gibbs free energy change equilibrium criteria and, 339-340 rate determining steps and, 148-149 Glucose, 116-117 Glucose isomerase, 117 Goddard, S. A., 139, 140,253 Gonzo, E., 18 Goodness of fit linear regression, 32-33, 345-347 nonlinear regression, 348 Goodwin, J. G., 125 Growth rate profiles, in chemical vapor deposition, 224-227 Guldberg-Waage rate expressions bimolecular reactions, 29 generalized form, 22-23 parallel reaction networks, 42 series reaction networks, 38 trimolecular reactions, 35 Gupta, V. E, 235 H Haag, W 0., 154, 170 Haber ammonia synthesis process, 2 Half-life, 27 Halpern, J., 240, 241, 242, 244, 257, 258 Hansen, E. W, 256 Hawley, M. c., 317 Heat flux expressions, 214 Heat of adsorption, 162-166 Heat of reactions for ammonia synthesis, 1 defined, 57 in nonisothermal operation, 289-290 Heat transfer effects criteria for kinetic analysis, 228-229 external transport, 189-190 internal transport, 212-217 vessel size and, 295-297 n-Heptanal, production of, 85 Heterogeneous catalysis, 133-177. See also Fixed-bed reactors; Fluidized-bed reactors; Multiphase reactors adsorption step, 140-147 bifunctional catalysts, 170-171 defined, 133-134 desorption step, 155-156 enthalpy of adsorption and, 162-166 kinetic sequences, evaluation of, 171-177 single crystal surfaces, 134-136 supported metal catalysts, 136-140 surface reactions, 147-155 two-step assumption for overall reactions, 157-162 zeolite catalysts, 165-170 Heterogeneous reactions vs. homogeneous, 133,315-317 reaction rates for, 17-18 Hexagonal crystal structure, 135 (figure) Hexane, cracking of, 169-170 2-Hexanol, formation of, 292-293 I-Hexene hydration of, 292-293 hydroformylation of, 85 isomerization of, 202-203 Hicks, J. S., 216 Hill, C. G., Jr., 54, 66, 78, 235, 299, 334 Hinrichsen, 0., 250, 251 Holies, J. H., 181 Homogeneous vs. heterogeneous catalysis, 133, 315-317 Horiuti, J., 252, 253, 256 Hot spots, 303, 309-310 Hudson, J. L, 96, 233 Hurwitz, H., 160 Hydration I-hexene, 292-293 isobutylene, 235 Hydrobromic acid, formation of, 131-132 Hydrocarbon fuels. See also Petrochemical industry combustion of, 131 reforming of, 170-171 Hydrocarbons, selective oxidation of, 183,332-333 Hydrochloric acid acetylene, reaction with, 47 formation of, 101 Hydroformylation commercial scale, 36-37 I-hexene, 85 mechanism, 36 (figure) propene, 23 (table), 68, 69 (figure) Hydrogen. See also Dihydrogen chemisorption, on nickel, 143 deuterium, exchange with, 109 Hydrogen peroxide, as decontaminator, 300 Hydrogenation benzene,210-212,236-237,318-319 cyclohexene, 18-19,230,238-239 dibromine, 4 ethylene, 99, 139-140,252-257 isopentene, 171 prochiral olefins, 240-246 propene, 23 (table) Hydrogenolysis of ethane, 125-126, 150-151 side products from, 42 Hydroxyl ions, reaction with ethyl acetate, 75-76 Ideal catalyst pellet geometries, 196-197, 201 (table) Ideal gas law, 15 Ideal limits of mixing, 76 Ideal reactors. See also Laboratory reactors; specific types, e.g., Plug flow reactors (PFRs) __ . ]odex.
361 defi ned, 64---65 equations for, 83 (table) Induced fit theory, 114-116 Induction period. See Relaxation time Industrial scale reactions. See Large scale reactions Inert species, effect on equilibrium, 342 Infinite cylinder catalyst pellets. See Cylindrical catalyst pellets Infinite slab catalyst pelIets, 196-197,201 (table) Inhibition by dihydrogen, 171, 250-251 in enzyme catalysis, 127-128 by product in feed, 173-174 Initial rate behavior. 173-174 Initiation steps defined, 1 styrene polymerization, 111 Instantaneous selectivity, 40 Integration factor method, 38, 106 Intermediates reactive, 5, 100-105, 102 (table), 242-245 transition state, 5, 6, 7 (figure) types of, 5, 6 Internal recycle reactors. See Berty reactors Internal transport effects, 190-218 adiabatic temperature rise, 217-218 cylindrical catalyst pellets, 203-206 cylindrical pores, ideal, 192-195 defined, 185 effectiveness factors, 199-202 flat plate catalyst pellets, 208-212 heat transfer effects, 212-217 ideal pellet geometries, 196-197,201 (table) pellet size, calculation of, 202-203 pore diffusion, influence on rate, 206-207 pore size and, 190-192 severe diffusional resistance, 202, 207-208 spherical catalyst pellets, 197-201 zig-zag pores, 195-196 Interphase effectiveness factors, 220 Interphase transport effects, See External transport effects Intraphase transport effects. See Internal transport effects Ionic strength, and rate constants, 61 Iron catalysts BET method and, 141-143 Haber process, 2 microkinetic analysis of, 246-250 partieIe size effect, 150 Irreversible elementary steps, notation for, xvi (table), 4 (table) In-eversible stoichiometric reactions defined, 14 notation for, xvi (table), 4 (table) rate expressions for, 21 Isoamyl alcohoL esterification of, 129-130 IsobutanoL dehydration of. 169 Isobutene, from I-butene, 28-29 Isobutylene dimerization of. 154-155 hydration of. 235 Isomerization I-butene, 28-29. 46 hexene. 202-203 and kinetie sequences, postulation of. 172-175 n-pentane, 170-171 rate expressions for, 26 (table) Isopentane, from n-pentane, 170-171 Isopropanol, oxidation of, 179 Isothermal batch reactors energy balance for, 291 procedure for solving, 66-67 Isothermal operation, 64 Isothennal plug flow reactors energy balance for, 298 temperature profile, exothermic reaction, 287 (figure) Isotopic transient kinetic analysis, 124-126 J .r factors, 189 Johnston, E. H., 96 K Kargi, E, 127 Kehoe,J.~G,,218,219, 236, 318 Kinetic coupling, 245 Kinetic sequences. See also Microkinetic analysis; Rates of reactions evaluation of, 171-177 types of, 100-101, 102 (table) zero intercepts for, 346 (table) Kladko. M" 295, 296, 297 Knudsen diffusion, 190-191 Komiyama, H.. 224, 225, 226, 227 Koshland, D. E., Jr.• 115 Koshland induced fit theory, 114-116 L Laboratory reactors. 82-95. See also Ideal reactors; specific types, e,g" Continuous flow stirred tank reactors (CSTRs) axial dispersion neglect of, 323 batch reactors. 84-87 ethylene oxidation kinetics, 92-94 limitations of, 95 purpose of. 82-83 stirred flow reactors. 88-92 tubular reactors, 87-88 Ladas, S" 149 Lago, R. M., 170 Laminar flow axial-dispersion coefficient for. 273 vs. turbulent flow. in tubular reactors, 260-262 Landis. C. R" 241, 242. 244. 257. 258 Langmuir adsorption isotherms. 144, 145 (figure) Langmuir-Hinshelwood steps defined. 152 rate and equilihrium constant calculation and, 174-175 vs. Rideal-Eley steps. 153-154, 176-177 Large scale reactions anl1110nia synthesis. 2-3 axial dispersion, neglect of, 323 362 Index Large scale reactions--COl1t.
commercial reactor, 328 (photograph) ethylene oxide production, 93-94 fructose, from enzyme catalysis, 117 hydroformylations, 36-37, 68, 69 (figure) maleic anhydride production, 333 transport phenomena and rates of, 184-185 Lauryl alcohol, 37 Law of Definitive Proportions defined, 9-10 fractional conversion and, 14 Law of Mass Action, 22 Least squares methods. See Linear regression; Nonlinear regression Lee, J. D., 48 Leininger, N., 51 Levodopa synthesis. See L-Dopa synthesis Lewis and Randall mixing rules, 3, 341 Limiting components, 13-14 Limiting conditions for reactors, 65 (table) Linear regression defined, 343-345 kinetic parameter evaluation and, 175 Lineweaver-Burk analysis, 119-121, 122 Lipase, 129-130 Liquids, diffusivity of, 186 Lock, c., 250 Long chain approximation, 112 Lotka-Volterra model, 51 Low-index surface planes, 135 (figure) M MAC, hydrogenation of, 241-246, 257-258 McClaine, B. c., 250 McKenzie, A. L., 59-60, 63 Macrocavities, in silicon wafers, 224-227 Macromixing, 272 Macropores, in catalyst pellets, 192 Madon, R. J., 229, 230 Madon-Boudart criterion, 229-230 Madsen, N. K., 327 Maleic anhydride, production of, 333 Mari (most abundant reaction intermediates), 158, 161-162 Mars-van Krevelen mechanism, 183,332-333 Masel, R., 135, 152, 153 Mass transfer coefficients, 187-189 Material balance expressions axial-dispersion model, 272-274 batch reactors, 65, 294 blood flow, 282-283 comonomers,72-73 CSTRs, 71, 73 cylindrical catalyst pellets, 204 first order reactions in series, 106 fixed-bed reactors, 317, 320, 321, 325, 335 flat plate catalyst pellets, 209, 212, 214, 222 ideal reactors, isothermal, 83 (table) isothermal batch reactors, 67 laminar flow reactors, 261 macrocavities, in silicon wafers, 225 nonisothermal operation, 286, 294, 299 PFRs, 76-77,286,299, 301 polymers, 72-73 radial dispersion model, 282 recycle reactors, 89-92 semibatch reactors, 68 spherical catalyst pellets, 197 Mean residence times, in CSTRs, 74 Mears criterion, 229 Mears, D. E., 20, 22, 228, 229, 231 Mechanisms ammonia synthesis, 247 (table), 250 (table) defined, 5 ethylene hydrogenation, 252, 253 hydroformylation, 36 (figure) induced fit, 115 (figure) Mars-van Krevelen, 183, 332-333 "ping pong bi bi," 129, 130 Mehta, D., 317 Mensah, P., 129 Metal catalysts. See also specific types, e.g., Platinum catalysts crystal structures of, 134-136 supports for, 18-19, 136-140 Metal oxide supports, in catalysis, 18-19, 136-140 Methane, oxidation of, 5-6 Methanol, 69-70 Methyl acrylate, formation of, 78-79 Methyl iodide, 34-35 Methyl tertiary butyl ether (MTBE), 28 Methyl-(Z)-a-acetamidocinnamate (MAC), 241-246, 257-258 Methylacetoxypropionate, 78-79 Methylcylohexane, dehydrogenation of, 160-161 2-Methylhexanal, production of, 85 MFRs (mixed flow reactors). See Continuous flow stirred tank reactors (CSTRs) Michaelis constant, 118 Michaelis-Menten form, 117-119 Michelsen, M. L., 301 Microcavities, in silicon wafers, 224-227 Microkinetic analysis ammonia synthesis, 246-252 defined, 240 ethylene hydrogenation, 246-252 olefin hydrogenation, 240-246 Micromixing, 272 Micropores, in catalyst pellets, 192 Mixed flow reactors (MFRs). See Continuous flow stirred tank reactors (CSTRs) Mixed parallel-series reaction networks, 43-45 Mixing limits, ideal, 76 Mobil Corporation, 331 Molar concentrations. See Concentrations Molar expansion factors, 14-16 Molar flux expressions cylindrical catalyst pellets, 204 derivation of, 349-353 dimensionless, 221, 302-303 flat plate catalyst pellets, 212, 220 spherical catalyst pellets, 197 Index 363 stagnant film diffusion, 187 zig-zag pores, 195-196 Mole balance. See Law of Definitive Proportions Molecular bromine. See Dibromine Molecular chlorine. See Dichlorine Molecular diffusion. See Diffusion Molecular hydrogen. See Dihydrogen Molecular nitrogen. See Dinitrogen Molecular sieves, zeolites as, 169 Molecularity defined,24 for elementary steps, 25 (table) Molybdenum, oxides of, 183 Momentum balance expressions derivation of, in porous media, 351-353 fixed-bed reactors, 317-318, 321, 335 Monsanto, 123 Monte Carlo algorithm, 256 Most abundant reaction intermediates (Mari), 158, 161-162 MTBE (methyl tertiary butyl ether), 28 Muhler, M., 250, 251 Multilayer adsorption, 140-143 Multiphase reactors, 329-330 Multiple reactions. See Reaction networks N Naphthalene, 333-334 Negative activation energies, 163-164 Networks. See Reaction networks Neurock, M., 254, 256 Nickel catalysts in benzene hydrogenation, 218-219, 236-237, 318-319 copper alloyed with, 150-151 hydrogen chemisorbed on, 143 thermal conductivity effect, 218-219 Nitration, of aromatic compounds, 303-305 Nitric oxide, as combustion byproduct, 131 Nitrogen. See Dinitrogen Nitrogen oxides, formation of, 12 Nitrous oxide, 181 Nomenclature list, xii-xvi Noncompetitive inhibitors, 128 Nonflow reactors. See Batch reactors Nonideal flow in reactors, 260-283 axial dispersion model, 272-277 Berty reactors, 268-269 conversion prediction, axial dispersion model, 277-280 conversion prediction, residence distribution times, 269-272 penicillin bioreactor, 280-281 perfectly mixed reactors, 264-265 radial dispersion model, 282 residence time distributions, 262-267, 263 (figure), 267 (figure) turbulent vs. laminar flow, 260-262 Nonisothermal effectiveness factors, 212-217 Nonisothermal reactors, 286-311 batch reactors. 288-294 CSTRs.303-305 energy balances for, 286-288 PFRs, 297-303 ratio of heat transfer area to reactor volume, 295-297 sensitivity of reactors, 309-311 small vs. large vessels, 295-297 stability of reactor steady-state, 305-309 Nonlinear regression described, 347-348 kinetic parameter evaluation and, 175 Lineweaver-Burk analysis and, 121-122 L-Norepinephrine, 123 Norskov, J. K., 246, 247 Numerical methods, use of, 321-322, 326-327 o Observed reaction kinetics criteria for kinetic analysis, 228-230 diffusional limitations and, 202, 207-208 Ogle, K. M., 109 Oil industry. See Hydrocarbon fuels; Petrochemical industry Olefins epoxidation of, 43-45 in ethylbenzene alkylation, 235-236 hydrogenation of, asymmetrical, 240-246 O'Neal, G., 49 One-dimensional fixed-bed reactor model, 317-325 One-point BET method, 142-143 One-way stoichiometric reactions. See Irreversible stoichiometric reactions Open reaction sequences, 101, 102 (table) Order of reaction, 21 Overall effectiveness factors fixed-bed reactors and, 316-317, 320-322 flat plate catalyst pellets, 221-223 Overall selectivity, 40 Oxidation. See also Partial oxidation of carbon monoxide, 10-11, 133-134, 149-150, 162-163 of dinitrogen, 131 of ethylene, 92-94 of hydrocarbons, 183,332-333 of isopropanol, 179 of methane, 5-6, 9 as series reaction network, 37-38 of sulfur dioxide, 234 of titanium tetrachloride, 98 Oxides of nitrogen, formation of, 12 Oxydehydrogenation, of propane, 46 Ozone decomposition of, 101, 103-105, 110-111 formation of, 12 loss of, 105 p Packed-bed reactors. See Fixed-bed reactors Palladium catalysts carbon monoxide oxidation, 149-150, 162-163 cyclohexene hydrogenation, 18-19 ethylene hydrogenation, 254-257 Para, G., 119 364 m~~~-----loLLdll:e",x _ Parallel reaetion networks, 42-43 Park, Y., 280, 281, 322, 323 Parkinson's disease, 122 m l24 Partial oxidation Mars~van Krevelen mechanism, 183,332-333 of naphthalene, 333-335 series reaction networks and, 37-40 Particle size, rates and, 316-317 Pauling, Linus, 116 PDECOL software, 327 Peclet number axial dispersion model and, 274-276 radial dispersion model, 282 Pellet size calculation of, 202-203 reaction rate, elfect on, 231 m 232 Peloso, A., 178 Penicillin bioreactors, 280, 281 (figure) n~Pentane, isomerization of, 170-171 Perfectly mixed reactors. See Continuous flow stirred tank reactors (CSTRs) PET (Positron emission tomography), 277 Petrochemical industry. See also Hydrocarbon fuels Iluidized~bed reactors in, 331-332 laboratory scale reactors for, 87-89 PFRs. See Plug flow reactors Pharmacokinetics, tracer experiments for, 265 Photocatalytic decomposition, 49 Phthalic anhydride, formation of, 333-334 Physisorption, 140-143 "Ping pong bi bi" mechanism, 129, 130 Plate catalyst pellets. See Flat plate catalyst pellets; Infinite slab catalyst pellets Platinum catalysts bifunctionalityof, 170-171 in carbon monoxide oxidation, 133-134 chemisorption of, 138-139 in ethylene hydrogenation, 139-140 in methylcyclohexane dehydrogenation, 160-161 in sulfur dioxide oxidation, 234 Plug flow reactors (PFRs) axial dispersion model and, 272-277, 278-279, 280 (figure) vs. CSTRs in series, 81-82, 322-323 defined. 76-78 flow rate limitations, 95 gas-phase reactions in, 78-79 homogeneous vs. heterogeneous reactions in, 315-317 nonisothermal operation, 297-303 as recycle reactor limit, 91 residence time distributions, 262, 263 (figure), 267 (figure), 270-272 space-time yield, vs. CSTRs, 79-80 temperature protlle, exothermic reaction, 287 (figure) vs. tubular reactor laminar flow, 260-262 Polanyi, M., 252, 253, 256 Pollution from hydrocarbon fuels, 131 ozone loss and, 105 standard index, 11-13 Polyn1erizatiOll, of styrene, II 1-112 Polymers, material balance for, 72-73 Polystyrene, production of, 111-112 Poly(styrene-sulfonic acid), as catalyst, 154-155 Poppa, H" 149 Pore size diffusion effects and, 190-192 rate, influence on, 206-207 in zeolite catalysts, 167-169 Porosity, 195 Porous media, derivation of flux in, 351-353 Positron emission tomography (PET), 277 Potential energy profiles elementary reactions, 57 (figure) hydrogen chemisorption on nickel, 143 (figure) Power law rate expressions defined,22-23 examples, 23 (table) Weisz-Prater criterion, 228 Prater, C. D., 228 Prater number, 215 Prater relation, 217 Predator-prey interactions, modeling of, 51 Prediction of reactor conversion axial-dispersion model, 277-280 residence time distribution function, 269-272 Pre-exponential factors in Arrhenius' Law, 21 calculating, 55 transition state theory vs. experiment, 152-153 unimolecular reactions, 62 Price, T. H., 318 Primary structure, of enzymes, 114 Prochiral olefins, hydrogenation of, 240-246 Product removal, 68 Propagation steps defined, 10I styrene polymerization, 111-112 Propane dehydrogenation of, 45 oxydehydrogenation of, 46 2-Propanol acetone from, 47-48 decomposition of, 59-60, 63 dehydrogenation of, 231 Propene hydroformylation of, 23, 68, 69 (figure) production of, 63 Propionic acid, esterification of, 129-130 n-Propylbromine, 31-33 Propylene, formation of, 45, 46 Protons, in zeolite catalysts, 169-170 Pseudo-equilibrium. See Quasi-equilibrium Pseudo mass action rate expressions. See Power law rate expressions PT (dimethyl-p-toludine), 34-35 Q Quantum chemical methods, 254-256 Index 365 Quasi-equilibrium vs. rate determining steps, 148-149 two-step assumption and, 157-158 R Radial dispersion, 282 Radial temperature gradient, effect of, 327 Radioactive decay, rate expressions for, 26-27 Rate constants, 53-62. See also Rates of reactions Arrhenius' Law and, 54-56 in CSTRs, 75-76 defined,21 equilibrium constant and, 24 from experimentation, 174-175 external transport effects and, 188 ionic strength and, 61 in nonideal systems, 60-61 transition state theory, 56-62 unimolecular reactions, 62 Rate-determining steps kinetic sequences and, 172-173 microkinetic analysis and, 248 notation for, xvi (table), 4 (table) surface reactions as, 148-149 Rate expressions. See also Rates of reactions ammonia synthesis, 158-160 bimolecular surface reactions, 152 carbon monoxide oxidation, 163 desorption, 156 dissociative adsorption, 145-147 enzyme catalysis, 117-119, 127-128 methylcyclohexane dehydrogenation, 161 two-step assumption, general form, 157 unimolecular reactions, 24, 25 (table), 26, 147 Rate of turnover. See Turnover frequency Rates of reactions. See also Rate constants; Rate expressions catalyzed vs. direct reactions, 101, 103, 104 (figure), 116 (figure), 134 (figure) defined, 16 diffusional limitations and, 207-208 external transport effects and, 188, 189 first order reactions, 24, 26 (table), 26-29 for heterogeneous reactions, 17-18 measurement of, by ideal reactors, 82-92, 83 (table) in polymerization, 111-112 pore diffusion, influence on, 206-207 rules for rate functions, 19-24 second order reactions, 29-35 steady-state approximation and, 110-111 trimolecular reactions, 35-37 vs. turnoverfrequency, 18, 19 volumic rates, 16-17 Reaction intermediates. See Intermediates Reaction networks defined,4 intermediates in,S mixed parallel-series, 43-45 parallel, 42-43 series, 37-40, 310-311 Reaction rates. See Rates of reactions Reaction sequences. See Kinetic sequences Reactive distillation, 68 Reactive intermediates. See also Steady-state approximation in coupled catalytic cycles, 242-245 defined,S in single reactions, 100-105, 102 (table) Reactors See also specific types, e.g, Batch reactors flow vs. nonflow, variables, for, 74 (table) ideal, 64-65, 83 (table) laboratory scale, for rate measurement, 82-95 limiting conditions, 65 (table) Recycle reactors, 88-92 Redox reactions, 183,333 Regeneration, in fluidized-bed reactors, 331-332 Regression analysis linear least squares method, 175,343-345 nonlinear least squares method, 121-122, 175,347-348 zero intercepts, correlation probability, 345-347, 346 (table) Rekoske,J.E., 139, 140,240,246,253 Relaxation time defined, 112-113 methods for determining, 124-126 Renouprez, A. J., 139 Residence time distribution function axial-dispersion model, 274 Berty reactors, 268-269 defined,262,264 impulse input, 263 (figure), 266 perfectly mixed reactors, 264-265 reactor conversion prediction and, 269-272 step input, 266, 267 (figure) Residence times in CSTRs, 73-74 limiting conditions, 65 PFR space-time analogy, 77-78 Reversible elementary steps, notation for, xvi (table), 4 (table) Reversible stoichiometric reactions defined, 13-14 notation for, xvi (table), 4 (table) rate expressions for, 24 Reynolds number axial-dispersion model and, 274-276 mass transfer coefficients and, 188-189 Rhodium catalysts in hexene isomerization, 202-203 in hydroformylation, 36-37 in nitrous oxide/carbon monoxide reaction, 181 in olefin hydrogenation, 240-246 propene and, 23 (table) supports for, 137 (figure) Rideal-Eley steps defined, 153-154 vs. Langmuir-Hinshelwood steps, 176-177 Rode, E., 23 Rosowski, E, 250, 251 Rowland, M., 265 R,R-l ,2-bis[(phenyl-o-anisol)phosphino]ethane (DIPAMP), 241-245 366 Index Rudd D. E, 240, 246 Rules I-V for reaction rate expressions, 19-24 Runaway reactor condition, 310 Ruthenium catalysts constant volume system, 49-50 microkinetic analysis of, 250-252 two-step assumption and, 159-160 s Schmidt number, 188-189 Second order reactions rate expressions for, 29-31 residence time distributions, 271-272 statistical analysis of, 31-35, 32 (figure), 33 (figure) Seinfeld, J. H., 25 (table) Selective oxidation. See Partial oxidation Selectivity. See also Stereoselectivity defined,40 in ethylene oxidation, 94 mixed parallel-series networks, 43-44 parallel networks, 43 Semibatch reactors exothermic heat of reaction removal and, 295-297 ideal, 66 (figure), 67-70 Sensible heat effects, 289-290 Sensitivity of reactors, 309-311 Separable reaction rates, 21 Sequences. See Kinetic sequences Series reaction networks reaction rates for, 37-40 reaction sensitivity and, 310-311 Severe diffusion limitations, 202, 207-208 Sharma, S. B., 253 Sherwood number, 188-189 Shuler, M. L., 127 Shulman, R. A, 160 Silicon wafers, in chemical vapor deposition, 224-227 Simple reactions, 4. See also Elementary steps; Single reactions Simulation, of ethylene hydrogenation, 256-257 Sincovec, R. E, 327 Sinfelt, J. H., ISO, lSI, 160 Single crystal surfaces microkinetic analysis and, 246 structure sensitive reactions and, lSI for transition metals, 134-136 Single tile diffusion, 191 Single reactions rate function rules I-V, 19-24 reactive intermediates in, 100-105, 102 (table) Site balance general form, 144 two competing species, 145 Slab catalyst pellets. See Flat plate catalyst pellets; Infinite slab catalyst pellets Slurry reactors, 329 (figure), 330 Smith, J. M., 234 Soaps, 37 Sodalite cage structures, 167 Sodium lauryl sulfate, production of, 37 Software for numerical solutions, 327 Solid acid catalysts, 154-155, 169 Space time as average exit time, 266 defined,73 in PFRs, 77 Space-time yield CSTR vs. PFR, 79-80 defined,72 Space-velocity, 73 Specific rate of reaction, 17 Spherical catalyst pellets characteristic length parameter, 201 (table) for cumene, cracking of, 206-207 diffusion/reaction in, 197-201 nonisothermal effectiveness factors, 216-217 reactor balance expressions for, 321-322 schematic of, 196 (figure) Stability of reactors, in steady-state, 305-309 Standard error, 345, 347 Statistical analysis. See also Experimentation L-dopa synthesis, 119-121 linear regression, 175, 343-345 nonlinear regression, 121-122, 175,347-348 second order reactions, 31-35, 32 (figure), 33(figure) Steady-state approximation, 105-113 coupled catalytic cycles and, 243 derivation of, 105-108 hydrogen-deuterium exchange and, 109 ozone decomposition rate and, 110-111 polymerization rate and, 111-112 relaxation time, 112-113, 124-126 surface reactions and, 148 two-step assumption and, 162 Steady-state operation, stability of, 305-309 Stefan-Maxwell equations cylindrical pores, ideal, 193 derivation of, 351 molecular diffusion, 185 multicomponent gas mixture, 211 Steps, elementary. See Elementary steps Stereoselectivity L-dopa synthesis, 123-124 olefin hydrogenation, 240-246 Stirred-flow reactors. See also Continuous flow stirred tank reactors (CSTRs) laboratory scale, 88-92 material balance for, 83 (table) Stirring speed, effect of, 231 Stoichiometric coefficients. 8 Stoichiometric numbers, 157 Stoichiometric reactions defined, 4, 8 vs. elementary steps, 8, 53-54 generalized equation, 9 notation for, xvi (table), 4 (table) reactive intermediates in, 100-102 Stolze, P., 246, 247 Structure sensitive/insensitive reactions, 149-151 Student Hest described, 345, 347 in reaction order example, 32-33 Stutzman, C F., 180 Styrene, polymerization of, 111-112 Substrates binding of, 114-116 as inhibitor, 128, 130 Sucrose, 116-1 i7 Sugar, I 16-1 i7 Sulfur dioxide, oxidation of, 234 Supported metal catalysts, 18-19, 136-140 Surface area of catalysts, 141-143 Surface concentrations Damkohler number and, 219 external transport effects and, 188 notation for, 144 Surface reactions, 147-155 bimolecular reactions, 151-152 kinetics of IE dimerization, 154-155 pre-exponential factors, 152-153 as rate-determining step, 172-173 Rideal-Eley steps, 153-154 structure sensitivity/insensitivity, 149-151 unimolecular reactions, 147-149 Surfactants, 37 Switzer, M. A., 181 Symbols list, xii-xvi T Tarmy, B. L., 328, 329, 332 Taylor, K. C, 13 Teller, Edward, 141, 142 Temperature gradients across mm, 190 criteria for importance of, 228-229 internal vs, external transport effects, 223 thermal conductivity effect, 218-219 Temperature profiles, in PFRs, 287 (figure) Termination steps defined, iO I styrene polymerization, 11 1-112 Tertiary structure, of enzymes, 114 Thermal conductivity, effect of, 218-219 Thermodynamic constraints importance of, 1-2 kinetic sequence postulation and, 177 Thiele modulus cylindrical catalyst pellets, 204 dellned. 194-195 effectiveness factors and, 200 (llgure), 20 I-202 fiat plate catalyst peiJets, 210 general fonn, 207 silicon wafer chemical vapor deposition, 226 spherical catalyst peiJets, ]98, 199 (figure) Thodor, G.. 180 Titanium tetrachloride, oxidation of. 98 Titration, 18-19 Toluene benzene from, 47 formation of, 160-16 I Topsoe, H., 150 Topsoe, N., 150 Tortuosity, 195-196 Total order of reaction, 21-22 Tozer, T. N., 265 Tracer experiments impulse input, 262-265, 263 (figure) step input, 266, 267 (figure) Transient material balance expressions axial-dispersion model, 272-274 radial-dispersion model, 282 Transition diffusivity, 191 Transition state intermediates defined, 5, 6, 7 (figure) ethyicne hydrogenation, 254 (figure) Transition state theory, 56-62 vs. Arrhenius form, 60 genera] equation of, 58-59 potential energy and, 56-57 pre-exponential factors, experiment vs., 152-153 Transmission eleetron microscopy, 138, 139 Transport limitations in solid-catalyzed reactions, 184-230 adiabatic temperature rise, 2i7-218 chemical vapor deposition, 223-228 eombined internal/external transport, 220-223 commercial rates, adjustment of, ]84- I85 cylindrical catalyst peiJets, 203-206 cylindrical pores, ideal, 192-I95 Damkohler number, 219-220 effectiveness factors, internal transport, 199-202 effectiveness factors, overall, 221-223 fixed-bed reactors and, 3] 7 flat plate catalyst pellets, 208-212, 213-215, 219-223 heat transfer effects, external, 189-190 ideal peiJet geometries, 196-197, 20i (table) kinetic analysis criteria, 228-230 mass transfer effects, extemal, 187-189 moiccu1ar diffusion, calculation of 185-] 88 nonisothermal reactions, 212-2 i7 pellet size, calculation of, 202-203 pore diffusion, iufluence on rate, 206-207 pore size, and intemal diffusion, ]90-192 rate data analysis, 228-232 severe diffusional resistance, 202, 207-208 spherical catalyst peiJets, 197-201 thermal conductivitiy effect, 2 I8-2 I9 zig-zag pores, 195-I96 Trevino, A. A., 240, 246, 247, 248, 249 Trickle-bed reactors, 329 (figure), 330 Trimethylamine, 31-33 Trimolecular reactions, 25 (table), 35-37 Triphenylmethylchloride, 69-70 Tubular flow reactors. See also Fixed-bed reactors: Plug flow reactors (PFRs) laboratory seale, 87-88 material balance for, 83 (table) Peelet number for, 275-276, 282 368 Index Tubular flow reactors-eO/u.
sensitivity of, 309-311 turbulent vs. laminar flow, 260--262 Turbulent flow axial-dispersion coefficient for, 273 axial-dispersion effects and, 276 vs. laminar flow, in tubular reactors, 260-262 Turnover frequency defined, 18 normalization of rates and, 139, 140 vs. reaction rate, 19 vs. relaxation time, 113 structure sensitivity/insensitivity and, 149-151 in surface reactions, 149 Two-dimensional fixed-bed reactor model, 325-327 Two-step assumption for overall reactions, 157-162 Two-way stoichiometric reactions. See Reversible stoichiometric reactions U Uncompetitive inhibitors, 127-128 l-Undecene, hydroformylation of, 37 Unimolecular reactions pre-exponential factors for, 152 rate constants for, 62 rate expressions for, 24, 25 (table), 26, 147 surface-catalyzed, 147-149 Uranium, radioactive decay of, 26-27 v Vaccines, bioreactors for, 300 van de Runstraat, A., 258 van der Waals forces, 140 van Grondelle, J., 258 van Krevelen-Mars mechanism, 183,332-333 van Santen, R. A., 254, 258 Vanadium catalysts, 183,333-334 Vessel size, effects of, 295-297 Villadsen, J., 301 Vinyl chloride, production of, 47 Voids, in catalyst pellets, 192 Volume of reactor, and heat transfer area, 295-297 Volumic rate. 16-17 VPO (vanadium phosphorus oxide) catalysts, 333 W Wallis, D. A., 280, 281, 322, 323 Wastewater, dyes in, 48-49 Watanabe, K., 224, 225, 226, 227 Water gas shift reaction (WGS), 182 Water, in enzyme catalysis, 114 Weekman, V. W., Jr., 83 Wei, J., 164, 165, 166 Weisz, P. B., 2,170,184,191,216,228 Weisz-Prater criterion, 228 White, J. M., 109 Wilhelm, R. H., 274, 276 Wood stoves, exhausting of, 331 Wu, L-W., 49 x X-ray diffraction, 138, 139 Xylene, production of, 47 y Yates, D. J. C., 150, 151 Yeh, C. Y, 235 Yield mixed parallel-series networks, 44-45 parallel networks, 43 series reaction networks, 41 Young, L. C., 323 z Zeolite catalysts axial-dispersion coefficients, 277 for catalytic cracking, 331-332 described, 164 (figure), 165 (figure), 166-170 in ethylbenzene alkylation, 235-236 Zero intercept, in linear regression, 32, 345-347, 346 (table) Zewail, A, 5, 57, 58 Zig-zag pore networks, 195-196 Zirconium tetrabenzyl, 111-112
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Snap Language
Getting Smarter through Language
esl b-level reading Using Antonyms and Synonyms as Context Clues: Intermediate Reading Course: Snap Language™
Intermediate Reading Course. Section 1: The Basics
Using Antonyms and Synonyms as Context Clues
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Antonyms as Context Clues
Antonyms are words that have opposite meanings (e.g., rich and poor, or tall and short). If you run into a new word while reading, sometimes you can tell what it means because the writer uses the word in contrast with another word or with the idea in the text.
Check out the examples below.
Anete Lusina | Pexels
Antonym or contrast clues
Example 1
It was hard to believe that such an erudite man had grown up in such an uneducated, uncultured home.
”Erudite” is contrasted with “uneducated” and “uncultured.”
Example 2
Pedro and Jenny are an odd couple. Pedro is very laconic while Jenny will talk for an hour just to tell you what she had for breakfast.
Pedro’s “laconic” nature is contrasted with the idea that Jenny is very talkative, perhaps talks too much. Being laconic means Pedro is a man of few words.
Example 3
The first was very original and kept the audience interested and entertained. On the other hand, the second was simply insipid.
From the contrast between the presenters, you can tell that “insipid” is the opposite of “original and interesting.”
Synonyms as Context Clues
Alena Darmel | Pexels
Synonyms are words that have the same or similar meanings (e.g., rich and wealthy, or big and large).
Sometimes you can figure out the meaning of an unknown word because the writer includes its synonymous meaning or concept. In essence, the writer repeats the idea using a synonymous word or expression.
Synonyms or repeat context clues
Example 1
Samira always approached her academic studies with verve. She showed equal enthusiasm for photography as a hobby.
“Verve” and “enthusiasm” have similar meanings.
Example 2
People were tired of his belligerent behavior at work. Being quick to argue and fight with everyone made him difficult to work with.
”Belligerence” refers to being eager or quick to argue and fight.
Example 3
The paucity of natural resources in the country accounts for its serious economic problems. Such a lack of resources is made worse by a history of political troubles.
“Paucity” of resources in the first sentence is restated as “lack” of resources in the second.
Video Activity
Watch a video to learn about using synonyms and antonyms in the
paragraph or passage to figure out the meaning of unknown words during reading. Take good notes.
Why Do Writers Do That?
Sometimes writers just try to find different ways to describe or explain something and end up providing synonyms and antonyms to avoid repetition. Other times, writers are actually aware that their readers may not know a word or expression, so they deliberately include a synonym or antonym to clarify its meaning.
As the readers, you should be aware that texts have such context clues. This way, you do not look up a word unnecessarily, which interrupts your reading and disrupts comprehension.
Important Caveats
Do not rely on context clues alone in the following situations:
When when it is important to know the exact meaning of a word.
When it is important to understand the meaning of a keyword exactly so that the text makes sense.
Note. A “caveat” is a detail that you must consider carefully when interpreting some information.
Up Next: Using Definitions and Examples as Context Clues
Go to the next lesson to learn about definition and example context clues.
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https://askfilo.com/user-question-answers-physics/a-uniform-rod-of-mass-and-length-hangs-from-a-smooth-hinge-35393439303335
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Question asked by Filo student
A uniform rod of mass M and length 2a hangs from a smooth hinge at one end. Find the length of the equivalent simple pendulum. It is struck sharply, with impulse X , at a point a distance b below the hinge. Use the angular momentum equation to find the initial value of the angular velocity. Find also the initial momentum. Determine the point at which the rod may be struck without producing any impulsive reaction at the hinge. Show that, if the rod is struck elsewhere, the direction of the impulsive reaction depends on whether the point of impact is above or below this point.
Views: 5,131 students
Updated on: Nov 2, 2023
Text SolutionText solutionverified iconVerified
Step 1: Find the length of the equivalent simple pendulum
The equivalent simple pendulum has the same period of oscillation as the given rod. The moment of inertia for a uniform rod of mass M and length 2a rotating about one end is I=31M(2a)2. For a simple pendulum with mass m and length L, the moment of inertia (around the pivot point) is I=mL2. The period of a simple pendulum is given by T=2πmglI. Equating the expression for the period of the given rod and simple pendulum gives 31M(2a)2=mL2. Solving for L gives L=34a2=32a. Therefore, the equivalent simple pendulum has a length of 32a.
Step 2: Find the initial angular velocity after the impulse
Given that an impulse X is applied at a distance b below the hinge, use the angular momentum equation, ΔL=IΔω, where ΔL is the change in angular momentum and Δω is the change in angular velocity. The change in angular momentum due to the impulse is given by ΔL=Xb. Since the rod is initially at rest, the change in angular velocity is simply the initial angular velocity, ω0. Therefore, Xb=31M(2a)2ω0. Solving for ω0 gives ω0=M(2a)23Xb. Therefore, the initial angular velocity is M(2a)23Xb.
Step 3: Find the initial momentum
The initial momentum can be found using the equation p=Mv, where v is the initial linear velocity of the rod's center of mass. Since we know the initial angular velocity, we can find the linear velocity by noting that v=rω0, where r is the distance from the pivot point to the center of mass. For a rod of length 2a, the center of mass is at a distance a from the pivot point. Thus, the initial momentum is p=Ma⋅M(2a)23Xb=2a3Xb. Therefore, the initial momentum is 2a3Xb.
Step 4: Determine the striking point for no impulsive reaction at the hinge
For no impulsive reaction at the hinge, the torque due to the impulse must be equal to the torque due to the reaction force. The torque due to the impulse is given by τ=Xb. The torque due to the reaction force is given by τ=Fh, where F is the reaction force and h is the perpendicular distance from the pivot point to the line of action of the impulse. To have no impulsive reaction at the hinge, set those torques equal: Xb=Fh. Solving for h gives h=XbF. Therefore, the rod can be struck at a point where the perpendicular distance from the pivot to the line of action of the impulse is XbF.
Step 5: Show that impulsive reaction depends on point of impact
If the point of impact is above the point determined in Step 4, the torque due to the impulse will be smaller, as b would be smaller in that case. Conversely, if the point of impact is below that point, the torque due to the impulse will be larger. As a result, the direction of the impulsive reaction depends on whether the point of impact is above or below the point found in Step 4. The direction will be opposite to that of the torque caused by the impulse.
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| Question Text | A uniform rod of mass M and length 2a hangs from a smooth hinge at one end. Find the length of the equivalent simple pendulum. It is struck sharply, with impulse X , at a point a distance b below the hinge. Use the angular momentum equation to find the initial value of the angular velocity. Find also the initial momentum. Determine the point at which the rod may be struck without producing any impulsive reaction at the hinge. Show that, if the rod is struck elsewhere, the direction of the impulsive reaction depends on whether the point of impact is above or below this point. |
| Updated On | Nov 2, 2023 |
| Topic | All topics |
| Subject | Physics |
| Class | High School |
| Answer Type | Text solution:1 |
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Adding 2-digit numbers | Mental math
KidsMathLogic
1420 subscribers
24 likes
Description
2198 views
Posted: 22 Feb 2024
In this video we talk about how to to add 2-digit numbers. This is mental math strategy.
Links to Mental math videos :
- Subtracting 2-digit numbers | Mental math
- Adding a mixed number to a whole number | Mental math
- Subtracting a mixed number from a whole number | Mental math
mentalmath #mentalmaths #addition #adding #twodigit #stepbystep #kidsmathlogic
Disclaimer:
The information provided in this video is for general informational purposes only. All information is provided in good faith, however we make no representation or warranty of any kind, expressed or implied, regarding the accuracy, adequacy, validity, reliability or completeness of this information.
Transcript:
Let’s practice mental math. How we can add
56 and 18 in an easy way? Numbers like 10, 20, 30 or 40 are friendly numbers meaning
they are numbers ending with 0s and they are the easier to add. Back to our example. Let’s
see what friendly number is closest to 56. It is 60. To turn 56 into 60 we have to add 4 to it
but if we are adding something to one number, we also have to subtract the same from the
other number so we also subtract 4 from 18. That way we have 60 + 14. It looks much
easier now, right? I am sure you already see the answer. It’s 74.
So, 56 + 18 is 74
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The number of non-empty equival ence relations on the s... - Solution & Explanation | iExplain | iExplain
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The number of non-empty equival ence relations on ...
The number of non-empty equival ence relations on the set {1,2,3} is : (1)6 (2)7 (3)5 (4)4
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Published July 8, 2025
Mathematics
Set Theory
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Equivalence Relations
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Detailed Explanation
An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive. Equivalence relations correspond exactly to partitions of the set. That means each equivalence relation divides the set into disjoint subsets called equivalence classes.
For the set {1, 2, 3}, the number of equivalence relations equals the number of ways to partition the set into non-empty subsets.
The number of partitions of a set with n elements is given by the Bell number B_n. For n=3, the Bell number B_3 = 5.
These partitions are:
{ {1}, {2}, {3} }
{ {1, 2}, {3} }
{ {1, 3}, {2} }
{ {2, 3}, {1} }
{ {1, 2, 3} }
Each partition corresponds to one equivalence relation. Since the question asks for non-empty equivalence relations, and equivalence relations are never empty (they always contain at least the identity pairs), the answer is 5.
Simple Explanation (ELI5)
Imagine you have three friends: 1, 2, and 3. You want to group them so that each group has friends who are 'equivalent' to each other. An equivalence relation is like making groups where everyone in a group is connected in a special way. The question is: how many different ways can you make these groups without leaving the whole set empty? To solve this, we need to find all possible ways to split the set {1, 2, 3} into groups where each group is connected properly.
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Step-by-Step Solution
Step 1: Understand that equivalence relations correspond to partitions of the set.
Step 2: Find the number of partitions of the set {1, 2, 3}.
Step 3: The number of partitions of a set with 3 elements is the Bell number B_3.
Step 4: Bell numbers for small n are:
B_1 = 1
B_2 = 2
B_3 = 5
Step 5: List the partitions of {1, 2, 3}:
{ {1}, {2}, {3} }
{ {1, 2}, {3} }
{ {1, 3}, {2} }
{ {2, 3}, {1} }
{ {1, 2, 3} }
Step 6: Each partition corresponds to one equivalence relation.
Step 7: Since equivalence relations are never empty, all 5 are valid.
Final answer: 5
So, the correct option is (3) 5.
Examples
Example 1
Grouping students in a class into teams where each team is an equivalence class.
Example 2
Classifying animals into species where each species is an equivalence class.
Example 3
Dividing tasks into categories where tasks in the same category are equivalent.
Example 4
Organizing books by genre where each genre forms an equivalence class.
Example 5
Partitioning a network into connected components where each component is an equivalence class.
Visual Representation
Failed to render diagram. Please check the syntax.
View diagram code
graph TD
A[Start with set {1,2,3}] --> B[Find all partitions of the set]
B --> C[Count number of partitions (Bell number B3)]
C --> D[Identify each partition as an equivalence relation]
D --> E[Count total equivalence relations]
E --> F[Answer is 5]
References
Discrete Mathematics and Its Applications by Kenneth H. Rosen
Introduction to Set Theory by Karel Hrbacek and Thomas Jech
Bell Numbers - Wikipedia
Equivalence Relations - Khan Academy
Partition of a Set - Brilliant.org
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The sum of all real values of ( x ) for which [ \frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12} ] is equal to _____.
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https://www.youtube.com/watch?v=dcmrJMguwBY
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How To Calculate Entropy Changes: Liquids, Solids, and Phase Changes
LearnChemE
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Posted: 20 Jul 2018
Organized by textbook: Derives equations to calculate entropy changes for liquids and solids and for phase changes. Made by faculty at the University of Colorado Boulder, Department of Chemical & Biological Engineering. Check out our various resources by topic: Thermodynamics playlists:
Our website for interactive thermodynamics simulations:
Our Physical Chemistry playlist:
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Transcript:
INSTRUCTOR: Here we're gonna look at how to calculate entropy
changes if we have a liquid or solid, temperature changes, or if we go through a phase change from say a liquid to a gas. And the way we're gonna do this is to start with our
definition, entropy change, which is the integral
dQ reversible over T, where T is absolute temperature. The idea is if we're at state one, and we go to state two, if we pick a reversible pathway and then the heat transfer at any point divided by the temperature
of that point integrated gives us the entropy change. And so to get the heat transfer, we're gonna use the first law, so let me write down
the differential form, first law for a closed system. So Q reversible and work reversible. So work reversible would mean pressure times the differential. It turns out changing pressure
has very little effect on a liquid or a solid, we
don't change the anthopy, we don't change the entropy, we don't change the volume, et cetera. So to our first approximation
for a liquid solid, we can write this as
minus the differential PV. Now, what I'm going to
do is take this term, bring it to the other
side of the equation, and we'll see that we have
on the left dU plus d[PV] and u plus PV is H, so this
is dH is dQ reversible. dH for a liquid or solid is
heat capacity, Cp times dT. So that means if we now want to calculate the entropy change, and again
let's rewrite our definition, and dQ reversible is CpdT over T. Heat capacity for liquid or solid does not change much with temperature, some reasonable temperature range, so if that's a constant, then the entropy change
starting temperature, our final temperature. Here's entropy change
as we change temperature for a liquid or a gas. Well, the other case
we want to look at now is entropy change for a phase change. So we want to look at entropy
change for a phase change, an important aspect of a phase change is it takes place at constant temperature and at constant pressure. So our equation for first law is identical for a constant pressure
reversible process, and so we have dH is dQ reversible. The difference is dH is the anthopy change for a phase change. So this is a number, right,
we're at one temperature. And so at a given temperature,
this is gonna be a constant, so when I go back to my
definition of dQ reversible over T for a phase change dH over T integrated, T is a constant so with delta H over T and so for a phase change, entropy change, and again this is per mole, and could we change per mole divided by absolute temperature. Again let's emphasize this is per mole.
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https://en.wikipedia.org/wiki/Tangent_half-angle_substitution
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Tangent half-angle substitution
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From Wikipedia, the free encyclopedia
Change of variable for integrals involving trigonometric functions
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In integral calculus, the tangent half-angle substitution is a change of variables used for evaluating integrals, which converts a rational function of trigonometric functions of into an ordinary rational function of by setting . This is the one-dimensional stereographic projection of the unit circle parametrized by angle measure onto the real line. The general transformation formula is:
The tangent of half an angle is important in spherical trigonometry and was sometimes known in the 17th century as the half tangent or semi-tangent. Leonhard Euler used it to evaluate the integral in his 1768 integral calculus textbook, and Adrien-Marie Legendre described the general method in 1817.
The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name. It is known in Russia as the universal trigonometric substitution, and also known by variant names such as half-tangent substitution or half-angle substitution. It is sometimes misattributed as the Weierstrass substitution. Michael Spivak called it the "world's sneakiest substitution".
The substitution
[edit]
Introducing a new variable sines and cosines can be expressed as rational functions of and can be expressed as the product of and a rational function of as follows:
Similar expressions can be written for tan x, cot x, sec x, and csc x.
Derivation
[edit]
Using the double-angle formulas and and introducing denominators equal to one by the Pythagorean identity results in
Finally, since , differentiation rules imply
and thus
Examples
[edit]
Antiderivative of cosecant
[edit]
We can confirm the above result using a standard method of evaluating the cosecant integral by multiplying the numerator and denominator by and performing the substitution .
These two answers are the same because
The secant integral may be evaluated in a similar manner.
A definite integral
[edit]
We wish to evaluate the integral:
A naïve approach splits the interval and applies the substitution . However, this substitution has a singularity at , which corresponds to a vertical asymptote. Therefore, the integral must be split at that point and handled carefully:
Note: The substitution maps to and to . The point corresponds to a vertical asymptote in , so the integral is evaluated as a limit around this point.
Alternatively, we can compute the indefinite integral first:
Using symmetry:
Thus, the value of the definite integral is:
Third example: both sine and cosine
[edit]
if
Geometry
[edit]
As x varies, the point (cos x, sin x) winds repeatedly around the unit circle centered at (0, 0). The point
goes only once around the circle as t goes from −∞ to +∞, and never reaches the point (−1, 0), which is approached as a limit as t approaches ±∞. As t goes from −∞ to −1, the point determined by t goes through the part of the circle in the third quadrant, from (−1, 0) to (0, −1). As t goes from −1 to 0, the point follows the part of the circle in the fourth quadrant from (0, −1) to (1, 0). As t goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1, 0) to (0, 1). Finally, as t goes from 1 to +∞, the point follows the part of the circle in the second quadrant from (0, 1) to (−1, 0).
Here is another geometric point of view. Draw the unit circle, and let P be the point (−1, 0). A line through P (except the vertical line) is determined by its slope. Furthermore, each of the lines (except the vertical line) intersects the unit circle in exactly two points, one of which is P. This determines a function from points on the unit circle to slopes. The trigonometric functions determine a function from angles to points on the unit circle, and by combining these two functions we have a function from angles to slopes.
Hyperbolic functions
[edit]
As with other properties shared between the trigonometric functions and the hyperbolic functions, it is possible to use hyperbolic identities to construct a similar form of the substitution, :
Similar expressions can be written for tanh x, coth x, sech x, and csch x. Geometrically, this change of variables is a one-dimensional stereographic projection of the hyperbolic line onto the real interval, analogous to the Poincaré disk model of the hyperbolic plane.
Alternatives
[edit]
There are other approaches to integrating trigonometric functions. For example, it can be helpful to rewrite trigonometric functions in terms of eix and e−ix using Euler's formula.
See also
[edit]
Mathematics portal
Rational curve
Stereographic projection
Tangent half-angle formula
Trigonometric substitution
Euler substitution
Further reading
[edit]
Courant, Richard (1937) . "1.4.6. Integration of Some Other Classes of Functions §1–3". Differential and Integral Calculus. Vol. 1. Blackie & Son. pp. 234–237.
Edwards, Joseph (1921). "§1.6.193". A Treatise on the Integral Calculus. Vol. 1. Macmillan. pp. 187–188.
Hardy, Godfrey Harold (1905). "VI. Transcendental functions". The integration of functions of a single variable. Cambridge. pp. 42–51. Second edition 1916, pp. 52–62
Hermite, Charles (1873). "Intégration des fonctions transcendentes" [Integration of transcendental functions]. Cours d'analyse de l'école polytechnique (in French). Vol. 1. Gauthier-Villars. pp. 320–380.
Stewart, Seán M. (2017). "14. Tangent Half-Angle Substitution". How to Integrate It. Cambridge. pp. 178–187. doi:10.1017/9781108291507.015. ISBN 978-1-108-41881-2.
Notes and references
[edit]
^ Other trigonometric functions can be written in terms of sine and cosine.
^ Gunter, Edmund (1673) . The Works of Edmund Gunter. Francis Eglesfield. p. 73
^ Euler, Leonhard (1768). "§1.1.5.261 Problema 29" (PDF). Institutiones calculi integralis [Foundations of Integral Calculus] (in Latin). Vol. I. Impensis Academiae Imperialis Scientiarum. pp. 148–150. E342, Translation by Ian Bruce.
Also see Lobatto, Rehuel (1832). "19. Note sur l'intégration de la fonction ∂z / (a + b cos z)". Crelle's Journal (in French). 9: 259–260.
4. ^ Legendre, Adrien-Marie (1817). Exercices de calcul intégral [Exercises in integral calculus] (in French). Vol. 2. Courcier. p. 245–246.
5. ^ For example, in chronological order,
Hermite, Charles (1873). Cours d'analyse de l'école polytechnique (in French). Vol. 1. Gauthier-Villars. p. 320.
Johnson, William Woolsey (1883). An Elementary Treatise on the Integral Calculus. MacMillan. p. 52.
Picard, Émile (1901) . Traité d'analyse (in French). Vol. 1 (2e ed.). Gauthier-Villars. p. 77.
Goursat, Édouard (1904) . A Course in Mathematical Analysis. Vol. 1. Translated by Hedrick, Earle Raymond. Ginn. pp. 236ff.
Wilson, Edwin Bidwell (1911). Advanced Calculus. Ginn. p. 21.
Edwards, Joseph (1921). A Treatise on the Integral Calculus. Vol. 1. MacMillan. pp. 187–188.
Courant, Richard (1937) . Differential and Integral Calculus. Vol. 1. Translated by McShane, E. J. (2nd ed.). Blackie & Son. pp. 234–238.
Peterson, Thurman S. (1950). Elements of Calculus. Harper & Brothers. pp. 201–202.
Apostol, Tom M. (1967) . Calculus. Vol. 1 (2nd ed.). Xerox. pp. 264–265.
Larson, Ron; Hostetler, Robert P.; Edwards, Bruce H. (1998) . Calculus of a Single Variable (6th ed.). Houghton Mifflin. p. 520. ISBN 9780395885789.
Rogawski, Jon (2011) . Calculus: Early Transcendentals (2nd ed.). Macmillan. p. 435. ISBN 9781429231848.
^ Piskunov, Nikolai (1969). Differential and Integral Calculus. Mir. p. 379. Zaitsev, V. V.; Ryzhkov, V. V.; Skanavi, M. I. (1978). Elementary Mathematics: A Review Course. Ėlementarnai͡a matematika.English. Mir. p. 388.
^
In 1966 William Eberlein attributed this substitution to Karl Weierstrass (1815–1897):
Eberlein, William Frederick (1966). "The Circular Function(s)". Mathematics Magazine. 39 (4): 197–201. doi:10.1080/0025570X.1966.11975715. JSTOR 2688079. (Equations (3) [], (4) [], (5) [] are, of course, the familiar half-angle substitutions introduced by Weierstrass to integrate rational functions of sine, cosine.)
Two decades later, James Stewart mentioned Weierstrass when discussing the substitution in his popular calculus textbook, first published in 1987:
Stewart, James (1987). "§7.5 Rationalizing substitutions". Calculus. Brooks/Cole. p. 431. ISBN 9780534066901. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution t = tan(x/2) will convert any rational function of sin x and cos x into an ordinary rational function.
Later authors, citing Stewart, have sometimes referred to this as the Weierstrass substitution, for instance:
Jeffrey, David J.; Rich, Albert D. (1994). "The evaluation of trigonometric integrals avoiding spurious discontinuities". Transactions on Mathematical Software. 20 (1): 124–135. doi:10.1145/174603.174409. S2CID 13891212.
Merlet, Jean-Pierre (2004). "A Note on the History of Trigonometric Functions" (PDF). In Ceccarelli, Marco (ed.). International Symposium on History of Machines and Mechanisms. Kluwer. pp. 195–200. doi:10.1007/1-4020-2204-2_16. ISBN 978-1-4020-2203-6.
Weisstein, Eric W. (2011). "Weierstrass Substitution". MathWorld. Retrieved 2020-04-01.
Neither Eberlein nor Stewart provided any evidence for the attribution to Weierstrass. A related substitution appears in Weierstrass’s Mathematical Works, from an 1875 lecture wherein Weierstrass credits Carl Gauss (1818) with the idea of solving an integral of the form by the substitution
Weierstrass, Karl (1915) . "8. Bestimmung des Integrals ...". Mathematische Werke von Karl Weierstrass (in German). Vol. 6. Mayer & Müller. pp. 89–99.
8. ^ Spivak, Michael (1967). "Ch. 9, problems 9–10". Calculus. Benjamin. pp. 325–326.
External links
[edit]
Weierstrass substitution formulas at PlanetMath
| v t e | |
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| Types of integrals | Riemann integral Lebesgue integral Burkill integral Bochner integral Daniell integral Darboux integral Henstock–Kurzweil integral Haar integral Hellinger integral Khinchin integral Kolmogorov integral Lebesgue–Stieltjes integral Pettis integral Pfeffer integral Riemann–Stieltjes integral Regulated integral |
| Integration techniques | Substitution + Trigonometric + Euler + Weierstrass By parts Partial fractions Euler's formula Inverse functions Changing order Reduction formulas Parametric derivatives Differentiation under the integral sign Laplace transform Contour integration Laplace's method Numerical integration + Simpson's rule + Trapezoidal rule Risch algorithm |
| Improper integrals | Gaussian integral Dirichlet integral Fermi–Dirac integral + complete + incomplete Bose–Einstein integral Frullani integral Common integrals in quantum field theory |
| Stochastic integrals | Itô integral Russo–Vallois integral Stratonovich integral Skorokhod integral |
| Miscellaneous | Basel problem Euler–Maclaurin formula Gabriel's horn Integration Bee Proof that 22/7 exceeds π Volumes + Washers + Shells |
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http://www.math.rwth-aachen.de/~Gabriele.Nebe/papers/GAUSS.pdf
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A parallel algorithm for Gaussian elimination over finite fields Stephen Linton, Gabriele Nebe, Alice C. Niemeyer, Richard Parker, Jon Thackray December 16, 2024 School of Computer Science, University of St. Andrews, St. Andrews, Fife KY169SX, Scotland E-mail address: steve.linton@st-andrews.ac.uk Lehrstuhl für Algebra und Zahlentheorie, RWTH Aachen University, 52056 Aachen, Germany E-mail address: nebe@math.rwth-aachen.de Lehrstuhl für Algebra und Darstellungstheorie, RWTH Aachen Uni-versity, 52056 Aachen, Germany E-mail address: alice.niemeyer@mathb.rwth-aachen.de Cambridge, UK www-address: Cambridge, UK E-mail address: jgt@pobox.com Abstract In this paper we describe a parallel Gaussian elimination algorithm for matrices with entries in a finite field. Our algorithm subdivides a very large input matrix into smaller submatrices by subdividing both rows and columns into roughly square blocks sized so that computing with individual blocks on individual processors provides adequate concurrency.
The algorithm also returns the transformation matrix, which encodes the row operations used. We go to some lengths to avoid storing any unnecessary data as we keep track of the row operations, such as block columns of the transformation matrix known to be zero. This helps the algorithm to scale to very large dimensions.
The algorithm is accompanied by a concurrency analysis which shows that the improvement in concurrency is of the same order of magnitude as the number of blocks. An implementation of the algorithm has been tested on matrices as large as 1500000 × 1500000 over GF(2).
Keywords: Gaussian elimination, parallel algorithm, finite fields MSC 2010:15A06, 68W10, 68W05 Dedication: This paper is dedicated to the memory of Richard Parker, who died on 26th January, 2024. The ideas expressed within reflect Richard’s relentless pursuit 1 for algorithmic excellence as well as his ceaseless desire to obtain the maximum possible performance from the available hardware. The original underlying context is modular representation theory, a field in which theoretically intractable problems abound, and for which he initially conceived the Meataxe, see [15, 21]. The Meataxe has been used to solve many problems, including the existence proof of the sporadic simple group J4, and the publication of the Modular Atlas . In particular, this paper addresses effective use of multi-processor CPUs, which became the norm once clock speeds of around 3GHz had been achieved. Not all such issues are mentioned here, eg vector extensions, but we do address concurrency, cache friendliness, synchronisation and memory footprint.
Richard was an original thinker (see ) and his loss has left us all poorer.
1 Introduction Already employed for solving equations by hand in China over 2000 years ago , the Gaussian elimination method has become an invaluable tool in many areas of science.
When computing machines became available, this algorithm was one of the first to be implemented on a computer. In general the algorithm takes as input a matrix with entries in a field (or a division ring) and transforms this matrix to a matrix in row echelon form.
It can be employed for several different purposes, and computer implementations can be tailored to suit the intended application. Different variants of the Gaussian elimination algorithm can be envisaged, for example computing the rank of a matrix, computing a row echelon form or a reduced row echelon form of a matrix, or computing one of these echelon forms together with the transformation matrix.
Often one of these versions of the Gaussian elimination algorithm lies at the heart of other algorithms for solving problems in a broad range of areas and their overall performance is often dictated by the performance of the underlying Gaussian elimination algorithm. Thus an implementation of a Gaussian elimination algorithm is required to display exceptional performance.
Since their invention, computers have become faster and more powerful every year.
Yet, for over a decade this increase in computing power is no longer primarily due to faster CPUs but rather to the number of different processors an individual computer has, paired with the increasingly sophisticated memory hierarchy. It is therefore paramount that modern algorithms are tailored to modern computers. In particular this means that they need to be able to perform computations in parallel and store the data for the current computations readily in cache.
With the advance of parallel computers comes the need to design a parallel algo-rithm to perform Gaussian elimination on a matrix. Such a parallel algorithm would immediately result in immense speedups of higher level algorithms calling the Gaus-sian elimination without having to introduce parallelism to these algorithms themselves.
When designing a parallel Gaussian elimination algorithm it is important to keep the ap-plications of the algorithms in mind. Several versions of a parallel Gaussian elimination algorithm have been described when working over the field of real or complex numbers, see for a survey and PLASMA for implementations. In this paper we describe a 2 parallel version of the Gaussian elimination algorithm which, given a matrix with entries in a finite field, computes a reduced row echelon form together with the transformation matrix. Several approaches to Gaussian elimination have been surveyed in .
We note that when working with dense matrices over finite fields we are not con-cerned with sparsity, the selection of suitable pivot elements nor numerical accuracy. In particular, we can always choose the first non-zero element of a given row as our pivot element, ensuring that all entries to the left of a pivot are known to be zero. Moreover, we need not be concerned with producing elements in a field which require more and more memory to store them. Avoiding producing very large field elements would again complicate pivot selection. Thus our main concern is to design a parallel algorithm which makes optimal use of modern parallel computers.
We assume an underlying shared memory computational model in which we have access to k different processors, each of which can run a job independently from any other. The processors communicate with each other through the shared memory system.
Our design must take account of the limited total memory bandwidth of the system. The aim of our parallel algorithm is to achieve adequate concurrency by dividing the necessary computational work into smaller jobs and scheduling these to run simultaneously on the k processors. This in turn calls for a very careful organization of the jobs so that different jobs do not interfere with each other. We will solve these issues in Section 2.1.
It is well known, see for example [5, Theorems 28.7, 28.8], that the asymptotic com-plexities of matrix inversion and matrix multiplication are equal. An algorithm that shows this reduces inversion of a 2n × 2n matrix to six multiplications of n × n matrices together with two inversions, also of n × n matrices. Applying this approach recursively, almost all of the run-time of inversion is spent in multiplications of various sizes. It is not difficult to see that this extends to our somewhat more general computation of a reduced row echelon form, with transformation matrix.
We envisage that we are given a very large matrix over a finite field for which we need to compute a reduced row echelon form together with a transformation matrix.
Several approaches to achieving this already exist for finite fields. One such takes advantage of the reduction to multiplication mentioned above, and delegates the problem of parallelising the computation primarily to the much easier problem of parallelising the larger multiplications. Another represents finite field elements as floating point real numbers of various sizes in such a way that (with care) the exact result over the finite field can still be recovered. The problem can then be delegated to any of a number of highly efficient parallel floating point linear algebra systems . A third method by Albrecht et al. works over small finite fields of characteristic 2 (see and ). Other algorithms subdivide the matrix horizontally into slabs as for example in [18, 16, 19].
Slab methods tend not to be cache friendly. Approaches subdividing the input matrix by both row and column are examined in [8, 11, 10, 12]. Our strategy improves on the real time performance of these by orders of magnitude.
Large modern computers typically have a large number of cores but may well have an order of magnitude less real memory bandwidth per core than a typical laptop or desktop computer. On such a large modern computer, at the lowest level, a core can 3 only be fully occupied if all its data is in the smallest, fastest level of cache memory (L1). At the next level out, this work can only be started if all its data is in the next level of cache (L2). A similar statement is true for L3 cache. To use a modern computer effectively for matrix operations, we therefore repeatedly subdivide matrices in both directions producing submatrices that are roughly square at a scale commensurate with the size of the cache at each level.
It is not too hard to design an algorithm for matrix multiplication with these prop-erties. Gaussian elimination for these submatrices is described in Section 2.3.2.
For the whole matrix, we also need to subdivide to achieve concurrency. We split our very large input matrix into smaller submatrices, called blocks, by subdividing both rows and columns into roughly square blocks, not necessarily of the same size, such that com-puting with individual blocks on individual processors provides adequate concurrency.
We will show that we gain a concurrency improvement in the same order of magnitude as the number of blocks, see Proposition 5.3 and Theorem 5.4.
As well as computing the reduced row echelon form of the input matrix, we compute the transformation matrix, which encodes the row operations used.
We go to some lengths to avoid storing any unnecessary data as we keep track of the row operations, such as block columns of the transformation matrix known to be zero. This helps the algorithm scale to very large dimensions. Our experiments show that the memory usage during the execution of the algorithm remains fairly stable, and is similar to storing the input matrix. The runtime is broadly comparable to multiplication of matrices to the same size as the input matrix. This gives evidence that we have succeeded in keeping the cost of computing the transformation matrix as small as possible and is in accordance with the theoretical analysis of Bürgisser et al. in particular [4, Theorem 16.12].
The parallel Gauss algorithm is designed with three distinct environments in view, although we have only implemented the first so far.
The first (and original) target is to use a single machine with multiple cores with the matrix in shared memory. Here the objective is to subdivide the overall task into many subtasks that can run concurrently.
The second target is to use a single machine where the matrix does not fit in memory, but does fit on disk. Here the objective is to subdivide the matrix so that each piece fits into memory.
The third target is where several computers (probably each with multiple cores) are to work on a single problem simultaneously. Here the objective is again to subdivide the overall task into many subtasks that can run concurrently on different computers with access to the same central disk.
2 Preliminaries 2.1 Computational Model Our general approach to parallel computing is to decompose the work to be done into relatively small units with explicit dependencies. Units all of whose dependencies are 4 met are called “runnable” and a fixed pool of worker threads carry out the runnable units, thereby discharging the dependencies of other units and making them in turn runnable.
A module, called the scheduler, is charged with keeping track of the dependencies and scheduling the execution of the individual units. Data describing the units of work and their input and output data reside in a shared data store, but we take considerable care to ensure that the speed of this store is not critical to overall performance. It can thus be the large amount of shared DRAM on a multicore server, or disk storage local to a single server (for the case where we only have one server, but data too large for its RAM) or shared disk provided sufficiently strong consistency can be guaranteed.
In our implementations, we have used two variations of this model. In one, the task-model, the units of work are tasks and are relatively coarse-grained. A task represents, roughly speaking, all the work that needs to be done in a particular part of the matrix at a particular stage of the algorithm. Dependencies are between tasks, so one task cannot execute until certain others have completed and it will find the data it needs where those previous tasks have stored it. To simplify understanding, we collect different data into data packages, the input and output of the tasks. For example a typical output of the task ClearDown is the data package A = (A, M, K, ρ′, E, λ) with six components which we refer to as A.A, A.M, etc.
In the other model, the job-model, the units of work are the jobs and are significantly more fine-grained and represent a single elementary computation such as a submatrix multiplication. More importantly, the dependencies are between the jobs and the data they produce and consume.
Each job requires zero or more input data objects and produces one or more output objects. A job is runnable when all of its input data has been produced. This finer grain approach allows more concurrency. Further gain in efficiency can be achieved by giving the scheduler guidance as to which jobs or tasks are urgent and which are less so. This aspect is mainly ignored in this paper. The implementation of Meataxe64 [22, 23] uses the job-model and identifies the urgent jobs.
An implementation in HPC-GAP by Jendrik Brachter and Sergio Siccha is based on the task-model.
The parallel Gaussian elimination algorithm is described in Section 3 as a program called the Chief charged with defining the tasks and the data packages they work on.
The Chief is described as a sequential program but the reader should be warned that the tasks are executed in an unpredictable order which may bear little resemblance to the order in which the tasks are submitted by the Chief. The result of running the Chief is a plan consisting of a list of tasks, respectively jobs, together with their inputs and outputs whose collective execution performs the Gaussian elimination.
Below we specify The Chief in terms of tasks, specified in turn by jobs, for which the reader will find a more or less specific description in Section 4.
2.2 Gaussian elimination This subsection describes the Gaussian elimination process in a way we hope is familiar to the reader, but in our notation. Given a matrix H with entries in a field F the output 5 of the Gauss algorithm consists of matrices M, K and R and permutation matrices Pρ and Pγ such that M 0 K 1 PρHPγ = −1 R 0 0 .
(1) The matrices Pρ and Pγ perform row, respectively column, permutations on the input matrix H such that the top left-hand part of the resulting matrix PρHPγ is invertible (with inverse −M) with the same rank as H. It should be noticed that the permutation matrices Pρ and Pγ in Equation (1) are not uniquely defined. All that matters is that they put the pivotal rows and columns into the top left-hand corner of the matrix organised as the negative of an identity matrix. We therefore choose to only specify the sets of row and column numbers containing pivotal elements. We choose to use negative echelon form in order to avoid negations during ClearDown, see Section 4.2.1. As we are chopping our matrix into blocks, we use the word selected to specify a row or column in which a pivot has been already been found. Hence we have to apply a permutation to the rows during the course of the algorithm to ensure that our pivots remain located in columns with increasing indices. We formalize how we store these permutation matrices in the Definition 2.1.
Definition 2.1. When we enumerate elements of a set, we always implicitly assume that these are in order. To a subset ρ = {ρ1, . . . , ρ|ρ|} ⊆{1, . . . , α} associate a 0/1 matrix ρ ∈F|ρ|×α with ρi,j = ( 1 j = ρi 0 else.
We call ρ the row-select matrix and ρ the row-nonselect matrix associated to the set ρ and its complement ρ = {1, . . . , α} \ ρ.
Remark 2.2. Note that the matrix Pρ = ρ ρ ∈Fα×α is a permutation matrix associated to the permutation pρ with pρ(i) = ρi for i ≤|ρ| and pρ(i) = ρi−|ρ| for i > |ρ|. Note that pρ = 1 if and only if ρ = {1, . . . , i} for some 0 ≤i ≤α but for the other subsets ρ we may recover ρ from pρ as the image {pρ(1), . . . , pρ(i)} if pρ(i + 1) < pρ(i).
In this sense we keep switching between permutations, subsets, and bitstrings in {0, 1}α which represent the characteristic function of the subset. In particular, this means that a subset of cardinality 2α −α of special permutations of all the α! permutations of {1, . . . , α} is sufficient for all our purposes.
We also note that for a matrix H ∈Fα×β and ρ ⊆{1, . . . , α} and γ ⊆{1, . . . , β} the matrix ρ × H ∈F|ρ|×β consists of those rows of H whose indices lie in ρ (retaining the ordering) and the matrix H × γtr ∈Fα×|γ| consists of those columns of H whose indices lie in γ. Therefore we also call γtr the column-select matrix and γtr the column-nonselect matrix associated to γ.
6 Remark 2.3. Let H ∈Fα×β be of rank r.
Then the echelonisation algorithm will produce sets ρ ⊆{1, . . . , α} and γ ⊆{1, . . . , β} of cardinality |γ| = |ρ| = r and matrices M ∈Fr×r, K ∈F(α−r)×r, R ∈Fr×(β−r) such that M 0 K 1 ρ ρ H γtr γtr = −1 R 0 0 We will refer to these matrices as (M, K, R, ρ, γ) := ECH(H).
Strictly speaking, the job ECH computes the negative row reduced echelon form of the input matrix H. However we will simply call this the echelon form of H.
We assume we have an implementation of ECH suitable for use on a single core. The goal of this paper is to show how to scale it up.
Example We now present an example to highlight some of the structure of the algorithm. Consider the matrix C ∈F6×6 3 given by 0 2 2 0 1 0 0 2 2 1 2 2 1 0 1 0 2 1 2 0 1 0 2 2 0 1 1 2 1 1 1 2 2 2 0 0 and divided into four blocks as shown. Our first step is to echelonise the top left block.
This yields 0 2 0 1 0 0 2 0 1 P{1,3} 0 2 2 0 2 2 1 0 1 P{1,2} = 2 0 2 0 2 2 0 0 0 matching Equation 1. Here P{1,3} = 1 0 0 0 0 1 0 1 0 and P{1,2} = 1 0 0 0 1 0 0 0 1 , where the bars separate selected from non-selected rows or columns. So the outputs from this step are the multiplier M = 0 2 1 0 , K = 2 0 , and R = 2 2 as well as the row-select set ρ, selecting the first and third row and the column-select set γ, selecting the first two columns.
After this, two further steps are available, our parallel implementation will do both concurrently. We mimic the row transformations, applied to the top left block, on the 7 top right block. We also use the echelonised top left block to clean out some columns in the block beneath it. We explain them in that order.
Mathematically, we need to left multiply the top right block by 0 2 0 1 0 0 2 0 1 P{1,3}.
We take advantage of the known structure of this matrix to speed up this computation as follows.
We divide the top right block into the selected rows, as described by ρ: 0 1 0 0 2 1 and the rest 1 2 2 . We add the product of K and the selected rows to the non-selected rows, giving 1 1 2 , and then multiply the selected rows by M giving 0 1 2 0 1 0 .
The other step requires us to add multiples of the pivot rows from the echelonised top left block 2 0 2 0 2 2 to the bottom left block, so as to clear the pivotal columns. We again take advantage of the known structure of this matrix to speed up this computation as follows. We divide the bottom left block into the selected columns, as described by γ: 2 0 0 1 1 2 and the non-selected columns 1 1 2 . Now we add the product of the selected columns and R to these non-selected columns and obtain 2 0 2 .
We must now mimic the row transformations used to clear the pivotal columns of the bottom left block by adding multiples of rows from the top right hand block to the bottom right hand block, i.e. we add the product of 2 0 0 1 1 2 and 0 1 2 0 1 0 to the bottom right hand block, which becomes 0 1 0 2 2 1 2 0 2 .
After all these steps, the overall matrix is 2 0 2 0 1 2 0 2 2 0 1 0 0 0 0 1 1 2 0 0 2 0 1 0 0 0 0 2 2 1 0 0 2 2 0 2 .
At this stage we have dealt with all the consequences of the pivots found in the top left block.
What remains to be done is to echelonise the top-right and bottom-left submatrices in the picture above and deal with the consequences of any pivots found.
Finally part of the bottom right hand block will need to be echelonised. This example, chopped 2 × 2 does not demonstrate pivotal row merging as described in Section 2.3.1 8 2.3 Some guiding points 2.3.1 Subdividing a huge matrix Our parallel version of the Gaussian elimination algorithm takes as input a huge matrix and subdivides it, both horizontally and vertically, into blocks. We do this partly to obtain concurrency and partly to reduce the row length for cache reasons. Once the top-left block has been echelonised, the same row operations can be applied simultaneously to all the blocks along the top row. Putting the rest of the left-most block column into echelon form requires addressing the blocks sequentially from the top down. However, in the common case where the co-rank of the top-left block is small, it is not a great deal of computation. Once the top block row and left-most block column are in echelon form, we can update each of the blocks in the other rows and columns concurrently.
It should be remarked that nothing else can happen until this first block is echelonised, suggesting that we should make this block smaller than the rest. A similar comment applies to the last (bottom-right) block.
Proceeding down the left-most block column sequentially enables us to merge into a single row of blocks all those rows whose pivots lie there. This merging is done to reduce the amount of data access. Usually, the work of doing this merging is not great, so that soon after the first block echelonisation is complete, a large block multiply-and-add can be done to every block of the matrix. Without the merging, this multiply-and-add would be done piecemeal, requiring multiple passes through the block.
2.3.2 Echelonisation of a single block The performance of the echelonisation of a single block (as defined in Remark 2.3) can have a considerable impact on the concurrency, as many later jobs may depend on each echelonisation.
The sequential algorithm used to echelonise individual blocks is recursive, combin-ing elements of the recursive inversion algorithm already mentioned, with the greater generality of the technique described in this paper. A block is divided into two, either horizontally or vertically, and the top (resp. left) part is echelonised. Using a simplified version of ClearDown (resp. UpdateRow), see Sections 4.2.1 and 4.2.2, the remain-der of the matrix and the transformation matrix are updated, producing a second block which must also be echelonised. The results of the two echelonisations can be combined to compute the data package consisting of M, K, R, Pρ and Pγ. Using this technique recursively for all matrices bigger than a threshold size (about 300 dimensions) and a simple direct Gaussian elimination algorithm below this size, echelonisation of a block takes essentially the same time as a block multiply.
2.3.3 Blocks change size and shape In our description of the algorithm, especially in Equations (3) and (4), we imagine that rows are permuted so that those containing pivots (in blocks to the left) are moved to 9 the top, and the rest are moved to the bottom. In the program, however, these two sets of rows are held in different matrices, but it seemed better to try to include all the information in one matrix, attempting to clarify the relationships between the parts.
As a consequence of moving rows about, the sizes and shapes of the blocks change during the course of the algorithm. We use a superscript to indicate the “stage” of a particular block, so that, for example, Cj ik is the matrix block at the (i, k)-position in its j’th stage. In some ways the original input matrix C (see Equation (2) below) is gradually converted into the matrix R. An intermediate matrix B collects the pivotal rows from C which are subsequently deleted from C.
2.3.4 Riffles Our subset-stored permutations are used both to pull matrices apart and to merge them together. We call the pulling apart an ‘extract’ where one matrix is separated into two matrices with the selected rows (or columns) going to one, and the remaining rows to the other. We call the merging a ‘riffle’ where one matrix is assembled from two inputs with the subset-stored permutation directing from which input each row (or column) comes.
2.3.5 Transformation matrix abbreviation To compute the transformation matrix, we could apply the same row operations we performed on the input matrix to the identity matrix. In practice this would be wasteful, both of memory and computational effort, since initially the identity matrix contains no information at all, and the early row operations would mainly be adding zeros to other zeros. Considerable effort has been expended in this paper to avoid storing any parts of matrices whose values are known a priori, thereby saving both memory, and work manipulating them. The graphs shown in Section 6 suggest that this has been successful.
The details of this optimisation are the subject of Section 4.2.3, which may be skipped on first reading.
Note that although the output matrix M in Equation 2 below is square, the number of blocks in each row may differ from the number of blocks in each column, since the block rows are indexed by the block column in which the pivot was found and vice versa.
3 A parallel Gauss algorithm 3.1 The structure of the algorithm We now describe a parallel version of the Gaussian elimination algorithm which takes as input a huge matrix and subdivides it, both horizontally and vertically, into blocks.
To distinguish between the huge matrix and its blocks we use different fonts.
Let F be a finite field and C ∈Fm×n a huge matrix of rank r.
We describe a parallel version of the well-known Gauss algorithm, which computes matrices R, a 10 transformation matrix T = M 0r×(m−r) K 1(m−r)×(m−r) and subsets ϱ ⊆{1, . . . m} and Υ ⊆ {1, . . . , n}, both of cardinality r, such that M 0r×(m−r) K 1(m−r)×(m−r) ϱ ϱ C Υtr Υ tr = −1r×r R 0 0 (2) is in (negative row reduced) echelon form.
Comparing to Remark 2.3, we see that our task is to chop our huge input matrix into smaller blocks and then, using ECH and other jobs on the blocks, to effect the same operation on the huge matrix as ECH does on a single block. We therefore choose positive integers ai, bj such that a X i=1 ai = m, b X j=1 bj = n and our algorithm copies the block-submatrices of the input matrix C into data packages Cij ∈Fai×bj, called blocks, and performs tasks (as described in Section 4) on these smaller blocks.
We call the a matrices C1j, . . . , Caj the j-th block column and the b matrices Ci1, . . . , Cib the i-th block row of C. Echelonising the overall matrix C is achieved by performing an echelonisation algorithm on individual blocks and using the resulting data to modify others.
The result of the Gaussian elimination as well as the intermediate matrices are par-titioned into blocks: When the Gauss algorithm has completed, the matrix R, which occurs in Equation (2), consists of blocks and has the form R = R1 R′ 12 . . .
. . .
R′ 1b 0 R2 R′ 23 . . .
R′ 2b .
.
.
...
...
...
.
.
.
0 . . .
0 Rb−1 R′ b−1,b 0 . . .
. . .
0 Rb (3) with Rj ∈Frj×(bj−rj) and R′ jk ∈Frj×(bk−rk). Here r = Pb j=1 rj is the rank of C and for k = 1, . . . , b the sum Pk j=1 rj is the rank of the submatrix of C consisting of the first k block columns.
The time-consuming parts of the Gaussian elimination algorithm consist of Step 1 and Step 3, whereas the intermediate Step 2 is not. After the first step the matrix C has been transformed into an upper triangular matrix and prior to permuting columns the matrix (−1r×r|R) ∈Fr×n has the shape −1 | R1 X12 | R12 . . .
. . .
X1b | R1b 0 | 0 −1 | R2 X23 | R23 . . .
X2b | R2b .
.
.
...
...
...
.
.
.
0 | 0 . . .
0 | 0 −1 | Rb−1 Xb−1,b | Rb−1,b 0 | 0 . . .
. . .
0 | 0 −1 | Rb , (4) 11 where Rjk ∈Frj×(bk−rk) and Xjk ∈Frj×rk.
To simplify notation in the algorithms below we define the data packages Bjk = (Xjk|Rjk) ∈Frj×bk for 1 ≤j < k ≤b and Dj = (Dj.R, Dj.γ), where Dj.R = Rj and Dj.γ ⊆{1, . . . , bj} is the set of indices of pivotal columns in the block column j.
3.1.1 Computing the transformation matrix During the course of the algorithm we also compute the transformation matrix T ∈ GLm(F) as given in Equation (2). Of course this could be achieved by simply performing the same row operations on an identity matrix that were performed on C. This involves considerable work on blocks known to be either zero or identity matrices. For example, if C is invertible, the work required to compute its inverse is needlessly increased by 50%. To avoid such extra work, we only store the relevant parts of the blocks of the transformation matrix.
During the computation, the data packages Mji (j = 1, . . . , b; i = 1, . . . , a) and Kih (i, h = 1, . . . , a) record the matching status of the transformation matrix. If ˜ Cik denotes the i, k-block of the original input matrix C, then initially Cik = ˜ Cik and Bjk has no rows. Likewise, initially Kih is the identity matrix if i = h and the zero matrix otherwise and the matrix Mji has no rows and no columns. When storing the data packages K and M, we omit the columns and rows in the blocks that are zero or still unchanged from the identity matrix.
To obtain the row select matrix we maintain further data packages Eij = (Eij.ρ, Eij.δ) (1 ≤i ≤a, 1 ≤j ≤b) with Eij.ρ ⊆{1, . . . , ai} is the set of indices of pivotal rows in block row i with pivot in some block column 1, . . . , j and Eij.δ ∈{0, 1}|Eij.ρ| records which indices already occurred up to block column j −1.
During the algorithm, having handled block row i in Step 1, we have a X h=1 Mjh × Ehi.ρ × ˜ Chk = Bjk and a X h=1 Kℓh × Ehi.ρ × ˜ Chk = Cℓk.
The final column select matrix is the block diagonal matrix Υ = diag(Da 1.γ, . . . , Da b.γ).
and the row select matrix is ϱ = diag(E1b.ρ, . . . , Eab.ρ).
3.2 Step 1 Step 1 loops over the block rows. For the j-th column of the i-th block row, the algorithm calls Task ClearDown with the two data packages Cij and Di−1 j as input.
Task 12 ClearDown amalgamates the pivots in Di−1 j .γ with the pivots in the matrix Cij to produce the enlarged set Di j.γ as well as a new matrix Di j.R of the (negative) echelon form (−1 | Di j.R) followed by 0 rows (up to column permutations which are remembered in Di j.γ). Moreover, the task ClearDown records in its output data package Aij the row operations performed. With the help of these data packages, the first step then propagates the same elementary row operations to the remaining blocks in block row i as well as to block row i of the transition matrix using Task UpdateRow.
Hence Step 1 assembles in the block row 0 . . .
0 −1|Dj.R Bj,j+1 . . .
Bjb the rows of the original input matrix whose pivotal entries lie in block column j for j ≤b.
These rows are then deleted from the data package C. Thus having treated the j-th block column the matrix C contains no rows whose pivots lie in block columns 1, . . . , j and the number of rows of C is m −Pj k=1 rk.
In particular, during the course of the entire algorithm the block rows Ci,−con-tain fewer and fewer rows, whereas the number of rows of the block row Bj,−increases accordingly. After completing the block column j the matrices Bjk remain stable.
Similarly for the transformation matrix, the matrix M gains rows whereas K loses rows. However, things here are slightly more complicated due to the fact that we do not store the full transformation matrix. As we only store columns of Kih that are not known a priori to be zero or columns of an identity matrix, we have to ensure that all the needed columns are present when calling UpdateRowTrafo. The columns that are not yet present are precisely the columns that correspond to the positions of the pivot rows and are stored in Ehj.δ. If i = h this means we need to insert into the correct positions columns of an identity matrix and if i ̸= h then columns of a zero matrix. To achieve this, and also to efficiently deal with the cases where the matrices Kih or Mjh are to be initialized we adapted the task UpdateRow to obtain UpdateRowTrafo.
3.3 Step 2 This intermediate step becomes necessary as we do only store the relevant parts of the transformation matrices Mji. Before the upwards cleaning in Step 3 we need to riffle in zero columns in Mji so that the number of columns in Mji is equal to the number of columns in Mbi for all j.
3.4 Step 3 Then back cleaning only performs upwards row operations on the matrix from Equation (4) to eliminate the Xjk. The matrices Rjk from Equation (4) are stored in the data packages Rjk in the Chief. Having cleaned block columns b, . . . , k −1 the algorithm adds the Xjk multiple of block row k to block row j for all j ≤k−1 to clear block column k. The same row operations are performed on the relevant part M of the transformation matrix.
13 Algorithm 1: The Chief Input : C =: (C1 ik)i=1,...,a,k=1,...,b, where C1 ik ∈Fai×bk Output: R = (Rk−j jk )j≤k=1,...,b, M = (M2a jh)j=1,...,b,h=1,...,a, K = (Ka ih)i,h=1,...,a, a row select matrix ϱ ⊆{1, . . . , m}, the concatenation of the Eib.ρ ⊆{1, . . . , ai} (i = 1, . . . , a), and a column select matrix Υ ⊆{1, . . . , n}, the concatenation of the Da j.γ ⊆{1, . . . , bj} (j = 1, . . . , b), such that M 0 K 1 ϱ ϱ C Υ Υ = −1 R 0 0 Step 1: for i from 1 to a do for j from 1 to b do (Di j; Aij) := ClearDown(Cj ij, Di−1 j , i); Eij :=Extend(Aij, Ei,j−1, j); for k from j + 1 to b do (Cj+1 ik , Bi jk) := UpdateRow(Aij, Cj ik, Bi−1 jk , i); for h from 1 to i do (Kj+1 ih , Mi jh) :=UpdateRowTrafo(Aij, Kj ih, Mi−1 jh , Ehj, i, h, j); Step 2: for j from 1 to b do for h from 1 to a do Ma+1 jh :=RowLengthen(Ma jh, Ehj, Ehb); Step 3: for k from 1 to b do R0 kk :=Copy(Da k); for k from b downto 1 do for j from 1 to k −1 do (Xjk, R0 jk) := PreClearUp(Ba jk, Da k); for ℓfrom k to b do Rℓ−k+1 jℓ := ClearUp(Rℓ−k jℓ, Xjk, Rℓ−k kℓ); for h from 1 to a do Ma+h jh := ClearUp(Ma+h−1 jh , Xjk, Ma+h−1 kh ); 14 4 Jobs and Tasks 4.1 The jobs In this section we describe the jobs. These are fundamental steps that are later used to define the tasks. Many of the jobs take as input one or more matrices. While the input and output matrices of the jobs within the global context of the parallel Gauss algorithm are blocks computed from a huge input matrix, the jobs described in this section work locally only on these matrices. In current implementations, each job can be performed by a single threaded computation, entirely in RAM and in a reasonable amount of time.
cpy This task simply copies the input matrix to the output.
mul This job performs a matrix multiplication. It takes as input two matrices A ∈Fα×β and B ∈Fβ×δ and returns as output the matrix A × B ∈Fα×δ.
mad This job performs a matrix multiplication followed by a matrix addition. It takes as input matrices A ∈Fα×δ and B ∈Fα×β and C ∈Fβ×δ and returns the matrix A + B × C ∈Fα×δ.
ech This job performs an echelonisation as described in Remark 2.3. We will refer to the job as (M, K, R, ρ, γ) := ECH(H).
cex This job performs two column extracts. It takes as input a matrix H ∈Fα×β and a subset γ ⊆{1, . . . , β} and returns the matrices H × γtr and H × γtr, consisting of all those columns of H whose indices lie in γ, respectively do not lie in γ, as described in Definition 2.1.
rex This job performs two row extracts. It takes as input a matrix H ∈Fα×β and a subset ρ ⊆{1, . . . , α} and returns the matrices ρ × H and ρ × H, consisting of all those rows of H whose indices lie in ρ, respectively do not lie in ρ, as described in Definition 2.1.
unh This job performs a union plus history. It takes as input a subset ρ1 ⊆{1, . . . , α} and, for α0 = α−|ρ1|, a subset ρ2 ⊆{1, . . . , α0} and returns a subset ρ ⊆{1, . . . , α} defined as follows. Write {1, . . . , α} \ ρ1 = {x1, . . . , xα0}. Define ρ = ρ1 ∪{xi | i ∈ρ2} =: {y1, . . . , yr} as an ordered set. Then u ∈{0, 1}r with uℓ= 0 if yℓ∈ρ1 and uℓ= 1 otherwise.
We refer to this job as (ρ, u) := UNH(ρ1, ρ2).
un0 This job does the same as unh except that the first input set of unh is omitted and assumed empty.
15 mkr It takes two sets ρ1 ⊆ρ2 and produces a bitstring λ ⊆{0, 1}|ρ2| with 1s corre-sponding to the elements in ρ1 and 0s corresponding to the elements in ρ2\ρ1.
rrf This job performs a row riffle. The input consists of a bit string u ∈{0, 1}r and two matrices B ∈Fα×β and C ∈Fγ×β with α + γ = r, where the number of 0s in u is α and the number of 1s in u is γ. The job returns the new matrix A ∈Fr×b whose rows are the rows of B and C combined according to u. In some sense this is the inverse of row extract.
crz Similarly to row riffles we also need column riffles, but we only need to riffle in zero columns.
adi It takes as input a matrix K ∈Fα×β and a bitstring δ ∈{0, 1}β and puts Ki,ji := 1 if ji is the position of the ith 0 in δ. Note that combining crz with adi allows us to riffle in columns of the identity matrix.
4.2 The tasks We now describe the tasks on which our Gaussian elimination algorithm depends. As mentioned above, a task receives data packages as input, which in turn may consist of several components, and returns data packages as output.
Definition 4.1. A data package is a record of one or several components. A data package is called ready (for a given scheduler) if the task that produces it as output has finished, regardless whether this task has computed all its components. If there is no task having this data package as an output, then we also consider it ready.
Example: Task UpdateRow.
If called with the parameter i = 1 the task UpdateRow can start, even though the com-ponent A.A of the data package A has not been computed, after the task ClearDown for i = 1 has completed. Note that for i = 1, no job in the task UpdateRow takes the component A.A as an input. Also the data package B0 jk is an input to UpdateRow but not computed by any task in the Chief. So therefore it is also considered ready.
Task 1: Extend Input : A = (A, M, K, ρ′, E, λ), E = (ρ, δ) with ρ ⊂{1, . . . , α}, δ a riffle, j Output: E.
j = 1 (UN0): (E.ρ, E.δ) := UN0(A.ρ′); j ̸= 1 (UNH): (E.ρ, E.δ) := UNH(E.ρ, A.ρ′); 16 Task 2: RowLengthen Input : M ∈Fα×g1, E1.ρ ⊆E2.ρ ⊆{1, . . . , α} of sizes g1, g2 with g1 ≤g2.
Output: M ∈Fα×g2.
(MKR): λ := MKR(E1.ρ, E2.ρ); (CRZ): M := CRZ(M, λ); Task 3: ClearUp Input : R ∈Fα×β, X ∈Fα×γ, M ∈Fγ×β.
Output: R ∈Fα×β.
(MAD): R := R + X × M; Task 4: PreClearUp Input : B ∈Fα×β, D, with D.γ ⊆{1, . . . , β} of cardinality g.
Output: X ∈Fα×g, R ∈Fα×(β−g).
(CEX): X := B × D.γtr; R := B × D.γ tr; Task 5: Copy Input : D, with D.R ∈Fα×β.
Output: R ∈Fα×β.
(CPY): R := D.R; 4.2.1 Task ClearDown Task ClearDown works on block columns. Suppose that j ∈{1, . . . , b} and ClearDown works on block column j which contains bj columns. Task ClearDown assumes that block column j truncated after row i −1 is in row echelon form and the aim of task ClearDown is to replace the block column j truncated after row i by its row echelon form.
Task ClearDown takes two data packages C and D as input. The first data package C is the block Cij which is the block in the i-th block row of block column j. The second data set D contains two data elements. The data element D.R is a matrix such that block column j truncated after block row i−1 is in row echelon form (−1 | D.R) followed by 0 rows. The data element D.γ ⊆{1, . . . , bj} contains indices of the pivots assembled in block column j truncated after block row i −1.
The task produces two data packages A and D as outputs. The data elements stored in the data package A are required to propagate row operations performed during the call to Task UpdateRow to other blocks in block row i. The data elements stored in 17 the data package D are required for a subsequent call to ClearDown for the block Ci+1,j in block column j.
We begin by partitioning the input block C according to D.γ into pivotal and non pivotal columns C = (A.A | A′). Using the rows of the matrix (−1 | D.R) we can reduce C to (0 | H′) where H′ = A′+A.A×D.R. The next step is to call job ECH to echelonise H′ and obtain (A.M, A.K, R, A.ρ′, γ′) := ECH(H′), where A.ρ′ is the set of pivotal rows of H′ and γ′ the set of pivotal columns.
As block column j truncated after block row i−1 is in row echelon form (−1r×r | D.R) followed by 0 rows, we now wish to determine a new remnant matrix ˆ R (which will become the new D.R) such that block column j truncated after block row i is in row echelon form (−1(r+r′)×(r+r′) | ˆ R) followed by 0 rows. To achieve this, the we have to add the r′ pivots of H′ to −1r×r and reduce D.R according to (−1r′×r′ | R). This amounts to first separating the columns of D.R into those containing pivot entries of H′ and those that do not, i.e. writing D.R = (A.E | R′) with the help of the row select and row non-select matrices γ′ and γ′. We then use the rows of the matrix (−1r′×r′ | R) to reduce D.R to (0 | R′+A.E×D.R). The new set D.γ of all pivotal columns of block column j truncated after block row i is now obtained by combining the old set D.γ and γ′. We record in A.λ the information which of these indices came from the r′ pivots of H′. Finally, the new remnant ˆ R is obtained by interleaving the rows of R′ + A.E × R with the rows of R according to A.λ and storing the resulting matrix as the new D.R.
The following pseudo code details Task ClearDown: Task 6: ClearDown Input : C ∈Fα×β, D.γ ⊆{1, . . . , β} of cardinality r, D.R ∈Fr×(β−r), i; Output: D.R ∈F(r+r′)×(β−r−r′), D.γ ⊆{1, . . . , β} of cardinality r + r′ and A = (A, M, K, ρ′, E, λ) where A ∈Fα×r, M ∈Fr′×r′, E ∈Fr×r′, K ∈F(α−r′)×r′, ρ′ ⊆{1, . . . , α −r} of cardinality r′, λ ∈{0, 1}r+r′.
if i = 1 then (ECH): (A.M, A.K, D.R, A.ρ′, D.γ) := ECH(C); else (CEX): A.A := C × D.γtr; A′ := C × D.γ tr; (MAD): H := A′ + A.A × D.R; (ECH): (A.M, A.K, R, A.ρ′, γ′) := ECH(H); (CEX): A.E := D.R × (γ′)tr, R′ := D.R × (γ′)tr; (MAD): R′ := R′ + A.E × R; (UNH): (D.γ, A.λ) := UNH(D.γ, γ′); (RRF): D.R := RRF(A.λ, R′, R); end 18 4.2.2 Task UpdateRow Given i ∈{1, . . . , a}, the Task UpdateRow works on block C = Cik in block row i and block column k. It takes as input data packages A, C and B, where the data package A encodes the necessary information computed by ClearDown when transforming an earlier block in the same block row i into echelon form.
The same row operations that were performed on this earlier block now need to be performed on C. This subroutine also assembles in the matrix B the rows in block column k whose pivotal entry lies in block column j for j + 1 ≤k ≤b.
The new data package C returned by Tasks UpdateRow then is equal to the transformed input matrix C with these rows deleted.
The following pseudo code details Task UpdateRow: Task 7: UpdateRow Input : A = (A, M, K, ρ′, E, λ), C ∈Fα×β, B ∈Fr×β, i.
Output: C ∈F(α−r′)×β, B ∈F(r+r′)×β.
(1) i ̸= 1 (MAD): Z := C + A.A × B; i = 1 (CPY): Z := C; (2) always (REX): V := A.ρ′ × Z; and W := A.ρ′ × Z; (3) always (MUL): X := A.M × V ; (4) i ̸= 1 (MAD): S := B + A.E × X; (5) i ̸= 1 (RRF): B := RRF(A.λ, S, X); i = 1 (CPY): B := X; (6) always (MAD): C := W + A.K × V ; Remark 4.2. In the case i = 1 in Task UpdateRow we work with the first block row.
Therefore we do not need to perform the upwards cleaning on the data package C and the data package B is initialized accordingly. Note that for i = 1 the task ClearDown did not compute the components A.A, A.E and A.λ and also the input data package B of UpdateRow is not present.
4.2.3 Task UpdateRowTrafo If one is not too concerned about performance, then it would be possible to generate an identity matrix K and apply the UpdateRow task replacing C by K and B by M to mimic the relevant row operations performed to obtain the transformation matrix M and the cleaner matrix K. This would result in a lot of needless work performed on zero or identity matrices. The main difference is that we never store any columns known to belong to an identity or a zero matrix. Instead we insert these columns just before they are needed. Moreover, we never add a matrix known to be zero or multiply by a matrix known to be the identity.
As a result, we require a separate procedure, UpdateRowTrafo, to mimic the row operations on the transformation matrix. UpdateRowTrafo still performs the 19 same steps as UpdateRow, identified by the same numbers, however it requires some additional steps, indicated by the symbol + in the first column and which insert some unstored columns into the matrix K. The various instances of a given step are due to the fact that we can often avoid unnecessary work. In particular, UpdateRowTrafo takes as an additional input the integers i, j, h, with h ≤i. The integer i indicates the current block row and h the current block column, on which to mimic the row operations performed during UpdateRow on block row i. If j > 1 then we already computed some input K into which we need to riffle in zero (if i ̸= h) or the relevant columns of the identity matrix (if i = h). It turns out that it is easier to always riffle in zero (the first line marked with +) and mimic the special case i = h by adding the correct 0/1 matrix to V later in the other line marked with +. If j = 1 then we should initialise K with zero (if i ̸= h) or the relevant columns of the identity matrix (if i = h). As we only need the input K to define V and W in line (2), we mimic this by remembering that W = 0 in this case and V is either 0 (if i ̸= h) or the identity matrix if i = h. So for j = 1 and h = i we omit the multiplication by the identity in lines (3) and (6).
If i = 1 again Remark 4.2 applies accordingly to line (4). Note that due to the fact that h ≤i, this only happens if h = i = 1.
In the following pseudo code describing Task UpdateRowTrafo we indicate in the last column which unstored matrices are implicitly known to be 0 or the identity 1.
20 Task 8: UpdateRowTrafo Input : A = (A, M, K, ρ′, E, λ), K ∈Fα×β, M ∈Fr×β′, E = (ρ, δ), i, h, j.
Output: K ∈F(α−r′)×(β+|δ|), M ∈F(r+r′)×β′.
case job command remark + j ̸= 1 (CRZ): K := CRZ(K, E.δ); j = 1 − K is 0 (1) h ̸= i, j ̸= 1 (MAD): Z := K + A.A × M; h ̸= i, j = 1 (MUL): Z := A.A × M; K is 0 h = i, j ̸= 1 (CPY): Z := K; M is 0 h = i, j = 1 − Z is 0 (2) ¬(j = 1 ∧h = i) (REX): V := A.ρ′ × Z; W := A.ρ′ × Z; j = 1 ∧h = i − V , W are 0 + j ̸= 1 ∧h = i (ADI): V := ADI(V, E.δ); j = 1 ∧h = i − V is 1 (3) ¬(j = 1 ∧h = i) (MUL): X := A.M × V ; j = 1 ∧h = i (CPY): X := A.M V is 1 (4) h ̸= i (MAD): S := M + A.E × X; h = i ̸= 1 (MUL): S := A.E × X; M is 0 h = i = 1 − S no rows (5) ¬(h = i = 1) (RRF): M := RRF(A.λ, S, X); h = i = 1 (CPY): M := X; S no rows (6) ¬(j = 1 ∧h = i) (MAD): K := W + A.K × V ; j = 1 ∧h = i (CPY): K := A.K; V is 1, W is 0 5 Concurrency analysis To measure the degree of concurrency of our algorithm we assign costs to each of the tasks. We perform a relative analysis, comparing the cost of a parallel Gauss algorithm with the cost of a sequential algorithm. Therefore, to simplify our analysis, we assume that the cost of a matrix multiplication of (α × β) · (β × γ) possibly followed by addition is αβγ and the cost of echelonising an α × β matrix of rank r is αβr. It seems plausible that when assuming that these costs are homogeneous functions of some degree, bounded below by ω (see [4, Chapter 16]), then in the results of Proposition 5.3 and Theorem 5.4 the degree of concurrency can be replaced by some constant times aω−1. We also assume that all blocks are square matrices of size α × α, where α is not too small and α = n a = m b .
Then the tasks Extend, RowLengthen, PreClearUp, and Copy do not perform any time consuming operations (compared to ClearDown and UpdateRow), so we assign cost 0 to these tasks.
21 ↓(i-1, j) →(i, j-1, j) ↓(i, j-1) →(i-1, j, k) →(i, j-1, k) ↓(i, j) →(i, j, k) Figure 1: Extract of task dependency graph Lemma 5.1. The cost of ClearDown is bounded above by α3 and the cost of Up-dateRow is bounded above by 1.25α3.
Proof. We start analysing the task ClearDown: For i = 1 only one job ECH is performed contributing cost α3. Otherwise the first call of MAD multiplies a matrix A.A of size α×r with a matrix D.R of size r ×(α−r), contributing cost αr(α−r). The echelonisation is done on an α×(α−r) matrix of rank r′ and the second MAD multiplies an r × r′ matrix by an r′ × (α −r −r′) matrix. So in total the cost of ClearDown is αr(α −r) + α(α −r)r′ + rr′(α −r −r′) = α2(r + r′) −rr′(α + r + r′) ≤α3 as r + r′ ≤α.
For the task UpdateRow we similarly obtain the cost αrα for the MAD in row (1), r′r′α for the MUL in row (3), rr′α for the MAD in row (4), and (α −r′)r′α for the MAD in row (6). Summing up we obtain α2(r + r′) + αrr′ ≤1.25α3 again since r + r′ ≤α.
Ignoring all tasks of cost 0 Step 1 only involves the tasks ClearDown and Up-dateRow. The graph of task dependencies decomposes naturally into layers accord-ing to the value of i + j. We abbreviate the call ClearDown(Cj ij, Di−1 j , i) with data packages depending on i, j by ↓(i, j) and similarly UpdateRow(Aij, Cj ik, Bi−1 jk , i) by →(i, j, k).Then Figure 1 displays the local task dependencies in Step 1 for layers i+j −1 and i + j, where k = j + 1, . . . , b.
Recall that a critical path in a task dependency graph is the longest directed path between any pair of start node and finish node. Its length is weighted by the cost of the nodes along the path.
22 Proposition 5.2.
1. The weighted length of a critical path in the task dependency graph of Step 1 is 2.25α3(a + b −1).
2. The weighted length of a critical path in Step 3 is max((b −1)α3, aα3).
Proof. 1.) The task dependency graph splits naturally into a + b −1 layers according to the value of i + j ∈{2, . . . , a + b}.
Within each layer, the critical paths involve ClearDown and UpdateRow exactly once. So in total the length of a critical path is (a + b −1)(α3 + 1.25α3).
2.) Step 3 only involves the task ClearUp of non-zero cost α3. The data package Rjh is only changed by the data packages below in the same column so h −j times.
The maximum of h −j is achieved at R1b contributing the cost α3(b −1).
For the transformation matrix, which can be done independently, each of the Mjh is touched a times contributing aα3.
To determine the average degree of concurrency we divide the cost of the sequential Gaussalgorithm (with transformation matrix) applied to the m×n-matrix ClearDown by the weighted length of a critical path. For simplicity we assume that m = n, a = b and that all blocks are of the same size α × β with α = β.
Proposition 5.3. Under the assumptions above and using Lemma 5.1 the average degree of concurrency of the Chief is 1 5.5a2.
Proof. By our assumptions n = m and the cost of the sequential Gaussalgorithm (with transformation matrix) applied to the huge matrix ClearDown is n3 = a3α3.
By Proposition 5.2 the weighted length of a critical path in the complete algorithm the Chief is (2.25(2a −1) + a)α3 = (5.5a −2.25)α3 ∼5.5aα3.
In practical examples the gain of performance is much better. This is partly due to the fact that we split our matrix into blocks that fit into the computer’s memory; an echelonisation of the huge matrix, however, would require to permanently read and write data to the hard disk. The other reason is that in random examples the length of a critical path is much shorter. To make this precise we assume, in addition to the assumptions above of starting with a square matrix partitioned into square blocks of equal size α, that our input matrix is well-conditioned, by which we mean that the a top-left square submatrices of the input matrix of size jα (j = 1, . . . , a) have full rank.
Then, properly implemented, the cost of ClearDown is α3, if it is called for i = j and 0 otherwise. Also in Update Row r′ = 0 and so the cost of Update Row is α3 (this can be shown without using the assumptions in Lemma 5.1). In particular in the dependency graph above, the weighted length of a critical path in any odd layer is α3 and in an even layer, this length is 2α3 (resp. α3 for the last layer). In total this shows the following Theorem 5.4. For a well-conditioned square matrix, the length of a critical path in the dependency graph is α3(3a −2) and hence the average degree of concurrency of the Chief is 1 3a2.
23 Remark 5.5. The concurrency analysis above is not sufficient to ensure the effective use of all processors throughout the entire run that we see in the experimental results below.
Although the steps are estimated at their worst case in practice many tasks early in the task dependency graph execute a lot faster. Provided there are sufficiently many blocks, enough work is available for execution considerably earlier than the task dependency graph might suggest. Assigning a priority i + j to a task pertaining to the i-th row and j-th column directs an appropriate scheduling.
6 Experimental results We give timings for the implementation of the code available through the Meataxe project, see . In order to demonstrate the power of this algorithm the initial tests were done on a machine with 64-piledriver cores with 512 GB of memory running at 2.7 gigaherz. Further tests were performed on an HEDT comprising an Intel i7-11700 (Rocket Lake) machine with 8 cores and 128 GB ram.
While runs on examples of up to dimension 1.5 million have been run by the fifth author, we chose as our first example a random 1, 000, 000×1, 000, 000-matrix with entries in the field of order 2. To put this matrix into reduced echelon form with transformation matrix we chose to chop the matrix into blocks of size 50, 000 × 50, 000. This run took 520min.
The following graph shows the progress of the calculation. The red shows that over 60 cores were used for the vast majority of the time. The blue shows that during Steps 1 and 2 the memory footprint was fairly constant but at the transition to Step 3 about 30% more memory was needed, due mainly to the expansion of the matrix M.
For a second example, on the same machine, we echelonised with transformation matrix a random 600, 000 × 600, 000-matrix with entries in the field of order 3 in 460 min, using a block size of 30, 000. We do not give a graph for this run, as it is almost indistinguishable from the one given above.
24 The above examples were done with a carefully chosen block size. The following two examples highlight the effect of too large a block size. We echelonised a random 300, 000 × 300, 000-matrix with entries in the field of order 3 using block sizes of 30, 000 and 15, 000, respectively. In the first graph we see that 15, 000 is again a good choice of block size. Almost all cores are used for most of the time. In the second graph we see that 30, 000 is too large a block size, so that the 64 available cores are seldomly utilised at once, leading to a run time which is about 50% longer.
Note that in general terms the necessary block size agrees with the concurrency analysis.
Our next three examples are intended to demonstrate that our methods are not restricted to tiny fields, nor to prime fields. We measured the wall clock time r (h:m:s) to run Gaussian elimination with transformation matrices on random matrices over a finite field F.
dim |F| r 200k 193 7:41:00 200k 1331 = 112 10:25:00 100k 50653 = 373 3:33:00 In order to demonstrate comparable performance of our Gaussian elimination al-gorithm with transformation matrix (Gauss) and matrix multiplication (Multiply), we compared these algorithms on several random matrices over F2 and F3 on the HEDT (see above).
For the multiplication we also list the parallel factor f on the 8 core machine with all cores made available to the algorithm. Here r is again the wall clock time and u denotes the CPU time over all processors used. The column labelled |G|/|M| records the quotient of the r-time of Gauss over the r-time of Multiply.
25 Multiply Gauss |G|/|M| dim |F| r u f r u f 100k 2 0:00:51 0:05:23 6.3 0:00:26 0:02:31 5.8 0.51 100k 3 0:02:02 0:14:41 7.2 0:01:25 0:10:18 7.3 0.70 200k 2 0:04:51 0:32:44 6.8 0:02:36 0:18:36 7.2 0.54 200k 3 0:12:34 1:32:22 7.4 0:9:39 1:12:42 7.5 0.77 300k 2 0:14:55 1:45:6 7.0 0:07:52 0:55:36 7.1 0.53 300k 3 0:42:44 5:21:9 7.5 0:35:49 4:31:43 7.6 0.84 500k 2 0:59:10 7:09:50 7.3 0:39:9 4:56:15 7.6 0.66 500k 3 2:54:5 22:12:32 7.7 3:02:13 18:17:4 6.0 1.05 7 Acknowledgements We thank Martin Albrecht for discussions in the early stages of the algorithm design.
The first full implementation of this algorithm was developed by Jendrik Brachter in GAP which was essential to getting the details of the algorithm right. A parallel version of this implementation, using HPC-GAP, was produced by Jendrik Brachter and Sergio Siccha.
The second and third author acknowledge support by the German Research Foun-dation (DFG) – Project-ID 286237555 – within the SFB-TRR 195 “Symbolic Tools in Mathematics and their Applications”.
References Martin R. Albrecht, The M4RIE library for dense linear algebra over small fields with even characteristic. ISSAC 2012-Proceedings of the 37th International Sympo-sium on Symbolic and Algebraic Computation, 28–34, ACM, New York, 2012.
Martin Albrecht,Gregory Bard and William Hart, Algorithm 898: Efficient multipli-cation of dense matrices over GF (2) ACM Transactions on Mathematical Software (TOMS) 37 (1), Article 9, 2010.
Martin Albrecht et al.,Linear Algebra over F2 (and F2e) Peter Bürgisser, Michael Clausen and M. Amin Shokrollahi, Algebraic complexity theory. With the collaboration of Thomas Lickteig. Grundlehren der Mathematis-chen Wissenschaften 315. Springer-Verlag, Berlin, 1997.
T.H. Cormen, C.E. Leiserson, R.L. Rivest, C. Stein, Introduction to Algorithms. (2.
ed.), The MIT Press, Cambridge, Massachusetts, London McGraw-Hill Book Com-26 pany, Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Montreal Toronto, (2001).
Simplice Donfack, Jack Dongarra, Mathieu Faverge, Mark Gates, Jakub Kurzak, Piotr Luszczek and Ichitaro Yamazaki, A survey of recent developments in parallel implementations of Gaussian elimination.
Concurrency and Computat.: Pract.
Exper. (2014). DOI: 10.1002/cpe.3306 Jean-Guillaume Dumas, Pascal Giorgi and Clément Pernet, Dense Linear Algebra over Word-Size Prime Fields: the FFLAS and FFPACK Packages. ACM Trans. on Mathematical Software (TOMS) 35(3), ACM Press, NY, USA, (2008), 1–42. DOI: 10.1145/1391989.1391992 Jean-Guillaume Dumas, Clément Pernet, Ziad Sultan, Simultaneous computation of the row and column rank profiles. ISSAC 2013 - 38th International Symposium on Symbolic and Algebraic Computation, Jun 2013, Boston, MA, United States., 181–188, 2013. DOI: 10.1145/2465506.2465517.
Jean-Guillaume Dumas, Thierry Gautier, Clément Pernet & Ziad Sultan, Parallel Computation of Echelon Forms. Euro-Par 2014 Parallel Processing, F. Silva, I.
Dutra, V. Santos Costa (eds), Lecture Notes in Computer Science, (8632), Springer, Cham., 2014. DOI: 10.1007/978-3-319-09873-9_42 Jean-Guillaume Dumas, Clément Pernet, Ziad Sultan, Computing the rank profile matrix, in ISSAC’15—Proceedings of the 2015 ACM International Symposium on Symbolic and Algebraic Computation, 149–156, ACM, New York, 2015.
Jean-Guillaume Dumas, Clément Pernet, Ziad Sultan, Recursion based paralleliza-tion of exact dense linear algebra routines for Gaussian elimination, Parallel Com-put. 57, 235–249, 2016.
Jean-Guillaume Dumas, Clément Pernet, Ziad Sultan, Fast computation of the rank profile matrix and the generalized Bruhat decomposition. J. Symbolic Comput. 83, 187–210, 2017.
The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.9.1; 2018. ( Joseph F. Grcar, Mathematicians of Gaussian elimination. Notices Amer. Math.
Soc. 58 (6), (2011), 782–792.
Derek F. Holt, The Meataxe as a tool in computational group theory. The At-las of Finite Groups - Ten Years On, R.T. Curtis RT, R.A. Wilson, eds. London Mathematical Society Lecture Note Series. Cambridge University Press; 1998:74–81.
27 Oscar H. Ibarra, Shlomo Moran and Roger Hui, A generalization of the fast LUP matrix decomposition algorithm and applications. J. Algorithms 3, no. 1, 45–56, 1982.
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(N.S.), 11 Oxford Sci. Publ. The Clarendon Press, Oxford University Press, New York, xviii+327, 1995.
Claude-Pierre Jeannerod, Clément Pernet and Arne Storjohann, Rank-profile re-vealing Gaussian elimination and the CUP matrix decomposition, J. Symbolic Com-put., 56, 46–68, 2013. DOI: 10.1016/j.jsc.2013.04.004 Walter Keller-Gehrig, Fast algorithms for the characteristic polynomial, Theoret.
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Equation for Distance of the Straight line from the Origin.
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By reduction of the equation ax+by+c=0 of a straight line to the normal form , we get
(−aa2+b2−−−−−−√)x+(−ba2+b2−−−−−−√)y=ca2+b2−−−−−−√
And,
p=∣c∣a2+b2−−−−−−√
And my textbook says that p is the distance of the straight line from the origin. I don't know why we are getting it as a distance from origin?
I know p=xcosθ+ysinθ ( where p is distance of line from origin).
Also I want to know how can we relate both equations?
geometry
analytic-geometry
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edited Jan 7, 2017 at 13:42
Jyrki Lahtonen
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asked Oct 8, 2016 at 3:46
FawadFawad
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Have you tried to check what's written about this on Wikipedia or in other posts on this site like this or this?
– Martin Sleziak
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Oct 8, 2016 at 10:09
@THE, you have made edits to about 30 questions in a matter of minutes, flooding the front page and driving other, newer questions off it. Please don't do that! Please edit three or four questions a day, not 30 in half an hour.
– Gerry Myerson
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Jan 4, 2017 at 3:47
@GerryMyerson, I knew that soon something like that will come to me, and I appologise for what I did. But II did not took them from previous pages, they were already on the front page (Someone else had edited them), I just revised small mistakes.
– Vidyanshu Mishra
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Jan 4, 2017 at 3:51
@THE, yes, so I see. I have also left a comment for the other editor who is, if I'm not mistaken, a repeat offender.
– Gerry Myerson
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Jan 4, 2017 at 3:52
Okay, @Garry, you do not need to worry shot it now. Nothing such will happen again from my side
– Vidyanshu Mishra
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Jan 4, 2017 at 4:23
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There is a simple derivation of what you want to know. Without going into details, let me introduce the result:
The perpendicular distance of a line (Ax+By+c=0)from a point is equal to |Ax′+By′+cA2+B2√|. Where (x′,y′) are coordinates of the point.
Since you want distance of line from origin, the coordinates become (0,0) and hence the perpendicular distance of a line from origin is |Ax′+By′+cA2+B2√|=|0+0+cA2+B2√|=|cA2+B2√|.
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answered Jan 4, 2017 at 3:25
Vidyanshu MishraVidyanshu Mishra
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I got that after few weeks I posted that. But I still don't know why cosθ and sinθ is getting when divided by a2+b2−−−−−−√
– Fawad
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Jan 4, 2017 at 11:22
Wait, Have you got the book, SL Loney ??
– Vidyanshu Mishra
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Jan 4, 2017 at 11:23
Of coordinate geometry??
– Vidyanshu Mishra
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Jan 4, 2017 at 11:23
No :( I only follow my Textbook
– Fawad
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Jan 4, 2017 at 11:24
Okay, then I m gonna send you pictures from that book of the section we are discussing here, I m sure it will help you, But where can I send you them ??
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Note that by Cauchy Schwarz, if (x,y) satisfies ax+by+c=0, then
c2=(−c)2=(ax+by)2≤(a2+b2)(x2+y2).
(The last inequality can be checked directly). Thus every points (x,y) on the line satisfies
x2+y2−−−−−−√≥|c|a2+b2−−−−−−√.(1)
On the other hand, the point
(x,y)=(−aca2+b2−−−−−−√,−bca2+b2−−−−−−√)
lies on the line and has distance
|c|a2+b2−−−−−−√
from the origin. Thus we are done.
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edited Oct 8, 2016 at 18:29
answered Oct 8, 2016 at 10:04
user99914user99914
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I can't get any thing from your answer, can you explain in any simple way?
– Fawad
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Oct 8, 2016 at 10:25
Which part do you not understand, @Ramanujan?
– user99914
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Oct 8, 2016 at 10:53
After −c=ax+by= to last
– Fawad
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Oct 8, 2016 at 10:58
Please see the edit @Ramanujan
– user99914
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Oct 8, 2016 at 11:01
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@JohnMa I suppose you wanted to write (ax+by)2≤(a2+b2)(x2+y2) rather than (ax+by)2=(a2+b2)(x2+y2)...?
– Martin Sleziak
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For this derivation, we won’t be needing any trigonometric equation nor you have to prior study trigonometry.
For this derivation we just need the following things:
Pythagoras Theorem
Algebra
Area of triangle
12×height×base
Let us take the line equation Ax+By+C=0.
So by putting first y=0 solving for x and then putting x=0 and solving for y.
By putting y=0 and solve for x:
Ax+By+CAx+B(0)+CAx+CAxx=0,y=0=0=0=−C=−CA
By putting x=0 and solve for y:
Ax+By+CA(0)+By+CBy+CByy=0,x=0=0=0=−C=−CB
By doing the above we get the base distance and perpendicular distance of the triangle formed by the line equation Ax+By+C=0.
Since the triangle formed is a right-angled triangle, we can get its area and hypotenuse with no problem.
So first, deriving the hypotenuse:
Rise=∣∣∣−CA∣∣∣, Run=∣∣∣−CB∣∣∣
Pythagoras Thm.: Hyp.2 = Rise2 + Run2:
Hyp.2Hyp.=∣∣∣−CA∣∣∣2+∣∣∣−CB∣∣∣2=C2A2+C2B2=C2(A2+B2)(AB)2;=C2(A2+B2)(AB)2−−−−−−−−−−−√=|C|A2+B2−−−−−−−√|AB|.
Deriving the Area:
BaseHeightArea=∣∣∣−CA∣∣∣,=∣∣∣−CB∣∣∣;=12×∣∣∣−CA∣∣∣×∣∣∣−CB∣∣∣=C22|AB|.
We can also write the area as half the hypotenuse times the height that is perpendicular to the hypotenuse (Note: the height perpendicular to the hypotenuse is the shortest distance to the origin, so let us call this height w).
AreaC22|AB|C2|AB||C||C|A2+B2−−−−−−−√=12×hypotenuse×perp. height=12×|C|A2+B2−−−−−−−√|AB|×w=|C|A2+B2−−−−−−−√|AB|×w=A2+B2−−−−−−−√×w=w.
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edited Sep 11, 2022 at 6:22
Rócherz
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answered Nov 10, 2021 at 6:21
daemondaemon
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By simple identification, the two equations are identical when
−aa2+b2−−−−−−√=cosθ,−ba2+b2−−−−−−√=sinθ,ca2+b2−−−−−−√=p.
This is coherent as you verify cos2θ+sin2θ=1. If c is negative, change all signs.
As one can verify by substitution,
x=pcosθ−tsinθ,y=psinθ+tcosθ.
describes any point along the line, by varying t.
The distance from the origin to this point is given by, after simplification,
d=x2+y2−−−−−−√=p2+t2−−−−−−√.
The minimum value is indeed p.
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answered Oct 8, 2016 at 10:11
user65203user65203
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How we getx=pcosθ−tsinθ,y=psinθ+tcosθ.
– Fawad
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Oct 8, 2016 at 10:19
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Question on counting electrons for identifying lone pairs and formal charge : r/chemhelp
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Question on counting electrons for identifying lone pairs and formal charge
In this image
For example: the top and bottom left structures that are circled
So, when counting the electrons around an atom to meet its octet and to also identify any lone pairs, we count each bond connected to the atom as 2 electrons each, correct?
Now If that's the case, what confuses me is how you count for "number of valence electrons in bonded atom" - using formal charge. You can see only 1 electron is being counted for each of the bonds connected to O in the top circled structure to get a total of 6 electrons for the formal charge being 6-6=0.
This was a misconception I've had and I think I may have somehow been mixing up how I should be accounting for electrons.
Again, I just thought that a single bond always represents 2 electrons, so I count like 2-4-6-8 but apparently we only account for 1 electron per bond when we're dealing with FC?
Sorry if my question is confusing or you dont understand what I'm asking. I'm basically just a bit confused with how to always know where the lone pairs are and what the charge will always be. It's a simple thing that I hope someone can clear up- like why only 1 electron is accounted for in FC instead of 2.
Thanks
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Half-plane intersection | Computational Geometry | Faculty of Natural Science at Masaryk University
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Computational Geometry Faculty of Natural Science at Masaryk University doc. RNDr. Martin Čadek, CSc.
česky
in English
Introduction1. Convex hull in the plane2. Line segment intersection via sweep algorithm3. Map overlay4. Polygon triangulation5. Half-plane intersection
IntroductionDescription of half-plane intersectionAlgorithm for half-plane intersectionPseudocodes and their running timeAnimation
6. Linear programming7. Orthogonal range searching8. Point location9. Voronoi Diagrams10. Delaunay TriangulationReferences
Half-plane intersection
Introduction
In this chapter, we will show you how to describe and find the intersection of n n half-planes in the plane effectively. We want to proceed recurrently according to the "divide and conquer" strategy. That is, we divide the set
H={h 1,h 2,…,h n}H={h 1,h 2,…,h n}
of given half-planes into two parts H 1 H 1 and H 2 H 2 of roughly same size and we compute the intersection
C=⋂h i∈H h i C=⋂h i∈H h i
as an intersection C 1∩C 2 C 1∩C 2, where
C 1=⋂h i∈H 1 h i,C 2=⋂h i∈H 2 h i.C 1=⋂h i∈H 1 h i,C 2=⋂h i∈H 2 h i.
A major role for the gradual computing of intersections C 1∩C 2 C 1∩C 2 has the way how we describe these intersections.
Description of half-plane intersection
Intersections of half-planes are convex sets that can be both bounded and unbounded. We exclude the case that among given half-planes there are two opposite to each other. In this case, the intersection would be a subset of their common line, and this would lead to a one-dimensional task of the intersection of half-lines on the line.
If the intersection is bounded and 2 2-dimensional, it is a convex polygon and we can describe it using a double-connected edge list with two faces, one bounded and one unbounded. In case the intersection is unbounded, we need to modify this description. For this purpose, we will use a lexicographic arrangement of points in the plane defined as follows:
p>q iff p y>q y or(p y=q y and p xq iff p y>q y or(p y=q y and p x<q x).
Let us consider a non-empty convex set C C, which is the intersection of the half-planes and is not a subset of a line. There is just one of the following options:
The set C has a maximum point p p and a minimum point q q in the lexicographic arrangement. They both lie on the boundary and divide it into the left and right parts, abbreviated to the left and right boundaries. Each of these is a polygonal chain determined by the sequence of vertices and segments. We denote these sequences L(C)L(C) and P(C)P(C), respectively.
Figure 5.1
The left boundary of C C is L(C)=(p=v 1,e 1,v 2,e 2,v 3,e 3,q=v 4)L(C)=(p=v 1,e 1,v 2,e 2,v 3,e 3,q=v 4), the right boundary is P(C)=(p=v 1,e 7,v 7,e 6,v 6,e 5,v 5,e 4,q=v 4)P(C)=(p=v 1,e 7,v 7,e 6,v 6,e 5,v 5,e 4,q=v 4).
2. The set C C contains the maximum point p p but does not contain the minimum point. In this case the point p p divides the boundary of C C again into two parts, left and right. Each of them is determined by a sequence of vertices and segments terminated by a half-line.
Figure 5.2
The left boundary L(C)=(p=v 1,e 1,v 2,e 2,v 3,e 3)L(C)=(p=v 1,e 1,v 2,e 2,v 3,e 3), the right boundary P(C)=(p=v 1,e 7,v 7,e 6,v 6,e 5,v 5,e 4)P(C)=(p=v 1,e 7,v 7,e 6,v 6,e 5,v 5,e 4).
3. C C contains a minimum point q q but does not have a maximum point. In this case q q divides the boundary again into two parts, left and right. These are sequences, starting with a half-line followed by vertices and segments.
Figure 5.3
L(C)=(e 1,v 2,e 2,v 3,e 3,q=v 4)L(C)=(e 1,v 2,e 2,v 3,e 3,q=v 4), P(C)=(e 7,v 7,e 6,v 6,e 5,v 5,e 4,q=v 4)P(C)=(e 7,v 7,e 6,v 6,e 5,v 5,e 4,q=v 4).
4. The set C C has neither a maximum point nor a minimum point. In this case, there are two possibilities. Either the boundary has left and right parts formed by parallel lines or the boundary has only one part, left or right. This is determined by a single line or a sequence beginning and ending with a half-line with a sequence of vertices and segments between them. We will speak about the other part of the boundary as empty.
Figure 5.4
Left: L(C)=(e 1)L(C)=(e 1), L(C)=(e 2)L(C)=(e 2). Right: L(C)=(e 1,v 2,e 2,v 3,e 3)L(C)=(e 1,v 2,e 2,v 3,e 3), P(C)=()P(C)=().
Algorithm for half-plane intersection
Let C 1 C 1 and C 2 C 2 be convex sets that have been created as intersections of disjoint sets of half-planes. Therefore they are determined by their left and right boundaries. We will describe an algorithm which will create two sequences describing the left and right boundary of the intersection C 1∩C 2 C 1∩C 2 out of the sequences L(C 1)L(C 1), P(C 1)P(C 1), L(C 2)L(C 2), P(C 2)P(C 2).
To do this, it is sufficient to realize that the vertices on the left boundary of C 1∩C 2 C 1∩C 2 are the following:
vertices of L(C 1)L(C 1), which lie inside C 2 C 2,
vertices of L(C 2)L(C 2), which lie inside C 1 C 1,
points of intersection of the left boundaries of C 1 C 1 and C 2 C 2,
points of intersection of the left boundary of C i C i and the right boundary of C j C j for i≠j i≠j. They form in the given lexicographic order a maximal or a minimal vertex in L(C 1∩C 2)L(C 1∩C 2).
Figure 5.5
The vertices on the left boundary of the intersections C 1∩C 2 C 1∩C 2 are p p -- the point of the intersection of the left boundary of C 2 C 2 and the right boundary of C 1 C 1, w 2∈L(C 2)w 2∈L(C 2) lying inside C 1 C 1, q q -- the point of the intersection of the left boundaries, v 3∈L(C 1)v 3∈L(C 1) lying inside C 2 C 2 and r r -- the point of the intersection of the left boundary of C 1 C 1 and the right boundary of C 2 C 2.
Analogously, the points of the right boundary of C 1∩C 2 C 1∩C 2 are:
vertices of P(C 1)P(C 1) lying inside C 2 C 2,
vertices of L(C 2)L(C 2) lying inside C 1 C 1,
points of intersection of the right boundaries of C 1 C 1 and C 2 C 2,
points of intersection of the left boundary of C i C i and the right boundary of C j C j for i≠j i≠j They are maximal or minimal vertices of P(C 1∩C 2)P(C 1∩C 2).
To create lists L(C 1∩C 2)L(C 1∩C 2) and P(C 1∩C 2)P(C 1∩C 2) we have to select the appropriate vertices of C 1 C 1 and C 2 C 2, to calculate the intersections of the boundaries and to organize chosen vertices lexicographically. Our algorithm will be similar to the algorithm for segment intersection in Chapter 1. So we use again sweep line method.
Events will be the vertices of C 1 C 1 and C 2 C 2 and the calculated intersections of boundaries. At the beginning of the algorithm, we set the boundary vertices lexicographically into the queue. Because the vertices are already arranged at both left and right boundaries, the creation of the queue takes time proportional to the number of vertices. Gradually, we will include the calculated intersections of boundaries into the event queue. Since we know on which segments or half-lines these intersections lie, their queuing will take only a constant time.
If neither of the convex sets C 1 C 1 and C 2 C 2 have a maximum point, we calculate the intersections of the half-lines that begin the left and right boundaries of both sets and we will place the obtained points into the queue. Since C 1 C 1 and C 2 C 2 are intersections of disjoint sets of half-planes, the set of the boundary intersections will be finite.
Horizontal sweep line l l moves from top to bottom. It starts at a position above the highest event. For every event we will record on which boundary it lies, and which lines, half-lines or segments pass through it. (For each of the three types of objects we will use a common notation edge.) We will also record the order in which the edges of each boundary cross the sweep line. Since there are no more than four edges crossing the sweep line, there is no need to keep the order in a binary balanced tree. When passing the event v v, we perform the following actions:
We include or do not include the event v v as a vertex in the sequence L(C 1∩C 2)L(C 1∩C 2) or P(C 1∩C 2)P(C 1∩C 2) using criteria given above.
If v v is the first event in the left or right boundary of the intersection, and if there are some half-lines going upwards from it, we decide which one will be in the right and the left boundary of the intersection.
Figure 5.6
The vertex v v is the first event in L(C 1∩C 2)L(C 1∩C 2). The half-line f 1 f 1 will be a part of L(C 1∩C 2)L(C 1∩C 2) above v v, and hence L(C 1∩C 2)=(f 1,v,…)L(C 1∩C 2)=(f 1,v,…).
We decide which of the edges going downwards from the event v v is in the left and which is in the right boundary of the intersection.
Figure 5.7
The edges e e and f f are going down the event v∈L(C 1∩C 2)v∈L(C 1∩C 2). The edge e e will be included in L(C 1∩C 2)L(C 1∩C 2) below the event v v, while f f is not any more a part of the boundary below v v.
We compute the intersection of the left edge e l e l originating from the event v v downwards with the adjacent left edge s l s l of the boundary of the second set, and the intersection of the right edge e p e p originating from the event v v downwards with the adjacent right edge s p s p of the boundary of the second set. If intersections exist and lie below v v, we put them in the queue.
Figure 5.8
The choice of edges e l e l, e p e p, s l s l, s p s p for the sweep line below the event v v. We put the point p=e l∩s l p=e l∩s l in the queue. The intersection e p∩s p e p∩s p is empty.
If the event v v is the minimum vertex of the left and right boundary of the intersection, we will remove all events from the queue as the description of both boundaries of the intersection is already complete. Otherwise, only the event v v is removed from the queue.
The algorithm ends when the queue is empty.
Pseudocodes and their running time
Algorithm 1:HalfplanesIntersection(H H)
Input. A set H={h 1,h 2,…,h n}H={h 1,h 2,…,h n} of n n half-planes in the plane.
Output. Lists L(C)L(C) and P(C)P(C) describing the left and the right boundaries of the intersection C C of all half-planes from H H.
if n=1 n=1 then
determine the left and the right boundaries.
else
put H 1={h 1,h 2,…,h[n/2]}H 1={h 1,h 2,…,h[n/2]}, H 2=H∖H 1 H 2=H∖H 1.
C 1←C 1←HalfplanesIntersection(H 1 H 1).
C 2←C 2←HalfplanesIntersection(H 2 H 2).
C←C←IntersectionOfTwo(C 1,C 2 C 1,C 2).
end
The intersection of two convex sets can be obtained in this way:
Algorithm 2:IntersectionOfTwo(C 1,C 2 C 1,C 2)
Input. Convex sets C 1 C 1 a C 2 C 2 described using the left and right boundaries.
Output. The intersection C=C 1∩C 2 C=C 1∩C 2 described using a list L(C)L(C) for the left boundary and a list P(C)P(C) for the right boundary.
Put L(C)=()L(C)=() a P(C)=()P(C)=().
if C 1 C 1 a C 2 C 2 are half-planes then
compute L(C)L(C) a P(C)P(C).
else
make an event queue out of vertices of C 1 C 1 and C 2 C 2.
end
if C 1 C 1 a C 2 C 2 are not bounded from above then
compute intersections of boundary half-planes or lines which are at the beginnings of boundaries of both sets, and insert them into the queue.
end
while the queue of events is not empty do
take its first vertex v v
HandleEvent(C 1,C 2,v C 1,C 2,v)
end
return L(C 1∩C 2)L(C 1∩C 2) a P(C 1∩C 2)P(C 1∩C 2).
The passage of the sweep line through an event is recorded in the pseudocode:
Algorithm 3:HandleEvent(C 1,C 2,v C 1,C 2,v)
Input. A vertex v v on the boundary of C 1 C 1 or C 2 C 2 and lists L(C 1∩C 2)L(C 1∩C 2) and P(C 1∩C 2)P(C 1∩C 2) for the left and right boundaries of the intersection above v v.
Output. Updated lists L(C 1∩C 2)L(C 1∩C 2) and P(C 1∩C 2)P(C 1∩C 2), updated queue Q Q.
if v∈L(C i)v∈L(C i) lies between the left and right boundaries of C j C j, i≠j i≠j then
insert v v into L(C 1∩C 2)L(C 1∩C 2).
end
if v∈P(C i)v∈P(C i) lies between the left and right boundaries of C j C j, i≠j i≠j then
insert v v into P(C 1∩C 2)P(C 1∩C 2).
end
if v v lies in the intersection of the left boundaries then
insert v v into L(C 1∩C 2)L(C 1∩C 2).
end
if v v lies in the intersection of the right boundaries then
insert v v into P(C 1∩C 2)P(C 1∩C 2).
end
if v v lies in the intersection of the left boundary of C i C i and the right boundary of C j C j, i≠j i≠j then
insert v v into L(C 1∩C 2)L(C 1∩C 2) and also into P(C 1∩C 2)P(C 1∩C 2).
end
if v v lies in L(C 1∩C 2)L(C 1∩C 2) or in P(C 1∩C 2)P(C 1∩C 2)then
if v v is the first vertex in L(C 1∩C 2)L(C 1∩C 2) or in P(C 1∩C 2)P(C 1∩C 2)then
find which halflines with lower vertex v v belong to L(C 1∩C 2)L(C 1∩C 2) or P(C 1∩C 2)P(C 1∩C 2).
end
Find which edges with the upper vertex v v belong to L(C 1∩C 2)L(C 1∩C 2) or P(C 1∩C 2)P(C 1∩C 2).
end
From the edges which go downwards from the vertex v v denote e l e l the mostleft one and e p e p the mostright one. To e l e l find a left adjacent edge s l s l from the boundary of the other convex set. To e p e p find a right adjacent edge s p s p from the boundary of the other set.
Compute intersections s l∩e l s l∩e l and s p∩e p s p∩e p below v v and insert them into the queue.
if v v is the last member of the sequence L(C 1∩C 2)L(C 1∩C 2) or/and P(C 1∩C 2)P(C 1∩C 2) (i.e. it is not followed by an edge) then
remove all events from the queue
else
remove v v from the queue
end
return L(C 1∩C 2)L(C 1∩C 2) and P(C 1∩C 2)P(C 1∩C 2).
The running time of the algorithm from the last pseudocode is constant. Let the convex sets C 1 C 1 and C 2 C 2 have n 1 n 1 and n 2 n 2 vertices, respectively. Then the running time of the algorithm from the pseudocode IntersectionOfTwo(C 1,C 2 C 1,C 2) is O(n 1+n 2)O(n 1+n 2).
The running time T(n)T(n) of the whole algorithm for the computation of the intersection of n n half-planes is given by the recurrent relation
T(n)=2 T(n 2)+O(n).T(n)=2 T(n 2)+O(n).
This leads to the resulting running time
T(n)=O(n log n).T(n)=O(n logn).
Animation
An example of the computation of the intersection of two convex sets C 1 C 1 and C 2 C 2 according to the algorithms described by the second and the third pseudocodes is captured in the following animation.
The algorithm creates the queue Q=(w 2,w 3,w 5,v 2,v 4,v 3,w 4)Q=(w 2,w 3,w 5,v 2,v 4,v 3,w 4). There are no intersections of half-lines e 1 e 1, e 4 e 4, f 1 f 1, f 5 f 5 above w 2 w 2. The queue remains unchanged. The position of sweep line is above w 2 w 2.
doc. RNDr. Martin Čadek, CSc.
Ústav matematiky a statistiky – Přírodovědecká fakulta Masarykovy univerzity Masarykovy univerzity
Technical cooperation:
Service Center for E-learning, Faculty of Informatics, Masaryk University, 2018
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Molecules
. 2020 Feb 7;25(3):711. doi: 10.3390/molecules25030711
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How do the Hückel and Baird Rules Fade away in Annulenes?
Irene Casademont-Reig
Irene Casademont-Reig
1 Donostia International Physics Center (DIPC), 20018 Donostia, Euskadi, Spain; irenecasre@gmail.com (I.C.-R.); eloy.raco@gmail.com (E.R.-C.); miqueltorrentsucarrat@gmail.com (M.T.-S.)
2 Kimika Fakultatea, Euskal Herriko Unibertsitatea (UPV/EHU), 20080 Donostia, Euskadi, Spain
Find articles by Irene Casademont-Reig
1,2, Eloy Ramos-Cordoba
Eloy Ramos-Cordoba
1 Donostia International Physics Center (DIPC), 20018 Donostia, Euskadi, Spain; irenecasre@gmail.com (I.C.-R.); eloy.raco@gmail.com (E.R.-C.); miqueltorrentsucarrat@gmail.com (M.T.-S.)
2 Kimika Fakultatea, Euskal Herriko Unibertsitatea (UPV/EHU), 20080 Donostia, Euskadi, Spain
Find articles by Eloy Ramos-Cordoba
1,2, Miquel Torrent-Sucarrat
Miquel Torrent-Sucarrat
1 Donostia International Physics Center (DIPC), 20018 Donostia, Euskadi, Spain; irenecasre@gmail.com (I.C.-R.); eloy.raco@gmail.com (E.R.-C.); miqueltorrentsucarrat@gmail.com (M.T.-S.)
2 Kimika Fakultatea, Euskal Herriko Unibertsitatea (UPV/EHU), 20080 Donostia, Euskadi, Spain
3 IKERBASQUE, Basque Foundation for Science, 48013 Bilbao, Euskadi, Spain
Find articles by Miquel Torrent-Sucarrat
1,2,3, Eduard Matito
Eduard Matito
1 Donostia International Physics Center (DIPC), 20018 Donostia, Euskadi, Spain; irenecasre@gmail.com (I.C.-R.); eloy.raco@gmail.com (E.R.-C.); miqueltorrentsucarrat@gmail.com (M.T.-S.)
3 IKERBASQUE, Basque Foundation for Science, 48013 Bilbao, Euskadi, Spain
Find articles by Eduard Matito
1,3,
Editors: Diego Andrada, Israel Fernández
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1 Donostia International Physics Center (DIPC), 20018 Donostia, Euskadi, Spain; irenecasre@gmail.com (I.C.-R.); eloy.raco@gmail.com (E.R.-C.); miqueltorrentsucarrat@gmail.com (M.T.-S.)
2 Kimika Fakultatea, Euskal Herriko Unibertsitatea (UPV/EHU), 20080 Donostia, Euskadi, Spain
3 IKERBASQUE, Basque Foundation for Science, 48013 Bilbao, Euskadi, Spain
Correspondence: ematito@gmail.com; Tel.: +34-943018513
Roles
Diego Andrada: Academic Editor
Israel Fernández: Academic Editor
Received 2019 Dec 20; Accepted 2020 Jan 29; Collection date 2020 Feb.
© 2020 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC7037833 PMID: 32045990
Abstract
Two of the most popular rules to characterize the aromaticity of molecules are those due to Hückel and Baird, which govern the aromaticity of singlet and triplet states. In this work, we study how these rules fade away as the ring structure increases and an optimal overlap between p orbitals is no longer possible due to geometrical restrictions. To this end, we study the lowest-lying singlet and triplet states of neutral annulenes with an even number of carbon atoms between four and eighteen. First of all, we analyze these rules from the Hückel molecular orbital method and, afterwards, we perform a geometry optimization of the annulenes with several density functional approximations in order to analyze the effect that the distortions from planarity produce on the aromaticity of annulenes. Finally, we analyze the performance of three density functional approximations that employ different percentages of Hartree-Fock exchange (B3LYP, CAM-B3LYP and M06-2X) and Hartree-Fock. Our results reveal that functionals with a low percentage of Hartree-Fock exchange at long ranges suffer from severe delocalization errors that result in wrong geometrical structures and the overestimation of the aromatic character of annulenes.
Keywords: annulenes, aromaticity, antiaromaticity, Hückel rule, Baird rule, density functional theory, delocalization error
1. Introduction
Aromaticity is one of the most important concepts in chemistry, and it is associated with cyclic electron delocalization (or conjugation) in closed circuits giving rise to bond-length equalization, energy stabilization, large magnetic anisotropies and abnormal chemical shifts, among other well-known effects [1,2,3]. Despite the multidimensional character of aromaticity and the recent proliferation of aromatic compounds that extend beyond the organic chemistry realm [4,5], several simple but predictive models have been designed to characterize aromaticity [6,7,8,9,10]. From these models, rules of aromaticity have been obtained that are commonly used in assessing the aromatic character of molecules. Among these rules, the Hückel rule was the first and remains the most widely employed [12,13,14,15].
According to the Hückel rule, annulenes with electrons (n being an integer number) are more stable than the corresponding open-chain polyenes and, therefore, considered aromatic. Although this prediction is met for the first elements of the annulenes series, the rule is soon broken due to out-of-plane geometrical distortions that are more energetically favorable but disrupt the optimal overlaps between p orbitals that give rise to conjugated circuits . These geometrical distortions cannot be considered by the Hückel rule which, inevitably, overestimates the aromaticity of annulenes that are described as planar structures having bond equalization. Conversely, the Hückel rule also predicts that electrons annulenes are unstable with respect to their open-chain analogues, exhibit bond length alternation, and are considered antiaromatic. The application of this rule has not been limited to annulenes and it is currently used in all sorts of molecules, including non-organic molecules [4,5].
Another important rule of aromaticity, complementary to Hückel’s, is the Baird rule . While the Hückel rule is limited to singlet ground states, the Baird rule applies to the lowest-lying triplet and excited singlet states and predicts () -electron molecules to be aromatic (antiaromatic). Aihara calculated the resonance energies of singlet and triplet annulenes, concluding that Baird’s rule was satisfied, i.e., that the lowest-lying excited state of annulenes presented the opposite aromatic character to that in the ground state . Baird’s rule has been called the photochemistry analogue of Hückel’s rule because it has been used to tailor molecules with enhanced photochemical activity [20,21,22].
The most (anti)aromatic molecules do not usually have many atoms in the ring. The optimal overlap between orbitals that favors electron conjugation calls for a particular arrangement of the atoms in the ring that cannot be sustained as the ring size increases. Therefore, both aromaticity and antiaromaticity are expected to decrease with the ring size. Although there is evidence of large aromatic nanorings , studies on annulenes have led to the conclusion that systems with more than 30 electrons are nonaromatic .
In his seminal paper, Baird studied the velocity at which the antiaromaticity (measured through Dewar resonance energy, DRE) diminished with increasing ring size . Based on a few examples, he found that triplets lost their antiaromaticity with the ring size as fast as singlets . To the best of our knowledge, thus far, there have been very few attempts to actually quantify how the aromaticity rules fade away with the ring size and they were limited to annulenes [25,26]. This is the goal of this paper. We will first perform an analysis of how these rules are expected to change with the ring size as predicted by the model from which they originated. Second, we will study how the geometry distorts these molecules from the optimal (planar) geometry assumed in the model and how it affects the aromaticity. Finally, we will analyze the effect of the computational method in the description of these systems by taking into account several density functional approximations (DFAs) with a varying percentage of Hartree Fock (HF) exchange.
2. Methodology
2.1. Aromaticity Indices
In this section, we will review the expressions of several aromaticity indices based on electron delocalization . In the following, we will assume a molecule having at least one ring structure, which consists of N atoms represented by the string = {,,...,}, whose elements are ordered according to the connectivity of the atoms in the ring. For convenience, we adopt and .
2.1.1. The Aromatic Fluctuation Index: FLU
The FLU index measures aromaticity by comparison with the cyclic electron delocalization of some reference aromatic molecules . Its expression depends on the delocalization index (DI) [28,29,30,31], , which measures the electron sharing between atoms A and B,
(1)
where is the atomic valence for a closed-shell system , and is a simple function to ensure that the first term in Equation (1) is always greater or equal to 1,
(2)
is the DI corresponding to an aromatic molecule which has the pattern of bonding . For example, the aromatic reference for C-C bonds is benzene. FLU is close to zero for aromatic species, and greater than zero for non-aromatic or antiaromatic species.
Aromaticity indices based on references, such as FLU or HOMA [33,34], do not measure aromaticity but the similarity with respect to some aromatic molecule. Therefore, they are not adequate to describe reactivity [35,36].
2.1.2. The Bond-Length and Bond-Order Alternation Indices
Two popular indicators of aromaticity are the bond-length (BLA) and the bond-order alternation (BOA), which compare the average of bond lengths and bond orders, respectively, of consecutive bonds in a ring:
(3)
where and , being the floor function of y, that returns the largest integer less than or equal to y, and x is either the bond length or the bond order. Unlike the atomic charges , the bond orders are much less dependent on the computational method and the basis set used and, therefore, aromaticity indices based on bond orders are not highly dependent on the level of theory employed. An exception to this rule is the delocalization error that is present in some DFAs with a low percentage of Hartree-Fock exchange, which causes an overestimation of the aromaticity of some compounds (see Section 3.4).
The definition of Equation (3) presents a serious drawback: it is not well defined for a ring of an odd number of members because its value depends on the order of the atoms in the ring. For the sake of generality, in this paper we adopt an alternative definition of BLA/BOA:
(4)
and we choose the delocalization index [28,29,30,31] as a measure of bond order.
2.1.3. A Many-Center Electron Delocalization Index:
Giambiagi and coworkers suggested to use the multicenter index, which was previously defined by them to account for the simultaneous electron sharing among various centers , as a measure of aromaticity . This index was named and its formulation for single-determinant wavefunctions reads as follows:
(5)
where is the overlap of molecular orbitals i and j in the atom A, i.e.,
(6)
will provide large values for aromatic molecules. Although it will not be considered in this paper, it is worth to mention that Bultinck and co-workers generalized considering also the delocalization of a non-Kekule arrangement of the atoms in the ring; the index is known as MCI . Some of us have shown that both and MCI are ring-size dependent and, therefore, for convenience, we will calculate the multicenter electron delocalization per atom that can be obtained as .
2.1.4. AV1245 and
Both and MCI are among the less fallible aromaticity indices available in the literature [2,36,45] and, therefore, they have been used in a plethora of cases involving a difficult assessment of aromaticity [46,47,48,49,50,51,52,53,54]. However, these indices present some drawbacks that prevent their use in large rings and, therefore, we have recently designed and tested [40,56] a new electronic aromaticity index, AV1245, based on MCI but free of the shortcomings of this index. AV1245 is defined as the average value of the four-atom MCI index between relative positions 1–2 and 4–5 constructed from each five-atom fragment (five consecutive atoms) along the perimeter of the ring . The latter gives an average picture of the electron delocalization among the atoms in the ring. However, for the purpose of measuring aromaticity, it has been found more adequate to use the minimum absolute MCI value evaluated along the perimeter, . identifies the weakest link, i.e., the fragment with the lowest electron delocalization which is usually responsible for the loss of aromaticity in the ring. A detailed study of all the MCI values along the perimeter has been recently shown to be useful in identifying electron delocalization patterns and it will be the topic of discussion in a forthcoming work. Aromatic molecules are thus identified by large values of () and non-aromatic molecules exhibit very low values. Antiaromatic molecules usually exhibit intermediate values and are more difficult to identify. To this end, the analysis is complemented by the examination of either four-atom MCI profiles along the perimeter of the ring or the BOA, which help differentiate between aromatic and antiaromatic compounds.
2.2. Hückel Molecular Orbital Method
Despite the drastic approximations inherent in the Hückel Molecular Orbital (HMO) approach [12,13,14,15], organic aromatic molecules are usually well described within the HMO method. It is thus usual to learn the HMO method at the same time as Hückel’s rule and other aromaticity measures given by the HMO method, such as the resonance energy (RE), the RE per electron (REPE) or the topological REPE (TREPE) . Some recent works have studied Hückel’s rule from the perspective of electron delocalization [58,59]. Since the studies of Hückel on organic molecules, the concept of aromaticity has extended importantly including all sorts of new aromatic molecules such as metalloaromatic molecules [4,51,52,60,61], fullerens , nanotubes , porphyrins [64,65] with a Möbius-like structure [66,67,68,69,70,71,72,73], and all-metal clusters [4,5], among others.
For a cyclic polyene of n carbon atoms and N electrons, the Hückel molecular orbital (HMO) method provides a general formulation for its orbitals:
(7)
where is the atomic orbital 2 of the carbon atom and . Excepting for the first () and the last (), these orbitals are complex and degenerate by pairs (). From these equations, one can easily calculate the atomic overlap matrices (AOMs), the bond orders, as well as many aromaticity indices (see Section 2.1), obtaining analytical expressions .
The typical way to assess the aromaticity within the framework of HMO theory is through the study of stabilization or resonance energy. Since the computational chemistry study of resonance energies puts severe limitations to analyze more complicated molecules, especially those containing atoms other than carbon, we prefer to study electron delocalization to assess the aromatic character of compounds [2,55]. In particular, we focus here on the study of which, in the case of HMO theory, takes a very simple form (compare to Equation (5)),
(8)
where is the bond-order between atoms A and B and it is related to the DI: .
In order to compare annulenes of different sizes among them, we will study the normalized quantity . The calculation of the bond orders takes a very simple form for singlet annulenes with electrons . The study of other multiplicities or number of electrons is less evident. The HMO theory does not distinguish between alpha and beta electrons (beyond the fact of permitting only one electron of each kind to populate each orbital) and, therefore, the study of triplets is not entirely satisfactory. However, to a reasonable extent, one can analyze the triplets as cases where one electron is promoted from a l orbital to a orbital. If the orbitals obtained from HMO are those of Equation (7), the results are the same regardless the electrons are promoted from l or orbitals (or whether they are promoted to or orbitals). Hence, the study of triplets can be done without ambiguity. However, the study of neutral singlets, which are expected to exhibit localized structures of symmetry , cannot be conducted because both l and orbitals contribute the same amount to all the bond orders in the perimeter of the ring. In order to enforce the appearance of mesomeric structures, we should obtain a different set of orbitals from the HMO theory. Since degenerate orbitals can be freely combined without altering the energy of the system, it is convenient to recombine each pair of degenerate orbitals among themselves to produce this set of orbitals:
(9)
for . Unlike those in Equation (7), these orbitals are real and produce one of the two mesomeric structures that one can expect in neutral annulenes, depending on whether the last two electrons occupy a l or orbital. Finally, we find that the study of neutral triplets within the HMO theory is far from obvious because both the singlet-open and the triplet states would be actually described by the same orbital occupancies. For the sake of completeness, we have also included this case in our study. To this end, we have employed the original set of orbitals (Equation (7)) and these results are labeled as . (We could have also chosen the second set of orbitals that lead to mesomeric structures. In such case, we would have obtained different values but the same qualitative trend (see Section 3.1)).
3. Results
3.1. Aromaticity from the HMO Method
First of all, we will study Hückel’s and Baird’s rules from the simple HMO theory [12,14]. Although we cannot expect this theory to provide a reliable description of annulenes (especially the large ones), the results we obtain will provide a high bound to the aromaticity/antiaromaticity expected in these species.
All the values of (Equation (8)) are collected in Figure 1. Interestingly, all the curves conform to the same general expression: , where N is the number of ring members. The values of a do not differ significantly among the four cases giving values very close to the theoretical limit for annulenes, , whereas the values of b present large differences among them: −2.1867, 1.0822, −5.1151, and −5.5910 for , singlet, triplet, and singlet structures, respectively. These results are in agreement with the chemical intuition: (i) for very large annulenes both Hückel and Baird rules break and all the species become equally aromatic, () the initial (anti)aromatic character decreases smoothly with the annulene size. We can also see that singlets, which are aromatic, display values above the limit, whereas triplets and singlets, which are expected to be antiaromatic, present values below this limit. On the other hand, annulenes also behave like antiaromatic molecules, which does not conform with the aromaticity expected in triplets and, therefore, these species are not well described from the delocalization measures we can obtain from the HMO theory. Finally, we can compare the velocity with which the (anti)aromaticity decreases according to Hückel’s and Baird’s rule. The values of b show that the velocity at which the aromaticity decreases with the ring size is about five times smaller for singlet annulenes than the corresponding decrease of antiaromaticity for triplets. Interestingly, the latter is very close to the decrease of antiaromaticity found in singlets, in agreement with the study of Baird on cyclic hydrocarbons using DRE . We have also studied the HMO method forcing bond-length alternation by employing two different resonance integrals for each bond type (single and double) [75,76], finding that the antiaromaticity of singlets decreases more rapidly when the annulenes are forced to exhibit bond alternation. Since we cannot compare singlets and triplets, we cannot extract any relevant conclusion about which rule, Hückel’s or Baird’s, is broken more quickly with the ring size. We can, however, conclude from this data that molecules are more resilient to the loss of aromaticity than to the loss of antiaromaticity, as one would expect from purely energetic grounds.
Figure 1.
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Values of for the annulenes series against the number of C atoms (N) for different singlets and triplets. The species have been divided according to the number of electrons (4 n and 4 n + 2) and the spin multiplicity (singlet and triplet).
3.2. Geometrical Relaxation
Thus far, we have studied ideally planar annulenes within the HMO method. In this section, we employ quantum chemistry methods to study the geometry of annulenes, which often do not attain a planar conformation in their ground state configuration (see Figure 2). We will employ HF and three DFAs: B3LYP, M06-2X and CAM-B3LYP. These results will be compared against the benchmark data available in the literature. We will split the results into two blocks: and -electron annulenes.
Figure 2.
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Geometrical structures of the studied annulenes.
3.2.1. Annulenes
First of all, we have studied the lowest-lying structures of singlet and triplet benzene. The ground state of benzene has been largely studied and its characterization does not present a challenge for DFAs. The molecule is identified by all the aromaticity measures as the most aromatic molecule among the studied annulenes. Conversely, the triplet state of benzene has a more elusive character, presenting two conformations which are very close in energy, the quinoidal (Q) and antiquinodial (AQ) one . The former is characterized by two clear double bounds separated by two radical C atoms, whereas the AQ shows two allylic structures merged by two single C-C bonds. The energy difference between the two conformations is below 1 kcal/mol and they are connected by a transition state with an energy barrier of 2.5 kcal/mol . Both conformers are expected to be antiaromatic, displaying an energy destabilization that is responsible for their prominent photochemical reactivity . The lowest-lying triplet state of benzene has been identified by HF, B3LYP, and CAM-B3LYP as AQ-like, whereas the Q structure could not be located with these methods. Conversely, M06-2X provides a Q-like structure. The aromaticity indices display values which are much smaller than those of benzene and the BOA index reveals a bond-order alternation that is characteristic of antiaromatic molecules. values are small, which in conjunction with the oscillating pattern displayed by the dissected index profile (see Supplementary Material), agrees with the antiaromatic character anticipated in this species.
Singlet annulene presents several low-energy isomers, which are difficult to sort energetically from the results of computational calculations. Most ab initio methods find the twist conformation to be the lowest in energy, the exception being B3LYP, which predicts the heart conformation to be the ground state . Our calculations agree with these results. The most delocalized structure, corresponding to a symmetry, lies more than 30kcal/mol above the ground state geometry and, hence, the isomer expected to be the most aromatic it is actually not stable. On the contrary, the twist isomer (which does not exhibit a planar conformation) presents a very modest aromatic/antiaromatic character according to all the aromaticity indices. In fact, the molecule could be easily classified as nonaromatic from the value, which is very close to zero and, as the dissected index profile shows (see Supplementary Material), the twisted bonds are those that prevent an optimal overlap of p orbitals. Conversely, the heart configuration (lying 0.6 and 2.3 kcal/mol above the ground state according to CAM-B3LYP and M06-2X) is an aromatic molecule as stated by all aromaticity indices. To the best of our knowledge, triplet annulene has been much less studied and there are only examples of triplet dicationic annulene structures on the literature [20,21]. The lowest-lying triplet state of annulene displays naphthalene (B3LYP, CAM-B3LYP and M06-2X) and twist (HF) configurations. Following the Baird rule, this molecule should be antiaromatic. The naphthalene conformation is indeed mildly antiaromatic but the twist conformation is rather nonaromatic (see Table 1).
Table 1.
Aromaticity indices for annulenes calculated with HF, B3LYP, CAM-B3LYP, and M06-2X and the 6-311G(d,p) basis set. values were too small for an accurate calculation of 1/N.
| Structure | Multiplicity | Functional | FLU | 1/N | BOA | BLA | 1/N | || |
:---: :---: :---: :---:
| C H | S | HF | 0.000 | 0.624 | 0.000 | 0.000 | 0.597 | 10.25 |
| B3LYP | 0.000 | 0.625 | 0.000 | 0.000 | 0.603 | 10.72 |
| CAM-B3LYP | 0.000 | 0.628 | 0.000 | 0.000 | 0.603 | 10.71 |
| M06-2X | 0.000 | 0.626 | 0.000 | 0.000 | 0.603 | 10.73 |
| C H | T | HF | 0.024 | 0.393 | 0.246 | 0.089 | 0.341 | 0.39 |
| B3LYP | 0.025 | 0.399 | 0.275 | 0.090 | 0.353 | 1.51 |
| CAM-B3LYP | 0.025 | 0.408 | 0.276 | 0.091 | 0.363 | 1.20 |
| M06-2X | 0.041 | 0.467 | 0.281 | 0.056 | 0.380 | 0.28 |
| C H (twist) | S | HF | 0.068 | 0.421 | 0.728 | 0.157 | 0.339 | 0.00 |
| B3LYP | 0.052 | 0.480 | 0.639 | 0.128 | 0.322 | 0.03 |
| CAM-B3LYP | 0.059 | 0.466 | 0.677 | 0.137 | 0.341 | 0.03 |
| M06-2X | 0.058 | 0.463 | 0.670 | 0.136 | 0.325 | 0.05 |
| C H (heart) | S | HF | 0.065 | 0.436 | 0.712 | 0.153 | 0.380 | 0.01 |
| B3LYP | 0.000 | 0.610 | 0.007 | 0.009 | 0.579 | 5.19 |
| CAM-B3LYP | 0.000 | 0.614 | 0.009 | 0.010 | 0.579 | 5.13 |
| M06-2X | 0.000 | 0.611 | 0.010 | 0.010 | 0.579 | 5.11 |
| C H (naphthalene) | T | HF | 0.030 | 0.470 | 0.364 | 0.085 | 0.353 | 0.14 |
| B3LYP | 0.023 | 0.531 | 0.328 | 0.064 | 0.460 | 0.91 |
| CAM-B3LYP | 0.027 | 0.524 | 0.364 | 0.072 | 0.446 | 0.68 |
| M06-2X | 0.028 | 0.522 | 0.367 | 0.073 | 0.444 | 0.63 |
| C H (twist) | T | HF | 0.020 | 0.478 | 0.266 | 0.068 | 0.295 | 0.05 |
| B3LYP | 0.020 | 0.517 | 0.305 | 0.066 | 0.352 | 0.08 |
| CAM-B3LYP | 0.022 | 0.513 | 0.324 | 0.071 | 0.347 | 0.07 |
| M06-2X | 0.022 | 0.511 | 0.328 | 0.072 | 0.337 | 0.10 |
| C H | S | HF | 0.050 | 0.497 | 0.626 | 0.136 0.49 |
| B3LYP | 0.001 | 0.605 | 0.010 | 0.008 4.24 |
| CAM-B3LYP | 0.026 | 0.561 | 0.449 | 0.091 1.80 |
| M06-2X | 0.025 | 0.561 | 0.438 | 0.088 1.89 |
| C H (TS) | S | CAM-B3LYP | 0.000 | 0.609 | 0.007 | 0.007 4.29 |
| M06-2X | 0.001 | 0.606 | 0.007 | 0.007 4.27 |
| C H | T | HF | 0.021 | 0.502 | 0.282 | 0.069 0.01 |
| B3LYP | 0.017 | 0.554 | 0.302 | 0.060 0.13 |
| CAM-B3LYP | 0.023 | 0.544 | 0.348 | 0.071 0.08 |
| M06-2X | 0.022 | 0.544 | 0.349 | 0.071 0.04 |
| C H | S | HF | 0.049 | 0.504 | 0.616 | 0.133 | 0.472 | 0.57 |
| B3LYP | 0.001 | 0.606 | 0.026 | 0.011 | 0.573 | 4.27 |
| CAM-B3LYP | 0.026 | 0.563 | 0.446 | 0.090 | 0.530 | 1.80 |
| M06-2X | 0.025 | 0.561 | 0.444 | 0.089 | 0.529 | 1.81 |
| C H (TS) | S | CAM-B3LYP | 0.001 | 0.609 | 0.022 | 0.010 | 0.572 | 4.29 |
| M06-2X | 0.001 | 0.607 | 0.022 | 0.010 | 0.572 | 4.28 |
| C H | T | HF | 0.018 | 0.514 | 0.257 | 0.060 | 0.411 | 0.12 |
| B3LYP | 0.013 | 0.570 | 0.265 | 0.053 | 0.533 | 0.58 |
| CAM-B3LYP | 0.019 | 0.559 | 0.324 | 0.066 | 0.513 | 0.30 |
| M06-2X | 0.019 | 0.559 | 0.324 | 0.065 | 0.517 | 0.46 |
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The ground-state annulene geometry corresponds to a distorted pyrene perimeter that can undergo a facile isomerization reaction through a Möbius antiraromatic transition state . At the B3LYP level of theory, we locate a minimal structure with symmetry that is aromatic according to all criteria of aromaticity. However, the CAM-B3LYP and M06-2X configurations correspond to a transition state (TS) that through a small energy barrier (ca. 2.5 kcal/mol) connects two -symmetry antiaromatic energy minima with complementary bond-length alternation. The triplet conformer leads to an energy minimum that is only mildly antiaromatic at all the levels of theory.
The Hückel rule predicts that annulene is an aromatic molecule but the situation is similar to annulene. B3LYP finds the molecule to be a highly symmetric () structure that is also aromatic from all aromaticity indices. However, both CAM-B3LYP and M06-2X predict that this structure is actually a transition state that connects two identical structures that exhibit bond-order alternation and, therefore, are antiaromatic. This discrepancy has been attributed to the delocalization error of DFAs with a low percentage of Hartree-Fock exchange, a fact that has been confirmed through the comparison of experimental proton chemical shifts . A detailed study of the delocalization error is done in Section 3.4 of this manuscript. The lowest-lying triplet displays a symmetry (HF) or a symmetry (CAM-B3LYP, B3LYP, and M06-2X). DFAs agree on the triplet annulene being weakly antiaromatic, whereas HF predicts it to be rather nonaromatic.
3.2.2. Annulenes
The photochemical formation of annulenes is very important in excited-state aromaticity . Cyclobutadiene is often used as the paradigmatic example of an antiaromatic molecule following the rule. Although (anti)aromaticity can be easily overestimated with a single-determinant wavefunction [82,83,84], cyclobutadiene is the molecule with the largest bond-length and bond-order alternation and all indicators of aromaticity clearly confirm its antiaromatic character. Baird was first to suggest that triplet -electron annulenes should be regarded as aromatic and confirm it through DRE calculations . In particular, the DRE confirmed that triplet cyclobutadiene is an aromatic molecule. A fact that is further substantiated by its symmetric geometry and the values of the aromaticity indices gathered in Table 2.
Table 2.
Aromaticity indices for the studied annulenes calculated with HF, B3LYP, CAM-B3LYP, and M06-2X and the 6-311G(d,p) basis set. values were too small for an accurate calculation of 1/N. cannot be calculated for rings with less than six members.
| Structure | Multiplicity | Method | FLU | 1/N | BOA | BLA | 1/N | || |
:---: :---: :---: :---:
| C H | S | HF | 0.101 | 0.391 | 0.888 | 0.249 | 0.262
| B3LYP | 0.104 | 0.416 | 0.900 | 0.247 | 0.266
| CAM-B3LYP | 0.103 | 0.398 | 0.898 | 0.245 | 0.264
| M06-2X | 0.103 | 0.405 | 0.898 | 0.242 | 0.268
| C H | T | HF | 0.010 | 0.507 | 0.000 | 0.000 | 0.433
| B3LYP | 0.012 | 0.499 | 0.000 | 0.000 | 0.440
| CAM-B3LYP | 0.011 | 0.504 | 0.000 | 0.000 | 0.439
| M06-2X | 0.011 | 0.505 | 0.000 | 0.000 | 0.438
| C H | S | HF | 0.067 | 0.436 | 0.726 | 0.156 | 0.406 | 0.30 |
| B3LYP | 0.056 | 0.477 | 0.664 | 0.134 | 0.441 | 0.72 |
| CAM-B3LYP | 0.061 | 0.468 | 0.693 | 0.140 | 0.428 | 0.52 |
| M06-2X | 0.062 | 0.460 | 0.694 | 0.141 | 0.427 | 0.51 |
| C H | T | HF | 0.001 | 0.590 | 0.000 | 0.000 | 0.534 | 4.07 |
| B3LYP | 0.001 | 0.589 | 0.000 | 0.000 | 0.540 | 4.31 |
| CAM-B3LYP | 0.001 | 0.593 | 0.000 | 0.000 | 0.539 | 4.29 |
| M06-2X | 0.001 | 0.591 | 0.000 | 0.000 | 0.539 | 4.29 |
| C H | S | HF | 0.063 | 0.445 | 0.698 | 0.153 0.04 |
| B3LYP | 0.042 | 0.511 | 0.565 | 0.115 0.01 |
| CAM-B3LYP | 0.050 | 0.494 | 0.624 | 0.128 0.02 |
| M06-2X | 0.050 | 0.488 | 0.624 | 0.128 0.06 |
| C H | T | HF | 0.021 | 0.487 | 0.280 | 0.067 0.07 |
| B3LYP | 0.002 | 0.590 | 0.033 | 0.012 0.07 |
| CAM-B3LYP | 0.015 | 0.562 | 0.288 | 0.056 0.21 |
| M06-2X | 0.008 | 0.577 | 0.208 | 0.039 0.13 |
| C H (S) | S | HF | 0.054 | 0.486 | 0.651 | 0.139 | 0.440 | 0.33 |
| B3LYP | 0.029 | 0.551 | 0.476 | 0.095 | 0.513 | 0.96 |
| CAM-B3LYP | 0.041 | 0.529 | 0.564 | 0.113 | 0.484 | 0.64 |
| M06-2X | 0.040 | 0.526 | 0.562 | 0.113 | 0.484 | 0.63 |
| C H (C) | S | HF | 0.053 | 0.488 | 0.643 | 0.139 | 0.452 | 0.25 |
| B3LYP | 0.029 | 0.548 | 0.474 | 0.096 | 0.512 | 0.78 |
| CAM-B3LYP | 0.040 | 0.530 | 0.555 | 0.113 | 0.487 | 0.55 |
| M06-2X | 0.039 | 0.526 | 0.553 | 0.112 | 0.487 | 0.58 |
| C H (C) | T | HF | 0.001 | 0.598 | 0.013 | 0.005 1.08 |
| B3LYP | 0.002 | 0.596 | 0.057 | 0.013 1.16 |
| CAM-B3LYP | 0.015 | 0.568 | 0.294 | 0.059 0.35 |
| M06-2X | 0.011 | 0.576 | 0.252 | 0.050 0.74 |
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Unlike cyclobutadiene, cycloocta-1,3,5,7-tetraene (COT) is not a planar molecule in its ground state. COT shows a boat-like geometry that, although presents bond-length and bond-order alternation, is not a very antiaromatic molecule [16,85]. The aromaticity indices reveal that this molecule could be classified as mildly antiaromatic. Conversely, the lowest-lying triplet state of COT presents symmetry and values of the aromaticity indices that confirm the aromatic character anticipated by the Baird rule. Interestingly, the realization of planar triplet COT in some substituted annulenes has been studied as an acceleration path for the photochemical inversion of the ring .
In contrast to smaller annulenes, annulene presents several energy minima in the potential energy surface, five of which lie within 5 kcal/mol according to CCSD(T) calculations . The instability and the easy isomerization of this species explain the difficulty in characterizing it experimentally. All DFAs and HF predict a CTCTCT () geometry with a large bond-order alternation, where C and T stand for the arrangement of C-C bonds that can be either cis (C) or trans (T). Nevertheless, the molecule presents very small values that indicate a nonaromatic character. The lowest-lying triplet state of annulene shows a very similar geometry in agreement with a CTCTCT arrangement of the atoms but corresponding to a CTCTCT () geometry, according to all DFAs. The HF lowest-lying triplet also presents a CTCTCT geometry () but it is severely distorted with respect to the latter one. Regardless of the method used, value is small, prompting us to also classify the triplet state as a nonaromatic species.
As in the previous case, a large number of annulene isomers have been found and, therefore, this species also easily undergoes isomerization even through quantum mechanical tunneling . According to all the methods, the ground state structure can be classified as CTCTCTCT (). Interestingly, there is another structure, which presents a distorted CTCTTCTT structure () lying 5-8 kcal/mol above the ground state structure. The lowest-lying triplet presents a conformation very close to the latter one. Although the singlet is less antiaromatic than COT or cyclobutadiene, we find that both and present a similar character and are more antiaromatic than annulene. Therefore, annulene could be classified as weakly antiaromatic. Likewise, the triplet provides values of the indices that indicate a more aromatic character than annulene triplet. Although values oscillate between 0.35 and 1.16, these values indicate a mild aromatic character in between triplet annulene and triplet cyclobutadiene, which agrees with the fact that this molecule displays a more symmetric structure () than its singlet counterpart ().
3.3. Aromaticity from DFAs
The HMO cannot take into account the geometrical relaxation from a planar structure (permitting an optimal overlap of p orbitals) and, therefore, the results obtained in the Section 3.1 are upper bounds to aromaticity and antiaromaticity in annulenes. In this section, we will consider the optimization of several singlet and triplet annulenes and how it affects the aromaticity of these compounds. As an improvement over HMO results, we will calculate an approximate version of that consists in using only the two-center bond orders of bonded atoms to estimate the N-center delocalization. In particular, we will employ the equivalent of Equation (8) in the framework of quantum mechanics. To this end, we will take and subtract 1.00 to (approximately) remove the sigma contribution to the C-C bond, and multiply all resulting DIs of the bonds around the perimeter of the ring,
(10)
where we have taken the square root to account for the fact that at the HMO level the DI corresponds to the square of the Hückel bond-order . One can consider the latter as a rough approximation that bridges the HMO definition (Equation (8)) and the actual value (Equation (5)). We have collected the results of in Figure 3.
Figure 3.
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Values of for the annulenes series in terms of the number of C atoms (N) for the lowest-lying singlets and triplets. The species have been divided according to the number of electrons (4 n and 4) and the spin multiplicity (singlet and triplet). Calculations were performed with CAM-B3LYP/6-311G(d,p).
The results show some differences with respect to the HMO values we have shown earlier. singlet compounds show a similar trend to the HMO results except for annulene, which exhibits a smaller aromaticity than expected. This molecule undergoes important geometrical distortions that disrupt the overlap of p orbitals and causes the apparent loss of aromaticity. singlets are expected to be antiaromatic, as is confirmed by the large BOA values (see Table 1 and Table 2), and they become less and less antiaromatic as they increase the ring size (see the increasing values of Figure 3). triplet annulenes follow a very similar trend in agreement with Baird’s rule. Finally, we examine the triplet annulenes that are expected to be aromatic and for which there is no clear trend. On one hand, COT, which displays a planar structure, is actually among the most aromatic compounds, even more aromatic than cyclobutadiene in its lowest-lying triplet state, which actually shows the smallest value among the molecules that are expected to be aromatic. On the other hand, neither nor annulene present a planar structure and, therefore, they exhibit larger BOA values, suggesting that these molecules are rather nonaromatic or slightly antiaromatic. Interestingly, all the aromaticity trends are similar to those found for pure HMO calculations for annulenes and singlets with the only mentioned exception of annulene.
Table 1 and Table 2 also include the actual values of but we have not discussed them because they mostly corroborate the trends we have found with and some of the values for large annulenes could not be obtained with enough precision. This is one of the main shortcomings of for large rings , the values become so small that they conflict with the numerical precision of the calculation and, consequently, we cannot obtain a reliable normalization () which permits the comparison among rings of different sizes.
In order to remedy this problem, was recently designed . values for the annulenes series are collected in Figure 4. Although the trends are not so clear as in HMO calculations or the approximate account of aromaticity with , the picture we can extract from does not differ entirely from that we obtained from . The four most aromatic molecules are the same according to both indices and each single-triplet pair shows the same order of aromaticity, e.g., singlet benzene is more aromatic than triplet benzene. The only apparent exception to this rule is annulene that has a larger value for the lowest-lying triplet than for the ground state singlet. The latter is due to the fact that the ground state configuration of annulene is found to be nonaromatic (the second lowest-lying structure, the heart configuration, is actually quite aromatic) whereas the triplet configuration is antiaromatic.
Figure 4.
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Values of for the annulenes series (lowest-lying singlets and triplets) in terms of the number of C atoms (N). The species have been divided according to the number of electrons (4 n and 4) and the spin multiplicity (singlet and triplet). Calculations were performed with CAM-B3LYP/6-311G(d,p).
Finally, we find that FLU, BOA, and BLA give a similar qualitative trend to (see Table 1 and Table 2, and also the Supplementary Material) with the exception of large annulenes (annulene and annulene) that are found to be more aromatic in their triplet state than in their ground state. One should bear in mind that these indices are based on pairwise interactions (FLU is also based on fixed ground-state aromatic references) and they are not as reliable as indices based on multicenter calculations .
3.4. The Delocalization Error in DFAs
DFAs suffer the so-called delocalization error [90,91,92], which tends to give delocalized electronic structures. This is the exact opposite behavior of HF, which tends to overestimate the localization of the electronic structures. DFAs with a low percentage of HF exchange are more prone to delocalization errors. This effect has been observed in conjugate systems and aromatic systems [40,81,94,95] including singlet annulenes . Recently, Contreras-García, et al. have suggested that the exact result is often bracketed between two extreme situations, the highly localized one (HF) and a highly delocalized one (LDA, which does not include HF exchange). They have used this result to construct error bars for the calculation of some solid-state properties. Likewise, Burke, et al. have defined the sensitivity of DFAs to the density as the energy difference between DFA calculations from LDA and HF densities.
In this section, we investigate how the delocalization error affects the aromaticity measures in the studied annulene series through the analysis of calculated with four methods that use a different amount of HF exchange. In our study, LDA is ruled out because it is not reliable for gas-phase calculations and B3LYP is used as the most delocalizing method (includes 19% of HF exchange). As the most localizing method we employ HF (that obviously uses 100% of HF exchange), and, as DFA with a large percentage of HF exchange, we have selected M06-2X (54%) and a range-separation functional , CAM-B3LYP, which presents a variable amount of HF exchange that goes from 19% at short ranges to 65% at large interelectronic separations.
values for different methods are collected in Figure 5. Our results confirm that in the large majority of cases B3LYP and HF are giving the most and the less aromatic species, respectively, among the DFA studied. In some cases (see, singlet benzene, triplet cyclobutadiene or triplet COT) there are no large differences among the DFAs because these species do not suffer from the delocalization error. However, in many other cases (see singlet annulenes larger than benzene) B3LYP clearly overestimates electron delocalization, sometimes even favoring a different geometrical arrangement of the atoms in the ring that permits larger electron delocalization (see Section 3.2). As expected, HF tends to overestimate electron localization whereas CAM-B3LYP and M06-2X provide very similar results lying between HF and B3LYP values. There is only two exceptions to this rule: the triplet states of benzene and annulene. Unlike other DFAs and HF, the M06-2X geometry optimization of triplet benzene leads to a non-planar structure that does not correspond to the AQ structure and, therefore, it is not as antiaromatic as one would expect. A careful examination of the potential energy surface of the triplet state of benzene confirms that M06-2X does not identify the AQ conformation as a minimum of energy at this level of theory. In the case of annulene, the exception is that HF value is actually larger than that of CAM-B3LYP or M06-2X because, according to HF, the structure is more planar than predicted by these DFAs.
Figure 5.
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Values of for the lowest-lying states of the studied annulenes obtained with methods using a different percentage of Hartree Fock (HF) exchange: HF (100%), M06-2X (54%), CAM-B3LYP (19–65%) and B3LYP (19%).
Interestingly, singlet molecules (with the exception of benzene) and singlets tend to suffer the most from electron delocalization errors, whereas triplet molecules barely exhibit differences between B3LYP and other DFAs with higher percentage of HF exchange. The fact that M06-2X and CAM-B3LYP present very similar values for most molecules corroborates that it is actually the long-range part of Hartree-Fock exchange the one which is relevant to decrease the delocalization error . In this sense, we recommend the use of a range-separation functional, such as CAM-B3LYP, to study aromatic and antiaromatic compounds.
4. Materials and Methods
The singlet ground state and the first triplet excited state of all the studied structures have been fully characterized. The optimizations have been performed with the Gaussian16 software package using B3LYP [100,101], CAM-B3LYP , M06-2X DFAs, and HF in combination with the 6-311G(d,p) basis set . The harmonic vibrational frequencies were calculated at the corresponding level of theory in order to verify the nature of the stationary points of their potential energy surface (minima or transition states).
The calculation of the electronic aromaticity indices (AV1245, , BOA and FLU) uses a QTAIM atomic partition performed by the AIMAll software . The AOM resulting from the QTAIM partition and the molecular geometries are the input for the in-house ESI-3D program [27,31,107], which provides AV1245 , , BLA, BOA, DIs [29,30], FLU , HOMA (included in the Supplementary Material), , and MCI values. The numerical accuracy of the QTAIM calculations has been assessed using two criteria: (i) The integration of the Laplacian of the electron density () within an atomic basin must be close to zero; () the number of electrons in a molecule must be equal to the sum of all the electron populations of the molecule, and also equal to the sum of all the localization indices and half of the delocalization indices in the molecule. For all atomic calculations, integrated absolute values of were always lower than 0.001 a.u. For all molecules, errors in the calculated number of electrons were always lower than 0.01 a.u. From our experience, these errors provide sufficient accuracy for all the indices here calculated except for the normalized multicenter indices of large rings, which require a numerical precision of the AOM well beyond the accuracy that one can obtain with AIMall or any other similar software available in the literature. For the latter and other reasons commented in Ref. , multicenter indices (MCI and ) cannot be used in large rings.
We have employed Mathematica to perform all Hückel calculations, fittings and extrapolations presented in this manuscript.
5. Conclusions
In this paper, we have studied how the Hückel and Baird rules fade away in cyclic polyenes. According to pure HMO calculations and the seminal Baird study , antiaromatic annulenes lose their antiaromatic character at the same speed as the ring size increases, regardless of their multiplicity. However, a two-resonance parameter HMO method, permitting bond-order alternation, shows that the antiaromaticy of singlets decreases more rapidly with the ring size. The conclusion is far less clear from calculations that consider the geometry relaxation because the potential energy surface of annulenes with more than eight carbon atoms often shows several configurations close in energy that display disparate aromatic characters. Nevertheless, density functional approximations reveal that the rules fade away much more quickly than it would be expected from the HMO method; they just do not follow a smooth trend.
The study of the level of theory employed in the calculation of annulenes reveals a clear tendency of density functional approximations with a low-percentage of HF exchange at long ranges to exhibit delocalization errors that lead to the overestimation of the aromatic character of the molecule and sometimes even to wrong geometries. Molecules with large ring structures are more prone to this kind of errors. These results are in line with previous findings [40,81,94] and suggest caution in choosing an appropriate density functional approximation to study aromatic and antiaromatic molecules. In particular, we recommend the use of a range-separation density functional approximation such as CAM-B3LYP.
Acknowledgments
The authors acknowledge the computational resources and technical and human support provided by DIPC and the SGI/IZO-SGIker UPV/EHU.
Abbreviations
The following abbreviations are used in this manuscript:
AOM Atomic overlaps matrix
AV1245 Aromaticity index for large rings
Minimal value of 12-45 delocalizations
BLA Bond-length alternation
BOA Bond-order alternation
DFA Density Functional Approximation
DI Delocalization index
FLU Fluctuation aromaticity index
HF Hartree-Fock
HMO Hückel Molecular Orbital
HOMA Harmonic Oscillator Model of Aromaticity
LDA Local density approximation
MCI Multicenter index
RE Resonance energy
TREPE Topological resonance energy per electron
Giambiagi’s multicenter index
Approximation to (Equation(8))
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Supplementary Materials
Click here for additional data file. (398.9KB, zip)
The Cartesian coordinates of the molecules as well as full tables and supplementary graphics are available online at
Author Contributions
Conceptualization, E.M.; methodology, E.R.-C. and E.M.; validation, E.R.-C., M.T.-S. and E.M.; formal analysis, I.C.-R., E.R.-C., M.T.-S. and E.M.; data curation, I.C.-R., E.R.-C., M.T.-S. and E.M.; writing—original draft preparation, E.M.; writing—review and editing, I.C.-R., E.R.-C., M.T.-S. and E.M.; supervision, E.R.-C., M.T.-S. and E.M.; funding acquisition, E.R.-C., M.T.-S. and E.M. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded by Spanish MINECO grant numbers PGC2018-098212-B-C21, CTQ2016-80375-P, EUIN2017-88605 and EUR2019-103825, the Basque Government/Eusko Jaurlaritza (GV/EJ) grant numbers IT1346-19, IT1254-19, PIBA19-0004, and 2019-CIEN-000092-01, and the doctoral grant PRE_2016_1_0159. E.R.C. acknowledges funding from the Juan de la Cierva program IJCI-2017-34658.
Conflicts of Interest
The authors declare no conflict of interest.
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https://www.quora.com/How-I-can-prove-that-the-volume-of-a-sphere-is-4-3%CF%80r-3
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3D Solid Modeling
The Sphere
Mathematics Homework Ques...
Proofs (mathematics)
Volume (mathematics)
Analytic Solid Geometry
3D Geometry
5
How I can prove that the volume of a sphere is 4/3πr^3?
Former Math Expert at Photomath (2021–2022)
·
Author has 330 answers and 761.7K answer views
·
6y
Related questions
How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
Why is the volume of a sphere 4/3 pi r^3?
Why is the volume of sphere
4
3
π
r
3
?
How can we prove that the volume of a sphere is equal to 4/3pir^3 by integration?
How do I prove the volume of a sphere and cone?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Upvoted by
Jean-Claude Matthys
, MSC & AESS Mathematics, Catholic University of Louvain (1964) · Author has 6.8K answers and 52.8M answer views
·
5y
Originally Answered: How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
·
I don’t think there is anything unusual in the 4/3 part of the formula. If we just work out the volume using integration the 4/3 is just part of it.
Just follow this…
Imagine the sphere is made up of countless thin discs as above.
The sum of the volumes of these d...
I don’t think there is anything unusual in the 4/3 part of the formula. If we just work out the volume using integration the 4/3 is just part of it.
Just follow this…
Imagine the sphere is made up of countless thin discs as above.
The sum of the volumes of these d...
Amitava Das
loves maths puzzles
·
8y
Originally Answered: Why is the volume of circle 4/3πr^3?
·
first of all , circle is a 2D concept .. so what we are talking of here is a sphere .
now , this problem needs to be traced from a long way back i.e. first comes a point, following it a hollow circle, then a hollow shell and finally a solid sphere .
so let me consider a solid sphere of radius R. it can be imagined to be formed of several hollow spheres each of negligible thickness whose radius r : 0<r<R.
now volume is basically summing up the net surface area of all those infintesemal spheres… what we mean here is simple integration.
so V=integration of 4pi(r^2) from 0 to R ie [4/3 r^2]0 to R = 4/
first of all , circle is a 2D concept .. so what we are talking of here is a sphere .
now , this problem needs to be traced from a long way back i.e. first comes a point, following it a hollow circle, then a hollow shell and finally a solid sphere .
so let me consider a solid sphere of radius R. it can be imagined to be formed of several hollow spheres each of negligible thickness whose radius r : 0<r<R.
now volume is basically summing up the net surface area of all those infintesemal spheres… what we mean here is simple integration.
so V=integration of 4pi(r^2) from 0 to R ie [4/3 r^2]0 to R = 4/3 r^3
Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010)
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Author has 10.2K answers and 15.2M answer views
·
5y
Originally Answered: How can you derive 4/3pir^3 (volume formula of a sphere) from a triple integral?
·
How can you derive the formula for volume of a sphere (43πr3) using a triple integral?
Consider the spherical coordinate system (r,θ,ϕ), where r,θ,ϕ are the radial distance, polar angle and azimuthal angle respectively.
In this coordinate system, a small volume inside the sphere is given by
dV=(dr)(rdθ)(rsinθdϕ)=r2sinθdrdθdϕ.
⇒ Volume of a sphere, V=2π∫0π∫0R∫0r2sinθdrdθdϕ
=2π∫0π∫0\l
How can you derive the formula for volume of a sphere (43πr3) using a triple integral?
Consider the spherical coordinate system (r,θ,ϕ), where r,θ,ϕ are the radial distance, polar angle and azimuthal angle respectively.
In this coordinate system, a small volume inside the sphere is given by
dV=(dr)(rdθ)(rsinθdϕ)=r2sinθdrdθdϕ.
⇒ Volume of a sphere, V=2π∫0π∫0R∫0r2sinθdrdθdϕ
=2π∫0π∫0[r33]R0sinθdθdϕ=2π∫0π∫0[R33]sinθdθdϕ
=R332π∫0[−cosθ]π0dϕ=R332π∫0dϕ
=2R33[ϕ]2π0=2R33[2π].
⇒ Volume of a sphere, V=43πr3.
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Related questions
Was the 4 in the volume formula for a sphere (4pir^3) /3 derived from pi/4?
We know volume of sphere is
V
=
4
π
3
R
3
. Then how do I prove that
d
V
V
=
3
d
R
R
?
How have we derived the formula (i.e. 4/3πr^3) of the volume of a solid sphere?
Why is the formula for surface area of a sphere
4
π
r
2
, and why is volume
4
3
π
r
3
?
How can we derive the volume of sphere (4÷3×Pi×r^3) for 10th class?
Assistant
Bot
·
1y
To prove that the volume V of a sphere is given by the formula V=43πr3, we can use several methods. One common approach is to use calculus, specifically integration. Here’s a step-by-step outline of how to derive this formula using integration.
Derivation Using Integration
Understanding the Sphere:
A sphere of radius r can be thought of as a set of circular disks stacked along an axis (usually the z-axis).
Equation of a Sphere:
The equation of a sphere centered at the origin is:
x2+y2+z2=r2
Volume Element:
To find the volume, we can consider a thin disk of
To prove that the volume V of a sphere is given by the formula V=43πr3, we can use several methods. One common approach is to use calculus, specifically integration. Here’s a step-by-step outline of how to derive this formula using integration.
Derivation Using Integration
Understanding the Sphere:
A sphere of radius r can be thought of as a set of circular disks stacked along an axis (usually the z-axis).
Equation of a Sphere:
The equation of a sphere centered at the origin is:
x2+y2+z2=r2
Volume Element:
To find the volume, we can consider a thin disk of thickness dz at a height z with radius √r2−z2 (from the equation of the sphere). The area A of this disk is:
A=π(√r2−z2)2=π(r2−z2)
Volume of the Disk:
The volume dV of the thin disk is:
dV=A⋅dz=π(r2−z2)dz
Integrating to Find Total Volume:
To find the total volume of the sphere, integrate dV from z=−r to z=r:
V=∫r−rπ(r2−z2)dz
Calculating the Integral:
This integral can be split into two parts:
V=π∫r−r(r2−z2)dz=π(∫r−rr2dz−∫r−rz2dz)
The first integral is:
∫r−rr2dz=r2[z]r−r=r2(r−(−r))=r2(2r)=2r3
The second integral is:
∫r−rz2dz=[z33]r−r=(r)33−(−r)33=r33+r33=2r33
Combining the Results:
Substitute back into the volume integral:
V=π(2r3−2r33)=π(6r33−2r33)=π(4r33)=43πr3
Conclusion
Thus, we have proven that the volume V of a sphere of radius r is:
V=43πr3
This derivation is a classic application of integral calculus to geometric shapes.
Adrian Lu
Former A-level maths student
·
Author has 133 answers and 321.7K answer views
·
4y
Related
Why is the volume of a sphere 4/3 pi r^3?
Imagine slicing a sphere with radius r into infinitesimal small pyramids as shown
:
The volume of the pyramid is given by:
Vpyramid=13×BA×r,
where BA is its base area.
If we sum up all the base areas of the sphere, we obtain the surface area of the sphere which is equal to 4πr2 .
Hence the volume of the sphere is equal to the sum of the volumes of the pyramids:
V=∑volumes of pyramids=13(4πr2)r=43πr3
Footnotes
Surface Area and Volume of Spheres
Imagine slicing a sphere with radius r into infinitesimal small pyramids as shown
:
The volume of the pyramid is given by:
Vpyramid=13×BA×r,
where BA is its base area.
If we sum up all the base areas of the sphere, we obtain the surface area of the sphere which is equal to 4πr2 .
Hence the volume of the sphere is equal to the sum of the volumes of the pyramids:
V=∑volumes of pyramids=13(4πr2)r=43πr3
Footnotes
Surface Area and Volume of Spheres
Paul Dunkley
Associate
·
Author has 993 answers and 2M answer views
·
Updated 2y
Originally Answered: How do you derive the volume of a sphere 4/3πr³?
·
Take a cylinder with radius R equal to height, by Pythagoras the area of a slice through the semisphere is equal to a corresponding area through the punctured cylinder.
That’s true of any and all corresponding slices, so take another slice to cut off the solids in the second figure. Now if the area of the top slices are equal, and the areas of the bottom slices are equal, and the areas of all the intermediate slices are equal, and since the solids share the same altitude, then the volumes are equal for all slices.
Therefore the volume of the whole semisphere must be equal to the whole volume of
Take a cylinder with radius R equal to height, by Pythagoras the area of a slice through the semisphere is equal to a corresponding area through the punctured cylinder.
That’s true of any and all corresponding slices, so take another slice to cut off the solids in the second figure. Now if the area of the top slices are equal, and the areas of the bottom slices are equal, and the areas of all the intermediate slices are equal, and since the solids share the same altitude, then the volumes are equal for all slices.
Therefore the volume of the whole semisphere must be equal to the whole volume of the punctured cylinder. But the punctured cylinder is 2/3 the volume of the cylinder, so the semisphere is 2/3 the volume of the cylinder. But the cylinder is (pi R³) so the semisphere is (2piR³)/3 and the whole sphere is (4piR³)/3
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Arpan Banerjee
4y
Some others have already proved this using knowledge of the formula for the surface area of a sphere, but it is also possible to prove without any prior formula.
Using spherical coordinates, let’s find the volume of a sphere with radius r
Then the bounds are
0≤ρ≤r0≤θ≤2π0≤ϕ≤π
This image has been removed for violating Quora's policy.
Then the volume of the sphere is ∫π0∫2π0∫r0ρ2sinϕdρdθdϕ which evaluates to 4πr33
This image has been removed for violating Quora's policy.
Some others have already proved this using knowledge of the formula for the surface area of a sphere, but it is also possible to prove without any prior formula.
Using spherical coordinates, let’s find the volume of a sphere with radius r
Then the bounds are
0≤ρ≤r0≤θ≤2π0≤ϕ≤π
This image has been removed for violating Quora's policy.
Then the volume of the sphere is ∫π0∫2π0∫r0ρ2sinϕdρdθdϕ which evaluates to 4πr33
This image has been removed for violating Quora's policy.
Harish Chandra Rajpoot
authored 'Advanced Geometry' on research articles in Mathematics & Radiometry
·
Author has 1.6K answers and 7.7M answer views
·
2y
Originally Answered: How do you derive the volume of a sphere 4/3πr³?
·
Method-1: Using Pappus’s Centroid Theorem
A sphere is a solid of revolution generated by revolving a semi-circular plane (i.e. semi-disk) about its diameter (as shown in the figure-1 below)
Figure-1
According to Pappus’s second theorem, the volume V of the sphere of radius say R will be equal to product of the area of semi-circular plane (A) and the distance (d) traveled by the centroid in one complete rotation which is mathematically given as
V=A⋅d=(12πR2)(2π⋅4R3π)
V=43πR3
Method-2: Using Calculus (spherical coordinate system)
Co
Method-1: Using Pappus’s Centroid Theorem
A sphere is a solid of revolution generated by revolving a semi-circular plane (i.e. semi-disk) about its diameter (as shown in the figure-1 below)
Figure-1
According to Pappus’s second theorem, the volume V of the sphere of radius say R will be equal to product of the area of semi-circular plane (A) and the distance (d) traveled by the centroid in one complete rotation which is mathematically given as
V=A⋅d=(12πR2)(2π⋅4R3π)
V=43πR3
Method-2: Using Calculus (spherical coordinate system)
Consider an element of solid sphere with radius R having arc length rdθ , arc width rsinθdϕ and thickness dr as shown in the figure-2 below.
Figure-2
The volume of spherical element (as shaded in the figure-2 above) will be equal to the volume of the cuboid having infinitesimal small length rdθ , width rsinθdϕ and thickness dr as follows
dV=(rdθ)(rsinθdϕ)(dr)=r2sinθdrdθdϕ
Now, using the proper limits of the variables i.e. r from r=0 to r=R , θ from θ=0 to θ=π and ϕ from ϕ=0 to ϕ=2π , the total volume of the sphere of radius R is determined as follows
V=∫dV
=∫ϕ2ϕ1∫θ2θ1∫r2r1r2sinθdrdθdϕ
=∫2π0(∫π0(∫R0r2dr)sinθdθ)dϕ
=∫2π0(∫π0([r33]R0)sinθdθ)dϕ
=∫2π0(∫π0R33sinθdθ)dϕ
=∫2π0(R33[−cosθ]π0)dϕ
=∫2π0R33(2)dϕ
=2R33[ϕ]2π0
=2R33(2π)
=43πR3
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Steve Dujmovic
ME, BE in Electrical Engineering & Telecommunications, University of New South Wales (Graduated 1984)
·
5y
Originally Answered: How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
·
I like the following approach to remembering and arriving at the volume of a sphere, which also answers the OP’s question of where the 4/3 comes from.
With the above diagram we represent a sphere of radius R. We’ll assume that we know (and can readily derive) the surface area of a sphere to be,
As=4πr2
Now if we assume a small sphere of radius dr, then the volume of the sphere with radius R is given by;
Vs=∫R04πr2dr
Vs=43πr3
And that’s it - short and sweet.
Now how did we get the surface area of the sphere? Well that’s another question.
I like the following approach to remembering and arriving at the volume of a sphere, which also answers the OP’s question of where the 4/3 comes from.
With the above diagram we represent a sphere of radius R. We’ll assume that we know (and can readily derive) the surface area of a sphere to be,
As=4πr2
Now if we assume a small sphere of radius dr, then the volume of the sphere with radius R is given by;
Vs=∫R04πr2dr
Vs=43πr3
And that’s it - short and sweet.
Now how did we get the surface area of the sphere? Well that’s another question.
David Townsend
Staff Software Engineer (1982–present)
·
Author has 29.5K answers and 11M answer views
·
5y
Originally Answered: How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
·
You could derive the formula for the volume of sphere from first principles, either through integration or using a simple quadrature approach.
Imagine you had a sphere which you could slice repeatedly through its diameters so that you have a lot of pyramid shaped wedges. You could arrange these wedges so that the curved surface is laid flat on a surface and the pointed end is vertically above. The volume of a wedge is 1/3 area of base x height. Now if the wedges were sufficiently small, the height would be approximately the radius of the sphere. The total volume would be the sum of each of the
You could derive the formula for the volume of sphere from first principles, either through integration or using a simple quadrature approach.
Imagine you had a sphere which you could slice repeatedly through its diameters so that you have a lot of pyramid shaped wedges. You could arrange these wedges so that the curved surface is laid flat on a surface and the pointed end is vertically above. The volume of a wedge is 1/3 area of base x height. Now if the wedges were sufficiently small, the height would be approximately the radius of the sphere. The total volume would be the sum of each of the volumes of the wedges, so we get volume = area of sphere 1/3 radius = 4pi r^2 x 1/3 r = 4/3 pi r^3.
You can also arrive at the same formula by integration of the formula for the area of a sphere 4pir^2
Gerrit Bernard
Studied at Karlsruhe Institute of Technology
·
Author has 1.7K answers and 1.6M answer views
·
5y
Originally Answered: How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
·
To understand the 4/3, you need to understand integration. The volume of a sphere 4/3πr3
is the definite integral from 0 to r of the surface area of the sphere 4πr2.
Ultimately, it comes from the radius, integrated three times. Here's one way to do it.
Analytically, integrate over circular discs.
A sphere with radius R positioned around the origin consists of circular discs. The radius of such a disc is sqrt(R^2 - x^2). The surface of such a disc pi (that radius) ^2 = pi (R^2 - x^2)
Integrating that from -R to +R gives your 4/3 pi R^2
Phil Scovis
I play guitar. And sing in the car.
·
Author has 6.9K answers and 13.3M answer views
·
Updated 5y
Originally Answered: How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
·
One way to determine the volume of a sphere is sum the volumes of concentric spherical shells of infinitesimal thickness. The volume of each shell is its area times its infinitesimal thickness. The area of a sphere is:
A=4πr2
Here, you can see the ‘4’ already making an appearance.
Adding up concentric shells from radius 0 to radius r, we get:
∫r04πx2dx
At this point, the power rule for integrals introduces the 1/3:
=4π13x3∣∣∣r0
=43πr3
An alternative way to find the volume (the way usually ta
One way to determine the volume of a sphere is sum the volumes of concentric spherical shells of infinitesimal thickness. The volume of each shell is its area times its infinitesimal thickness. The area of a sphere is:
A=4πr2
Here, you can see the ‘4’ already making an appearance.
Adding up concentric shells from radius 0 to radius r, we get:
∫r04πx2dx
At this point, the power rule for integrals introduces the 1/3:
=4π13x3∣∣∣r0
=43πr3
An alternative way to find the volume (the way usually taught in elementary Calculus classes) comes from using the “disk method” on y=√r2−x2 revolved around the x-axis. The resulting integral is:
∫r−rπ(√r2−x2)2dx
The same answer is obtained, although the appearance of ‘4’ is more subtle.
Related questions
How does one acquire the 4/3 in the sphere-volume formula, V=4/3πr^3?
Why is the volume of a sphere 4/3 pi r^3?
Why is the volume of sphere
4
3
π
r
3
?
How can we prove that the volume of a sphere is equal to 4/3pir^3 by integration?
How do I prove the volume of a sphere and cone?
Was the 4 in the volume formula for a sphere (4pir^3) /3 derived from pi/4?
We know volume of sphere is
V
=
4
π
3
R
3
. Then how do I prove that
d
V
V
=
3
d
R
R
?
How have we derived the formula (i.e. 4/3πr^3) of the volume of a solid sphere?
Why is the formula for surface area of a sphere
4
π
r
2
, and why is volume
4
3
π
r
3
?
How can we derive the volume of sphere (4÷3×Pi×r^3) for 10th class?
Why is the volume of the sphere V=4/3 pi r3? By my calculations, it should not be 4/3.
The largest sphere is carved out of a cube of side 7 cm. What is the volume of the sphere (takeπ=3.14..)?
How do we get the factor of 4/3 in the formula for volume of a sphere?
The volume of a sphere is 288π mm^3. What is the diameter of the sphere?
V^(2/3) ∝ A is a formula for how the volume of a sphere is proportional to the surface area. Can anyone prove how this works or cite the name of the equation?
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454
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https://www.cuemath.com/questions/what-is-1-16-as-a-percent/
|
What is 1/16 as a percent
Converting a fraction to percent means converting the whole into 100 and find how many are considered out of 100.
Answer: 1/16 is equivalent to 6.25%.
Let's understand the conversion of a fraction to a percentage.
Explanation:
To convert a fraction to percent, we multiply it by 100.
Therefore, 1/16 = 1/16 × 100 %
= 100/16 %
= 6.25 %
In short, to convert a fraction to percent, multiply it by 100 followed by the symbol %.
Thus, 1/16 is equivalent to 6.25%.
|
455
|
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/6%3A_Relations/6.3%3A_Equivalence_Relations_and_Partitions
|
Skip to main content
6.3: Equivalence Relations and Partitions
Last updated
: May 5, 2020
Save as PDF
6.2: Properties of Relations
7: Combinatorics
Page ID
: 26444
Harris Kwong
State University of New York at Fredonia via OpenSUNY
( \newcommand{\kernel}{\mathrm{null}\,})
Recall:
A relation on a set is an equivalence relation if it is reflexive, symmetric, and transitive. We often use the tilde notation to denote a relation.
Also, when we specify just one set, such as is a relation on set , that means the domain & codomain are both set .
For an equivalence relation, due to transitivity and symmetry, all the elements related to a fixed element must be related to each other. Thus, if we know one element in the group, we essentially know all its “relatives.”
Definition: Equivalence Class
Let be an equivalence relation on set . For each we denote the equivalence class of as defined as:
Example
Define a relation on by Find the equivalence classes of .
Answer
: Two integers will be related by if they have the same remainder after dividing by 4. The possible remainders are 0, 1, 2, 3.
hands-on exercise
Define a relation on by Find the equivalence classes of .
example
Let
For convenience, label
Define this equivalence relation on by
Find the equivalence classes of .
Answer
: Two sets will be related by if they have the same number of elements.
The element in the brackets, [ ] is called the representative of the equivalence class. An equivalence class can be represented by any element in that equivalence class. So, in Example 6.3.2, This equality of equivalence classes will be formalized in Lemma 6.3.1.
Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets.
Take a closer look at Example 6.3.1.All the integers having the same remainder when divided by 4 are related to each other. The equivalence classes are the sets It is clear that every integer belongs to exactly one of these four sets. Hence, These four sets are pairwise disjoint. From this we see that is a partition of .
Equivalence Classes form a partition (idea of Theorem 6.3.3)
The overall idea in this section is that given an equivalence relation on set , the collection of equivalence classes forms a partition of set (Theorem 6.3.3).
The converse is also true: given a partition on set , the relation "induced by the partition" is an equivalence relation (Theorem 6.3.4).
As another illustration of Theorem 6.3.3, look at Example 6.3.2.
Thus, is a partition of set .
In order to prove Theorem 6.3.3, we will first prove two lemmas.
Lemma
If is an equivalence relation on , then .
Proof
: Let be an equivalence relation on with
First we will show
Let by definition of equivalence class. Now we have
thus by transitivity (since is an equivalence relation). Since by definition of equivalence classes.
We have shown if , thus by definition of subset.
Next we will show
Let by definition of equivalence class. Since , we also have by symmetry.
Now we have thus by transitivity. Since by definition of equivalence classes.
We have shown if , thus by definition of subset.
by the definition of set equality.
One may regard equivalence classes as objects with many aliases. Every element in an equivalence class can serve as its representative. So we have to take extra care when we deal with equivalence classes. Do not be fooled by the representatives, and consider two apparently different equivalence classes to be distinct when in reality they may be identical.
Example
Define on a set of individuals in a community according to We can easily show that is an equivalence relation. Each equivalence class consists of all the individuals with the same last name in the community. Hence, for example, Jacob Smith, Liz Smith, and Keyi Smith all belong to the same equivalence class. Any Smith can serve as its representative, so we can denote it as, for example, Liz Smith.
Example
Define on according to Hence, two positive real numbers are related if and only if they have the same decimal parts. It is easy to verify that is an equivalence relation, and each equivalence class consists of all the positive real numbers having the same decimal parts as has. Notice that which means that the equivalence classes , where , form a partition of .
hands-on exercise
Prove that the relation in Example 6.3.4 is indeed an equivalence relation.
Lemma
Given an equivalence relation on set , if then either or
Proof
: Let be an equivalence relation on set with
Case 1:
In this case or is true.
Case 2:
by definition of empty set & intersection.
and by definition of equivalence classes. Also since , by symmetry.
We have and , so by transitivity. Since , by Lemma 6.3.1.
In this case or is true.
These are the only possible cases. So, if then either or
Theorem 6.3.3 and Theorem 6.3.4 together are known as the Fundamental Theorem on Equivalence Relations.
Theorem
If is an equivalence relation on any non-empty set , then the distinct set of equivalence classes of forms a partition of .
Proof
: Suppose is an equivalence relation on any non-empty set . Denote the equivalence classes as .
WMST
First we will show
If then belongs to at least one equivalence class, by definition of union.
By the definition of equivalence class, . Thus
Next we show .
If , then since is reflexive. Thus .
for some since is an equivalence class of .
So, by definition of subset. And so, by the definition of equality of sets.
Now WMST is pairwise disjoint.
For any , either or by Lemma 6.3.2. So, is mutually disjoint by definition of mutually disjoint.
We have demonstrated both conditions for a collection of sets to be a partition and we can conclude
if is an equivalence relation on any non-empty set , then the distinct set of equivalence classes of forms a partition of .
Conversely, given a partition , we could define a relation that relates all members in the same component. This relation turns out to be an equivalence relation, with each component forming an equivalence class. This equivalence relation is referred to as the equivalence relation induced by .
Definition
Given is a partition of set , the relation, , induced by the partition, ,is defined as follows:
Example
Consider set with this partition:
Find the ordered pairs for the relation , induced by the partition.
Proof
Theorem
If is a set with partition and is a relation induced by partition then is an equivalence relation.
Proof
: Let be a set with partition and be a relation induced by partition WMST is an equivalence relation.
Reflexive
Let Since the union of the sets in the partition must belong to at least one set in
Since by the definition of a relation induced by a partition.
is reflexive.
Symmetric
Suppose by the definition of a relation induced by a partition.
Since
is symmetric.
Transitive
Suppose
and by the definition of a relation induced by a partition.
Because the sets in a partition are pairwise disjoint, either or
Since belongs to both these sets, thus
Both and belong to the same set, so by the definition of a relation induced by a partition.
is transitive.
We have shown is reflexive, symmetric and transitive, so is an equivalence relation on set
if is a set with partition and is a relation induced by partition then is an equivalence relation.
Example
Over , define It is not difficult to verify that is an equivalence relation. There are only two equivalence classes: and , where contains all the positive integers, and all the negative integers. It is obvious that .
hands-on exercise
The relation defined on the set is known to be Confirm that is an equivalence relation by studying its ordered pairs. Determine the contents of its equivalence classes.
Example
Consider the equivalence relation induced by the partition of .
(a) Write the equivalence classes for this equivalence relation. (b) Write the equivalence relation as a set of ordered pairs.
Answer
: (a)
(b) From the two 1-element equivalence classes and , we find two ordered pairs and that belong to . From the equivalence class , any pair of elements produce an ordered pair that belongs to . Therefore,
Summary Review
A relation on a set is an equivalence relation if it is reflexive, symmetric, and transitive.
If is an equivalence relation on the set , its equivalence classes form a partition of .
In each equivalence class, all the elements are related and every element in belongs to one and only one equivalence class.
The relation determines the membership in each equivalence class, and every element in the equivalence class can be used to represent that equivalence class.
In a sense, if you know one member within an equivalence class, you also know all the other elements in the equivalence class because they are all related according to .
Conversely, given a partition of , we can use it to define an equivalence relation by declaring two elements to be related if they belong to the same component in the partition.
Exercises
Exercise
Find the equivalence classes for each of the following equivalence relations on .
a)
b)
Answer
: (a) The equivalence classes are of the form for some integer . For instance, , , , and .
(b) There are two equivalence classes: , and .
Exercise
For this relation on defined by :
a) show is an equivalence relation.
b) find the equivalence classes for .
Exercise
Let be a fixed subset of a nonempty set . Define the relation on by Show that is an equivalence relation. In particular, let and .
a) True or false: ?
b) How about ?
c) Find
d) Describe for any .
Answer
: (a) True
(b) False
(c)
(d) . In other words, if contains the same element in , plus possibly some elements not in .
Exercise
For each of the following relations on , determine whether it is an equivalence relation. For those that are, describe geometrically the equivalence class .
.
Exercise
For each of the following relations on , determine whether it is an equivalence relation. For those that are, describe geometrically the equivalence class .
Answer
: (a) Yes, with . In other words, the equivalence classes are the straight lines of the form for some constant .
(b) No. For example, and , but . Hence, the relation is not transitive.
Exercise
For each of the following relations on , determine whether it is an equivalence relation. For those that are, describe geometrically the equivalence class .
Exercise
Define the relation on by is an equivalence relation. Describe the equivalence classes and .
Answer
: We find , and .
Exercise
Define the relation on by Show that is an equivalence relation. Describe the equivalence classes , and .
Exercise
Consider the following relation on : This is an equivalence relation. Describe its equivalence classes.
Answer
Exercise
Each part below gives a partition of . Find the equivalence relation (as a set of ordered pairs) on induced by each partition.
(a)
(b)
(c)
(d)
Exercise
Write out the relation, induced by the partition below on the set
Answer
Exercise
Consider the relation, induced by the partition on the set shown in exercises 6.3.11 (above).
Answer these questions True or False.
(a) Every element in set is related to every other element in set
(b)
(c)
(d) Every element in set is related to itself.
(e) The relation, , is transitive.
(f)
(g)
(h)
(i)
(j)
6.2: Properties of Relations
7: Combinatorics
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https://www.asc.ohio-state.edu/kagan.1/AS1138/Lectures/coherence-and-stability-e894l15l_1-9.pdf
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Holography & Coherence • For Holography need coherent beams • Two waves coherent if fixed phase relationship between them for some period of time Coherence • Coherence appear in two ways Spatial Coherence • Waves in phase in time, but at different points in space • Required for interference and diffraction • Before lasers need to place slits far from source or pass light through slit so only part of source seen Temporal Coherence • Correlation of phase at the same point but at different times • Regular sources rapidly change phase relationships • Single atom on 10-8 sec coherent lifetime of atom in an excited state • Much shorter for groups of atoms • For lasers in single mode much longer time Coherence Length and Time • Time of coherence given by τcoh • Coherence time about time taken for photon to pass a given distance (Coherence length) in space • Coherence length is coh coh c L τ = • Best seen in Michelson-Morley experiment • Beam is split into two beam paths, reflected and combine • If get interference pattern then within Coherence lengths • Before lasers paths needed to be nearly equal • With lasers only require ( ) coh 2 1 L L L 2 < − • Coherence last 50 - 100 m with lasers Coherence Length and Lasers • It can be shown Coherence time related to laser frequency width Δν (linewidth) ν τ Δ = 1 coh • As the coherence length is coh coh c L τ = • For holography setup distances must be < coherence length • Long coherence lasers have small linewidth, thus high stability Example of Coherence Length Sodium vapour lamp yellow "D" line • λ = 589 nm and linewidth 5.1x1011 Hz • Thus coherence time and length is sec 10 x 96 .
1 10 x 1 .
5 1 1 12 11 coh − = = Δ = ν τ ( ) mm 59 .
0 m 10 x 88 .
5 10 x 96 .
1 10 x 98 .
2 c L 4 12 8 coh coh = = = = − − τ • Coherence small hence hard to create holograms • HeNe laser in multimode operation • λ = 632.8 nm and linewidth 1500 MHz • Thus coherence time and length is sec 10 x 67 .
6 10 x 5 .
1 1 1 10 9 coh − = = Δ = ν τ ( ) m 2 .
0 10 x 67 .
6 10 x 98 .
2 c L 10 8 coh coh = = = − τ • If single mode HeNe operation linewidth goes to 1 Mz and cohrence time is 1 microsec, cohrence length 300 m Interferometers • Can use interference effects to precisely measure distance • First example Michelson Interferometer • Have 2 mirrors (M1 & M2) placed on arms at 90 degrees • Splitting mirror O (half silvered mirror) at intersection • Splitter mirror – reflects part (~50%) of light 90o Lets part pass directly through • Of a thin film of Aluminium ~100 nm, not full absorbing • Monochormatic & coherent light source along path of one arm • Detector at other arm • Light to M1 is reduced, reflected by M1 then by splitter to detector • Light at splitter reduced but reflected to M2 • The passed through splitter O to detector Michelson Interferometer • At detector two beams combine to create interference • Let path length difference be Δl • Then if Δl = Nλ/2 get constructive interference – bright • Dark if λ 4 1 2 + = Δ N l • Now can measure very small distance changes • Eg if put glass plate C in can see small defects in glass • Interferometers used in measuring distance • Digitize light level and measure changes – can get λ/64 or 256 • Measure 2 nm distance • Need extremely stable laser Michelson Interferometer • Actually see circular interference at the detector • Reason distance from detector to splitter is d • The angle θ at the detector when destructive interference is ( ) λ θ m cos d m = 2 • Result is rings of interference Michelson Morley Experiment • Use Michelson interferometer floating on mercury • Align so path along direction of earth around sun • Other path at along radius to sun • Then rotate by 90 degrees • Classic physics: Along the path in direction of motion • light should arrive sooner to addition of velocities • But no difference found – first indication of relativity
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https://economy-finance.ec.europa.eu/economic-research-and-databases/economic-databases/tax-and-benefits-indicators-database/methodology-tax-and-benefits-indicators-database_en
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Methodology of tax and benefits indicators database - European Commission
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Methodology of tax and benefits indicators database
The Tax and benefits database is used by to examine the likely impacts of tax policies on economic growth and employment. The following definitions form part of the analysis.
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Tax wedge
Marginal effective tax rate - traps
Net increase in disposable income
Net replacement rate
Tax wedge
The tax wedge measures the difference between the total labour cost of employing a worker and the worker’s net earnings. A high tax wedge can exert a negative impact both on labour supply and demand, weakening incentives to look for work, to work additional hours and to hire new staff.
The tax wedge is defined as the sum of personal income taxes and employee and employer social security contributions net of family allowances, expressed as a percentage of total labour costs (the sum of the gross wage and social security contributions paid by the employer).
Marginal effective tax rate - traps
Marginal effective tax rates measure what part of an increase in earnings, due for instance to an increase in the number of hours worked or to a change in employment situation, is "taxed away" by the imposition of personal income taxes and employee social security contributions, and the possible withdrawal of earnings-related benefits.
Depending on the initial situation, such benefits can include:
social assistance (SA)
housing benefits (HB)
family benefits (FB)
unemployment benefits (UB)
in-work benefits (IWB)
The higher the marginal effective tax rate, the lower the incentives to look for work or to work additional hours. Because high marginal effective tax rates may discourage a change in employment situation, they are often referred to as traps.
The unemployment trap measures the short-term financial incentive for an unemployed person receiving unemployment benefits to move to paid employment. It is defined as the share of additional gross income of such a transition that is taxed away.
Example
A person received EUR 500 in unemployment benefits and does not pay taxes. Taking up work provides a gross salary of EUR 1000 while personal income taxes and employee social security contributions amount to EUR 200 and unemployment benefits are fully withdrawn. His new net income will be EUR 800. Out of his increase in the gross salary (EUR 1000), 70% will be "taxed away" and his net income increase will be EUR 300.
The inactivity trap measures the short-term financial incentive for an inactive person not entitled to unemployment benefits (but potentially receiving other benefits such as social assistance) to move from inactivity to paid employment. It is defined as the rate at which the additional gross income of such a transition is taxed.
The low-wage trap measures the financial incentive to increase a low level of earnings by working additional hours. It is defined as the rate at which the additional gross income of such a move is taxed.
Net increase in disposable income
Like the inactivity and unemployment traps, the net increase in disposable income indicator provides a measure of the financial incentives related to the transition from unemployment or inactivity into work. It is defined as the increase in disposable income when moving from unemployment or inactivity to employment, expressed as a percentage of the initial disposable income when unemployed or inactive. The higher the rate, the higher the incentive to look for work.
Example
An inactive person receives EUR 500 in social assistance, net of taxes. Taking up work provides a gross salary of EUR 800 per month, with income taxes and employee social security contributions amounting to EUR 150, and social assistance benefits are fully withdrawn. The additional net income is of EUR 150, corresponding to an increase in disposable income of 30 %.
Net replacement rate
The net replacement rate provides a measure of the generosity of unemployment and other social benefits. It is defined as the net income of an unemployed person receiving unemployment and possibly other benefits, expressed as a share of the income earned previously in the job before becoming unemployed. The higher the rate is, the more generous the benefits system, and consequently, the lower the financial incentives to return to work. The indicator can be measured at different points in time, as the duration of unemployment benefits is generally limited, and their generosity can decline over the unemployment spell.
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http://galton.uchicago.edu/~lalley/Courses/385/BrownianMotion.pdf
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Brownian Motion 1 Brownian motion: existence and first properties 1.1 Definition of the Wiener process According to the De Moivre-Laplace theorem (the first and simplest case of the cen-tral limit theorem), the standard normal distribution arises as the limit of scaled and centered Binomial distributions, in the following sense. Let ξ1, ξ2, . . . be inde-pendent, identically distributed Rademacher random variables, that is, independent random variables with distribution P{ξi = +1} = P{ξi = −1} = 1 2, (1) and for each n = 0, 1, 2, . . . let Sn = Pn i=1 ξi. The discrete-time stochastic process {Sn}n≥0 is the simple random walk on the integers. The De Moivre-Laplace theorem states that for each x ∈R, lim n→∞P{Sn/√n ≤x} = Φ(x) := Z x −∞ ϕ1(y) dy where (2) ϕt(y) = 1 √ 2πte−y2/2t (3) is the normal (Gaussian) density with mean 0 and variance t.
The De Moivre-Laplace theorem has an important corollary: the family {ϕt}t≥0 of normal densities is closed under convolution. To see this, observe that for any 0 < t < 1 the sum Sn = S[nt] + (Sn −S[nt]) is obtained by adding two independent Rademacher sums; the De Moivre-Laplace theorem applies to each sum separately, and so by an elementary scaling we must have ϕ1 = ϕt ∗ϕ1−t. More generally, for any s, t ≥0, ϕs ∗ϕt = ϕs+t.
(4) This law can, of course, be proved without reference to the central limit theo-rem, either by direct calculation (“completing the square”) or by Fourier transform.
1 However, our argument suggests a “dynamical” interpretation of the equation (4) that the more direct proofs obscure. For any finite set of times 0 = t0 < t1 < · · · < tm < ∞there exist (on some probability space) independent, mean-zero Gaussian random variables Wti+1 −Wti with variances ti+1 −ti. The De Moivre-Laplace theorem implies that as n →∞, 1 √n(S[nt1], S[nt2], . . . , S[ntm]) D − →(Wt1, Wt2, . . . , Wtm).
(5) The convolution law (4) guarantees that the joint distributions of these limiting ran-dom vectors are mutually consistent, that is, if the set of times {ti}i≤m is enlarged by adding more time points, the joint distribution of Wt1, Wt2, . . . , Wtm will not be changed. This suggests the possibility of defining a continuous-time stochastic pro-cess {Wt}t≥0 in which all of the random vectors (Wt1, Wt2, . . . , Wtm) are embedded.
Definition 1. A standard (one-dimensional) Wiener process (also called Brownian mo-tion) is a continuous-time stochastic process {Wt}t≥0 (i.e., a family of real random variables indexed by the set of nonnegative real numbers t) with the following properties: (A) W0 = 0.
(B) With probability 1, the function t →Wt is continuous in t.
(C) The process {Wt}t≥0 has stationary, independent increments.
(D) For each t the random variable Wt has the NORMAL(0, t) distribution.
A continuous-time stochastic process {Xt}t≥0 is said to have independent increments if for all 0 ≤t0 < t1 < · · · < tm the random variables (Xti+1 −Xti) are mutually independent; it is said to have stationary increments if for any s, t ≥0 the distribution of Xt+s −Xs is the same as that of Xt −X0. Processes with stationary, independent increments are known as Lévy processes.
Properties (C) and (D) are mutually consistent, by the convolution law (4), but it is by no means clear that there exists a stochastic process satisfying (C) and (D) that has continuous sample paths. That such a process does exist was first proved by N. WIENER in about 1920.
Theorem 1. (Wiener) On any probability space (Ω, F, P) that supports an infinite se-quence of independent, identically distributed Normal−(0, 1) random variables there exists a standard Brownian motion.
We will give at least one proof, due to P. LÉVY, later in the course. Lévy’s proof shows that exhibits the Wiener process as the sum of an almost surely uniformly convergent series. Lévy’s construction, like that of Wiener, gives some insight into the mathematical structure of theorem Wiener process but obscures the connection with random walk. There is another approach, however, that makes direct use 2 of the central limit theorem (5). Unfortunately, this approach requires a bit more technology (specifically, weak convergence theory for measures on metric space; see he book Weak Convergence of Probability Measures by P. Billingsley for details) than we have time for in this course. However, it has the advantage that it also explains, at least in part, why the Wiener process is useful in the modeling of natural processes. Many stochastic processes behave, at least for long stretches of time, like random walks with small but frequent jumps. The weak convergence theory shows that such processes will look, at least approximately, and on the appropriate time scale, like Brownian motion.
Notation and Terminology. A Brownian motion with initial point x is a stochastic process {Wt}t≥0 such that {Wt −x}t≥0 is a standard Brownian motion. Unless other-wise specified, Brownian motion means standard Brownian motion. To ease eyestrain, we will adopt the convention that whenever convenient the index t will be written as a functional argument instead of as a subscript, that is, W(t) = Wt.
1.2 Brownian motion and diffusion The mathematical study of Brownian motion arose out of the recognition by Ein-stein that the random motion of molecules was responsible for the macroscopic phenomenon of diffusion. Thus, it should be no surprise that there are deep con-nections between the theory of Brownian motion and parabolic partial differential equations such as the heat and diffusion equations. At the root of the connection is the Gauss kernel, which is the transition probability function for Brownian motion: P(Wt+s ∈dy | Ws = x) ∆ = pt(x, y)dy = 1 √ 2πt exp{−(y −x)2/2t}dy.
(6) This equation follows directly from properties (3)–(4) in the definition of a standard Brownian motion, and the definition of the normal distribution. The function pt(y|x) = pt(x, y) is called the Gauss kernel, or sometimes the heat kernel. (In the parlance of the PDE folks, it is the fundamental solution of the heat equation). Here is why: Theorem 2. Let f : R →R be a continuous, bounded function. Then the unique (contin-uous) solution ut(x) to the initial value problem ∂u ∂t = 1 2 ∂2u ∂x2 (7) u0(x) = f(x) (8) is given by ut(x) = Ef(W x t ) = Z ∞ y=−∞ pt(x, y)f(y) dy.
(9) 3 Here W x t is a Brownian motion started at x.
The equation (7) is called the heat equation. That the PDE (7) has only one solution that satisfies the initial condition (8) follows from the maximum principle: see a PDE text for details. More important (for us) is that the solution is given by the expectation formula (9). To see that the right side of (9) actually does solve (7), take the partial derivatives in the PDE (7) under the integral in (9). You then see that the issue boils down to showing that ∂pt(x, y) ∂t = 1 2 ∂2pt(x, y) ∂x2 .
(10) Exercise: Verify this.
1.3 Brownian motion in higher dimensions Definition 2. A standard d−dimensional Brownian motion is an Rd−valued continuous-time stochastic process {Wt}t≥0 (i.e., a family of d−dimensional random vectors Wt indexed by the set of nonnegative real numbers t) with the following properties.
(A)’ W0 = 0.
(B)’ With probability 1, the function t →Wt is continuous in t.
(C)’ The process {Wt}t≥0 has stationary, independent increments.
(D)’ The increment Wt+s −Ws has the d−dimensional normal distribution with mean vector 0 and covariance matrix tI.
The d−dimensional normal distribution with mean vector 0 and (positive definite) covariance matrix Σ is the Borel probability measure on Rd with density ϕΣ(x) = ((2π)d det(Σ))−1/2 exp{−xTΣ−1x/2}; if Σ = tI then this is just the product of d one-dimensional Gaussian distributions with mean 0 and variance t. Thus, the existence of d−dimensional Brownian motion follows directly from the existence of 1−dimensional Brownian motion: if {W (i)}t≥0 are independent 1−dimensional Brownian motions then Wt = (W (1) t , W (2) t , . . . , W (d) t ) is a d-dimensional Brownian motion.
One of the important properties of the d−dimensional normal distribution with mean zero and covariance matrix tI pro-portional to the identity is its invariance under orthogonal transformations. This im-plies that if {Wt}t≥0 is a d−dimensional Brownian motion then for any orthogonal transformation U of Rd the process {UWt}t≥0 is also a d−dimensional Brownian motion.
4 1.4 Symmetries and Scaling Laws Proposition 1. Let {W(t)}t≥0 be a standard Brownian motion. Then each of the following processes is also a standard Brownian motion: {−W(t)}t≥0 (11) {W(t + s) −W(s)}t≥0 (12) {aW(t/a2)}t≥0 (13) {tW(1/t)}t≥0.
(14) Exercise: Prove this.
The scaling law (13) is especially important. It is often advisable, when con-fronted with a problem about Wiener processes, to begin by reflecting on how scaling might affect the answer. Consider,as a first example, the maximum and minimum random variables M(t) := max{W(s) : 0 ≤s ≤t} and (15) M −(t) := min{W(s) : 0 ≤s ≤t}.
(16) These are well-defined, because the Wiener process has continuous paths, and continuous functions always attain their maximal and minimal values on compact intervals. Now observe that if the path W(s) is replaced by its reflection −W(s) then the maximum and the minimum are interchanged and negated. But since −W(s) is again a Wiener process, it follows that M(t) and −M −(t) have the same distribution: M(t) D = −M −(t).
(17) Next, consider the implications of Brownian scaling. Fix a > 0, and define W ∗(t) = aW(t/a2) and M ∗(t) = max 0≤s≤t W ∗(s) = max 0≤s≤t aW(s/a2) = aM(t/a2).
By the Brownian scaling property, W ∗(s) is a standard Brownian motion, and so the random variable M ∗(t) has the same distribution as M(t). Therefore, M(t) D = aM(t/a2).
(18) Exercise: Use Brownian scaling to deduce a scaling law for the first-passage time random variables τ(a) defined as follows: τ(a) = min{t : W(t) = a} (19) or τ(a) = ∞on the event that the process W(t) never attains the value a.
5 1.5 The Wiener Isometry The existence theorem (Theorem 1) was first proved by N. Wiener around 1920.
Simpler proofs have since been found, but Wiener’s argument contains the germ of an extremely useful insight, which is now known as the Wiener isometry (or, in some of the older literature, the Wiener integral). Following is an account of Wiener’s line of thought.
Suppose that Brownian motion exists, that is, suppose that on some probability space (Ω, F, P) there is a centered Gaussian process {Wt}t∈[0,1] with covariance EWtWs = min(s, t). The random variables Wt are all elements of the space L2(P) consisting of the real random variables defined on (Ω, F, P) with finite second moments. This space is a Hilbert space with inner product ⟨X, Y ⟩= (X, Y ).
Now consider the Hilbert space L2[0, 1], consisting of all real-valued square-integrable functions on the unit interval, with inner product ⟨f, g⟩= Z 1 0 f(x)g(x) dx.
Indicator functions 1[0,t] of intervals [0, t] are elements of L2[0, 1], and obviously ⟨1[0,t], 1[0,s]⟩= min(s, t).
Thus, the indicators 1[0,t] have exactly the same inner products as do the random variables Wt in the Hilbert space L2(P). Wiener’s key insight was that this identity between inner products implies that there is a linear isometry IW from L2[0, 1] into L2(P) mapping each indicator 1[0,t] to the corresponding random variable Wt.
Theorem 3. (Wiener’s Isometry) Let {Wt}t≥0 be a standard Wiener process defined on a probability space (Ω, F, P). Then for any nonempty interval J ⊆R+ the mapping 1(s,t] 7→Wt −Ws extends to a linear isometry IW : L2(J) →L2(Ω, F, P). For every function ϕ ∈L2(J), the random variable IW(ϕ) is mean-zero Gaussian.
Proof. Given the identity ⟨1[0,t], 1[0,s]⟩= ⟨Ws, Wt⟩, the theorem is a straightforward use of standard results in Hilbert space theory. Let H0 be the set of all finite linear combinations of interval indicator functions 1A. Then H0 is a dense, linear subspace of L2(J), that is, every function f ∈L2(J) can be approximated arbitrarily closely in the L2−metric by elements of H0. Since IW is a linear isometry of H0, it extends uniquely to a linear isometry of L2(J), by standard results in Hilbert space theory.
We claim that for any ϕ ∈L2(J), the random variable IW(ϕ) must be (mean-zero) Gaussian (or identically zero, in case ϕ = 0 a.e.). This can be seen as follows: Since H0 is dense in L2, there exists a sequence ϕn in H0 such that ϕn →ϕ in L2.
Since IW is an isometry, IW(ϕn) →IW(ϕ) in L2. Now convergence in L2 implies 6 convergence in measure, which in turn implies convergence in distribution: in particular, for any real θ, lim n→∞E exp iθIW(ϕn) = E exp iθIW(ϕ).
Each of the random variables IW(ϕn) has a centered Gaussian distribution, and consequently has a characteristic function (Fourier transform) E exp iθIW(ϕn) = exp{−θ2σ2 n/2}.
The weak convergence of these random variables now implies that the variances σ2 n converge to a nonnegative limit σ2, and that the limit random variable IW(ϕ) has characteristic function E exp iθIW(ϕ) = exp{−θ2σ2/2}.
But this implies that IW(ϕ) has a (possibly degenerate) Gaussian distribution with mean 0 and variance 2.
The Hilbert space isometry IW suggests a natural approach to explicit represen-tations of the Wiener process, via orthonormal bases. The idea is this: if {ψn}n∈N is an orthonormal basis of L2[0, 1], then {IW(ψn)}n∈N must be an orthonormal set in L2(P). Since uncorrelated Gaussian random variables are necessarily independent, it follows that the random variables ξn := IW(ψn) must be i.i.d. standard normals.
Finally, since IW is a linear isometry, it must map the L2−series expansion of 1[0,t] in the basis ψn to the series expansion of Wt in the basis ξn. Thus, with almost no further work we conclude the following.
Corollary 1. Assume that the probability space (Ω, F, P) supports an infinite sequence ξn of independent, identically distributed N(0, 1) random variables, and let {ψn}n∈N be any orthonormal basis of L2[0, 1]. Then for every t ∈[0, 1] the infinite series Wt := ∞ X n=1 ξn⟨1[0,t], ψn⟩ (20) converges in the L2−metric, and the resulting stochastic process {Wt}t∈[0,1] is a mean-zero Gaussian process with covariance function EWtWs = min(s, t).
Remark 1. Theorem 3 implies not only that the random variables Wt individually have Gaussian distributions, but that for any finite collection 0 < t1 < t2 < · · · < tm ≤1 the random vector (Wti)1≤i≤m has a (multivariate) Gaussian distribution. To prove this, first observe that it suffices to show that the joint characteristic function E exp{i m X j=1 θjWtj} 7 is the characteristic function of the appropriate multivariate Gaussian distribution.
But this follows from Theorem 3, because for each choice of real constants θj, m X j=1 θjWtj = IW m X j=1 θj1[0,tj] !
.
Because the convergence in (20) is in the L2−metric, rather than the sup-norm, there is no way to conclude directly that the process so constructed has a version with continuous paths. Wiener was able to show by brute force that for the particu-lar basis ψn(x) = √ 2 cos πnx the series (20) converges (along an appropriate subsequence) not only in L2 but also uniformly in t, and therefore gives a version of the Wiener process with continuous paths: Wt = ξ0t + ∞ X k=1 2k−1 X n=2k−1 n−1ξn √ 2 sin πnt.
(21) The proof of this uniform convergence is somewhat technical, though, and more-over, it is in many ways unnatural. Thus, rather than following Wiener’s construc-tion, we will describe a different construction, due to P. Lévy.
1.6 Lévy’s Construction Lévy discovered that a more natural orthonormal basis for the construction of the Wiener process is the Haar wavelet basis. The Haar basis is defined as follows: first, let ψ : R →{−1, 1} be the “mother wavelet” function ψ(t) = 1 if 0 ≤t ≤1 2; −1 if 1 2 < t ≤1; 0 otherwise.
(22) Then for any integers n ≥0 and 0 ≤k < 2n define the (n, k)th Haar function by ψn,k(t) = 2n/2ψ(2nt −k) (23) The function ψn,k has support [k/2n, (k + 1)/2n], and has absolute value equal to 2n/2 on this interval, so its L2−norm is 1. Note that ψ0,0 = 1 on [0, 1]. Moreover, the functions ψn,k are mutually orthogonal: ⟨ψn,k, ψm,l⟩= ( 1 if n = m and k = l; 0 otherwise.
(24) 8 Exercise 1. Prove that the Haar functions {ψn,k}n≥0,0≤k<2n form a complete or-thonormal basis of L2[0, 1]. HINT: Linear combinations of the indicator functions of dyadic intervals [k/2n, (k + 1)/2n] are dense in L2[0, 1].
In certain senses the Haar basis is better suited to Brownian motion than the Fourier basis, in part because the functions ψn,k are “localized” (which fits with the independent increments property), and in part because the normalization of the functions forces the scale factor 2n/2 in (23) (which fits with the Brownian scaling law).
The series expansion (20) involves the inner products of the basis functions with the indicators 10,t. For the Haar basis, these inner products define the Schauder functions Gn,k, which are defined as the indefinite integrals of the Haar functions: Gn,k(t) = ⟨1[0,t], ψn,k⟩= Z t 0 ψn,k(s) ds (25) The graphs of these functions are steeply peaked “hats” sitting on the dyadic inter-vals [k/2n, (k + 1)/2n], with heights 2−n/2 and slopes ±2n/2. Note that G0,0(t) = t.
Theorem 4. (Lévy) If the random variables ξm,k are independent, identically distributed with common distribution N(0, 1), then with probability one, the infinite series W(t) := ξ0,1t + ∞ X m=1 2m−1 X k=0 ξm,kGm,k(t) (26) converges uniformly for 0 ≤t ≤1 and the limit function W(t) is a standard Wiener process.
Lemma 1. If Z is a standard normal random variable then for every x > 0, P{Z > x} ≤2e−x2/2 √ 2πx .
(27) Proof.
P{Z > x} = 1 √ 2π Z ∞ x e−y2/2 dy ≤ 1 √ 2π Z ∞ x e−xy/2 dy = 2e−x2/2 √ 2πx .
Proof of Theorem 4. By definition of the Schauder functions Gn,k, the series (26) is a particular case of (20), so the random variables W(t) defined by (26) are centered Gaussian with covariances that agree with the covariances of a Wiener process.
9 Hence, to prove that (26) defines a Brownian motion, it suffices to prove that with probability one the series converges uniformly for t ∈[0, 1].
The Schauder function Gm,k has maximum value 2−m/2, so to prove that the series (26) converges uniformly it is enough to show that ∞ X m=1 2m X k=1 |ξm,k|/2m/2 < ∞ with probability 1. To do this we will use the Borel-Cantelli Lemma and the tail estimate of Lemma 1 for the normal distribution to show that with probability one there is a (possibly random) m∗such that max k |ξm,k| ≤2m/4 for all m ≥m∗.
(28) This will imply that almost surely the series is eventually dominated by a multiple of the geometric series P 2−(m+2)/4, and consequently converges uniformly in t.
To prove that (28) holds eventually, it suffices (by Borel-Cantelli) to show that the probabilities of the complementary events are summable. By Lemma 1, P{|ξm,k| ≥2m/4} ≤ 4 2m/4√ 2πe−2m/2.
Hence, by the Bonferroni inequality (i.e., the crude union bound), P{ max 1≤k≤2m |ξm,k| ≥2m/4} ≤2m2−m/4p 2/πe−2m−1.
Since this bound is summable in m, Borel-Cantelli implies that with probability 1, eventually (28) must hold. This proves that w.p.1 the series (26) converges uni-formly, and therefore W(t) is continuous.
2 The Markov and Strong Markov Properties 2.1 The Markov Property Property (12) is a rudimentary form of the Markov property of Brownian motion.
The Markov property asserts something more: not only is the process {W(t + s) −W(s)}t≥0 a standard Brownian motion, but it is independent of the path {W(r)}0≤r≤s up to time s. This may be stated more precisely using the language of σ−algebras. Define FW t := σ({Ws}0≤s≤t) (29) 10 to be the smallest σ−algebra containing all events of the form {Ws ∈B} where 0 ≤s ≤t and B ⊂R is a Borel set. The indexed collection of σ−algebra {FW t }t≥0 is called the standard filtration associated with the Brownian motion.
Example: For each t > 0 and for every a ∈R, the event {M(t) > a} is an element of FW t . To see this, observe that by path-continuity, {M(t) > a} = [ s∈Q:0≤s≤t {W(s) > a}.
(30) Here Q denotes the set of rational numbers. Because Q is a countable set, the union in (30) is a countable union. Since each of the events {W(s) > a} in the union is an element of the σ−algebra FW t , the event {M(t) > a} must also be an element of FW t .
In general, a filtration is an increasing family of σ−algebras {Gt}t≥0 indexed by time t. A stochastic process X(t) is said to be adapted to a filtration {Gt}t≥0 if the random variable X(t) is measurable relative to Gt for every t ≥0. It is sometimes necessary to consider filtrations other than the standard one, because in some situations there are several sources of randomness (additional independent Brownian motions, Poisson processes, coin tosses, etc.).
Definition 3. Let {Wt}t≥0 be a Wiener process and {Gt}t≥0 a filtration of the prob-ability space on which the Wiener process is defined. The filtration is said to be admissible1 for the Wiener process if (a) the Wiener process is adapted to the filtra-tion, and (b) for every t ≥0, the post-t process {Wt+s −Wt}s≥0 is independent of the σ−algebra Gt.
Of course, we have not yet defined what it means for a stochastic process {Xt}t∈J to be independent of a σ−algebra G, but the correct definition is easy to guess: the process {Xt}t∈J is independent of G if the σ−algebra σ(Xt)t∈J generated by the random variables Xt (i.e., the smallest σ−algebra containing all events of the form Xt ∈B, where t ∈J and B is a Borel set) is independent of G.
Proposition 2. (Markov Property) If {W(t)}t≥0 is a standard Brownian motion, then the standard filtration {FW t }t≥0 is admissible.
Proof of the Markov Property. This is nothing more than a sophisticated restatement of the independent increments property of Brownian motion. Fix s ≥0, and consider two events of the form A = ∩n j=1{W(sj) −W(sj−1) ≤xj} ∈Fs and B = ∩m j=1{W(tj + s) −W(tj−1 + s) ≤yj}.
1This is not standard terminology. Some authors (for instance, Karatzas and Shreve) call such filtrations Brownian filtrations.
11 By the independent increments hypothesis, events A and B are independent. Events of type A generate the σ−algebra Fs, and events of type B generate the smallest σ−algebra with respect to which the post-s Brownian motion W(t + s) −W(s) is measurable. Consequently, the post-s Brownian motion is independent of the σ−algebra Fs.
2.2 Sufficient conditions for independence The Markov property is a straightforward consequence of the independent incre-ments property of the Wiener process, and the proof extends without change to an arbitrary Lévy process, as neither the continuity of paths nor the Gaussian distribution of the increments played any role in the argument. Establishing the independence of a stochastic process from a σ−algebra is not always so easy, how-ever.
In general, two σ−algebras G and H are independent if for every G ∈G and every H ∈H the events G, H are independent, that is, if P(G ∩H) = P(G)P(H).
Equivalently, G and H are independent if for every H ∈H the conditional proba-bility of H given G coincides with the unconditional probability: E(1H | G) = E1H = P(H) almost surely.
(31) When H = σ(Xt)t∈J is the σ−algebra generated by a (real) stochastic process Xt, other sufficient conditions for independence from G are sometimes useful. Since any event in H is the almost sure limit of a sequence of cylinder events (events of the form ∩m i=1{Xti ∈Bi}), to prove (31) it suffices to consider cylinder events.
Similarly, for any cylinder event A = ∩m i=1{Xti ∈Bi} there is a sequence of bounded, continuous functions gn : Rm →[0, 1] such that lim n→ε gn(Xt1, Xt2, . . . , Xtm) = 1A almost surely; thus,H is independent of G if and only if for every finite subset {ti}i≤m ⊂J and every bounded, continuous function g : Rm →R, E(g(Xt1, Xt2, . . . , Xtm) | G) = Eg(Xt1, Xt2, . . . , Xtm) almost surely.
(32) Finally, since any bounded, continuous function g can be arbitrarily well approxi-mated by finite trig polynomials, to prove (32) it suffices to show that the conditional joint characteristic function (Fourier transform) of Xt1, Xt2, . . . , Xtm given G coin-cides (almost surely) with the unconditional joint characteristic function, that is, E(exp{i m X j=1 θjXtj} | G) = E exp{i m X j=1 θjXtj} almost surely.
(33) 12 Exercise 2. Prove the approximation statements used in the arguments above.
Specifically, (a) prove that any event in H is the almost sure limit of cylinder events; and (b) the indicator function of any cylinder event is the almost sure limit of continuous functions.
The sufficient conditions (32) and (33) have an easy consequence that is some-times useful.
Lemma 2. Let {Xn t }n≥1,t∈J be a sequence of stochastic processes indexed by t ∈J, all defined on the same underlying probability space (Ω, F, P). Let G be a σ−algebra contained in F, and suppose that each process {Xn t }t∈J is independent of G. If lim n→∞Xn t = Xt a.s. for each t ∈J (34) then the stochastic process {Xt}t∈J is independent of G.
Proof. It suffices to establish the identity (32) for every finite subset {ti}i≤m of the index set J and every bounded continuous function g on Rm. Because each process {Xn t }t∈J is independent of G, equation (32) holds when X is replaced by Xn. Since g is continuous, the hypothesis of the lemma implies that lim n→∞g(Xn t1, Xn t2, . . . , Xn tm) = g(Xt1, Xt2, . . . , Xtm) a.s.
Consequently, by the dominated convergence theorem (applied twice, once for ordinary expectations and once for conditional expectations), equation (32) holds for the limit process {Xt}t∈J 2.3 Blumenthal’s 0-1 Law Recall that the intersection of any collection of σ−algebras is a σ−algebra. Thus, if F = {Ft}t≥0 is a filtration, then so is the collection F+ = {Ft+}t≥0, where for each t, Ft+ = ∩s>tFs (35) Call this the right-continuous augmentation of the filtration F. The term right-continuous is used because the enlarged filtration has the property that Ft+ = ∩s>tFs+.
Clearly, Ft ⊂Ft+, because the σ−algebra Ft is contained in Ft+ε for every ε > 0.
One might wonder what (if any) additional “information” there is in Ft+ that is not already contained in Ft.
Example 1. Since the random variable W0 = 0 is constant, the σ−algebra FW 0 is trivial: it contains only the events ∅and Ω= ∅c. The augmentation FW 0+, however, contains interesting events, for instance, the events A+ = ∩n≥1 ∪q∈Q;q≤2−n {Wq > 0} and (36) A−= ∩n≥1 ∪q∈Q;q≤2−n {Wq < 0}.
13 Because Brownian paths are continuous, the event A+ (respectively, A−) is the event that the path enters the positive (respectively, negative) halfline at arbitrarily small times. These events are elements of FW 0+, because the inner union in each definition is an event in FW 2−n.
Proposition 3. If {Wt}t≥0 is a standard Brownian motion and {FW t }t≥0 is the standard filtration then the augmented filtration {FW t+}t≥0 is admissible.
Proof. We must show that for each s ≥0, the post-s process {∆W(s, t) := Wt+s − Ws}t≥0 is independent of the σ−algebra FW s+. By the Markov property (Proposi-tion 2), for any m ≥1 the post-(s + 2−m) process {∆W(s + 2m, t)}t≥0 is indepen-dent of FW s+2−n, and since FW s+ is contained in FW s+2−n, it follows that the process {∆W(s + 2m, t)}t≥0 is independent of FW s+. But the path-continuity of the Wiener process guarantees that for each t ≥0, lim m→∞∆W(s + 2−m, t) = ∆W(s, t), and so Lemma 2 implies that the process {∆W(s, t)}t≥0 is independent of FW s+.
Corollary 2. For every t ≥0 and every integrable random variable Y that is measurable with respect to the σ−algebra FW ∞generated by a standard Wiener process, E(Y | FW t+) = E(Y | FW t ).
(37) Consequently, for any event F ∈FW t+ there is an event F ′ ∈FW t such that P(F∆F ′) = 0.
Proof. By the usual approximation arguments, to prove (37) it suffices to consider random variables of the form Y = exp{i m X j=1 θj(Wtj −Wtj−1)} where 0 ≤t1 < t2 < · · · < tk = t < tk+1 < · · · < tm. For such tj, the random variable P j≤k θj(Wtj−Wtj−1) is measurable with respect to FW t , and hence also with respect to FW t+, while by Proposition 3 the random variable P k 0} has probability at least 1/2, because the chance that W2−n > 0 is 1/2. Consequently, the intersection must have probability at least 1/2, and so by Blumenthal’s Law, P(A+) = 1. A similar argument shows that P(A−) = 1. Hence, P(A+ ∩A−) = 1.
(38) On the event A+, the path Wt takes positive values at indefinitely small positive times; similarly, on A−the path Wt takes negative values at indefinitely small positive time. Therefore, on the event A+ ∩A−, the path Wt must cross and recross the origin 0 infinitely often in each time interval (0, 2−n). Blumenthal’s 0-1 Law implies that this must be the case with probability one.
2.4 Stopping times and stopping fields A stopping time for a filtration F = {Ft}t≥0 is defined to be a nonnegative (possibly infinite)2 random variable τ such that for each (nonrandom) t ∈[0, ∞] the event {τ ≤t} is in the σ−algebra Ft.
Example: τ(a) := min{t : W(t) = a} is a stopping time relative to the standard filtration. To see this, observe that, because the paths of the Wiener process are continuous, the event {τ(a) ≤t} is identical to the event {M(t) ≥a}. We have already shown that this event is an element of Ft.
2A stopping time that takes only finite values will be called a finite stopping time.
15 Exercise 3. (a) Prove that if τ, σ are stopping times then so are σ ∧τ and σ ∨τ.
(b) Prove that if τ is a finite stopping time then so is τn = min{k/2n ≥τ}.
Thus, any finite stopping time is a decreasing limit of stopping times each of which takes values in a discrete set.
The stopping field Fτ (more accurately, the stopping σ−algebra) associated with a stopping time τ is the collection of all events B such that B ∩{τ ≤t} ∈Ft for every nonrandom t. Informally, Fτ consists of all events that are “observable” by time τ.
Exercise 4. (a) Show that Fτ is a σ−algebra. (b) Show that if σ ≤τ are both stopping times then Fσ ⊂Fτ.
Exercise 5. Let τ be a finite stopping time with respect to the standard filtration of a Brownian motion {Wt}t≥0 and let FW τ be the associated stopping field. Show that each of the random variables τ, Wτ, and Mτ is measurable relative to FW τ .
2.5 Strong Markov Property Theorem 5. (Strong Markov Property) Let {W(t)}t≥0 be a standard Brownian motion, and let τ be a finite stopping time relative to the standard filtration, with associated stop-ping σ−algebra Fτ. For t ≥0, define the post-τ process W ∗(t) = W(t + τ) −W(τ), (39) and let {F∗ t }t≥0 be the standard filtration for this process. Then (a) {W ∗(t)}t≥0 is a standard Brownian motion; and (b) for each t > 0, the σ−algebra F∗ t is independent of Fτ.
The following variant of the strong Markov property is also useful.
Theorem 6. (Splicing) Let {Wt}t≥0 be a Brownian motion with admissible filtration {F}t, and let τ be a stopping time for this filtration. Let {W ∗ s }s≥0 be a second Brownian motion on the same probability space that is independent of the stopping field Fτ. Then the spliced process ˜ Wt = Wt for t ≤τ, (40) = Wτ + W ∗ t−τ for t ≥τ is also a Brownian motion.
16 The hypothesis that τ is a stopping time is essential for the truth of the strong Markov Property. Consider the following example. Let T be the first time that the Wiener path reaches its maximum value up to time 1, that is, T = min{t : W(t) = M(1)}.
The random variable T is well-defined, by path-continuity, as this assures that the set of times t ≤1 such that W(t) = M(1) is closed and nonempty. Since M(1) is the maximum value attained by the Wiener path up to time 1, the post-T path W ∗(s) = W(T + s) −W(T) cannot enter the positive half-line (0, ∞) for s ≤1 −T.
Later we will show that T < 1 almost surely; thus, almost surely, W ∗(s) does not immediately enter (0, ∞). Now if the strong Markov Property were true for the random time T, then it would follow that, almost surely, W(s) does not immediately enter (0, ∞). Since −W(s) is also a Wiener process, we may infer that, almost surely, W(s) does not immediately enter (−∞, 0), and so W(s) = 0 for all s in a (random) time interval of positive duration beginning at 0. But we have already shown, using Blumenthal’s Law, that with probability 1, Brownian paths immediately enter both the positive and negative halflines. This is a contradiction, so we conclude that the strong Markov property fails for the random variable T.
Proof of the strong Markov property. First, we will show that it suffices to prove the theorem for stopping times that take values in the discrete set Dm = {k/2m}k≥0, for some m ≥1; then we will prove that the strong Markov property holds for stopping times that take values in Dm.
Step 1. Suppose that the theorem is true for all stopping times that take values in one of the sets Dm; we will show that it is then true for any finite stopping time. By Exercise 3, if τ is a finite stopping time then so is τm = min{k/2m ≥τ}. Clearly, the sequence τm is non-increasing in m, and τm →τ as m →∞. Hence, by Exercise 4, the stopping fields Fτm are reverse-ordered, and Fτ ⊂Fτm for every m ≥1.
By hypothesis, for each m ≥1 the post-τm process ∆W(τm, t) = W(t + τm) − W(τm) is a standard Brownian motion independent of Fτm. Since Fτ ⊂Fτm, the pro-cess ∆W(τm,t) is also independent of Fτ. But Brownian paths are right-continuous, so the convergence τm ↓τ implies that for each t ≥0, lim m→∞∆W(τm, t) = ∆W(τ, t) = W ∗(t).
(41) Consequently, by Lemma 2, the post-τ process W ∗(t) is independent of Fτ.
To complete the first step of the proof we must show that the process W ∗(t) is a Brownian motion. The sample paths are obviously continuous, since the paths of the original Brownian motion W(t) are continuous, so we need only verify 17 properties (C) and (D) of Definition 1. For this is suffices to show that for any 0 = t0 < t1 < · · · < tk, E exp ( i k X j=1 θj(W ∗ tj −W ∗ tj−1) ) = E exp ( i k X j=1 θj(Wtj −Wtj−1) ) (42) But by hypothesis, each of the processes {∆W(τm, t)}t≥0 is a Brownian motion, so for each m the equation (42) holds when W ∗ t is replaced by ∆W(τm, t) in the expectation on the left side. Consequently, the equality (42) follows by (41) and the dominated convergence theorem.
Step 2: We must prove that if τ is a stopping time that takes values in Dm, for some m ≥0, then the post-τ process {W ∗ t }t≥0 is a Brownian motion and is independent of Fτ. For ease of notation, we will assume that m = 0; the general case can be proved by replacing each integer n in the following argument by n/2m.
It will suffice to show that if B is any event in Fτ then for any 0 = t0 < t1 < · · · < tk and θ1, θ2, . . . , θk, E1B exp ( i k X j=1 θj(W ∗ tj −W ∗ tj−1) ) = P(B)E exp ( i k X j=1 θj(Wtj −Wtj−1) ) .
(43) Since τ takes only nonnegative integer values, the event B can be partitioned as B = ∪n≥0(B ∩{τ = n}). Since τ is a stopping time, the event B ∩{τ = n} is in the σ−algebra Fn; moreover, on {τ = n}, the post-τ process coincides with the post-n process ∆W(n, t) = Wt+n −Wn. Now the Markov property implies that the post-n process ∆W(n, t) is a Brownian motion independent of Fn, and hence independent of B ∩{τ = n}. Consequently, E1B1{τ=n} exp {i k X j=1 θj(W ∗ tj −W ∗ tj−1) = E1B1{τ=n} exp ( i k X j=1 θj(∆W(n, tj) −∆W(n, tk−1)) ) = P(B ∩{τ = n})E exp ( i k X j=1 θj(Wtj −Wtj−1) ) .
Summing over n = 0, 1, 2, . . . (using the dominated convergence theorem on the left and the monotone convergence theorem on the right), we obtain (43).
Proof of the Splicing Theorem 6. This is done using the same strategy as used in prov-ing the strong Markov property: first, one proves that it suffices to consider stop-ping times τ that take values in Dm, for some m; then one proves the result for this 18 restricted class of stopping times. Both steps are very similar to the corresponding steps in the proof of the strong Markov property, so we omit the details.
3 Embedded Simple Random Walks Lemma 3. Define τ = min{t > 0 : |Wt| = 1}. Then with probability 1, τ < ∞.
Proof. This should be a familiar argument: I’ll define a sequence of independent Bernoulli trials Gn in such a way that if any of them results in a success, then the path Wt must escape from the interval [−1, 1]. Set Gn = {Wn+1 −Wn > 2}. These events are independent, and each has probability p := 1 −Φ(2) > 0. Since p > 0, infinitely many of the events Gn will occur (and in fact the number N of trials until the first success will have the geometric distribution with parameter p). Clearly, if Gn occurs, then τ ≤n + 1.
The lemma guarantees that there will be a first time τ1 = τ when the Wiener process has traveled ±1 from its initial point. Since this time is a stopping time, the post-τ process Wt+τ −Wτ is an independent Wiener process, by the strong Markov property, and so there will be a first time when it has traveled ±1 from its starting point, and so on. Because the post-τ process is independent of the stopping field Fτ, it is, in particular, independent of the random variable W(τ), and so the sequence of future ±1 jumps is independent of the first. By an easy induction argument, the sequence of ±1 jumps made in this sequence are independent and identically distributed. Similarly, the sequence of elapsed times are i.i.d. copies of τ. Formally, define τ0 = 0 and τn+1 := min{t > τn : |Wt+τn −Wτn| = 1}.
(44) The arguments above imply the following.
Proposition 4. The sequence Yn := W(τn) is a simple random walk started at Y0 = W0 = 0. Furthermore, the sequence of random vectors (W(τn+1) −W(τn), τn+1 −τn) is independent and identically distributed.
Corollary 4. With probability one, the Wiener process visits every real number.
Proof. The recurrence of simple random walk implies that Wt must visit every integer, in fact infinitely many times. Path-continuity and the intermediate value theorem therefore imply that the path must travel through every real number.
19 There isn’t anything special about the values ±1 for the Wiener process — in fact, Brownian scaling implies that there is an embedded simple random walk on each discrete lattice (i.e., discrete additive subgroup) of R. It isn’t hard to see (or to prove, for that matter) that the embedded simple random walks on the lattices 2−mZ “fill out” the Brownian path in such a way that as m →∞the polygonal paths gotten by connecting the dots in the embedded simple random walks converge uniformly (on compact time intervals) to the path Wt. This can be used to provide a precise meaning for the assertion made earlier that Brownian motion is, in some sense, a continuum limit of random walks. We’ll come back to this later in the course (maybe).
The embedding of simple random walks in Brownian motion has other, more subtle ramifications that have to do with Brownian local time. We’ll discuss this when we have a few more tools (in particular, the Itô formula) available. For now I’ll just remark that the key is the way that the embedded simple random walks on the nested lattices 2−kZ fit together. It is clear that the embedded SRW on 2−k−1Z is a subsequence of the embedded SRW on 2−kZ. Furthermore, the way that it fits in as a subsequence is exactly the same (statistically speaking) as the way that the embedded SRW on 2−1Z fits into the embedded SRW on Z, by Brownian scaling.
Thus, there is an infinite sequence of nested simple random walks on the lattices 2−kZ, for k ∈Z, that fill out (and hence, by path-continuity, determine) the Wiener path. OK, enough for now.
One last remark in connection with Proposition 4: There is a more general — and less obvious — theorem of Skorohod to the effect that every mean zero, finite variance random walk on R is embedded in standard Brownian motion. See sec. ??
below for more.
4 The Reflection Principle Denote by Mt = M(t) the maximum of the Wiener process up to time t, and by τa = τ(a) the first passage time to the value a.
Proposition 5.
P{M(t) ≥a} = P{τa ≤t} = 2P{W(t) > a} = 2 −2Φ(a/ √ t).
(45) The argument will be based on a symmetry principle that may be traced back to the French mathematician D. ANDRÉ. This is often referred to as the reflection principle. The essential point of the argument is this: if τ(a) < t, then W(t) is just as likely to be above the level a as to be below the level a. Justification of this claim requires the use of the Strong Markov Property. Write τ = τ(a). By Corollary 4 20 above, τ < ∞almost surely. Since τ is a stopping time, the post-τ process W ∗(t) := W(τ + t) −W(τ) (46) is a Wiener process, and is independent of the stopping field Fτ. Consequently, the reflection {−W ∗(t)}t≥0 is also a Wiener process, and is independent of the stopping field Fτ. Thus, if we were to run the original Wiener process W(s) until the time τ of first passage to the value a and then attach not W ∗but instead its reflection −W ∗, we would again obtain a Wiener process. This new process is formally defined as follows: ˜ W(s) = W(s) for s ≤τ, (47) = 2a −W(s) for s ≥τ.
Proposition 6. (Reflection Principle) If {W(t)}t≥0 is a Wiener process, then so is { ˜ W(t)}t≥0.
Proof. This is just a special case of the Splicing Theorem 6.
Proof of Proposition 5. The reflected process ˜ W is a Brownian motion that agrees with the original Brownian motion W up until the first time τ = τ(a) that the path(s) reaches the level a. In particular, τ is the first passage time to the level a for the Brownian motion ˜ W. Hence, P{τ < t and W(t) < a} = P{τ < t and ˜ W(t) < a}.
After time τ, the path ˜ W is gotten by reflecting the path W in the line w = a.
Consequently, on the event τ < t, W(t) < a if and only if ˜ W(t) > a, and so P{τ < t and ˜ W(t) < a} = P{τ < t and W(t) > a}.
Combining the last two displayed equalities, and using the fact that P{W(t) = a} = 0, we obtain P{τ < a} = 2P{τ < t and W(t) > a} = 2P{W(t) > a}.
Corollary 5. The first-passage time random variable τ(a) is almost surely finite, and has the one-sided stable probability density function of index 1/2: f(t) = ae−a2/2t √ 2πt3 .
(48) Proof. Differentiate the equation (45) with respect to the variable t.
21 Essentially the same arguments prove the following.
Corollary 6.
P{M(t) ∈da and W(t) ∈a −db} = 2(a + b) exp{−(a + b)2/2t} (2π)1/2t3/2 dadb (49) It follows, by an easy calculation, that for every t the random variables |Wt| and Mt −Wt have the same distribution. In fact, the processes |Wt| and Mt −Wt have the same joint distributions: Proposition 7. (P. Lévy) The processes {Mt −Wt}t≥0 and {|Wt|}t≥0 have the same distri-butions.
Exercise 6. Prove this. Hints: (A) It is enough to show that the two processes have the same finite-dimensional distributions, that is, that for any finite set of time points t1, t2, . . . , tk the joint distributions of the two processes at the time points ti are the same. (B) By the Markov property for the Wiener process, to prove equality of finite-dimensional distributions it is enough to show that the two-dimensional distributions are the same. (C) For this, use the Reflection Principle.
Remark 2. The reflection principle and its use in determining the distributions of the max Mt and the first-passage time τ(a) are really no different from their analogues for simple random walks, about which you learned in 312. In fact, we could have obtained the results for Brownian motion directly from the corresponding results for simple random walk, by using embedding.
Exercise 7. Brownian motion with absorption.
(A) Define Brownian motion with absorption at 0 by Yt = Wt∧τ(0), that is, Yt is the process that follows the Brownian path until the first visit to 0, then sticks at 0 forever after. Calculate the transition probability densities p0 t(x, y) of Yt.
(B) Define Brownian motion with absorption on [0, 1] by Zt = Wt∧T, where T = min{t : Wt = 0 or 1}. Calculate the transition probability densities qt(x, y) for x, y ∈(0, 1).
5 Wald Identities for Brownian Motion 5.1 Doob’s optional sampling formula A continuous-time martingale with respect to a filtration {Ft}t≥0 is an adapted process {Xt}t≥0 such that each Xt is integrable, and such that for any 0 ≤s ≤t < ∞, E(Xt | Fs) = Xs almost surely.
(50) 22 Equivalently, for every finite set of times 0 ≤t1 < t2 < · · · < tn, the finite sequence {Xti}i≤n is a discrete-time martingale with respect to the discrete filtration {Fti}i≤n.
Observe that the definition does not require that the sample paths t 7→Xt be continuous, or even measurable.
Theorem 7. (Doob) If {Xt}t≥0 is a martingale with respect to the filtration {Ft}t≥0 such that {Xt}t≥0 has right-continuous paths and if τ is a bounded stopping time for this filtration, then Xτ is an integrable random variable and EXτ = EX0.
(51) Proof. By hypothesis there is a finite constant C such that τ ≤C. By Exercise 3, the random variables τm := min{k/2m ≥τ} are stopping times. Clearly, each τm takes values in a finite set , C + 1 ≥τ1 ≥τ2 ≥· · · ≥· · · , and τ = lim m→∞τm.
Since τm takes values in a discrete set, Doob’s identity for discrete-time martingales implies that for each m, EXτm = EX0.
Furthermore, because the sample paths of {Xt}t≥0 are right-continuous, Xτ = lim Xτm pointwise. Hence, by the dominated convergence theorem, to prove the identity (51) it suffices to show that the sequence Xτm is uniformly integrable. For this, observe that the ordering C + 1 ≥τ1 ≥τ2 ≥· · · of the stopping times implies that the sequence {Xτm}m≥1 is a reverse martingale with respect to the backward filtration {Fτm}. Any reverse martingale is uniformly integrable.
Remark 3. (a) If we knew a priori that supt≤C+1 |Xt| were integrable then there would be no need to show that the sequence {Xτm}m≥1 is uniformly integrable, and so the use of results concerning reverse martingales could be avoided. All of the martingales that will figure into the results of section 5.2 will have this property.
(b) It can be shown that every continuous-time martingale has a version with right-continuous paths. (In fact, this can be done by using only the upcrossings and maximal inequalities for discrete-time martingales, together with the fact that the dyadic rationals are dense in [0, ∞).) We will not need this result, though.
5.2 The Wald identities You should recall (see the notes on discrete-time martingales) that there are several discrete-time martingales associated with the simple random walk on Z that are quite useful in first-passage problems. If Sn = Pn i=1 ξi is a simple random walk 23 on Z (that is, the random variables {ξi}i≥1 are independent, identically distributed Rademacher, as in section 1) and Fn = σ(ξi)i≤n is the σ−algebra generated by the first n steps of the random walk, then each of the following sequences is a martingale relative to the discrete filtration (Fn)n≥0: (a) Sn; (b) S2 n −n; (c) exp{θSn}/(cosh θ)n.
There are corresponding continuous-time martingales associated with the Wiener process that can be used to obtain analogues of the Wald identities for simple random walk.
Proposition 8. Let {W(t)}t≥0 be a standard Wiener process and let G = {Gt}t≥0 be an admissible filtration. Then each of the following is a continuous martingale relative to G (with θ ∈R): (a) {Wt}t≥0 (b) {W 2 t −t}t≥0 (c) {exp{θWt −θ2t/2}}t≥0 (d) {exp{iθWt + θ2t/2}}t≥0 Consequently, for any bounded stopping time τ, each fo the following holds: EW(τ) = 0; (52) EW(τ)2 = Eτ; (53) E exp{θW(τ) −θ2τ/2} = 1 ∀θ ∈R; and (54) E exp{iθW(τ) + θ2τ/2} = 1 ∀θ ∈R.
(55) Observe that for nonrandom times τ = t, these identities follow from elementary properties of the normal distribution. Notice also that if τ is an unbounded stopping time, then the identities may fail to be true: for example, if τ = τ(1) is the first passage time to the value 1, then W(τ) = 1, and so EW(τ) ̸= 0. Finally, it is crucial that τ should be a stopping time: if, for instance, τ = min{t ≤1 : W(t) = M(1)}, then EW(τ) = EM(1) > 0.
Proof. The martingale property follows immediately from the independent incre-ments property. In particular, the definition of an admissible filtration implies that W(t + s) −W(s) is independent of Gs; hence (for instance), E(exp{θWt+s −θ2(t + s)/2} | Gs) = exp{θWs −θ2s/2}E(exp{θWt+s −Ws −θ2t/2} | Gs) = exp{θWs −θ2s/2}E exp{θWt+s −Ws −θ2t/2} = exp{θWs −θ2s/2}.
The Wald identities follow from Doob’s optional sampling formula.
24 5.3 Wald identities and first-passage problems for Brownian mo-tion Example 1: Fix constants a, b > 0, and define T = T−a,b to be the first time t such that W(t) = −a or +b. The random variable T is a finite, but unbounded, stopping time, and so the Wald identities may not be applied directly. However, for each integer n ≥1, the random variable T ∧n is a bounded stopping time. Consequently, EW(T ∧n) = 0 and EW(T ∧n)2 = ET ∧n.
Now until time T, the Wiener path remains between the values −a and +b, so the random variables |W(T ∧n)| are uniformly bounded by a + b. Furthermore, by path-continuity, W(T ∧n) →W(T) as n →∞. Therefore, by the dominated convergence theorem, EW(T) = −aP{W(T) = −a} + bP{W(T) = b} = 0.
Since P{W(T) = −a} + P{W(T) = b} = 1, it follows that P{W(T) = b} = a a + b.
(56) The dominated convergence theorem also guarantees that EW(T ∧n)2 →EW(T)2, and the monotone convergence theorem that ET ∧n ↑ET. Thus, EW(T)2 = ET.
Using (56), one may now easily obtain ET = ab.
(57) Example 2: Let τ = τ(a) be the first passage time to the value a > 0 by the Wiener path W(t). As we have seen, τ is a stopping time and τ < ∞with probability one, but τ is not bounded. Nevertheless, for any n < ∞, the truncation τ ∧n is a bounded stopping time, and so by the third Wald identity, for any θ > 0, E exp{θW(τ ∧n) −θ2(τ ∧n)} = 1.
(58) Because the path W(t) does not assume a value larger than a until after time τ, the random variables W(τ ∧n) are uniformly bounded by a, and so the random variables in equation (58) are dominated by the constant exp{θa}. Since τ < ∞with probability one, τ ∧n →τ as n →∞, and by path-continuity, the random variables 25 W(τ ∧n) converge to a as n →∞. Therefore, by the dominated convergence theorem, E exp{θa −θ2(τ)} = 1.
Thus, setting λ = θ2/2, we have E exp{−λτa} = exp{− √ 2λa}.
(59) The only density with this Laplace transform3 is the one–sided stable density given in equation (48). Thus, the Optional Sampling Formula gives us a second proof of (45).
Exercise 8. First Passage to a Tilted Line. Let Wt be a standard Wiener process, and define τ = min{t > 0 : W(t) = a −bt} where a, b > 0 are positive constants.
Find the Laplace transform and/or the probability density function of τ.
Exercise 9. Two-dimensional Brownian motion: First-passage distribution. Let Zt = (Xt, Yt) be a two-dimensional Brownian motion started at the origin (0, 0) (that is, the coordinate processes Xt and Yt are independent standard one-dimensional Wiener processes).
(A) Prove that for each real θ, the process exp{θXt + iθYt} is a martingale relative to any admissible filtration.
(B) Deduce the corresponding Wald identity for the first passage time τ(a) = min{t : Wt = a}, for a > 0.
(C) What does this tell you about the distribution of Yτ(a)?
Exercise 10. Eigenfunction expansions. These exercises show how to use Wald identities to obtain eigenfunction expansions (in this case, Fourier expansions) of the transition probability densities of Brownian motion with absorption on the unit interval (0, 1). You will need to know that the functions { √ 2 sin kπx}k≥1 (together with) constitute an orthonormal basis of L2[0, 1]. Let Wt be a Brownian motion started at x ∈[0, 1] under P x, and let T = T[0,1] be the first time that Wt = 0 or 1.
(A) Use the appropriate martingale (Wald) identity to check that Ex sin(kπWt)ek2π2t/21{T > t} = sin(kπx).
(B) Deduce that for every C∞function u which vanishes at the endpoints x = 0, 1 of the interval, Exu(Wt∧T) = ∞ X k=1 e−k2π2t/2( √ 2 sin(kπx))ˆ u(k) 3Check a table of Laplace transforms.
26 where ˆ u(k) = √ 2 R 1 0 u(y) sin(kπy) dy is the kth Fourier coefficient of u.
(C) Conclude that the sub-probability measure P x{Wt ∈dy ; T > t} has density qt(x, y) = ∞ X k=1 e−k2π2t/22 sin(kπx) sin(kπy).
6 Brownian Paths In the latter half of the nineteenth century, mathematicians began to encounter (and invent) some rather strange objects. Weierstrass produced a continuous function that is nowhere differentiable. Cantor constructed a subset C (the “Cantor set”) of the unit interval with zero area (Lebesgue measure) that is nevertheless in one-to-one correspondence with the unit interval, and has the further disconcerting property that between any two points of C lies an interval of positive length totally contained in the complement of C. Not all mathematicians were pleased by these new objects. Hermite, for one, remarked that he was “revolted” by this plethora of nondifferentiable functions and bizarre sets.
With Brownian motion, the strange becomes commonplace. With probability one, the sample paths are nowhere differentiable, and the zero set Z = {t ≤1 : W(t) = 0}) is a homeomorphic image of the Cantor set. These facts may be estab-lished using only the formula (45), Brownian scaling, the strong Markov property, and elementary arguments.
6.1 Zero Set of a Brownian Path The zero set is Z = {t ≥0 : W(t) = 0}.
(60) Because the path W(t) is continuous in t, the set Z is closed. Furthermore, with probability one the Lebesgue measure of Z is 0, because Fubini’s theorem implies that the expected Lebesgue measure of Z is 0: E|Z| = E Z ∞ 0 1{0}(Wt) dt = Z ∞ 0 E1{0}(Wt) dt = Z ∞ 0 P{Wt = 0} dt = 0, 27 where |Z| denotes the Lebesgue measure of Z. Observe that for any fixed (non-random) t > 0, the probability that t ∈Z is 0, because P{W(t) = 0} = 0. Hence, because Q+ (the set of positive rationals) is countable, P{Q+ ∩Z ̸= ∅} = 0.
(61) Proposition 9. With probability one, the Brownian path W(t) has infinitely many zeros in every time interval (0, ε), where ε > 0.
Proof. We have already seen that this is a consequence of the Blumenthal 0-1 Law, but we will now give a different proof using what we have learned about the distribution of M(t). First we show that for every ε > 0 there is, with probability one, at least one zero in the time interval (0, ε). Recall (equation (11)) that the distribution of M −(t), the minimum up to time t, is the same as that of −M(t). By formula (45), the probability that M(ε) > 0 is one; consequently, the probability that M −(ε) < 0 is also one. Thus, with probability one, W(t) assumes both negative and positive values in the time interval (0, ε). Since the path W(t) is continuous, it follows, by the Intermediate Value theorem, that it must assume the value 0 at some time between the times it takes on its minimum and maximum values in (0, ε].
We now show that, almost surely, W(t) has infinitely many zeros in the time interval (0, ε). By the preceding paragraph, for each k ∈N the probability that there is at least one zero in (0, 1/k) is one, and so with probability one there is at least one zero in every (0, 1/k). This implies that, with probability one, there is an infinite sequence tn of zeros converging to zero: Take any zero t1 ∈(0, 1); choose k so large that 1/k < t1; take any zero t2 ∈(0, 1/k); and so on.
Proposition 10. With probability one, the zero set Z of a Brownian path is a perfect set, that is, Z is closed, and for every t ∈Z there is a sequence of distinct elements tn ∈Z such that limn→∞tn = t.
Proof. That Z is closed follows from path-continuity, as noted earlier. Fix a rational number q > 0 (nonrandom), and define ν = νq to be the first time t ≥such that W(t) = 0. Because W(q) ̸= 0 almost surely, the random variable νq is well-defined and is almost surely strictly greater than q. By the Strong Markov Property, the post-νq process W(νq + t) −W(νq) is, conditional on the stopping field Fν, a Wiener process, and consequently, by Proposition 9, it has infinitely many zeros in every time interval (0, ε), with probability one. Since W(νq) = 0, and since the set of rationals is countable, it follows that, almost surely, the Wiener path W(t) has infinitely many zeros in every interval (νq, νq + ε), where q ∈Q and ε > 0.
Now let t be any zero of the path. Then either there is an increasing sequence tn of zeros such that tn →t, or there is a real number ε > 0 such that the interval 28 (t −ε, t) is free of zeros. In the latter case, there is a rational number q ∈(t −ε, t), and t = νq. In this case, by the preceding paragraph, there must be a decreasing sequence tn of zeros such that tn →t.
It can be shown (this is not especially difficult) that every compact perfect set of Lebesgue measure zero is homeomorphic to the Cantor set. Thus, with prob-ability one, the set of zeros of the Brownian path W(t) in the unit interval is a homeomorphic image of the Cantor set.
6.2 Nondifferentiability of Paths Proposition 11. With probability one, the Brownian path Wt is nowhere differentiable.
Proof. This is an adaptation of an argument of DVORETSKY, ERDÖS, & KAKUTANI 1961. The theorem itself was first proved by PALEY, WIENER & ZYGMUND in 1931.
It suffices to prove that the path Wt is not differentiable at any t ∈(0, 1) (why?).
Suppose to the contrary that for some t∗∈(0, 1) the path were differentiable at t = t∗; then for some ε > 0 and some C < ∞it would be the case that |Wt −Wt∗| ≤C|t −t∗| for all t ∈(t∗−ε, t∗+ ε), (62) that is, the graph of Wt would lie between two intersecting lines of finite slope in some neighborhood of their intersection. This in turn would imply, by the triangle inequality, that for infinitely many k ∈N there would be some 0 ≤m ≤4k such that4 |W((m + i + 1)/4k) −W((m + i)/4k)| ≤16C/4k for each i = 0, 1, 2.
(63) I’ll show that the probability of this event is 0. Let Bm,k = Bk,m(C) be the event that (63) holds, and set Bk = ∪m≤4kBm,k; then by the Borel-Cantelli lemma it is enough to show that (for each C < ∞) ∞ X k=1 P(Bm) < ∞.
(64) The trick is Brownian scaling: in particular, for all s, t ≥0 the increment Wt+s−Wt is Gaussian with mean 0 and standard deviation √s. Consequently, since the three increments in (63) are independent, each with standard deviation 2−k, and since the standard normal density is bounded above by 1/ √ 2π, P(Bm,k) = P{|Z| ≤16C/2k}3 ≤(32C/2k√ 2π)3.
4The constant 16 might really be 32, possibly even 64.
29 where Z is standard normal. Since Bk is the union of 4k such events Bm,k, it follows that P(Bk) ≤4k(32C/2k√ 2π)3) ≤(32C/ √ 2π)3/2k.
This is obviously summable in k.
Exercise 11. Local Maxima of the Brownian Path. A continuous function f(t) is said to have a local maximum at t = s if there exists ε > 0 such that f(t) ≤f(s) for all t ∈(s −ε, s + ε).
(A) Prove that if the Brownian path W(t) has a local maximum w at some time s > 0 then, with probability one, it cannot have a local maximum at some later time s∗ with the same value w. HINT: Use the Strong Markov Property and the fact that the rational numbers are countable and dense in [0, ∞).
(B) Prove that, with probability one, the times of local maxima of the Brownian path W(t) are dense in [0, ∞) (C) Prove that, with probability one, the set of local maxima of the Brownian path W(t) is countable. HINT: Use the result of part (A) to show that for each local maximum (s, Ws) there is an interval (s −ε, s + ε) such that Wt < Ws for all t ∈(s −ε, s + ε), t ̸= s.
7 Quadratic Variation Fix t > 0, and let Π = {t0, t1, t2, . . . , tn} be a partition of the interval [0, t], that is, an increasing sequence 0 = t0 < t1 < t2 < · · · < tn = t. The mesh of a partition Π is the length of its longest interval ti −ti−1. If Π is a partition of [0, t] and if 0 < s < t, then the restriction of Π to [0, s] (or the restriction to [s, t]) is defined in the obvious way: just terminate the sequence tj at the largest entry before s, and append s. Say that a partition Π′ is a refinement of the partition Π if the sequence of points ti that defines Π is a subsequence of the sequence t′ j that defines Π′. A nested sequence of partitions is a sequence Πn such that each is a refinement of its predecessor. For any partition Π and any continuous-time stochastic process Xt, define the quadratic variation of X relative to Π by QV (X; Π) = n X j=1 (X(tj) −X(tj−1))2.
(65) 30 Theorem 8. Let Πn be a nested sequence of partitions of the unit interval [0, 1] with mesh →0 as n →∞. Let Wt be a standard Wiener process. Then with probability one, lim n→∞QV (W; Πn) = 1.
(66) Note 1. It can be shown, without too much additional difficulty, that if Πt n is the restriction of Πn to [0, t] then with probability one, for all t ∈[0, 1], lim n→∞QV (W; Πt n) = t.
Before giving the proof of Theorem 8, I’ll discuss a much simpler special case5, where the reason for the convergence is more transparent. For each natural number n, define the nth dyadic partition Dn[0, t] to be the partition consisting of the dyadic rationals k/2n of depth n (here k is an integer) that are between 0 and t (with t added if it is not a dyadic rational of depth n). Let X(s) be any process indexed by s.
Proposition 12. Let {W(t)}t≥0 be a standard Brownian motion. For each t > 0, with probability one, lim n→∞QV (W; Dn[0, t]) = t.
(67) Proof. Proof of Proposition 12. First let’s prove convergence in probability. To simplify things, assume that t = 1. Then for each n ≥1, the random variables ξn,k ∆ = 2n(W(k/2n) −W((k −1)/2n))2, k = 1, 2, . . . , 2n are independent, identically distributed χ2 with one degree of freedom (that is, they are distributed as the square of a standard normal random variable). Observe that Eξn,k = 1. Now QV (W; Dn[0, 1]) = 2−n 2n X k=1 ξn,k.
The right side of this equation is the average of 2n independent, identically dis-tributed random variables, and so the Weak Law of Large Numbers implies conver-gence in probability to the mean of the χ2 distribution with one degree of freedom, which equals 1.
The stronger statement that the convergence holds with probability one can easily be deduced from the Chebyshev inequality and the Borel–Cantelli lemma.
The Chebyshev inequality and Brownian scaling implies that P{|QV (W; Dn[0, 1]) −1| ≥ε} = P{| 2n X k=1 (ξn,k −1)| ≥2nε} ≤Eξ2 1,1 4nε2 .
5Only the special case will be needed for the Itô calculus. However, it will be of crucial impor-tance — it is, in essence the basis for the Itô formula.
31 Since P∞ n=1 1/4n < ∞, the Borel–Cantelli lemma implies that, with probability one, the event |QV (W; Dn[0, 1]) −1| ≥ε occurs for at most finitely many n. Since ε > 0 can be chosen arbitrarily small, it follows that limn→∞QV (W; Dn[0, 1]) = 1 almost surely. The same argument shows that for any dyadic rational t ∈[0, 1], the convergence (67) holds a.s.
Exercise 12. Prove that if (67) holds a.s. for each dyadic rational in the unit interval, then with probability one it holds for all t.
In general, when the partition Π is not a dyadic partition, the summands in the formula (66) for the quadratic variation are (when X = W is a Wiener process) still independent χ2 random variables, but they are no longer identically distributed, and so Chebyshev’s inequality by itself won’t always be good enough to prove a.s. convergence. The route we’ll take is completely different: we’ll show that for nested partitions Πn the sequence QV (W; Πn) is a reverse martingale relative to an appropriate filtration Gn.
Lemma 4. Let ξ, ζ be independent Gaussian random variables with means 0 and variances σ2 ξ, σ2 ζ, respectively. Let G be any σ−algebra such that the random variables ξ2 and ζ2 are G−measurable, but such that sgn(ζ) and sgn(ζ) are independent of G. Then E((ξ + ζ)2 | G) = ξ2 + ζ2.
(68) Proof. Expand the square, and use the fact that ξ2 and ζ2 are G−measurable to extract them from the conditional expectation. What’s left is 2E(ξζ | G) = 2E(sgn(ξ)sgn(ζ)|ξ| |ζ| | G) = 2|ξ| |ζ|E(sgn(ξ)sgn(ζ) | G) = 0, because sgn(ξ) and sgn(ζ) are independent of G.
Proof of Theorem 8. Without loss of generality, we may assume that each partition Πn+1 is gotten by splitting one interval of Πn, and that Π0 is the trivial partition of [0, 1] (consisting of the single interval [0, 1]). Thus, Πn consists of n + 1 nonoverlap-ping intervals Jn k = [tn k−1, tn k]. Define ξn,k = W(tn k) −W(tn k−1), and for each n ≥0 let Gn be the σ−algebra generated by the random variables ξ2 m,k, where m ≥n and k ∈[m + 1]. The σ−algebras Gn are decreasing in n, so they form 32 a reverse filtration. By Lemma 4, the random variables QV (W; Πn) form a reverse martingale relative to the reverse filtration Gn, that is, for each n, E(QV (W; Πn) | Gn+1) = QV (W; Πn+1).
By the reverse martingale convergence theorem, lim n→∞QV (W; Πn) = E(QV (W; Π0) | ∩n≥0 Gn) = E(W 2 1 | ∩n≥0 Gn) almost surely.
Exercise 13. Prove that the limit is constant a.s., and that the constant is 1.
8 Skorohod’s Theorem In section 3 we showed that there are simple random walks embedded in the Wiener path. Skorohod discovered that any mean zero, finite variance random walk is also embedded, in a certain sense.
Theorem 9. (Skorohod Embedding I) Let F be any probability distribution on the real line with mean 0 and variance σ2 < ∞. Then on some probability space there exist (i) a standard Wiener process {Wt}t≥0; (ii) an admissible filtration {Ft}t≥0; and (iii) a sequence of finite stopping times 0 = T0 ≤T1 ≤T2 ≤· · · such that (A) the random vectors (Tn+1 −Tn, WTn+1 −WTn) are independent, identically dis-tributed; (B) each random variable , WTn+1 −WTn has distribution F; and (C) ETn+1 −ETn = σ2.
Thus, in particular, the sequence {WTn}n≥0 has the same joint distribution as a random walk with step distribution F, and Tn/n →σ2 as n →∞. For most applications of the theorem, this is all that is needed. However, it is natural to wonder about the choice of filtration: is it true that there are stopping times 0 = T0 ≤T1 ≤T2 ≤· · · with respect to the standard filtration such that (A), (B), (C) hold? The answer is yes, but the proof is more subtle.
Theorem 10. (Skorohod Embedding II) Let F be any probability distribution on the real line with mean 0 and variance σ2 < ∞, and let W(t) be a standard Wiener process. Then there exist stopping times 0 = T0 ≤T1 ≤T2 ≤· · · with respect to the standard filtration such that the conclusions (A), (B), (C) of Theorem 9 hold.
I will only prove Theorem 9. The proof will hinge on a representation of an arbitrary mean-zero probability distribution as a mixture of mean-zero two-point 33 distributions. A two-point distribution is, by definition, a probability measure whose support has only two points. For any two points −a < 0 < b there is a unique two-point distribution Fa,b with support {−a, b} and mean 0, to wit, Fa,b({b}) = a a + b and Fa,b({−a}) = b a + b.
The variance of Fa,b is ab.
Proposition 13. For any Borel probability distribution F on R with mean zero there exists a Borel probability distribution G on R2 + such that F = Z Fa,b dG(a, b).
(69) Hence, Z x2 dF(x) = Z ab dG(a, b).
(70) Proof for compactly supported F. First we will prove this for distributions F with finite support. To do this, we induct on the number of support points. If F is sup-ported by only two points then F = Fa,b for some a, b > 0, and so the representation (69) holds with G concentrated at the single point (a, b). Suppose, then, that the result is true for mean-zero distributions supported by fewer than m points, and let F be a mean-zero distribution supported by m points. Then among the m support points there must be two satisfying −a < 0 < b, both with positive probabilities F(−a) and F(b). There are three possibilities: −aF(−a) + bF(b) = 0, or −aF(−a) + bF(b) > 0, or −aF(−a) + bF(b) < 0.
In the first case, F can obviously be decomposed as a convex combination of Fa,b and a probability distribution F ′ supported by the remaining m −2 points in the support of FIn the second case, where −aF(−a) + bF(b) > 0, there is some value 0 < β < F(b) such that −aF(−a) + βF(b) = 0, and so F can be decomposed as a convex combination of Fa,b and a distribution F ′ supported by b and the remaining m −2 support points. Similarly, in the third case F is a convex combination of Fa,b and a distribution F ′ supported by −a and the remaining m −2 supported points of F. Thus, in all three cases (69) holds by the induction hypothesis. The formula (70) follows from (69) by Fubini’s theorem.
Next, let F be a Borel probability distribution on R with mean zero and com-pact support [−A, A]. Clearly, there is a sequence of probability distributions Fn 34 converging to F in distribution such that Fn is supported by finitely many points.
In particular, if X is a random variable with distribution F, then set Xn = max{k/2n : k/2n ≤X} and let Fn be the distribution of Xn. Clearly, Xn →X pointwise as n →∞. The distributions Fn need not have mean zero, but by the dominated convergence theo-rem, EXn →EX = 0. Thus, if we replace Xn be Xn −EXn we will have a sequence of mean-zero distributions converging to F, each supported by only finitely many points, all contained in [−A −1, A + 1]. Therefore, since the representation (69) holds for distributions with finite support, there are Borel probability distributions Gn such that Fn = Z Fa,b dGn(a, b) and σ2 n := Z x2 dFn(x) = Z ab dGn(a, b).
Moreover, since each Fn has supports contained in [−A −1, A + 1], the mixing distributions Gn have supports contained in [0, A + 1]2. Hence, by Helly’s selection principle (i.e., Banach-Alaoglu to those of you from Planet Math), there is a subse-quence Gk that converges weakly to a Borel probability distribution G. It follows routinely sthat F = Z Fa,b dG(a, b) and σ2 := Z x2 dF(x) = Z ab dG(a, b).
Exercise 14. Finish the proof: Show that the representation (69) holds for all mean-zero, finite variance distributions.
Remark 4. Proposition 13 can also be deduced from CHOQUET’S theorem. The set of mean-zero, probability distributions on R with variances bounded by C is a convex, weak-∗compact subset of the space of finite Borel measures on R. It is not difficult to see that the extreme points are the two-point distributions Fa,b. Therefore, the representation (69) is a special case of Choquet’s theorem.
Proof of Theorem 9. Consider first the case where F is supported by only two points.
Since F has mean zero, the two support points must satisfy a < 0 < b, with p = F({b}) = 1 −F({a}). Let {Wt}t≥0 be a standard Wiener process, and let T be the first time that the Wiener process visits either a or b. Then T < ∞almost surely, and T is a stopping time with respect to any admissible filtration. Equations (56)–(57) imply that the distribution of WT is F, and ET = σ2. (Exercise: Fill in the details.) To complete the proof in the case of a two-point distribution, we now use the strong Markov property and an induction argument. Assume that stopping times 35 0 = T0 ≤T1 ≤T2 ≤· · · ≤Tm have been defined in such a way that properties (A), (B), (C) hold for n < m. Define Tm+1 to be the first time after Tm that WTm+1 − WTm = a or b; then by the strong Markov property, the random vector (Tm+1 − Tm, WTm+1 −WTm) is independent of the random vectors (Tn+1 −Tn, WTn+1 −WTn) for n < m (since these are all measurable with respect to the stopping field FTm), and (Tm+1−Tm, WTm+1−WTm) has the same distribution as does (T1−T0, WT1−WT0).
This completes the induction for two-point distributions. It should be noted that the random variables Tn are stopping times with respect to any admissible filtration.
Now let F be any mean-zero distribution with finite variance σ2. By Proposi-tion 13, F has a representation as a mixture of two-point distributions Fa,b, with mixing measure G. Let (Ω, F, P) be a probability space that supports a sequence {(An, Bn)}n≥1 of independent, identically distributed random vectors each with distribution G, and an independent Wiener process {Wt}t≥0. (Such a probability space can always be realized as a product space.) Let F0 be the σ−algebra gener-ated by the random vectors (An, Bn), and for each t ≥0 let Ft be the σ−algebra generated by F0 and the random variables Ws, for s ≤t; since the random vectors (An, Bn) are independent of the Wiener process, the filtration {Ft}t≥0 is admissible.
Define stopping times Tn inductively as follows: T0 = 0, and Tn+1 = min{t ≥0 : WTn+t −WTn = −An or Bn}.
Conditional on F0, the random vectors (Tn+1−Tn, WTn+1−WTn) are independent, by the strong Markov property. Furthermore, conditional on F0, the random variable WTn+1 −WTn has the two-point distribution Fa,b with a = An and b = Bn. Since the unconditional distribution of (An, Bn) is G, it follows that unconditionally the random vectors (Tn+1 −Tn, WTn+1 −WTn) are independent, identically distributed, and WTn+1 −WTn has distribution F. That ETn+1 −ETn = σ2 follows from the variance formula in the mixture representation (69).
Proof of Theorem 10 for the uniform distribution on (-1,1). The general case is proved by showing directly (without using the Choquet representation (69)) that a mean-zero probability distribution F is a limit of finitely supported mean-zero distribu-tions. I will consider only the special case where F is the uniform distribution (normalized Lebesgue measure) on [−1, 1]. Define a sequence of stopping times τn as follows: τ1 = min{t > 0 : W(t) = ±1/2} τn+1 = min{t > τn : W(t) −W(τn) = ±1/2n+1}.
By symmetry, the random variable W(τ1) takes the values ±1/2 with probabilities 1/2 each. Similarly, by the Strong Markov Property and induction on n, the random 36 variable W(τn) takes each of the values k/2n, where k is an odd number between −2n and +2n, with probability 1/2n. Notice that these values are equally spaced in the interval [−1, 1], and that as →∞the values fill the interval. Consequently, the distribution of W(τn) converges to the uniform distribution on [−1, 1] as n →∞.
The stopping times τn are clearly increasing with n. Do they converge to a finite value? Yes, because they are all bounded by T−1,1, the first passage time to one of the values ±1. (Exercise: Why?) Consequently, τ := lim τn = sup τn is finite with probability one. By path-continuity, W(τn) →W(τ) almost surely. As we have seen, the distributions of the random variables W(τn) approach the uniform distribution on [−1, 1] as n →∞, so it follows that the random variable W(τ) is uniformly distributed on [−1, 1].
Exercise 15. Show that if τn is an increasing sequence of stopping times such that τ = lim τn is finite with probability one, then τ is a stopping time.
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组合数
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数学概念
本词条由《中国科技信息》杂志社参与编辑并审核，经科普中国·科学百科认证 。
从n个不同元素中，任取m(m≤n)个元素并成一组，叫做从n个不同元素中取出m个元素的一个组合；从n个不同元素中取出m(m≤n)个元素的所有组合的个数，叫做从n个不同元素中取出m个元素的组合数。
中文名 组合数
外文名 combinatorial number
所属学科 数学
公 式 C(n,m)=n!/((n-m)!m!)（m≤n）
性质1 C(n,m)= C(n,n-m)
性质2 C(n,m)=C(n-1,m-1)+C(n-1,m)
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高中组合数公式
目录
1定义
2计算公式
3性质
定义
播报
组合是数学的重要概念之一。从 n 个不同元素中每次取出 m 个不同元素
，不管其顺序合成一组，称为从 n 个元素中不重复地选取 m 个元素的一个组合。所有这样的组合的种数称为组合数。
计算公式
播报
在线性写法中被写作C(n,m)。
组合数的计算公式为
n 元集合 A 中不重复地抽取 m 个元素作成的一个组合实质上是 A 的一个 m 元子集合。如果给集 A 编序
成为一个序集，那么 A 中抽取 m 个元素的一个组合对应于数段
到序集 A 的一个确定的严格保序映射。组合数
的常用符号还有
实际上是自然数集上的二元运算，这种运算既不满足交换律也不满足结合律。
性质
播报
1.互补性质
组合数基本公式
即从 n 个不同元素中取出 m 个元素的组合数=从 n 个不同元素中取出 (n-m) 个元素的组合数；
这个性质很容易理解，例如C(9,2)=C(9,7)，即从9个元素里选择2个元素的方法与从9个元素里选择7个元素的方法是相等的。
规定：C(n,0)=1 C(n,n)=1 C(0,0)=1
2.组合恒等式
若表示在 n 个物品中选取 m 个物品，则如存在下述公式：C(n,m)=C(n,n-m)=C(n-1,m-1)+C(n-1,m)。
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参考资料
1 《数学辞海》总编辑委员会．《数学辞海》第1卷．南京．东南大学出版社．2002.8
2 饶克勇.组合数公式的由来及演变[J].昭通师范高等专科学校学报,2002,24(5):12-15
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齐登记.无符号第一类Stirling数的组合数表示．《CNKI;WanFang》，2011
顾江民.正切数与集合的纯偶组合数．《湖州师范学院学报》，2015
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Mathematical Principles in Vision and Graphics: Projective Geometry – Part 3 Ass.Prof. Friedrich Fraundorfer SS 2018 1 Learning goals Understand the concept of vanishing points and vanishing lines Understand the calculation of vanishing points Unterstand the relation between vanishing points and camera orientation and calibration 2 Outline Vanishing points and lines Effects of geometric transformations 3 Vanishing points 4 [Source: Flickr] Vanishing points Ideal points in the projective 3-space are located at infinity, and have homogeneous coordinates of the form (x, y, z, 0).
These points are also called points at infinity.
The image of an ideal point under a projective mapping is called a vanishing point.
Recall that two parallel lines in the projective 3- space meets at an ideal point.
Thus the images of two or more parallel world lines converge at a vanishing point.
5 [Source: Flickr] Vanishing points The vanishing point v is the projection of a point at infinity.
Think of extending the line on the ground plane further and further into infinity.
6 v C line on ground plane Vanishing points • Any two parallel lines have the same vanishing point v • The vanishing point is the image of the intersection point of the two parallel lines.
• The ray from C through v is parallel to the lines • An image may have more than one vanishing point 7 v C lines on ground plane parallel to each other Vanishing lines Multiple vanishing points ▫Any set of parallel lines on the plane define a vanishing point ▫Lines at different orientation result in a different vanishing point ▫The union of all of vanishing points from lines on the same plane is the vanishing line 8 v1 v2 Vanishing lines Different planes define different vanishing lines.
9v1v2 Vanishing lines A set of parallel planes that are not parallel to the image plane intersect the image plane at a vanishing line. The horizon is a special vanishing line when the set of parallel planes are parallel to the ground reference. Anything in the scene that is above the camera will be projected above the horizon in the image. 10 C vanishing line 0 / 1 / / / 1 ) ( Z Y X Z Z Y Y X X Z Z Y Y X X D D D t t D t P D t P D t P tD P tD P tD P t P P Computing vanishing points Properties ▫ Pis a point at infinity, v is its projection ▫ They depend only on line direction ▫ Parallel lines P0 + tD, P1 + tD intersect at P D P P t t 0 ) ( ΠP v P0 D 11 v C Computing vanishing points (from lines) Intersect p1q1 with p2q2 12 v p1 q1 p2 q2 𝑣= (𝑝1 × 𝑞1) × (𝑝2 × 𝑞2) Computing vanishing points by projection Let P = K[ I | 0 ] be a camera matrix. The vanishing point of lines with direction d in 3-space is the intersection v of the image plane with a ray through camera center with direction d. This vanishing point v is given by v = Kd.
Example: Computing vanishing points of lines on a XZ plane (1) parallel to the Z axis, (2) at 45 deg to the Z axis (3) parallel to the X axis 𝑣= 1 0 0 1 0 0 0 0 0 0 1 0 𝑫 0 = 𝐷𝑥 𝐷𝑦 𝐷𝑧 𝐷1: 0 0 1 0 0 0 1 𝐷2: 1 0 1 0 1 0 1 𝐷3: 1 0 0 0 1 0 0 𝐷3 = 1 0 0 𝐷2 = 1 0 1 𝐷1 = 0 0 1 𝑋 𝑌 𝑍 x y z 13 Vanishing points and projection matrix P 4 3 2 1 p p p p 1 p 2 p 3 p 4 p T P p 0 0 0 1 1 = vx (X vanishing point) Z 3 Y 2 , similarly, v p v p = = O origin world of projection 1 0 0 0 4 T P π o v v v Z Y X P 14 Vanishing point of a line parallel to a plane lies on the vanishing line of the plane a b c Real vanishing points 15 [Image source: Richard Hartley and Andrew Zisserman] Image rectification using vanishing points before rectification after rectification 16 X Y Z x y z X Y Z x y z ϴ Camera rotation from vanishing points Two images of a scene obtained by the same camera from different position and orientation.
The images of the points at infinity, the vanishing points, are not affected by the camera translation, but are affected only by the camera rotation Vanishing points vi and vi’ have the following directions di, di’ The directions are related by a rotation matrix: If the directions are known, the rotation matrix can be computed from two directions 𝑑𝑖 ′ = 𝑅𝑑𝑖 𝑑𝑖= 𝐾−1𝑣𝑖/ 𝐾−1𝑣𝑖 𝑑𝑖 ′ = 𝐾−1𝑣𝑖 ′/ 𝐾−1𝑣𝑖 ′ 17 Camera rotation from vanishing points First camera is aligned with world coordinate system. di = [0 0 1] Second camera deviates, di’ = [dx,dy,dz] and can be computed from the image coordinates of vi’ The rotation that aligns the second image with the first image can be computed from di and di’ and 1 more direction.
(0,0,0) image plane P’ y x vi’ di’ z vi 18 Measuring heights using vanishing points 19 d1 d2 reference object column Height column = height of reference objectd2/d1 Camera calibration from orthogonal vanishing points 20 [Image source: Richard Hartley and Andrew Zisserman] Recap - Learning goals Understand the concept of vanishing points and vanishing lines Understand the calculation of vanishing points Unterstand the relation between vanishing points and camera orientation and calibration 21
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Convert kiloohm to ohm - Conversion of Measurement Units
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The answer is 0.001.
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1 kiloohm is equal to 1000 ohm.
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1 kiloohm to ohm = 1000 ohm
2 kiloohm to ohm = 2000 ohm
3 kiloohm to ohm = 3000 ohm
4 kiloohm to ohm = 4000 ohm
5 kiloohm to ohm = 5000 ohm
6 kiloohm to ohm = 6000 ohm
7 kiloohm to ohm = 7000 ohm
8 kiloohm to ohm = 8000 ohm
9 kiloohm to ohm = 9000 ohm
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Definition: Ohm
The ohm (symbol: Ω) is the SI unit of electrical impedance or, in the direct current case, electrical resistance, named after Georg Ohm. It is defined as the resistance between two points of a conductor when a constant potential difference of 1 volt, applied to these points, produces in the conductor a current of 1 ampere, the conductor not being the seat of any electromotive force.
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https://courses.lumenlearning.com/calculus2/chapter/population-growth-and-carrying-capacity/
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Population Growth and Carrying Capacity
Learning Outcomes
Describe the concept of environmental carrying capacity in the logistic model of population growth
To model population growth using a differential equation, we first need to introduce some variables and relevant terms. The variable t. will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem must specify the units used in that particular problem. The variable P will represent population. Since the population varies over time, it is understood to be a function of time. Therefore we use the notation P(t) for the population as a function of time. If P(t) is a differentiable function, then the first derivative dPdt represents the instantaneous rate of change of the population as a function of time.
In Exponential Growth and Decay, we studied the exponential growth and decay of populations and radioactive substances. An example of an exponential growth function is P(t)=P0ert. In this function, P(t) represents the population at time t,P0 represents the initial population (population at time t=0), and the constant r>0 is called the growth rate. Figure 1 shows a graph of P(t)=100e0.03t. Here P0=100 and r=0.03.
We can verify that the function P(t)=P0ert satisfies the initial-value problem
dPdt=rP,P(0)=P0.
This differential equation has an interesting interpretation. The left-hand side represents the rate at which the population increases (or decreases). The right-hand side is equal to a positive constant multiplied by the current population. Therefore the differential equation states that the rate at which the population increases is proportional to the population at that point in time. Furthermore, it states that the constant of proportionality never changes.
One problem with this function is its prediction that as time goes on, the population grows without bound. This is unrealistic in a real-world setting. Various factors limit the rate of growth of a particular population, including birth rate, death rate, food supply, predators, and so on. The growth constant r usually takes into consideration the birth and death rates but none of the other factors, and it can be interpreted as a net (birth minus death) percent growth rate per unit time. A natural question to ask is whether the population growth rate stays constant, or whether it changes over time. Biologists have found that in many biological systems, the population grows until a certain steady-state population is reached. This possibility is not taken into account with exponential growth. However, the concept of carrying capacity allows for the possibility that in a given area, only a certain number of a given organism or animal can thrive without running into resource issues.
Definition
The carrying capacity of an organism in a given environment is defined to be the maximum population of that organism that the environment can sustain indefinitely.
We use the variable K to denote the carrying capacity. The growth rate is represented by the variable r. Using these variables, we can define the logistic differential equation.
Definition
Let K represent the carrying capacity for a particular organism in a given environment, and let r be a real number that represents the growth rate. The function P(t) represents the population of this organism as a function of time t, and the constant P0 represents the initial population (population of the organism at time t=0). Then the logistic differential equation is
dPdt=rP(1−PK)
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Visit this website for more information on logistic growth.
The logistic equation was first published by Pierre Verhulst in 1845. This differential equation can be coupled with the initial condition P(0)=P0 to form an initial-value problem for P(t).
Suppose that the initial population is small relative to the carrying capacity. Then PK is small, possibly close to zero. Thus, the quantity in parentheses on the right-hand side of the definition is close to 1, and the right-hand side of this equation is close to rP. If r>0, then the population grows rapidly, resembling exponential growth.
However, as the population grows, the ratio PK also grows, because K is constant. If the population remains below the carrying capacity, then PK is less than 1, so 1−PK>0. Therefore the right-hand side of the definition is still positive, but the quantity in parentheses gets smaller, and the growth rate decreases as a result. If P=K then the right-hand side is equal to zero, and the population does not change.
Now suppose that the population starts at a value higher than the carrying capacity. Then PK>1, and 1−PK<0. Then the right-hand side of the definition is negative, and the population decreases. As long as P>K, the population decreases. It never actually reaches K because dPdt will get smaller and smaller, but the population approaches the carrying capacity as t approaches infinity. This analysis can be represented visually by way of a phase line. A phase line describes the general behavior of a solution to an autonomous differential equation, depending on the initial condition. For the case of a carrying capacity in the logistic equation, the phase line is as shown in Figure 2.
Figure 2. A phase line for the differential equation dPdt=rP(1−PK).
This phase line shows that when P is less than zero or greater than K, the population decreases over time. When P is between 0 and K, the population increases over time.
Example: Examining the Carrying Capacity of a Woodpecker Population
There were an estimated 150 pileated woodpeckers (Dryocopus pileatus) in a forest at the beginning of 2020. At the beginning of 2021, the population had increased to about 175. Based on the size of the forest, the carrying capacity is estimated to be about 400 pileated woodpeckers.
Write a logistic differential equation and initial condition to model this population. Use t=0 for the beginning of 2020.
Draw a direction field for this logistic differential equation, and sketch the solution curve corresponding to the initial condition.
Solve the initial-value problem for P(t).
According to this model, what will the population be at the beginning of 2025?
How long will it take the population to reach 75 of the carrying capacity?
Show Solution
The logistic differential equation formula is dPdt=rP(1−PK). We are told that the carrying capacity is 400 so we substitute this value in for K. To approximate r, we can use the fact that the population increased from 150 to 175 in one year. Thus, the growth rate is r=175−150150=25150=16≈16.67.To find the initial condition, we use the fact that there are 150 pileated woodpeckers at the beginning of 2020, so P0=150. Putting this all together gives us the following initial-value problem.
dPdt=0.1667P(1−P400), P(0)=150
3. We can solve the differential equation using separation of variables.
dPdt=0.1667P(1−P400)dPdt=0.1667P(400−P400)Rewrite the expression inside the parenthesesas one combined fraction.dPP(400−P)=0.1667400dtDivide both sides by the factors P and 400−P.Multiply both sides by dt.∫1400(1P−1400−P)dP=∫0.1667400dtUse partial fraction decomposition to rewrite the left sideand integrate both sides.1400(ln|P|−ln|400−P|)=0.1667400t+C1ln∣∣P400−P∣∣=0.1667t+C2Multiply both sides by 400 and use the quotient rule to combine the logarightms.∣∣P400−P∣∣=e0.1667t+C2Rewrite the equation using the definition of the natural logarithm.∣∣P400−P∣∣=Ce0.1667tRewrite the right side using the properties of exponents.P400−P=Ce0.1667tIf we allow the constant C to be positive or negative, we can remove the absolute value.
At this point, we can substitute the values from the initial condition to find C.
150400−150=Ce0.1667(0)t=0 and P=150.150250=C(1)C=35
Substitute C=35 and solve for P.
P400−P=35e0.1667tP=35e0.1667t(400−P)Multiply both sides by 400−P.P=400(35)e0.1667t–P(35)e0.1667tDistribute.P+35Pe0.1667t=240e0.1667tSimplify and bring all terms with P to the left side.P(1+35e0.1667t)=240e0.1667tFactor P from the left side.P=240e0.1667t1+35e0.1667tDivide to isolate P on the left side.P(t)=1200e0.1667t5+3e0.1667tMultiply numerator and denominator by 5 to simplify.
4. The beginning of 2025 corresponds to t=5 and P(5)≈232. There will be approximately 232 pileated woodpeckers at the beginning of 2025.
5. Note that 75 of the carrying capacity is 0.75(400)=300. We set P(t)=300 and solve for t.
1200e0.1667t5+3e0.1667t=3001200e0.1667t=300(5+3e0.1667t)Multiply both sides by the denominator.1200e0.1667t=1500+900e0.1667tDistribute.300e0.1667t=1500Subtract 900e0.1667t from both sides.t=ln(5)0.1667Divide by 300, take the natural logarithm, and divide by 5 on both sides.t≈9.65
The population will reach 300 by the end of 2029.
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Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at
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What is an Isoline Map - Isoline Map Definition
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Chest Pain Syndromes - RCEMLearning
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Chest Pain Syndromes
Author:Jason M Kendall, Isabelle Hancock /Editor:Taj Hassan /Reviewer:Lauren Fraser /Codes:CC1, CC11, CC2, CC7, CP1, ResC10, ResP1, SLO1, SLO3 / Published: 16/07/2019
Introduction
Acute chest pain is a common presenting complaint: it accounts for approximately 700 000 presentations to the emergency department (ED) per year in England and Wales and for 25% of emergency medical admissions .
Chest pain is caused by a spectrum of pathology ranging from the innocent to the extremely serious (see Table 1); amongst the latter are a number of conditions which are potentially catastrophic and can cause death within minutes or hours. The large volume of patients presenting with a potentially serious condition places chest pain at the very core of emergency medicine work.
Emergency physicians are responsible for robustly identifying and treating a significant minority of patients with serious pathologies whilst also avoiding unnecessary investigation and admission for the majority of patients who can be safely discharged. This is a difficult challenge: it has been reported that 6% of patients discharged from a UK ED have subsequently been proven to have prognostically significant myocardial damage .
Table 1: The spectrum of pathology presenting with chest pain.
| System | Life-threatening | Urgent | Non-urgent |
--- --- |
| Cardiovascular | Acute myocardial infarction Aortic dissection Pulmonary embolism | Unstable angina Coronary vasospasm Pericarditis Myocarditis | Stable angina Valvular heart disease Hypertrophic cardiomyopathy |
| Pulmonary | Tension pneumothorax | Simple pneumothorax | Viral pleurisy Pneumonia |
| Musculoskeletal | | | Costochondritis Chest wall injury |
| Gastrointestinal | Oesophageal rupture | Pancreatitis | Cholecystitis Oesophageal reflux Biliary colic Peptic ulcer |
| Other | | Mediastinitis | Postherpetic neuralgia Herpes zoster Malignancy Psychological/anxiety |
The diagnostic approach to chest pain in the emergency department
Patients attending the ED with chest pain present a classic diagnostic challenge.
A minority will have an imminently or potentially life-threatening condition and need to be quickly and robustly identified and treated.
A significant minority will have benign pathology or no significant cause identifiable for their pain and these also need to be identified, reassured and discharged.
The rest will have a specific identifiable cause for their pain that will require a diagnosis and treatment plan during their ED attendance.
A logical and systematic approach to patients will achieve a diagnosis in the vast majority of cases whilst they are in the ED (see algorithm in the chart click the image to see a larger version).
A focused history and physical examination supported by an ECG and chest x-ray in most, with some specific ancillary investigations in a few, will allow a firm diagnosis to be made or confident reassurance to be given.
All of the assessments and investigations discussed in this session are within the remit of the emergency physician and should be available within the ED.
This module is about achieving a diagnosis in a patient presenting with chest pain, not about the treatment of a condition once a diagnosis has been made.
When a diagnosis has been reached with the required degree of certainty, management of that condition is covered in the relevantlearning sessionand the patient “drops” out of the algorithm.
Figure 1: The diagnostic approach to patients with chest pain
Note – According to Chest Pain NICE guideline 95, anginal pain is (1) a constricting discomfort in the front of the chest, or in the neck, shoulders, jaw, or arms that is (2) precipitated by physical exertion, and (3) relieved by rest or GTN within 5 minutes.
NICE defines chest pain as being atypical if only two of these three features are present, or non-anginal if one or none of these exist. (3)
The European Society of Cardiology describes atypical presentations of MI include epigastric pain, indigestion-like symptoms and isolated dyspnoea. (4)
History
One of the main purposes of early and rapid assessment of a patient with chest pain is to identify life-threatening conditions and, in particular, to “rule-in” or “rule-out” an acute coronary syndrome (ACS). There are various features in the history that are traditionally associated with cardiac ischaemic chest pain and various features that make it less likely.
A thorough description of the pain is the first step in the diagnostic process and will help to make the initial differentiation between cardiac and non-cardiac pain and ischaemic and non-ischaemic pain (See Figure 2 and Table 2). Symptom evaluation will include a description of the character of the pain, the location, severity and radiation of the pain, onset and duration of the pain, relieving and aggravating factors, and associated symptoms. Other important features of the history will include risk factor determination, previous episodes and relevant past medical history.
Table 2: Characteristic description of symptoms associated with major causes of chest pain(3-10)
ConditionDescription of symptoms
Ischaemic cardiac pain Retrosternal pressure, tightness, constricting
Radiation to shoulders/arms/neck/jaw
Crescendo in nature, related to exertion
Associated with diaphoresis, sweating, nausea, pallor
Pericarditis Atypical, retrosternal, sometimes pleuritic
Positional relieved on sitting forward
Gastro-oesophageal Retrosternal, burning
Associated with ingestion
Aortic dissection Tearing pain, sudden in onset, severe or worst ever pain, radiation to back
Pulmonary embolism Atypical, may be pleuritic
Associated with breathlessness; occasional haemoptysis
Pneumothorax Pleuritic, sharp, positional; sudden in onset
Associated with breathlessness
Pneumonia Atypical, may be pleuritic
Associated with cough, sputum, fever
Musculoskeletal Sharp, positional, pleuritic
Aggravated by movement, deep inspiration and coughing
Figure 2: Cardiac and non-cardiac causes of chest pain
Likelihood Ratios
Likelihood ratios have been calculated in various studies linking features of the history with acute myocardial infarction (AMI)(11-14)(see Table 3). It numerically indicates the probability of a patient having an AMI based on their description of the pain. Conventionally, a likelihood ratio greater than 10 provides very strong evidence to rule in a diagnosis whilst a ratio less than 0.1 provides very strong evidence to rule it out; ratios greater than 5 and less than 0.2 provide good evidence, and ratios greater than 2 and less than 0.5 provide moderate evidence to rule a diagnosis in or out respectively(15).
Table 3: Value of specific components of the chest pain history for the diagnosis of acute myocardial infarction(11-14)
| Historical factor | Likelihood factor | |
---
| Increased likelihood of AMI:Radiation to right arm/shoulder Radiation to both arms/shoulders Associated with exertion Radiation to left arm Associated with diaphoresis Associated with nausea/vomiting Worse than previous angina/similar to previous MI Described as a pressure | [Ref 11] 4.7 4.1 2.4 2.3 2.0 1.9 1.8 1.3 | [Ref 12] 2.9 7.1 – 2.3 2.0 1.9 – – | [Ref 13] 2.6 1.4 – 1.5 2.1 1.9 1.3 1.4 | [Ref 14] 1.3 2.6 1.5-1.8 1.3 1.3-1.4 0.9-1.1 2.2 – |
| Decreased likelihood of AMI:Described as pleuritic Described as positional Described as sharp Reproducible with palpation Inframammary location Not associated with exertion Associated Palpitations Associated Syncope | 0.2 0.3 0.3 0.3 0.8 0.8 – – | 0.2 0.3 0.3 0.2-0.4 – – – – | 0.2 0.3 0.3 0.2 – – – – | 0.4-0.6 0.2 0.3 0.3 0.2 0.6-0.8 0.7 0.4-0.8 |
| In heterogenous studies, likelihood ratios expressed as a range. | | | |
‘Burning pain’ (LR 1-1.4), ‘improvement with GTN’ (LR 0.93-1.3) and ‘abrupt onset of pain’ (LR 1-1.2) have no meaningful effect on the likelihood ratio of AMI.(14)
Therefore, based on these analyses, chest pain history is a helpful, but not diagnostic, first step in the assessment of these patients. Specifically, no single factor in the history carries with it a consistently powerful enough likelihood ratio to allow the emergency physician to safely discharge a patient without further diagnostic testing. Studies have reiterated that clinician gestalt although suggestive, cannot be used to rule in or rule out ACS.(16)
Learning Bite
No single factor in the history alone can confidently rule in or rule out AMI.
Chest Pain History – A Starting Point
The history does, however, form a start point in the diagnostic process, broadly establishing whether pain is likely to be cardiac ischaemic (or not) in origin; it provides information to add to baseline risk factors (see Table 4) which makes the diagnosis of ACS significantly more or less likely. Specifically, radiation of the pain to the arms or shoulders, and its association with exertion or diaphoresis will make the diagnosis more likely.
Table 4: Risk factors associated with major life-threatening causes of chest pain
ConditionRisk Factors
Acute coronary syndromes Previous known coronary artery disease (previous myocardial infarction, angioplasty, etc)Positive family history
Advanced age, male gender
Diabetes, hypertension, hypercholesterolaemia
Active smoker, obesity, sedentary lifestyle
Aspirin usage
Aortic dissection(6,7)Chronic hypertensionInherited connective tissue disorder e.g. Marfans Syndrome, Ehlers-Danlos Syndrome
Bicuspid aortic valve
Atheroscleroisis
Iatrogenic related to cardiac catheterisation
Coarctation of the aorta
Pregnancy
Cocaine use
Inflammatory aortic disease e.g. Giant Cell Arteritis
Pulmonary embolism(8,9)Previous history of venous thromboembolic diseasePregnancy particularly 6 weeks post partum
Positive family history of venous thromboembolic disease (two or more family members)
Recent prolonged immobilisation (> 3 days)
Major surgery within previous 12 weeks
Fracture of lower limb within previous 12 weeks
Active cancer (within previous 6 months, recent treatment, palliation)
Lower extremity paralysis
Likelihood of ACS
Chest pain that is pleuritic, sharp, positional and reproducible on palpation makes an alternative diagnosis much more likely. This information, in addition to other rapidly available clinical features (e.g. examination findings and the ECG), will determine the continued direction of investigation and initial management.
To refresh your memory on the diagnostic approach to patients with chest pain, click this thumbnail image:
Learning Bite
Characteristics of the pain with the highest likelihood for AMI are radiation of the pain to the right arm/shoulder or to both arms/shoulders.
Chest Pain Score
Various studies have used a “chest pain score” (based upon ascribing positive or negative points to typical or atypical aspects of chest pain location, character, radiation, onset, and associated symptoms(17)) and then combined this with historical risk factors to generate positive predictive values for ruling in or ruling out an acute coronary syndrome(18,19). However, this approach has not led to sufficiently robust positive likelihood ratios to definitively rule in ACS (i.e. commit to specific therapies) or rule out ACS (i.e. allow safe discharge).
Other life-threatening conditions present with chest pain other than ACS. The above discussion of likelihood applies specifically to AMI. The clinician will have to consider the presenting features in the history that are typical of other conditions (e.g. a “tearing” feeling and radiation to the back in aortic dissection (see Table 5), or haemoptysis and shortness of breath in pulmonary embolism) alongside risk factors specific to these conditions (see Table 4) in order to narrow the differential diagnosis and guide subsequent investigation and management.
Table 5: Presenting features of aortic dissection as per the international Registry of Acute Aortic Dissection(5)
SymptomType A (n=617)Type B (n=384)Overall (n=1001 except n=381)
Chest or Back Pain 507 (85%)328 (86%)835 (85%)
Severe or worst-ever pain 211 (90%)135 (90%)346 (90%)
Abrupt onset of pain 453 (91%)332 (89%)785 (90%)
Migrating pain 85 (15%)80 (25%)175 (19%)
Pain presenting within 6hrs of symptoms 334 (79%)....
Physical Examination
The history will have established whether a patient’s chest pain is ischaemic in nature and likely to be related to an ACS, or whether it is pleuritic or atypical in nature, and unlikely to be related to an ACS. Examination findings will further refine the differential diagnosis generated from the history.
Physical findings associated with ACS are generally non-specific and include pallor, anxiety, sweating, tachycardia and tachypnoea.
Generally, specific physical findings are associated with other (non-ischaemic) causes for chest pain or are associated with the complications of AMI: for example, a third heart sound occurring in heart failure, a pan-systolic murmur from mitral valve regurgitation, hypotension related to cardiogenic shock, or pulmonary crepitations secondary to left ventricular failure. These physical findings make AMI more likely [17-20].
Table 6: Value of specific components of the physical examination for the diagnosis of acute myocardial infarction(19-21)
| Examination finding | Likelihood ratio |
--- |
| | Ref12 | Ref13 | Ref14 |
| Increased likelihood of AMI: | | | |
| Third heart sound | 3.2 | 3.2 | |
| Hypotension (systolic BP <80 mmHg) | 3.1 | 2.1 | 0.98-15 |
| Pulmonary crepitations | 52.1 | 2.1 | 1.0-4.0 |
| Decreased likelihood of AMI: | | | |
| Chest pain reproducible by palpation | 0.3 | 0.2-0.4 | 0.14-0.54 |
Learning Bite
The finding of a third heart sound, hypotension or pulmonary crepitations makes AMI more likely.
Table 7 shows the key physical findings associated with conditions presenting with chest pain in the emergency department. Pivotal physical signs, or combinations of signs, which are highly suggestive of the relevant diagnoses, are shown in bold. Aortic dissection is renowned for being very difficult to diagnose, with the classic clinical signs occurring infrequently, see percentage of occurrence within the brackets.(6) Diagnosing PE is also a challenge because the symptoms and signs are common and not specific, Table 8(9)
Learning Bite
Various combinations of physical findings can be pathognomonic for non-ischaemic chest pain conditions
Table 7:Key physical findings associated with conditions causing chest pain
| Diagnosis | Physical findings |
--- |
| Acute coronary syndrome | Diaphoresis, tachycardia, tachypnoea, pallor |
| Complications of acute MI | Hypotension,thirdheart sound,pulmonary crepitations, elevated JVP, bradycardia, new murmur |
| Aortic dissection | Diaphoresis, hypotension, hypertension, tachycardia,differential blood pressures and/or pulses (27%),new murmur(aortic regurgitation 32%),focal neurological findings (12%) |
| Pulmonary embolism | Acute respiratory distress, diaphoresis, hypotension, tachycardia, hypoxaemia, elevated JVP, pleural rub |
| Pneumonia | Fever, signs of pulmonary collapse / consolidation,tachycardia, tachypnoea |
| Oesophageal rupture | Diaphoresis, hypotension, tachycardia,fever,Hammans sign,subcutaneous emphysema, epigastric tenderness |
| Simple pneumothorax | Tachypnoea, tachycardia,unilateral diminished air entry and breath sounds,subcutaneous emphysema |
| Tension pneumothorax | Tachypnoea, hypotension, tachycardia, hypoxaemia,elevated JVP, unilateral diminished air entry and breath sounds, subcutaneous emphysema,tracheal deviation |
| Pericarditis | Tachycardia,fever,pericardial rub |
| Myocarditis | Hypotension, tachycardia,fever,third heart sound, pulmonary crepitations,displaced apex beat |
| Mediastinitis | Tachycardia,fever,Hammans sign, subcutaneous emphysema, hypotension |
| Cholecystitis | Diaphoresis, fever, tachycardia,right upper quadrant tenderness |
| Hammans sign: audible systolic noise on cardiac auscultation |
Certain physical signs, or combinations of signs, are highly suggestive of certain diagnoses and are highlightedin bold.
Hamman’s sign: audible systolic noise on cardiac auscultation
Table 8: Prevalence of Symptoms and Signs in Patients with suspected PE according to final diagnosis.(9)
Symptoms of PEPE confirmed (n=219)PE Excluded (n=546)
Dyspnoea 80%59%
Pleuritic Chest Pain 52%43%
Substernal Chest Pain 12%8%
Cough 20%25%
Haemoptysis 11%7%
Syncope 19%11%
Signs of PE
Tachypnoea (>20/min)70%68%
Tachycardia >100 26%23%
Signs of DVT 15%10%
Fever (>38.5 o c)7%17%
Cyanosis 11%9%
Investigations
The Electrocardiogram (ECG)
After taking a history and performing a physical examination, the most commonly and rapidly performed investigation for a patient with chest pain in the ED is an ECG (see Figure 1). An ECG should be performed as soon as possible in all patients presenting with chest pain, particularly if cardiac ischaemia is suspected from the history.
Table 10 presents the likelihood ratios for the association of various ECG changes and AMI(12-14). The presence of ST segment elevation, new Q wave formation, or a new conduction deficit (eg. left bundle branch block) in the context of acute ischaemic chest pain is associated with such significantly positive likelihood ratios for AMI that the diagnosis can be made with confidence and appropriate therapy commenced.
The presence of ST segment depression and/or T wave changes in the context of acute ischaemic chest pain normally indicates myocardial ischaemia (i.e. unstable angina) but is also associated with a positive likelihood ratio for AMI (i.e. non-ST elevation AMI see Table 9). Approximately 50% of patients with ST depression and 33% of patients with T wave inversion will subsequently be shown to have myocardial infarction as defined by an elevated cardiac marker(20,21). This group of patients are presenting with an ACS (i.e. unstable angina or non-ST elevation myocardial infarction).
Learning Bite
ST segment elevation is associated with the highest likelihood of AMI followed, in order, by new Q waves, new conduction deficit, ST depression and T wave changes
A normal ECG significantly reduces the probability of AMI(12-14). It does not, however, reduce this probability enough to allow confident safe discharge based upon the history and ECG alone(2,3,4,5,16). Therefore, patients who present with chest pain in whom cardiac ischaemia is suspected and who have a normal ECG should undergo further diagnostic testing (ie. delayed cardiac markers, before they can be confidently ascribed to a low risk group.
Learning Bite
A normal ECG in a patient with chest pain does not allow safe discharge without further investigation
ECG Likelihood Ratios
Table 9:Value of specific components of the ECG for the diagnosis of acute myocardial infarction(12-14)
| ECG finding | Likelihood ratio |
--- |
| | Ref12 | Ref13 | Ref14 |
| Increased likelihood of AMI: | | | |
| New ST segment elevation | 5.7-53.9 | 13.1 | 2.1-8.6 |
| New Q wave formation | 5.3-24.8 | 5.0 | |
| New conduction deficit | 6.3 | | |
| ischaemic ECG | | | 1.6-5.7 |
| New ST segment depression | 3.0-5.2 | 3.13 | |
| T-wave peaking and/or inversion | 3.1 | 1.9 | |
| Decreased likelihood of AMI: | | | |
| Normal ECG | 0.1-0.3 | 0.1 | |
In heterogenous studies likelihood ratios expressed as a range
Ischemic ECG defined as any T wave inversion, ST depression, Q waves
ECG Results
The ECG must be considered in the context of the history and physical examination. The discussion above refers to the pivotal role of an ECG in the patient with a history of ischaemic cardiac chest pain. However, the ECG will also be useful in patients who have non-ischaemic pain (see Figure 1 and Table 10).
Normal ECG
A normal ECG significantly reduces the probability of AMI [4,5]. It does not, however, reduce this probability enough to allow confident safe discharge based upon the history and ECG alone .
Therefore, patients who present with chest pain in whom cardiac ischaemia is suspected and who have a normal ECG should undergo further diagnostic testing (i.e. delayed cardiac markers, exercise testing, etc.) before they can be confidently ascribed to a low risk group.
Table 10: ECG findings associated with non-ischaemic chest pain conditions(4,6)
| ECG finding | Context | Diagnosis |
---
| |
| Diffuse concave-upward ST segment elevation | Positional pain Pericardial rub | Pericarditis |
| Right ventricular strain Pattern | Pleuritic pain Hypoxia Pleural rub | Pulmonary embolus |
| Diffuse ST/T wave changes | Atypical pain Heart failure | Myocarditis |
| Inferior ST elevation Normal (30%) LVH (26%) | Tearing chest pain Radiation to back Differential pulses Differential blood pressures New diastolic murmur | Aortic dissection |
Learning Bite
A normal ECG in a patient with chest pain does not allow safe discharge without further investigation.
Chest X-ray (CXR)
The CXR is the next investigation commonly performed in the ED for patients presenting with chest pain following initial clinical assessment and ECG (see Figure 1). Table 11 shows the radiographic findings in conditions presenting with chest pain.
The CXR is particularly useful in patients presenting with non-cardiac chest pain and can definitively confirm a diagnosis suspected on clinical grounds (e.g. pneumothorax or pneumonia) or contribute significantly to the diagnostic process (e.g. widened mediastinum in aortic dissection or pneumomediastinum from oesophageal rupture). A normal CXR will also be helpful in making a diagnosis by excluding other potential causes for a certain clinical presentation: for example, a normal CXR in a patient with respiratory distress, pleuritic pain and hypoxia will exclude pneumothorax, make pneumonia unlikely, and increase the probability of pulmonary embolism.
There are no specific diagnostic findings on chest radiography associated with ACS; the usefulness of the CXR in this setting is to exclude other (non-cardiac) causes of chest pain or to evaluate complications of AMI (e.g. pulmonary oedema).
Table 11:Radiographic findings in conditions presenting with chest pain| Condition | Radiocraphic Finding | Comment |
---
| Acute coronary syndrome | No specific radiographic finding | |
| Aortic dissection | Mediastinal widening Abnormal aortic contour Globular heart shadow (haemopericardium) Pleural effusion (haemothorax) | Suggestive in context Unusual finding Rare finding Rare finding |
| Pneumothorax | Absence of pulmonary vascular markings | Diagnostic |
| Tension pneumothorax | Absence of pulmonary vascular markings Mediastinal displacement | Diagnostic Diagnostic |
| Pneumonia | Localised or diffuse pulmonary infiltration Localised pulmonary atelectasis/consolidation | Diagnostic in context |
| Pulmonary embolism | Normal chest radiograph Localised pulmonary atelectasis Small pleural effusion | Suggestive in context Rare finding Rare finding |
| Oesophageal rupture | Pneumomediastinum | Diagnostic in context |
| Mediastinitis | Pneumomediastinum | Diagnostic in context |
| Pericarditis | Globular heart shadow | Pericardial effusion |
| Myocarditis | Enlarged cardiac shadow | Dilated cardiomyopathy |
Clinical Decision Rules and Ancillary Investigations
The history, physical examination, ECG and CXR will normally allow the emergency physician to be fairly confident to achieve a diagnosis in a patient with chest pain presenting to the ED. There will still be a significant number of patients in whom the diagnosis is not clear or who require further investigations to allow safe discharge or definitive treatment.
Various studies have used a chest pain score (based upon ascribing positive or negative points to typical or atypical aspects of chest pain location, character, radiation, onset, and associated symptoms(17)) and then combined this with historical risk factors to generate predictive values for ruling in or ruling out an acute coronary syndrome(18,19). This approach alone has not led to sufficiently robust likelihood ratios to definitively rule in ACS (i.e. commit to specific therapies) or rule out ACS (i.e. allow safe discharge).
However, more robust scoring systems have been generated to assess risk by looking at components of the initial presentation AND by including additional factors, in particular biochemical cardiac markers. The most robustly validated risk scores are the TIMI, HEART and GRACE risk scores, the latter providing a score of future mortality(3,4,5). These scores help clinicians to identify those who are at high risk and requiring admission and treatment, and those who are at low risk and safe to discharge (with or without appropriate follow up).
Patients who present with a history of ischaemic chest pain who have a normal examination, ECG and CXR will require further risk stratification in order to allow safe discharge from the emergency department. Such a rule-out strategy will involve the use of cardiac markers (e.g. troponin) and possible exercise testing. Clinical pathways which combine clinician gestalt with the admission ECG and subsequent troponin assay(s) have shown a 100% sensitivity.(16)
Pulmonary Embolism
Pulmonary embolism is a relatively common condition that needs to be excluded with confidence due to its’ significant associated mortality if undiagnosed. Unfortunately, the history is variable, and there are no common diagnostic findings on examination, ECG or CXR. However, these findings, considered in association with historical risk factors for venous thromboembolic disorders, will allow the emergency physician to confidently ascribe the patient to a “likely” or “unlikely” risk category (two level PE Wells score is endorsed by NICE- see Table 12)(9).
Table 12: Two-level PE Wells score(9)
Clinical featurePoints
Clinical signs and symptoms of DVT (minimum of leg swelling and pain with palpation of the deep veins)3
An alternative diagnosis is less likely than PE 3
Heart rate > 100 beats per minute 1.5
Immobilisation for more than 3 days or surgery in the previous 4 weeks 1.5
Previous DVT/PE 1.5
Haemoptysis 1
Malignancy (on treatment, treated in the last 6 months, or palliative)1
Clinical probability simplified scores
PE likely More than 4 points
PE unlikely 4 points or less
Without further investigation, however, patients at low risk cannot be confidently reassured and discharged, and those at high risk should not be committed to prolonged anticoagulation. For patients at low risk, the diagnosis can be confidently excluded with a negative D-dimer assay(9,10,21). For patients at intermediate or high risk the diagnosis can be confidently confirmed or excluded with a ventilation perfusion (V/Q) scan or CT pulmonary angiogram (CTPA)(9,10,21).
Learning Bite
Pulmonary embolism will rarely be definitively diagnosed without ancillary investigations (D-Dimer, V/Q scan, or CTPA)
Aortic Dissection
Aortic dissection is a diagnosis that should be strongly suspected if the appropriate features are present upon clinical assessment: the history (tearing pain), examination (new murmur of aortic regurgitation, differential blood pressures), ECG (inferior ischaemic changes) and CXR (widened mediastinum) will, when present in combination, be pathognomonic of aortic dissection. However, due to the potentially catastrophic nature of aortic dissection if undiagnosed, this condition will need to be definitively excluded even if the index of suspicion is low (e.g. if only one of the characteristic clinical features is present). In patients in whom the diagnosis is virtually certain from the clinical presentation, the anatomical extent of the dissection will need to be defined. In either case, a CT angiogram of the aorta will need to be performed and this will be diagnostic and define the anatomical extent.
Learning Bite
CT angiogram of the aorta will be required to definitively exclude aortic dissection and/or to define its anatomical extent
Key Learning Points
No single factor in the history is associated with a high enough likelihood ratio to confidently diagnose or exclude AMI (Grade of evidence 1b)
The features of the chest pain history most likely to be associated with AMI are radiation to the right arm/shoulder or radiation to both arms/shoulders (Grade of evidence 1b)
The features of the chest pain history least likely to be associated with AMI are sharp nature, positional, pleuritic and reproducible with palpation (Grade of evidence 1b)
Examination findings which increase the likelihood of AMI in the context of chest pain are the presence of a third heart sound, hypotension and pulmonary crackles (Grade of evidence 1b)
ECG findings diagnostic of AMI in the context of ischaemic chest pain are new ST segment elevation and new Q waves (Grade of evidence 1b)
A normal ECG in the context of ischaemic chest pain does not rule out prognostically significant myocardial damage and further diagnostic testing will be required (Grade of evidence 1b)
A normal examination ECG and CXR cannot rule out an Aortic dissection
Shortness of breath and pleuritic chest pain are sensitive but not specific for Pulmonary emboli. Use of clinical predictive rules are essential to assess the likelihood of the diagnosis and aid ongoing investigations.
A logical diagnostic approach to patients presenting with chest pain will achieve a diagnosis in the majority based on taking a history, performing an examination, ECG and CXR in the ED
Ancillary investigations will be required for the remainder and, specifically:
A rule-out strategy for patients with ischaemic pain and a normal ECG (Grade of evidence 1b)
D-dimer, V/Q scanning or CTPA for patients in whom pulmonary embolism is a possibility (Grade of evidence 1b)
CT angiogram of the aorta for patients suspected of having an acute aortic dissection
References
Technol Assess 2013;17(1).
Collinson PO, Premachandram S, Hashemi K. Prospective audit of incidence of prognostically important myocardial damage in patients discharged from emergency department.BMJ 2000;320:1702-5.
National Institute for Health and Care Excellence (NICE).Chest Pain of recent onset: assessment and diagnosis CG95 London: 2010. Updated 2016.
Task Force for the Management of Acute Coronary Syndromes in Patients Presenting without Persistent ST-Segment Elevation of the European Society of Cardiology (ESC)Guidelines for the management of acute coronary syndromesin patients presenting without persistent ST-segment elevation. European Heart Journal (2016) 37, 267315
The Task Force for the Diagnosis and Treatment of Non-ST-Segment Elevation Acute Coronary Syndromes of the European Society of Cardiology.Guidelines for the Diagnosis and treatment of non-ST-segment elevation acute coronary syndromes.Eur Heart Journal 2007;28:1598-1660.
Larson E, Edwards W. Risk factors for aortic dissection: a necropsy study of 161 cases.Am J Cardiol 1984;53:849-55.
Golledge J, Eagle K. Acute Aortic Dissection. Lancet 2008:372;55-66
Wells P, Ginsberg J, Anderson D, Kearon C, Gent M, Turpie A et al. Use of a clinical model for safe management of patients with suspected pulmonary embolism.Ann Intern Med 1998;129:997-1005.
National Institute for Health and Care Excellence (NICE)Venous thromboembolic diseases: the management of venous thromboembolic diseases and the role of thrombophilia testing; CG114 London 2012.
The Task Force for the Diagnosis and Management of Acute Pulmonary Embolism of the European Society of Cardiology (ESC)Guidelines on the diagnosis and management of acute pulmonary embolism. European Heart Journal (2008) 29, 22762315
Swap CJ and Nagurney JT. Value and limitations of chest pain history in the evaluation of patients with suspected acute coronary syndromes.JAMA 2005;294(20):2623-29.
Panju A, Hemmelgarn B, Guyatt G, and Simel D. Is this patient having a myocardial infarction?JAMA 1998;280:1256-63.
Fanaroff A, Rymer J, Goldstein S, Simel D, Newby K, Does This Patient with Chest Pain Have Acute Coronary Syndrome? JAMA 2015;314(18):1955-1965.
Mant J, McManus RJ, Oakes RAL, Delaney BC, Barton PM, Deeks JJ et al. Systematic review and modelling of the investigation of acute and chronic chest pain presenting in primary care.Health Technology Assessment 2004;Vol 8:No. 2.
Jaeschke R, Guyatt GH, Sackett DL. Users guide to the medical literature. III. How to use an article about a diagnostic test. B. What are the results and will they help me cure my patients? The Evidence Based Medicine Working Group.JAMA 1994;271:703-7.
Body R, Cook G, Burrows G, Carley S, Lewis PS. Can emergency physicians rule in and rule out acute myocardial infarction with clinical judgement? Emerg Med J. 2014:31(11):872-6
Geleijnse M, Elhendy J, Kasprzak J, Rambaldi R, van Domburg R, Cornel J et al. Safety and prognostic value of early dobutamine-atropine stress echocardiography in patients with spontaneous chest pain and a non-diagnostic electrocardiogram.Eur Heart J 2000;21:397-406.
Conti A, Paladini B, Toccafondi S, Magazzini S, Olivotto I, Galassi F et al. Effectiveness of a multidisciplinary chest pain unit for the assessment of coronary syndromes and risk stratification in the Florence area.Am Heart J 2002;144(4):630-5.
Sanchis J, Bodi V, Llacer A, Nunez J, Consuegra M, Bosch M et al. Risk stratification of patients with chest pain and normal troponin concentrations.Heart 2005;91(8):1013-8.
Myocardial infarction redefined A consensus document of The Joint European Society of Cardiology / American College of Cardiology Committee for the Redefiniton of Myocardial Infarction. The Joint European Socitey of Cardiology / American College of Cardiology Committee.Eur Heart J 2000;21:1502-1513.
Karlson B, Herlitz J, Wiklund O, Richter A, Hjalmarson A. Early prediction of acute myocardial infarction from clinical history, examination and electrocardiogram in the emergency room.Am J Cardiol 1991;68:171-5.
British Thoracic Societyguidelines for the management of suspected pulmonary embolism.Thorax 2003;58:470-484.
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#### Acute Coronary Syndromes
This session is about the pathophysiology of acute coronary syndromes, defining acute myocardial infarction and recognising the various presentations and clinical features associated with acute coronary syndromes.
#### Acute Coronary Syndromes
This session is about the pathophysiology of acute coronary syndromes, and recognising the various presentations and clinical features associated with them.
#### Chest Pain: Low Risk ‘Rule-Out’ Pathways
Radiation of pain to the arms or shoulders, and association with exertion, diaphoresis, nausea or vomiting are useful for ruling in the diagnosis of AMI.
35 responses
Matthew McKeesays: May 28, 2018 at 10:22 pm good overview Log in to Reply
Charlotte Walkersays: August 21, 2018 at 2:04 pm Useful resource Log in to Reply
Muhammad Sayyedsays: September 19, 2018 at 4:47 pm A quick and useful briefing on chest pain Log in to Reply
Michael Daveysays: September 20, 2018 at 10:42 am Excellent article. Puts an overall scheme onto my existing knowledge. Also prompted me to double-check my knowledge of some ECG patterns (including right ventricular strain pattern) Log in to Reply
kandaratnami5783says: January 31, 2020 at 8:15 pm Good topic Log in to Reply
Dr. Yusuf Dala Galisays: April 28, 2020 at 2:47 pm Thank you, great revision. Log in to Reply
iqbaljsays: May 20, 2020 at 11:58 am brilliant recapitulation of cardiac emergency medicine Log in to Reply
Miss Michelle Anne Maundsays: May 30, 2020 at 5:33 pm Useful resource and refresh of knowledge. Log in to Reply
Fatima Hamidisays: June 25, 2020 at 10:57 am Concise and informative. Log in to Reply
borakatssays: October 25, 2020 at 6:33 pm Good Log in to Reply
Chukwunonso Ibenegbusays: February 28, 2021 at 11:43 pm Very succinct but detailed.thanks Log in to Reply
Dr. Muralidhar Srima Phallesays: May 19, 2021 at 9:53 pm Good Revision Log in to Reply
Dr Kamran Syed Muhammadsays: May 21, 2021 at 10:47 am Well written with concise points for chest pain differential Log in to Reply
Mr. Salman Ahmed Abbasisays: August 8, 2021 at 3:58 am very nice Log in to Reply
Denise Marie Wilkinsonsays: October 5, 2021 at 2:06 pm Interesting article. Good revision Log in to Reply
Mrs. Elaine E. Burnettsays: October 23, 2021 at 6:12 pm Really good article, it will help me as I move forward with my ACP training, thank you Log in to Reply
Miss Helen Laura Wilsonsays: January 8, 2022 at 6:20 pm Good revision and confirmation of practice Log in to Reply
burrowst6671says: January 19, 2022 at 9:05 pm Concise revision of the differentials of chest pain. Log in to Reply
Dr John Oluwasegun Ajisafesays: March 5, 2022 at 9:21 pm comprehensive discussion on chest pain. Log in to Reply
Dr Ibrahim Abdo Mohammed Husseinsays: March 13, 2022 at 9:38 pm informative and fundamental review of chest pain DDx and approach to chest pain in ED Log in to Reply
Dr Syed Shahbaz Ahmersays: June 21, 2022 at 7:55 pm Excellent update on chest pain Log in to Reply
Dr Sartaj Khansays: July 7, 2022 at 11:41 am Very informative and fundamental indeed. Log in to Reply
Dr. Mihaela-Nicoleta Ionsays: August 12, 2022 at 6:22 pm Very good Log in to Reply
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chrislucy21@googlemail.comsays: December 20, 2022 at 1:32 pm Really good learning tool and very informative Log in to Reply
Arnold Tonye Wilcoxsays: January 3, 2023 at 4:49 pm good refresher Log in to Reply
Miss Sue Elizabeth Listersays: February 16, 2023 at 10:09 am informative direction in diagnostic thought process Log in to Reply
Dr Ellen Sheris Umohsays: June 20, 2023 at 11:50 pm very eduvative article Log in to Reply
Mr Salim Kadhim Shubbersays: September 19, 2023 at 9:08 am It was interesting to read about the likelihood ratios in relation to different parameters of the assessment from the history to the various investigative tools used for each possible diagnosis Log in to Reply
Dr Aatif Masaud Buttsays: November 18, 2023 at 5:56 pm Useful resource and good overview Log in to Reply
Dr Mohamed Abdelgadir Elshakh Mohamedsays: March 14, 2024 at 12:25 am Very useful and informative Log in to Reply
Dr Adnan Quddussays: August 4, 2024 at 4:03 pm described the best approach for chest pain Log in to Reply
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When carboxylic acid reacts with PCl5, what do we obtain, POCCl3 or POCl3? - Quora
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Chemistry
POCl3
Carboxilic Acid
PCL5 in Chemistry
Phosphorus Pentachloride
Organic Reactions
Chemical Synthesis
Organic Chemistry
Carboxyl Group
5
When carboxylic acid reacts with PCl5, what do we obtain, POCCl3 or POCl3?
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Ayush Kumar
Intern at Adesignguy.co (2019–present)
·7y
Phosphorus(V) chloride is a solid which reacts with carboxylic acids in the cold to give steamy acidic fumes of hydrogen chloride. It leaves a liquid mixture of the acyl chloride and a phosphorus compound, phosphorus trichloride oxide (phosphorus oxychloride) - POCL3
The acyl chloride can be separated by fractional distillation.
For example:
Continue Reading
Phosphorus(V) chloride is a solid which reacts with carboxylic acids in the cold to give steamy acidic fumes of hydrogen chloride. It leaves a liquid mixture of the acyl chloride and a phosphorus compound, phosphorus trichloride oxide (phosphorus oxychloride) - POCL3
The acyl chloride can be separated by fractional distillation.
For example:
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Mayur
7y
The answer is pocl3 bcz whenever oxygen containing compound react with pcl5 form by product pocl3
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Purna Chandra Sahu
M. Sc. in Pure Chemistry&Inorganic Chemistry, Calcutta Uniiversity (Graduated 1987) · Author has 2.9K answers and 6M answer views
·4y
Related
What product is formed when the same type of carboxylic acid reacts together?
When a carboxylic acid is heated in presence of an dehydrating agent, anhydride of the carboxylic acid is formed. From each two molecules...
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Related questions
More answers below
What is the mechanism of PCl5 and carboxylic acid?
When PCl5 reacts with carboxylic acid, is it possible to obtain 2 products like ROCl and RCl?
What happens when benzoic acid reacts with PCl5?
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Why are carboxylic acids stronger acids than alcohols?
Kurt VdB
Ph.D. in Organic Chemistry, KU Leuven · Author has 1.3K answers and 13M answer views
·8y
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Can a carboxylic acid react with another carboxylic acid? If it can, what are the products of this reaction?
dehydradation of two (simple) carboxylic acids provides the (symmetric) anhydride.
You can do this by simple heating e.g with acetic acid:
e.g. 2 CH3COOH → CH3COOCH3 + H2O
but depending on the conditions you could get competing reactions.
e.g.decarboxylation
CH3COOH → CO2 + CH4 or even
CH3COOH → H20 + H2C=C=0 (ketene)
(the last two reactions can not be considered as a carboxylic acid reaction with another carboxylic acid.)
Besides heating you could use a dehydration agents like P2O5 etc. this will make the reaction to proceed at lower temperature.
see e.g:
Mechanism of carboxylic acid and amide dehydrat
Continue Reading
dehydradation of two (simple) carboxylic acids provides the (symmetric) anhydride.
You can do this by simple heating e.g with acetic acid:
e.g. 2 CH3COOH → CH3COOCH3 + H2O
but depending on the conditions you could get competing reactions.
e.g.decarboxylation
CH3COOH → CO2 + CH4 or even
CH3COOH → H20 + H2C=C=0 (ketene)
(the last two reactions can not be considered as a carboxylic acid reaction with another carboxylic acid.)
Besides heating you could use a dehydration agents like P2O5 etc. this will make the reaction to proceed at lower temperature.
see e.g:
Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide
You could also do the reaction inter-molecular. e.g. phthalic acid that forms phthalic anhydride and I guess you could make mixed anhydrides as well although you would probably convert one of them to e.g. the acid-chloride first.
e.g. in Yamaguchi esterification
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Yashas Anilkumar
Electron donating group! ;)
·Updated 8y
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PCl5 is added to an organic compound and it becomes warm with the evolution of HCl gas. What is the compound?
This is a test used to identify the functional group ‘alcohol’ in an organic compound.
When PCl5 is added to an alcohol, we observe that HCl gas is liberated. The evolution of HCl gas can be easily verified by holding a glass rod dipped in Ammonium hydroxide solution near the mouth of the test tube, and if dense white fumes are observed, we may conclude that it is indeed HCl gas, and thus conclude that the organic compound falls under the category of alcohols.
The equations associated are:-
PCl5 + ROH — RCl + POCl3 + HCl
HCl + NH4OH — NH4Cl ( white solid) + H2O
Hope this helps! :)
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Why are carboxylic acids acidic in nature?
How can we convert a Nitro compound into a carboxylic acid?
Are all acids carboxylic acids like lactic acid?
What is the difference between a carboxylic acid and a stronger acid?
Is lactic acid a carboxylic acid?
Ernest Leung
M.Phil. in Chemistry, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·Jun 26
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PCl5 is added to an organic compound and it becomes warm with the evolution of HCl gas. What is the compound?
PCl₅ is added to an organic compound and it becomes warm with the evolution of HCl gas. What is the compound?
The compound may be an alcohol (ROH) or a carboxylic acid (RCOOH). They can react exothermically with phosphorus pentachloride to give alkyl/acyl chloride, phosphorus oxychloride and fumes of hydrogen chloride.
ROH + PCl₅ → RCl + POCl₃ + HCl
RCOOH + PCl₅ → RCOCl + POCl₃ + HCl
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Adianadi Adi
Chemistry Guru - running a famous website: www.adichemistry.com
·8y
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Can a carboxylic acid react with another carboxylic acid? If it can, what are the products of this reaction?
They can react by losing a water molecule to give an anhydride in presence of a dehydrating agent.
E.g. Two molecules of acetic acid will form acetic anhydride in presence of P2O5 by losing a water molecule.
2CH3COOH ————→ (CH3CO)2O + H2O
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Ernest Leung
M.Phil. in Chemistry, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·Mar 7
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How are KP and KC related in PCL5=PCL3+CL2?
Suppose the question is:
"How are Kp and Kc related in PCl₅ ⇌ PCl₃ + Cl₂?"
The answer is as follows.
Continue Reading
Suppose the question is:
"How are Kp and Kc related in PCl₅ ⇌ PCl₃ + Cl₂?"
The answer is as follows.
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Stuart Herring
Author has 11.7K answers and 8.2M answer views
·7y
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Are all organic acids carboxylic?
Are all organic acids carboxylic?
Most are, but there are important exceptions.
From the Wikipedia article on Organic acid:
“An organic acid is an organic compound with acidic properties. The most common organic acids are the carboxylic acids, whose acidity is associated with their carboxyl group –COOH. Sulfonic acids, containing the group –SO₂OH, are relatively stronger acids. Alcohols, with –OH, can act as acids but they are usually very weak. The relative stability of the conjugate base of the acid determines its acidity. Other groups can also confer acidity, usually weakly: the thiol group
Continue Reading
Are all organic acids carboxylic?
Most are, but there are important exceptions.
From the Wikipedia article on Organic acid:
“An organic acid is an organic compound with acidic properties. The most common organic acids are the carboxylic acids, whose acidity is associated with their carboxyl group –COOH. Sulfonic acids, containing the group –SO₂OH, are relatively stronger acids. Alcohols, with –OH, can act as acids but they are usually very weak. The relative stability of the conjugate base of the acid determines its acidity. Other groups can also confer acidity, usually weakly: the thiol group –SH, the enol group, and the phenol group. In biological systems, organic compounds containing these groups are generally referred to as organic acids.”
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Colin John Cook
Industrial Chemist in commercial chemicals · Author has 3.4K answers and 3.7M answer views
·1y
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Why does carboxylic acid give OH during esterification?
Carboxylic acid + alcohol gives ester + water. The hydroxyl of the alcohol reacts with the hydrogen of the acid (or vice versa) to make water . This is normally removed by dehydrating mineral acids (e.g. sulphuric), or by a catalyst, and/or by heat and distillation.
To check whether it is the hydroxy part of the acid or the alcohol that makes the water, I suggest you look up the mechanism. Using an isotope of oxygen to make versions of the acid and alcohol, one will find that either the ester contains it or the water. This will have enabled researchers to determine the mechanism.
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Purna Chandra Sahu
M. Sc. in Pure Chemistry&Inorganic Chemistry, Calcutta Uniiversity (Graduated 1987) · Author has 2.9K answers and 6M answer views
·4y
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What is added to RMgX to produce carboxylic acid?
RMgX reacts with dry ice (solid CO2), followed by hydrolysis gives...
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Abhijit Singh
iOS Developer at Noon Academy (2019–present) · Author has 349 answers and 1.3M answer views
·8y
Related
Which is more stable, PCL3 or PCL5?
The actual reason for unstability of PCl5 is because of presence of two axial bonds which form an angle of 90 deg with the triangular plane consisting of 1 P and 3 Cl atoms. Thus due to greater repulsions at axial positions in PCl5, it dissociates to give PCl3 and Cl2. See the figure below,
Continue Reading
The actual reason for unstability of PCl5 is because of presence of two axial bonds which form an angle of 90 deg with the triangular plane consisting of 1 P and 3 Cl atoms. Thus due to greater repulsions at axial positions in PCl5, it dissociates to give PCl3 and Cl2. See the figure below,
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What is the mechanism of PCl5 and carboxylic acid?
When PCl5 reacts with carboxylic acid, is it possible to obtain 2 products like ROCl and RCl?
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Why are carboxylic acids stronger acids than alcohols?
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What is the mechanism of PCl5 and carboxylic acid?
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What happens when benzoic acid reacts with PCl5?
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Why are carboxylic acids stronger acids than alcohols?
Why are carboxylic acids acidic in nature?
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Can a Cubic Equation have 2 Roots?
Last Updated :
23 Jul, 2025
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Can a Cubic Equation have 2 Roots?
Answer: Yes, a cubic equation can have two roots, but one of them will be repeated (a double root).
Explanation:
A cubic equation, typically of the form ax3+bx2+cx+d=0, can indeed have two roots, but the nature of these roots depends on the discriminant and coefficients. If the discriminant is positive, the cubic equation will have three distinct real roots. However, if the discriminant is zero, it will have two roots, and one of these roots will be repeated, known as a double root.
Possibility of Two Roots
Contrary to common perception, a cubic equation can indeed have two roots. However, it's important to note that one of these roots will be repeated, resulting in what is known as a double root. We explore the conditions under which a cubic equation exhibits this behavior and provide an illustrative example for clarity.
Example - Can a Cubic Equation have 2 Roots?
consider the cubic equation x3−6x2+9x=0. Factoring it, we get x(x−3)2=0, revealing that x=0 and x=3 are the roots. Here, x=3is a repeated root, indicating a double root.
So, while a cubic equation can have two roots, it does so when one of those roots is repeated, resulting in a total of two distinct solutions.
Conclusion
In conclusion, while it may seem counterintuitive, a cubic equation can indeed have two roots. However, this occurs when one of the roots is repeated, resulting in a total of two distinct solutions. Understanding the nature of roots in cubic equations enhances our comprehension of polynomial functions and their properties, contributing to the broader landscape of algebra and mathematics.
Related Resources:
Solving Cubic Equations: Definitions, Methods and Examples
Real-Life Applications of Cube Root
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https://intro.chem.okstate.edu/1515SP01/Database/VPWater.html
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Vapor Pressure of Water
| | | | |
--- --- |
| Temperature | Pressure | Temperature | Pressure |
| (degrees C) | (mmHg) | (degrees C) | (mmHg) |
| 0 | 4.6 | 50 | 92.5 |
| 1 | 4.9 | 51 | 97.2 |
| 2 | 5.3 | 52 | 102.1 |
| 3 | 5.7 | 53 | 107.2 |
| 4 | 6.1 | 54 | 112.5 |
| 5 | 6.5 | 55 | 118 |
| 6 | 7 | 56 | 123.8 |
| 7 | 7.5 | 57 | 129.8 |
| 8 | 8 | 58 | 136.1 |
| 9 | 8.6 | 59 | 142.6 |
| 10 | 9.2 | 60 | 149.4 |
| 11 | 9.8 | 61 | 156.4 |
| 12 | 10.5 | 62 | 163.8 |
| 13 | 11.2 | 63 | 171.4 |
| 14 | 12 | 64 | 179.3 |
| 15 | 12.8 | 65 | 187.5 |
| 16 | 13.6 | 66 | 196.1 |
| 17 | 14.5 | 67 | 205 |
| 18 | 15.5 | 68 | 214.2 |
| 19 | 16.5 | 69 | 223.7 |
| 20 | 17.5 | 70 | 233.7 |
| 21 | 18.7 | 71 | 243.9 |
| 22 | 19.8 | 72 | 254.6 |
| 23 | 21.1 | 73 | 265.7 |
| 24 | 22.4 | 74 | 277.2 |
| 25 | 23.8 | 75 | 289.1 |
| 26 | 25.2 | 76 | 301.4 |
| 27 | 26.7 | 77 | 314.1 |
| 28 | 28.3 | 78 | 327.3 |
| 29 | 30 | 79 | 341 |
| 30 | 31.8 | 80 | 355.1 |
| 31 | 33.7 | 81 | 369.7 |
| 32 | 35.7 | 82 | 384.9 |
| 33 | 37.7 | 83 | 400.6 |
| 34 | 39.9 | 84 | 416.8 |
| 35 | 42.2 | 85 | 433.6 |
| 36 | 44.6 | 86 | 450.9 |
| 37 | 47.1 | 87 | 468.7 |
| 38 | 49.7 | 88 | 487.1 |
| 39 | 52.4 | 89 | 506.1 |
| 40 | 55.3 | 90 | 525.8 |
| 41 | 58.3 | 91 | 546 |
| 42 | 61.5 | 92 | 567 |
| 43 | 64.8 | 93 | 588.6 |
| 44 | 68.3 | 94 | 610.9 |
| 45 | 71.9 | 95 | 633.9 |
| 46 | 75.7 | 96 | 657.6 |
| 47 | 79.6 | 97 | 682.1 |
| 48 | 83.7 | 98 | 707.3 |
| 49 | 88 | 99 | 733.2 |
| | | 100 | 760 |
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What Is Breadth-First Search (BFS)?
Breadth-First Search (BFS) is one of the simplest and most widely used algorithms for searching a graph. It systematically explores all nodes in the graph to find a solution, starting from a given starting point (or root node) and moving outward layer by layer. The algorithm does not make assumptions about the possible location of the solution but explores all nodes equally, ensuring that it checks every possibility until it finds the target node or exhausts all options.
Key Terminology in BFS:
Graph: A structure consisting of nodes (also called vertices) and edges connecting them. In BFS, the graph can be directed or undirected.
Node (Vertex): The individual elements or points in a graph where edges meet. BFS begins at a specific node and explores outward from it.
Edge: A connection between two nodes. In BFS, edges define how nodes are connected and dictate the traversal path.
Queue: A First-In-First-Out (FIFO) data structure used in BFS to track nodes that need to be explored. Nodes are added to the queue when discovered and removed when explored.
Visited Set: A collection of nodes that have already been explored to avoid revisiting the same node multiple times.
Root/Starting Node: The node where the BFS algorithm begins. From this node, the search proceeds to explore the graph.
Level/Layer: Refers to the distance from the starting node. BFS explores all nodes at the same level before moving on to nodes further away.
How BFS Works
BFS operates by exploring a graph level by level, starting from a given node and expanding outward based on distance. The detailed steps are:
Mark the start node as visited and place it in a queue.
Remove a node from the queue, visit it, and perform necessary actions.
Add all unvisited neighboring nodes of the current node to the queue and mark them as visited.
Repeat steps 2 and 3 until the queue is empty, indicating that all reachable nodes have been processed.
This process continues level by level, visiting each node's neighbors, until all nodes have been explored or the target is found. BFS is often used as the foundation for more advanced algorithms, such as Dijkstra's algorithm for finding the shortest path and Prim's algorithm for constructing minimum spanning trees.
Breadth-First Search (BFS) Example
Example Graph
Consider the following graph structure:
A / \ B C / \ \D E F
In this graph, nodes (also known as vertices) are represented by letters, and edges are the connections between these nodes.
BFS Traversal on the Example Graph
BFS explores the graph level by level, starting from a chosen root node (vertex A in this case). Here's how the traversal proceeds:
Initialization:
Enqueue the starting node A into a queue.
Mark A as visited to avoid revisiting it.
Exploration:
Step 1: Dequeue A from the queue (current node) and visit it. Then enqueue its neighbors B and C, and mark them as visited.
Step 2: Dequeue B and visit it. Then enqueue its neighbors D and E, marking them as visited.
Step 3: Dequeue C and visit it. Then enqueue its neighbor F, and mark F as visited.
Step 4: Dequeue D and visit it. Since D has no neighbors, move to the next node.
Step 5: Dequeue E and visit it. Similarly, E has no neighbors.
Step 6: Dequeue F and visit it. F has no neighbors.
Final BFS Traversal Order:
The nodes are visited in the following order:A -> B -> C -> D -> E -> F
Complexity Analysis
Time Complexity
Explanation: The time complexity of BFS is O(V + E), where:
V represents the number of vertices (nodes) in the graph.
E represents the number of edges (connections between nodes).
BFS explores every vertex and examines each edge once, resulting in this linear time complexity. This makes it efficient for traversing graphs, whether large or small.
Scenarios:
Best Case: BFS quickly finds the target node close to the start.
Worst Case: BFS needs to explore the entire graph before reaching the target node.
Average Case: BFS typically explores multiple levels of the graph.
Space Complexity
Explanation: The space complexity of BFS is O(V), where V is the number of vertices. This is because BFS uses:
A queue to store vertices that are waiting to be explored.
A visited set to track which nodes have already been visited.
In the worst case, the size of the queue could be as large as the number of vertices at the largest level of the graph, leading to O(V) space usage.
Memory Usage: The memory needed for BFS depends on the graph size. In addition to the queue, BFS needs space for the visited set or array, which has a size of O(V).
Applications of Breadth-First Search (BFS)
Real-World Applications
Real-World Applications
Shortest Path in Unweighted Graphs:BFS guarantees the shortest path in an unweighted graph by exploring nodes in increasing order of distance from the start. This is particularly useful in network routing, such as finding the most efficient path in road networks (e.g., Google Maps uses BFS for unweighted roads).
Web Crawlers:Web crawlers use BFS to systematically explore web pages. Starting from a seed URL, the crawler visits all links on that page, then proceeds to follow links from those pages, layer by layer. This method ensures a comprehensive search across a defined depth.
Theoretical Applications
Bipartite Graph Checking:BFS helps check if a graph is bipartite by attempting to color nodes with two colors. If BFS detects any adjacent nodes having the same color, the graph is not bipartite. This is used in scheduling problems and matching algorithms.
Finding Connected Components:In an undirected graph, BFS can identify connected components by exploring all vertices reachable from a starting vertex. Once a component is fully explored, BFS moves to the next unvisited node to find the next component. This is important for clustering and social network analysis.
Breadth-First Search (BFS) in Different Programming Languages
BFS in Python
Python Code Example
The following Python code demonstrates the implementation of the Breadth-First Search (BFS) algorithm:
from collections import dequedef bfs(graph, start_vertex): visited = set() queue = deque([start_vertex]) visited.add(start_vertex) while queue: vertex = queue.popleft() print(vertex, end=" ") for neighbor in graph[vertex]: if neighbor not in visited: visited.add(neighbor) queue.append(neighbor)# Example graph represented as an adjacency listgraph = { 'A': ['B', 'C'], 'B': ['D', 'E'], 'C': ['F'], 'D': [], 'E': [], 'F': []}bfs(graph, 'A')
Explanation:
A queue is used to track the vertices to be explored next.
The visited set ensures that vertices are not revisited.
As the queue processes each vertex, its unvisited neighbors are enqueued, and the process repeats.
BFS in Java
Java Code Example
The following Java code demonstrates the implementation of the Breadth-First Search (BFS) algorithm:
import java.util.;public class BFSExample { public static void bfs(Map> graph, String startVertex) { Set visited = new HashSet<>(); Queue queue = new LinkedList<>(); visited.add(startVertex); queue.add(startVertex); while (!queue.isEmpty()) { String vertex = queue.poll(); System.out.print(vertex + " "); for (String neighbor : graph.get(vertex)) { if (!visited.contains(neighbor)) { visited.add(neighbor); queue.add(neighbor); } } } } public static void main(String[] args) { Map> graph = new HashMap<>(); graph.put("A", Arrays.asList("B", "C")); graph.put("B", Arrays.asList("D", "E")); graph.put("C", Arrays.asList("F")); graph.put("D", Collections.emptyList()); graph.put("E", Collections.emptyList()); graph.put("F", Collections.emptyList()); bfs(graph, "A"); }}
Explanation of Java Code
The Java code defines a bfs method that takes a graph and a starting vertex as inputs. The method uses a HashSet to track visited vertices and a LinkedList as a queue.
Initialization: The code adds the starting vertex to the visited set and enqueues it.
Exploration: While the queue is not empty, the code dequeues a vertex, prints it, and enqueues its unvisited neighbors.
Termination: The process continues until the queue becomes empty.
The main method creates an example graph using a HashMap and initiates the BFS traversal from vertex A.
BFS in C++
C++ Code Example
The following C++ code demonstrates the implementation of the Breadth-First Search (BFS) algorithm:
```
include #include #include #include #include using namespace std;void bfs(unordered_map>& graph, char startVertex) { unordered_set visited; queue q; visited.insert(startVertex); q.push(startVertex); while (!q.empty()) { char vertex = q.front(); q.pop(); cout << vertex << " "; for (char neighbor : graph[vertex]) { if (visited.find(neighbor) == visited.end()) { visited.insert(neighbor); q.push(neighbor); } } }}int main() { unordered_map> graph; graph['A'] = {'B', 'C'}; graph['B'] = {'D', 'E'}; graph['C'] = {'F'}; graph['D'] = {}; graph['E'] = {}; graph['F'] = {}; bfs(graph, 'A'); return 0;}
```
Explanation of C++ Code
The C++ code defines a bfs function that takes a graph and a starting vertex as inputs. The function uses an unordered_set to track visited vertices and a queue to manage the order of exploration.
Initialization: The code inserts the starting vertex into the visited set and enqueues it.
Exploration: While the queue is not empty, the code dequeues a vertex, prints it, and enqueues its unvisited neighbors.
Termination: The process continues until the queue becomes empty.
The main function creates an example graph using an unordered_map and initiates the BFS traversal from vertex A.
Comparing Breadth-First Search (BFS) with Other Algorithms
BFS vs. Depth-First Search (DFS)
Traversal Strategy:
BFS explores nodes level by level using a queue.
DFS explores as far down a branch as possible using a stack (or recursion) before backtracking.
Use Cases:
BFS is ideal for finding the shortest path in unweighted graphs.
DFS is more suited for tasks like topological sorting and solving mazes.
BFS vs. Dijkstra's Algorithm
Graph Type:
BFS works well for unweighted graphs.
Dijkstra’s Algorithm is designed for weighted graphs, finding the shortest path based on cumulative edge weights.
Use Cases:
BFS is useful in social networks and bipartite graph checking.
Dijkstra's Algorithm is used in GPS systems and network routing where edge weights represent distances or costs.
Practice Problems
Find the Shortest Path in an Unweighted Graph:
Problem: Given a graph and two vertices, the goal is to find the shortest path between them using BFS. BFS is optimal for unweighted graphs because it explores all vertices at the current depth before moving to the next, ensuring that the first time a target node is reached, it's via the shortest path in terms of the number of edges.
Solution Explanation:
Start by initializing a queue with the starting vertex.
Perform BFS, recording each vertex’s parent node as you traverse.
Once you reach the target vertex, backtrack using the parent nodes to construct the shortest path.
Return the path.
Additional Note: BFS guarantees the shortest path in terms of the number of edges, making it ideal for this type of problem in unweighted graphs.
Check if a Graph is Bipartite:
Problem: Determine if a given graph can be colored using two colors such that no two adjacent vertices share the same color. BFS can be used to alternate colors between nodes as it traverses the graph.
Solution Explanation:
Start BFS from any unvisited node, assigning it the first color (say, "color 1").
For each neighbor of the current node, assign the opposite color ("color 2").
Continue traversing, ensuring no two adjacent nodes share the same color.
If a conflict is found (i.e., two adjacent vertices have the same color), the graph is not bipartite.
Additional Note: This problem is a variation of a graph coloring problem and is commonly applied in scheduling or partitioning tasks. BFS is preferred over DFS for this problem because it naturally lends itself to checking levels, which corresponds to color alternation.
Find Connected Components in a Graph:
Problem: Identify all connected components in an undirected graph. A connected component is a set of vertices such that there is a path between any two vertices in the set.
Solution Explanation:
Initialize BFS from any unvisited vertex and explore all reachable vertices, marking them as visited and part of the same component.
After completing one BFS, if any unvisited vertices remain, start BFS again from one of them to find another connected component.
Repeat the process until all vertices have been visited, identifying all connected components.
Additional Note: This is particularly useful in social network analysis and clustering, where identifying distinct subgroups of connected nodes is necessary. BFS ensures a complete traversal of each component.
Web Crawler Simulation:
Problem: Create a basic web crawler that uses BFS to systematically explore web pages starting from a given URL.
Solution Explanation:
Initialize a queue with the starting URL.
While the queue is not empty, dequeue a URL, visit the page, and enqueue any new links found on the page.
Use a set to track visited URLs to avoid revisiting the same page.
Continue until you reach the desired depth or number of pages.
Additional Note: This problem demonstrates how BFS can be applied to real-world tasks like web crawling, where the crawler explores links in layers (level by level). BFS ensures that all pages linked from the same "depth" (e.g., the home page) are visited before moving deeper into the site.
Solutions and Explanations
Shortest Path Problem:
Steps:
Initialize a queue with the starting vertex.
Perform BFS, visiting each vertex, and recording the path using a parent map.
Once the target is reached, backtrack from the target to the start using the parent map to construct the path.
Return the shortest path.
Key Insight: Since BFS explores all nodes at the same depth before moving to the next, the first time it encounters the target, it guarantees the shortest path (in terms of the number of edges).
Bipartite Graph Checking:
Steps:
Initialize a queue with any unvisited vertex and assign it the first color.
For each neighbor, assign the opposite color and enqueue them.
If two adjacent vertices have the same color, the graph is not bipartite.
If BFS completes without conflicts, the graph is bipartite.
Key Insight: BFS’s level-by-level nature makes it ideal for alternately coloring nodes and checking for coloring conflicts.
Connected Components:
Steps:
Start BFS from any unvisited node and mark all reachable nodes as part of the same component.
Repeat for all unvisited nodes until the entire graph is covered.
Key Insight: BFS naturally divides the graph into its connected subgraphs, making it straightforward to find all connected components.
Web Crawler Simulation:
Steps:
Initialize a queue with the starting URL.
Visit the URL, enqueue new links, and track visited URLs.
Continue until all reachable URLs are explored or a stopping condition (e.g., depth limit) is met.
Key Insight: BFS’s layer-by-layer exploration closely mirrors how a web crawler can explore all links from one page before moving on to deeper links.
The provided problems and their explanations are technically accurate and represent common scenarios where BFS is used. The elaboration helps clarify why BFS is suitable for these problems and how it ensures optimal solutions, particularly in unweighted graphs and scenarios requiring level-based exploration. Understanding BFS through these practical examples will solidify knowledge and enhance problem-solving abilities in both theoretical and real-world applications.
Additional Resources
Recommended Books and Articles
Books on Graph Algorithms
Books provide a comprehensive understanding of graph algorithms. The following books offer in-depth knowledge of Breadth-First Search (BFS) and related topics:
"Introduction to Algorithms" by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein: This book covers a wide range of algorithms, including BFS. The authors explain the concepts with clarity and provide detailed pseudocode.
"Algorithms" by Robert Sedgewick and Kevin Wayne: This book includes a thorough discussion of graph algorithms. The authors present BFS with practical examples and applications.
"Graph Theory with Applications to Engineering and Computer Science" by Narsingh Deo: This book focuses on graph theory and its applications. The author provides a detailed explanation of BFS and its uses in various fields.
Online Articles and Tutorials
Online resources offer quick and accessible information on BFS. The following articles and tutorials provide valuable insights:
GeeksforGeeks: The article on BFS explains the algorithm with examples and pseudocode. The website also offers practice problems to reinforce learning.
TutorialsPoint: This tutorial covers BFS with step-by-step explanations and code examples in different programming languages.
Khan Academy: The video tutorials on Khan Academy provide a visual explanation of BFS. The instructors break down the algorithm into simple steps for easy understanding.
Conclusion
BFS is a systematic and efficient graph traversal algorithm, providing a foundation for solving various problems, such as finding the shortest path, analyzing network structures, and more. Its simplicity and guaranteed exploration of all reachable nodes make it a powerful tool in computer science.
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Solved: y is inversely proportional to x^2 (b) Complete this table of values. Table 1: ["columnLi [Math]
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y is inversely proportional to x^2 (b) Complete this table of values. Table 1: ["columnList":["x","1","2","5",""],"lines":1,"columnList":["y","","50","","2"],"lines":2]
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Expert Verified Solution
97%(984 rated)
Answer
Table 1:
$$\begin{array}{cc} x & y \ \hline 1 & 200 \ 2 & 50 \ 5 & 8 \ 10 & 2 \ \end{array}$$x 1 2 5 10y 200 50 8 2
Explanation
Write the inverse proportionality equation for $$y$$y and $$x^{2}$$x 2, which is $$y = \frac{k}{x^{2}}$$y=x 2 k, where $$k$$k is the constant of proportionality.
Use the given condition that when $$x = 2$$x=2, $$y = 50$$y=50 to find the value of $$k$$k. Substitute $$x = 2$$x=2 and $$y = 50$$y=50 into the equation to get $$50 = \frac{k}{4}$$50=4 k
Solve for $$k$$k by multiplying both sides of the equation by 4 to get $$k = 200$$k=200
Now that we have the value of $$k$$k, we can write the complete inverse proportionality equation as $$y = \frac{200}{x^{2}}$$y=x 2 200
Use the equation to find the values of $$y$$y for the given values of $$x$$x
For $$x = 1$$x=1, $$y = \frac{200}{1^{2}} = 200$$y=1 2 200=200
For $$x = 5$$x=5, $$y = \frac{200}{5^{2}} = 8$$y=5 2 200=8
For $$x = 10$$x=10, $$y = \frac{200}{10^{2}} = 2$$y=1 0 2 200=2
Fill in the completed table with the calculated values:
Table 1:
$$\begin{array}{cc} x & y \ \hline 1 & 200 \ 2 & 50 \ 5 & 8 \ 10 & 2 \ \end{array}$$x 1 2 5 10y 200 50 8 2
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A company is using linear programming to decide how many units of each of its two products to make each week. Weekly production will be x units of Product X and y units of Product Y. At least 50 units of X must be produced each week, and at least twice as many units of Y as of X must be produced each week. Each urt of X requires 30 minutes of labour, and each unit of Y requires two hours of labour. There are 5,000 hours of labour available each week. Which of the following is the correct set of constraints? Submit your answer to view the feedback. 0.5x+2y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≤ 2x y ≥ 100 x+4y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≥ 2x y ≥ 2x
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5
How can I show that the modulus value of a complex number is the distance from the origin to the point representing the complex number?
4 Answers
Sort
David Gillies
BSc. Physics, MSc. RF Engineering
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 760 answers and 786.2K answer views
·
8y
I assume you know Pythagoras’ Theorem. The modulus of a complex number
z
,
z
=
a
+
i
b
is
defined
as
|
z
|
=
√
a
2
+
b
2
. If you represent this number on an Argand diagram then the distance from the origin to the point corresponding to the number will
necessarily
be the modulus.
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Master's Degree from Harvard University (Graduated 2011)
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Related questions
What is the modulus of a complex number?
How do I find modulus of a complex number?
Which complex number has a distance of from the origin on the complex plane?
How do I find the modulus and the principal argument of the complex number, 2/√ (3 - i)?
What is the use of modulus of a complex number?
Hunter Johnson
PhD in Mathematics, University of Maryland, College Park (Graduated 2008)
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 610 answers and 2.5M answer views
·
8y
You know that the distance of the point
a
+
b
i
in the complex plane from the origin should be given by the Euclidean distance formula
√
a
2
+
b
2
. Show that the modulus amounts to the same thing.
Donald Hartig
PhD in Mathematics, University of California, Santa Barbara (Graduated 1970)
·
Author has 7.4K answers and 2.8M answer views
·
7y
By definition, the modulus of z=a+ib is the distance from z to the origin, 0.
The fact that |z|=√a2+b2 follows immediately from the Pythagorean Theorem.
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1y
To show that the modulus of a complex number represents the distance from the origin to the point representing that complex number, we can follow these steps:
Definition of a Complex Number
: A complex number
z
can be expressed in the form:
z
=
a
+
b
i
where
a
is the real part and
b
is the imaginary part.
2. Modulus of the Complex Number
: The modulus (or absolute value) of the complex number
z
is defined as:
|
z
|
=
√
a
2
+
b
2
3. Geometric Interpretation
: In the complex plane (often referred to as the Argand plane), the real part
a
is represented on the x-axis, and the imaginary part
b
is re
To show that the modulus of a complex number represents the distance from the origin to the point representing that complex number, we can follow these steps:
Definition of a Complex Number
: A complex number
z
can be expressed in the form:
z
=
a
+
b
i
where
a
is the real part and
b
is the imaginary part.
2. Modulus of the Complex Number
: The modulus (or absolute value) of the complex number
z
is defined as:
|
z
|
=
√
a
2
+
b
2
3. Geometric Interpretation
: In the complex plane (often referred to as the Argand plane), the real part
a
is represented on the x-axis, and the imaginary part
b
is represented on the y-axis. The complex number
z
=
a
+
b
i
corresponds to the point
(
a
,
b
)
.
4. Distance Formula
: The distance
d
from the origin
(
0
,
0
)
to the point
(
a
,
b
)
can be calculated using the distance formula:
d
=
√
(
a
−
0
)
2
+
(
b
−
0
)
2
=
√
a
2
+
b
2
5. Conclusion
: From the definitions and geometric interpretations above, we see that the modulus of the complex number
|
z
|
=
√
a
2
+
b
2
is exactly equal to the distance
d
from the origin to the point
(
a
,
b
)
in the complex plane.
Thus, the modulus of a complex number indeed represents the distance from the origin to the point representing that complex number.
Related questions
What is the modulus of a complex number?
How do I find modulus of a complex number?
Which complex number has a distance of from the origin on the complex plane?
How do I find the modulus and the principal argument of the complex number, 2/√ (3 - i)?
What is the use of modulus of a complex number?
Can a complex number be represented graphically?
What are the applications of complex numbers?
What does a modulus of a complex number represent?
What are the units of a complex number?
What are the modulus and argument of the following complex numbers 1+I?
How do I understand a complex number raised to a complex number?
How do I find the complex number with the given modulus and argument in the form of Xtiy?
What is the modulus and argument of a complex number?
Are the complex numbers an ordered field?
How can a complex number show a 3D point?
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What Is a Linear Pair of Angles in Geometry?
By: Talon Homer
|
Aug 12, 2024
In the subjects of geometry and trigonometry, a linear pair of angles is any two adjacent angles formed together to add up to 180°, or π (pi) radians.
This rule is also known as the linear pair postulate. This arrangement of two adjacent angles is easy to spot in a math textbook or on a piece of paper because the bottom section of both the angles will form a perfectly straight line.
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Consequently, any straight line that has a second line segment branching from it will form a linear pair. Linear pairs of angles are helpful to us while studying geometry because knowing one half of the linear pair will allow us to easily solve the other angle's value.
Contents
Linear Pair vs. Straight Angle
Linear Pair vs. Supplementary Angles
Solving a Linear Pair of Angles
How Other Adjacent Angles Form
Linear Pair vs. Straight Angle
In geometry, a straight angle has a value of 180° or π radians. However, it doesn't appear to be an angle at all because the lines or opposite rays that meet at the common vertex form a straight line. In this way, a linear pair is very closely related to a straight angle.
A linear pair can also be thought of as straight angle which has a single line segment added to its common vertex. In the process, a straight angle is effectively split into two angles.
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Linear Pair vs. Supplementary Angles
If two angles add up to 180°, we call them supplementary angles. At first, this naming seems redundant because linear pairs add up the same value.
However, there is a subtle difference between the two: Supplementary angles do not necessarily have to be adjacent to one another. They can have non-common arms, or not be attached to each other at all.
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Linear pairs of angles are always adjacent to one another, and always sharing a common arm. In this way, a linear pair can alternatively be defined as a pair of supplementary adjacent angles. Depending on the author, your geometry textbook may also use supplementary angles as a synonym for a linear pair.
Solving a Linear Pair of Angles
Since we know that a linear pair is made up of exactly two adjacent angles, and we know that these angles add up to a specific value, we can quickly find the value of a second adjacent angle as long as one of the values is already known.
For example, let's we have a linear pair of angles and one of them has a given value of 73°:
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Angle AB = 73°, Linear Pair ABC = 180°
Angle BC = 180° – 73°
BC = 107°
As all linear pairs add up to 180°, any geometry problem involving a linear pair of angles can be solved just by subtracting the known angle value from our total of 180 and getting a final value for the linear pair angle where our lines intersect.
In cases where one part of our linear pair is a right angle equal to 90°, solving this problem for the other angle will be even easier. Since 90 is exactly half of 180, our linear pair will be congruent angles with the exact same angle measurement of 90°.
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How Other Adjacent Angles Form
Adjacent angles can be thought of as any two or more angles which meet up at a single point. In the case of a linear pair, these angles add up to 180°, but they can also equal 90°, 360° or some arbitrary angle measurement.
In this way, all linear pairs are adjacent angles, but not all adjacent angles form a linear pair.
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Let's go over some different types of adjacent angles, how they differ from a linear pair, and how they can help us find unknown angle values.
Complementary Adjacent Angles
Similarly to how supplementary angles and linear pairs make a 180° value, complementary adjacent angles are two angles that share a common vertex and a common arm, adding up to a 90° value.
You may also think of complementary angles as a single right angle split into two angles by a dividing line segment. Instead of forming a straight line, the non-common arms form a square edge.
Complementary angles can be solved in much the same way that a linear pair of angles can. Instead of subtracting our known angle value from 180° to find the other angle value, we must subtract it from 90°.
Angle AB = 73°, Complementary Pair ABC = 90°
Angle BC = 90° – 73°
BC = 17°
If the complementary angles are split perfectly down the middle, two congruent angles form with values of 45° each.
Vertical Angles
Vertical angles are created by crossing two lines or rays over each other. The result is four adjacent angles that all add up to a value of 360°, or 2π radians.
Vertical angles can also be thought of as two linear pairs which are adjacent and congruent to each other. Knowing these rules, we can solve three unknown angle values using only one known angle value.
Solving Vertical Angles
There are a few ways we can go about solving vertical angles. In this example, angles AB and BC form a linear pair, but AB and AD could also be considered to form linear pairs. Since AB and CD are opposite angles, we also know them to be congruent to each other. The same goes for BC compared to AD.
Let's once again assume that angle AB is 73°. Consequently, angle CD must have an equal value of 73°. By subtracting 73° from linear pair value of 180, we can now find the value of both BC and AD.
Angle AB = 73°, Linear Pair ABC = 180°
Angle CD = AB, CD = 73°
Angle BC = 180° – 73°
BC = 107°
AD = BC
AD = 107°
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Inclination of a Line In Section 1.3, you learned that the graph of the linear equation is a nonvertical line with slope and -intercept There, the slope of a line was described as the rate of change in with respect to In this section, you will look at the slope of a line in terms of the angle of inclination of the line.
Every nonhorizontal line must intersect the -axis. The angle formed by such an intersection determines the inclination of the line, as specified in the following definition.
Horizontal Line Vertical Line Acute Angle Obtuse Angle FIGURE 10.1 The inclination of a line is related to its slope in the following manner.
For a proof of the relation between inclination and slope, see Proofs in Mathematics on page 806.
θ x y θ x y = θ π 2 x y x = 0 θ y x x.
y 0, b.
y m y mx b 728 Chapter 10 Topics in Analytic Geometry What you should learn • Find the inclination of a line.
• Find the angle between two lines.
• Find the distance between a point and a line.
Why you should learn it The inclination of a line can be used to measure heights indirectly. For instance, in Exercise 56 on page 734, the inclination of a line can be used to determine the change in elevation from the base to the top of the Johnstown Inclined Plane.
Lines AP/Wide World Photos 10.1 Definition of Inclination The inclination of a nonhorizontal line is the positive angle (less than measured counterclockwise from the -axis to the line. (See Figure 10.1.) x ) Inclination and Slope If a nonvertical line has inclination and slope then m tan .
m, The HM mathSpace® CD-ROM and Eduspace® for this text contain additional resources related to the concepts discussed in this chapter.
333202_1001.qxd 12/8/05 8:54 AM Page 728 Finding the Inclination of a Line Find the inclination of the line Solution The slope of this line is So, its inclination is determined from the equation From Figure 10.2, it follows that This means that The angle of inclination is about 2.554 radians or about Now try Exercise 19.
The Angle Between Two Lines Two distinct lines in a plane are either parallel or intersecting. If they intersect and are nonperpendicular, their intersection forms two pairs of opposite angles.
One pair is acute and the other pair is obtuse. The smaller of these angles is called the angle between the two lines. As shown in Figure 10.3, you can use the incli-nations of the two lines to find the angle between the two lines. If two lines have inclinations where and the angle between the two lines is You can use the formula for the tangent of the difference of two angles to obtain the formula for the angle between two lines. tan 2 tan 1 1 tan 1 tan 2 tan tan2 1 2 1.
2 1 < 2, 1 < 2 1 and 2, 146.3. 2.554. 0.588 0.588 arctan 2 3 2 < < .
tan 2 3.
m 2 3.
2x 3y 6.
Section 10.1 Lines 729 Example 1 Angle Between Two Lines If two nonperpendicular lines have slopes and the angle between the two lines is tan m2 m1 1 m1m2 .
m2, m1 1 2 3 2x + 3y = 6 ≈146.3° θ 1 3 x y FIGURE 10.2 θ θ θ θ 2 θ1 1 θ2 = − x y FIGURE 10.3 333202_1001.qxd 12/8/05 8:54 AM Page 729 Finding the Angle Between Two Lines Find the angle between the two lines.
Line 1: Line 2: Solution The two lines have slopes of and respectively. So, the tangent of the angle between the two lines is Finally, you can conclude that the angle is as shown in Figure 10.4.
Now try Exercise 27.
The Distance Between a Point and a Line Finding the distance between a line and a point not on the line is an application of perpendicular lines. This distance is defined as the length of the perpendicular line segment joining the point and the line, as shown in Figure 10.5.
Remember that the values of and in this distance formula correspond to the general equation of a line, For a proof of the distance between a point and a line, see Proofs in Mathematics on page 806.
Finding the Distance Between a Point and a Line Find the distance between the point and the line Solution The general form of the equation is So, the distance between the point and the line is The line and the point are shown in Figure 10.6.
Now try Exercise 39.
3.58 units.
8 5 d 24 11 1 22 12 2x y 1 0.
y 2x 1.
4, 1 Ax By C 0.
C B, A, 79.70 1.391 radians arctan 11 2 11 2 .
114 24 34 2 1 234 tan m2 m1 1 m1m2 m2 3 4, m1 2 3x 4y 12 0 2x y 4 0 730 Chapter 10 Topics in Analytic Geometry 1 4 3 ≈79.70° θ 1 2 4 2x −y −4 = 0 3x + 4y −12 = 0 x y FIGURE 10.4 d (x1, y1) (x2, y2) x y FIGURE 10.5 −3 −2 1 2 3 4 5 −2 −3 −4 1 2 3 4 (4, 1) y = 2x + 1 x y FIGURE 10.6 Distance Between a Point and a Line The distance between the point and the line is d Ax1 By1 C A2 B2 .
Ax By C 0 x1, y1 Example 2 Example 3 333202_1001.qxd 12/8/05 8:54 AM Page 730 An Application of Two Distance Formulas Figure 10.7 shows a triangle with vertices and a. Find the altitude from vertex to side b. Find the area of the triangle.
Solution a. To find the altitude, use the formula for the distance between line and the point The equation of line is obtained as follows.
Slope: Equation: Point-slope form Multiply each side by 4.
General form So, the distance between this line and the point is b. Using the formula for the distance between two points, you can find the length of the base to be Distance Formula Simplify.
Simplify.
Simplify.
Finally, the area of the triangle in Figure 10.7 is Formula for the area of a triangle Substitute for b and h.
Simplify.
Now try Exercise 45. 13 square units. 1 2217 13 17 A 1 2bh 217 units. 68 82 22 b 5 3 2 2 02 AC Altitude h 10 44 3 12 42 13 17 units.
0, 4 x 4y 3 0 4y x 3 y 0 1 4x 3 m 2 0 5 3 2 8 1 4 AC 0, 4.
AC AC.
B h C5, 2.
A3, 0, B0, 4, Section 10.1 Lines 731 Example 4 W RITING ABOUT MATHEMATICS Inclination and the Angle Between Two Lines Discuss why the inclination of a line can be an angle that is larger than but the angle between two lines cannot be larger than Decide whether the following statement is true or false: “The inclination of a line is the angle between the line and the -axis.” Explain.
x 2.
2, Activities 1. Find the inclination of the line Answer: radian or 51.34 2. Find the angle between the lines and Answer: radians or 81.87 3. Find the distance between the point and the line Answer: unit d 3 5 4x 3y 12 0.
3, 1 1.429 3x y 6.
x 2y 5 0.896 5x 4y 20.
Group Activity: Graphing Utility Put your students in groups of two.
Ask each group to write a graphing calculator program that finds the angle between two lines based on user input.
Ask each group to demonstrate the program on an example or exercise in this section.
x 1 2 3 4 5 −2 1 2 4 5 6 C (5, 2) B (0, 4) A (−3, 0) h y FIGURE 10.7 333202_1001.qxd 12/8/05 8:54 AM Page 731 In Exercises 1–8,find the slope of the line with inclination 1.
2.
3.
4.
5.
radians 6.
radians 7.
radians 8.
radians In Exercises 9–14, find the inclination (in radians and degrees) of the line with a slope of 9.
10.
11.
12.
13.
14.
In Exercises 15–18, find the inclination (in radians and degrees) of the line passing through the points.
15.
16.
17.
18.
In Exercises 19–22, find the inclination (in radians and degrees) of the line.
19.
20.
21.
22.
In Exercises 23–32,find the angle (in radians and degrees) between the lines.
23.
24.
25.
26.
27.
28.
29.
30.
3x 5y 12 x 2y 2 3x 5y 3 x 2y 8 3x 5y 1 6x 2y 5 5x 2y 16 x 2y 7 θ 1 2 3 4 1 2 3 4 y x y x −1 −2 1 2 −1 −2 1 2 θ 4x 3y 24 3x 2y 1 2x 3y 22 3x 2y 0 θ −1 −2 −3 2 3 1 y x θ −2 −1 1 2 3 4 2 y x x 2y 3 2x y 2 x 3y 2 3x y 3 x y 10 0 5x 3y 0 4x 5y 9 0 6x 2y 8 0 0, 100, 50, 0 2, 20, 10, 0 4, 3 12, 8, 6, 1, 10, 8 m 5 2 m 3 4 m 2 m 1 m 2 m 1 m.
2.88 1.27 5 6 3 = θ 2 3 π y x x = θ 3 4 π y = θ π 4 y x = θ π 6 y x .
732 Chapter 10 Topics in Analytic Geometry Exercises 10.1 The HM mathSpace® CD-ROM and Eduspace® for this text contain step-by-step solutions to all odd-numbered exercises.They also provide Tutorial Exercises for additional help.
VOCABULARY CHECK: Fill in the blanks.
1. The _ of a nonhorizontal line is the positive angle (less than measured counterclockwise from the axis to the line.
2. If a nonvertical line has inclination and slope then 3. If two nonperpendicular lines have slopes and the angle between the two lines is 4. The distance between the point and the line is given by PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.
d _ .
Ax By C 0 x1, y1 tan _ .
m2, m1 m _ .
m, x-) 333202_1001.qxd 12/8/05 11:05 AM Page 732 Section 10.1 Lines 733 31.
32.
Angle Measurement In Exercises 33–36, find the slope of each side of the triangle and use the slopes to find the measures of the interior angles.
33.
34.
35.
36.
In Exercises 37–44,find the distance between the point and the line.
Point Line 37.
38.
39.
40.
41.
42.
43.
44.
In Exercises 45–48, the points represent the vertices of a triangle. (a) Draw triangle in the coordinate plane, (b) find the altitude from vertex of the triangle to side and (c) find the area of the triangle.
45.
46.
47.
48.
In Exercises 49 and 50, find the distance between the parallel lines.
49.
50.
51. Road Grade A straight road rises with an inclination of 0.10 radian from the horizontal (see figure). Find the slope of the road and the change in elevation over a two-mile stretch of the road.
52. Road Grade A straight road rises with an inclination of 0.20 radian from the horizontal. Find the slope of the road and the change in elevation over a one-mile stretch of the road.
53. Pitch of a Roof A roof has a rise of 3 feet for every horizontal change of 5 feet (see figure). Find the inclina-tion of the roof.
5 ft 3 ft 0.1 radian 2 mi 4 −2 −2 −4 −4 2 y x 4 −2 −2 4 y x 3x 4y 10 x y 5 3x 4y 1 x y 1 C 6, 12 B 3, 10, A 4, 5, C 5 2, 0 B 2, 3, A 1 2, 1 2, C 5, 2 B 4, 5, A 0, 0, C 4, 0 B 1, 4, A 0, 0, AC, B ABC x y 20 4, 2 6x y 0 0, 8 y 4 0 10, 8 x 1 0 6, 2 x y 2 2, 1 4x 3y 10 2, 3 2x y 4 0, 0 4x 3y 0 0, 0 4 2 4 (2, 1) ( 3, 4) − ( 2, 2) − −2 −2 −4 y x 4 2 4 (1, 0) (3, 2) (−4, −1) y x 2 −2 −2 4 −4 2 4 6 (2, 0) (1, 3) (−3, 2) y x 2 4 6 2 4 6 (4, 4) (6, 2) (2, 1) y x 0.03x 0.04y 0.52 0.02x 0.05y 0.19 0.07x 0.02y 0.16 0.05x 0.03y 0.21 333202_1001.qxd 12/8/05 8:54 AM Page 733 54. Conveyor Design A moving conveyor is built so that it rises 1 meter for each 3 meters of horizontal travel.
(a) Draw a diagram that gives a visual representation of the problem.
(b) Find the inclination of the conveyor.
(c) The conveyor runs between two floors in a factory. The distance between the floors is 5 meters. Find the length of the conveyor.
55. Truss Find the angles and shown in the drawing of the roof truss.
Synthesis True or False?
In Exercises 57 and 58,determine whether the statement is true or false. Justify your answer.
57. A line that has an inclination greater than radians has a negative slope.
58. To find the angle between two lines whose angles of inclination and are known, substitute and for and , respectively, in the formula for the angle between two lines.
59. Exploration Consider a line with slope and -intercept (a) Write the distance between the origin and the line as a function of (b) Graph the function in part (a).
(c) Find the slope that yields the maximum distance between the origin and the line.
(d) Find the asymptote of the graph in part (b) and interpret its meaning in the context of the problem.
60. Exploration Consider a line with slope and -intercept (a) Write the distance between the point and the line as a function of (b) Graph the function in part (a).
(c) Find the slope that yields the maximum distance between the point and the line.
(d) Is it possible for the distance to be 0? If so, what is the slope of the line that yields a distance of 0?
(e) Find the asymptote of the graph in part (b) and interpret its meaning in the context of the problem.
Skills Review In Exercises 61–66, find all -intercepts and -intercepts of the graph of the quadratic function.
61.
62.
63.
64.
65.
66.
In Exercises 67–72,write the quadratic function in standard form by completing the square. Identify the vertex of the function.
67.
68.
69.
70.
71.
72.
In Exercises 73–76, graph the quadratic function.
73.
74.
75.
76. gx x2 6x 8 gx 2x2 3x 1 f x 6 x 12 f x x 42 3 f x 8x2 34x 21 f x 6x2 x 12 f x x2 8x 15 f x 5x2 34x 7 f x 2x2 x 21 f x 3x2 2x 16 f x x2 9x 22 f x x2 7x 1 f x x 112 12 f x x 52 5 f x x 92 f x x 72 y x m.
3, 1 d 0, 4.
y m m.
d 0, 4.
y m m2 m1 2 1 2 1 2 36 ft 9 ft 6 ft 6 ft α β 734 Chapter 10 Topics in Analytic Geometry 56. Inclined Plane The Johnstown Inclined Plane in Johnstown, Pennsylvania is an inclined railway that was designed to carry people to the hilltop community of Westmont. It also proved useful in carrying people and vehicles to safety during severe floods. The railway is 896.5 feet long with a 70.9% uphill grade (see figure).
(a) Find the inclination of the railway.
(b) Find the change in elevation from the base to the top of the railway.
(c) Using the origin of a rectangular coordinate system as the base of the inclined plane, find the equation of the line that models the railway track.
(d) Sketch a graph of the equation you found in part (c).
896.5 ft Not drawn to scale θ Model It 333202_1001.qxd 12/8/05 8:54 AM Page 734
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29 Instructional days (29 hours)
Ratios and Unit Rates
Investigate and relate ratios, rates, and percents, applying reasoning in real world problem solving.
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Quantum chemistry ; Donald A. McQuarrie. - UC San Diego
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; Quantum chemistry ; Donald A. McQuarrie.
; McQuarrie, Donald A. (Donald Allan), author.;
; Sausalito, Calif. : University Science Books; ; 2008;
Quantum chemistry Available at Geisel Library Geisel Floor1 East, Books(QD462 .M4 2008)
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Title
Quantum chemistry
Quantum chemistry
Quantum chemistry
Donald A. McQuarrie.
Donald A. McQuarrie.
Donald A. McQuarrie.
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Creator
McQuarrie, Donald A. (Donald Allan), author.
McQuarrie, Donald A. (Donald Allan), author.
McQuarrie, Donald A. (Donald Allan), author.
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Edition
2nd ed.
2nd ed.
2nd ed.
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Publisher
Sausalito, Calif. : University Science Books
Sausalito, Calif. : University Science Books
Sausalito, Calif. : University Science Books
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Creation Date
2008
2008
2008
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Physical Description
xiii, 690 pages : illustrations (some color) ; 26 cm
xiii, 690 pages : illustrations (some color) ; 26 cm
xiii, 690 pages : illustrations (some color) ; 26 cm
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Copyright
©2008
©2008
©2008
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Contents
The dawn of the quantum theory -- A. Complex numbers -- The classical wave equation -- B. Probability and statistics -- The Schrodinger equation and a particle in a box -- C. Vectors -- The postulates and general principles of quantum mechanics -- D. Series and limits -- The harmonic oscillator and vibrational spectroscopy -- E. Spherical coordinates -- The rigid rotator and rotational spectroscopy -- F. Determinants -- The hydrogen atom -- G. Matrices -- Approximation methods -- H. Matrix Eigenvalue problems -- Many-electron atoms -- The chemical bond: one- and two-electron molecules -- Qualitative theory of chemical bonding -- The Hartree-Fock-Roothaan method.
The dawn of the quantum theory -- A. Complex numbers -- The classical wave equation -- B. Probability and statistics -- The Schrodinger equation and a particle in a box -- C. Vectors -- The postulates and general principles of quantum mechanics -- D. Series and limits -- The harmonic oscillator and vibrational spectroscopy -- E. Spherical coordinates -- The rigid rotator and rotational spectroscopy -- F. Determinants -- The hydrogen atom -- G. Matrices -- Approximation methods -- H. Matrix Eigenvalue problems -- Many-electron atoms -- The chemical bond: one- and two-electron molecules -- Qualitative theory of chemical bonding -- The Hartree-Fock-Roothaan method.
The dawn of the quantum theory -- A. Complex numbers -- The classical wave equation -- B. Probability and statistics -- The Schrodinger equation and a particle in a box -- C. Vectors -- The postulates and general principles of quantum mechanics -- D. Series and limits -- The harmonic oscillator and vibrational spectroscopy -- E. Spherical coordinates -- The rigid rotator and rotational spectroscopy -- F. Determinants -- The hydrogen atom -- G. Matrices -- Approximation methods -- H. Matrix Eigenvalue problems -- Many-electron atoms -- The chemical bond: one- and two-electron molecules -- Qualitative theory of chemical bonding -- The Hartree-Fock-Roothaan method.
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Subject
Quantum chemistry.
Quantum chemistry.
Quantum chemistry.
Quantum chemistry
Quantum chemistry
Quantum chemistry
Quantenchemie
Quantenchemie
Quantenchemie
Química cuántica.
Química cuántica.
Química cuántica.
Kvantkemi.
Kvantkemi.
Kvantkemi.
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Notes
Includes bibliographical references and index.
Includes bibliographical references and index.
Includes bibliographical references and index.
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Related titles
Available in other form: Online version: McQuarrie, Donald A. (Donald Allan) Quantum chemistry. 2nd ed. Sausalito, Calif. : University Science Books, ©2008
Available in other form: Online version: McQuarrie, Donald A. (Donald Allan) Quantum chemistry. 2nd ed. Sausalito, Calif. : University Science Books, ©2008
Available in other form: Online version: McQuarrie, Donald A. (Donald Allan) Quantum chemistry. 2nd ed. Sausalito, Calif. : University Science Books, ©2008
Available in other form: Quantum Chemistry
Available in other form: Quantum Chemistry
Available in other form: Quantum Chemistry
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Identifier
LC : 2007023879
LC : 2007023879
LC : 2007023879
ISBN : 9781891389504
ISBN : 9781891389504
ISBN : 9781891389504
ISBN : 1891389505
ISBN : 1891389505
ISBN : 1891389505
ISBN : 9781891389528
ISBN : 9781891389528
ISBN : 9781891389528
ISBN : 1891389521
ISBN : 1891389521
ISBN : 1891389521
OCLC : (OCoLC)141484672
OCLC : (OCoLC)141484672
OCLC : (OCoLC)141484672
OCLC : (OCoLC)ocn141484672
OCLC : (OCoLC)ocn141484672
OCLC : (OCoLC)ocn141484672
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text
text
text
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volume
volume
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Library Catalog
Library Catalog
Library Catalog
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MMS ID
991002745379706535
991002745379706535
991002745379706535
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Virtual Browse
Methods of molecular quantum mechanics ©1989
Elementary methods of molecular quantum mechanics 2007
Methods of molecular quantum mechanics : an introduction to electronic molecular structure 2009
Simple theorems, proofs, and derivations in quantum chemistry ©2003
Quantum chemistry ©1983
Quantum chemistry 2008
Quantum chemistry of organic compounds : mechanisms of reactions ©1990
Molecular electrostatic potentials : concepts and applications 1996
A New dimension to quantum chemistry : analytic derivative methods in ab initio molecular electronic structure theory 1994
The construction of spin eigenfunctions : an exercise book ©2000
Foundations of quantum chemistry 1968
Ideas of quantum chemistry 2007
Ideas of quantum chemistry 2014
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https://www.youtube.com/watch?v=bcyX7el3Eo4
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Context Clues: Definition and Exemplification Clues | TeacherBethClassTV
Teacher Beth Class TV
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Posted: 6 Aug 2022
In this video, you will be able to learn and understand the two types of Context Clues, the Definition and Exemplification Clues.
Teacher Beth Class TV
Let's LEARN, PLAY, and GROW . . . TOGETHER!
english #grammar #englishlesson #englishlearning #englishgrammar #sentences #simplesentences #compoundsentence #complexsentence #kids #education #educational #context #definition #exemplification #clues #contextual
Transcript:
[Music] hi kids welcome to teach your birth class tv [Music] english [Music] today we're going to discuss about the context clues with its two types the definition and exemplification clues are you ready in this lesson you will be able to identify the context clues used and use context clues to find the meaning of unfamiliar words through definition and exemplification [Music] do you always consult a dictionary to look for some unfamiliar words have you experienced that some of the words that you are looking for are not in the dictionary that you have don't worry because in this lesson we will learn about context clues [Music] what are context clues these are clues found within the context [Music] context clues are hints about a word phrase sentence or paragraph that help readers understand its meaning the words that surround a difficult word in a sentence may give clues to the meaning of the focus word we call these words context clues context clues are one of the basic skills in enhancing reading strategies it also promotes critical reading for young readers like you it is a reading strategy that helps in aiding readers it provides hands on the meaning of unfamiliar words [Music] these are the types of context clues to easily remember the types of context clues let's remember the acronym ideas i for inference d for definition e for exemplification or example a for antonyms and s for synonyms in this lesson we will focus on the two types of context clues the definition clues and the exemplification or example tools [Music] definition clues the meaning of the unfamiliar word is directly defined or given in the sentence itself some writers use single words or phrases to help the readers understand the meaning of the unfamiliar word or the difficult word signal words such as are that is as or is defined as means refers to and in other words they also use punctuation such as commas dashes and parentheses examples of definition clues it's amazing that is its skeleton-like appearance was frightening to see in this sentence the unfamiliar word is amitation and it gives the definition which is it's a skeleton-like appearance the asian chinese use the abacus a device with movable beads that can be used as a calculator in this example the unfamiliar word is abacus and the definition of the word abacus is a device with movable beads that can be used as a calculator exemplification or example clues it uses examples to help the reader infer the meaning of the unfamiliar word take note of some signal words used for example such as like including for instance and types or kinds [Music] examples of exemplification or example clues celestial bodies including the sun moon and stars have fascinated men through the centuries in this sentence the unfamiliar word is celestial bodies so it means the examples of celestial objects are those in the sky or heaven such as sun moon stars and planets piscatorial creatures such as flounder salmon and trout live in the coldest parts of the ocean in this sentence the unfamiliar word is piscatorial creatures this factorial refers to fish so the examples of these creatures are flounder salmon and trout [Music] let's have a recap context clues are hints about a word phrase sentence or paragraph that help readers understand its meaning definition clues the meaning of the information word is directly defined or given in the sentence itself exemplification or example clues it uses examples to help the reader infer the meaning of the unfamiliar world [Music] identify the context clues used in each statement if it is definition or exemplification then underline the given clues to actively participate in this activity you may write your answer in the comment section [Music] are you ready [Music] last sunday we opened our new business cafe a small coffee restaurant near the call center building [Music] the correct answer in this sentence is definition and the definition of the word cafe is a small coffee restaurant [Music] school stakeholders such as teachers parents students and other non-teaching personnel support the school's tree planting project the correct answer in this sentence is exemplification and the examples of stakeholders are teachers parents students and other non-teaching personnel [Music] yesterday we experienced different types of precipitation like rain snow sleet and hill the correct answer in this sentence is exemplification and the examples of the word precipitation are rain snow sleet and hill [Music] russell enjoys organizing a music fest she likes celebrations and other festive gatherings the correct answer in this sentence is definition and of course the definition of the word fast is celebrations and other fasting [Music] in 2020 our town was placed under the enhanced community quarantine a restraint on activities to prevent the spread of the virus [Music] the correct answer in this sentence is definition and the definition of the word quarantine is a restraint on activities to prevent the spread of the virus [Music] the decor including the fancy lights beautiful paintings and unique furniture made the house look stunning [Music] the correct answer is exemplification and the examples of decor are fancy lights beautiful paintings and unique furniture [Music] [Music] thank you for joining me let's learn play and grow together teacher bath class tv subscribe like and share thank you [Music]
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https://www.osmosis.org/answers/eczema-herpeticum
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Eczema Herpeticum
What Is It, Causes, Diagnosis, and More
Author: Nikol Natalia Armata, MD
Editor: Alyssa Haag, MD
Editor: Kelsey LaFayette, DNP, ARNP, FNP-C
Editor: Lily Guo, MD
Illustrator: Jillian Dunbar
Copyeditor: Joy Mapes
Modified: Sep 24, 2025
What is eczema herpeticum?
Eczema herpeticum is a
skin infection
caused by herpes simplex virus (HSV) 1 or 2 in individuals who have atopic dermatitis (i.e., eczema), an inflammatory skin condition characterized by skin dryness and itching.
Individuals with
atopic dermatitis
are more prone to developing viral
skin infection
s, such as eczema herpeticum, due to reduced immunity and their skin’s impaired ability to form a barrier against infections.
The skin lesions of eczema herpeticum typically present as painful clusters of vesicles and pustules (i.e., fluid-filled bumps of the skin) with hemorrhagic crust. The vesicles appear widely over the body but are most common on the face, neck, and trunk. Older lesions that have burst and dried commonly form “punched-out” erosions, which are circular breaks in the skin with sharply defined borders.
Some individuals develop additional symptoms, like fever, swollen lymph nodes (i.e., lymphadenopathy), and
cold sores
(i.e., vesicles around the lips), while others with the infection are asymptomatic. Frequently, the skin lesions of eczema herpeticum become infected with bacteria, commonly
Staphylococcus aureus
and
Streptococcus pyogenes
, causing a
secondary bacterial infection
. In such cases, impetiginization (i.e., honey-colored crusting) and increased redness, swelling, and pain may occur.
Eczema herpeticum most often affects infants and children, but people of all ages can develop this infection.
What causes eczema herpeticum?
Direct contact with the
HSV
is the most common cause of eczema herpeticum. HSV is categorized into two types: type 1 and type 2. Type 1 is highly contagious and is the leading cause of eczema herpeticum. Often, eczema herpeticum presents 5 to 12 days after the first contact with
herpes simplex
.
Non-eczematous conditions that disrupt the continuity of the skin are also considered risk factors for eczema herpeticum, as they can increase an individual’s susceptibility to localized herpes infections. Such skin conditions include burns; irritant contact dermatitis (e.g., physical or chemical injury to the skin); Hailey-Hailey disease; and Darier disease, both rare genetic disorders. Hailey-Hailey disease typically presents with eroded vesicles and blisters in the folds of the skin (e.g., armpits, groin), while Darier disease generally causes dry, scaly papules (i.e., small, raised bump) on the chest and face.
Is eczema herpeticum contagious?
Eczema herpeticum is a contagious infection that can spread through direct skin-to-skin contact with an infected individual, even if the infected individual does not have a current outbreak. Infected individuals are also able to infect other parts of their body through what is called “self-infection,” or autoinoculation.
How is eczema herpeticum diagnosed and treated?
Eczema herpeticum is often diagnosed clinically based on the individual's history (e.g., of
atopic dermatitis
) and the characteristic appearance of the skin lesions. Diagnosis can be confirmed via viral swabs taken from the base of fresh vesicles, which are typically analyzed through viral culture or polymerase chain reaction (PCR). A Tzanck smear, a diagnostic method that can rapidly detect a
herpes simplex
infection, can also be used to confirm diagnosis of eczema herpeticum and rule out other potential diagnoses. Other viruses and bacteria can cause similar-appearing skin lesions, so careful examination and proper diagnostic testing of the individual is necessary to obtain an accurate diagnosis.
In healthy adults eczema herpeticum is typically self-limited, lasting 2-4 weeks, and treatment can shorten the course. However, in children and young infants treatment should be initiated as soon as possible. Oral antiviral medication, such as acyclovir or valacyclovir, is often prescribed to minimize the risk of complications and prevent progression to severe disease.
Additional treatment will depend on the severity of the lesions. Most mild cases can be treated with oral
antiviral medication
for a range of 7-21 days. Cool compresses and lotions can be used for symptomatic relief of itching and pain. In more severe cases, specifically when the individual’s immune system is weakened due to another condition (e.g., HIV, prolonged diabetes, leukemia, undernutrition) or certain treatments (e.g., corticosteroids and other immunosuppressant medications, chemotherapy, radiation therapy), hospitalization should be considered for intravenous (IV) administration of
antiviral medication
and supportive care of other symptoms. Critically ill individuals may also need additional fluids, pain relievers, and wound care to assist with healing. Furthermore, individuals should be monitored for the development of
secondary bacterial infection,
and if infection occurs, they will most likely need treatment with systemic antibiotics.
To prevent spread of the infection, all individuals with eczema herpteticum should be educated about self-infection. Contact precautions, such as wearing gloves and frequently washing hands, are often recommended when
herpes simplex
infections are suspected.
What are the most important facts to know about eczema herpeticum?
Eczema herpeticum is a
skin infection
of
herpes simplex
virus (HSV) that can develop in individuals with
atopic dermatitis
. It is more common in children, where it presents with painful blisters and “punched out” erosions. Eczema herpeticum is caused by the
herpes simplex
virus, especially type 1 (HSV-1). Risk factors include pre-existing
atopic dermatitis
and other non-eczematous skin conditions that
disrupt the continuity of the skin. Eczema herpeticum is highly contagious among both symptomatic and asymptomatic individuals. Therefore, prompt and proper diagnosis is important and should be based on clinical presentation and, if necessary, viral swabs. Treatment typically involves
antiviral medication
, administered in doses that depend on the severity of the case. In cases of
secondary bacterial infection
,
antibiotics may also be prescribed.
Key Takeaways
| | |
--- |
| Definition | Eczema herpeticum is a skin infection caused by herpes simplex virus (HSV) 1 or 2 in individuals who have atopic dermatitis. |
| Causes | - Direct contact with HSV - Type 1 more common - Risk factors: - Atopic dermatitis (eczema) - Non-eczematous conditions that result in breaks in the skin (burns, contact dermatitis) |
| Transmissibility | - Contagious through direct skin-to-skin contact with infected individual (even if no current outbreak is present) - Autoinoculation (self-infection to other parts of their body) |
| Diagnosis and Treatment | - Diagnosis - Medical history - Skin assessment - Viral swabs, Tzanck smear - Treatment - Oral antiviral medications (IV antivirals if severe) - Cool compresses and lotions - Contact precautions |
Related videos and concepts
6:14
Atopic dermatitis
10:00
Herpes simplex virus
12:03
Herpesvirus medications
8:20
Impetigo: Nursing
References
AlAlhareth I, Alomer H, M, et al. Extensive eczema herpeticum in a previously well child. Int J Emerg Med. 2022;15(1):21. doi:10.1186/s12245-022-00425-5. PMID: 35597913.
Downing C, Mendoza N, Sra K, et al. Human herpesviruses. In: Bolognia JL, Schaffer JV, Cerroni L, eds. Dermatology. 4th ed. Philadelphia, PA: Elsevier; 2017:1400–1424.
National Organization for Rare Disorders (NORD). Hailey-Hailey disease. In: Rare Disease Database. Published 2018. Accessed March 30, 2021.
Traidl S, Roesner L, Zeitvogel J, et al. Eczema herpeticum in atopic dermatitis. Allergy.2021;76(10):3017–3027. doi:10.1111/all.14853. PMID: 33844308.
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Chemistry Lesson: Polyatomic Ions
[Download the PDF of Chemical Nomenclature Rules here.]
What Are Polyatomic Ions?
Unlike monoatomic ions, which are composed of a single atom, polyatomic ions are composed of many atoms. (The prefix poly- means “many,” as in polygon or polytheism.)
There are some rules and patterns which can help us to deduce the structures of these ions, but before we can learn those, there are a handful that don’t follow the patterns that you just need to memorize. Depending on your course or instructor, you may need to memorize some of the following. When it comes to memorizing the, flashcards are your friends!
| | | | | | |
--- --- --- |
| Cation | Formula | Anion | Formula | Anion | Formula |
| NH4+ | ammonium ion | C2H3O2– | acetate ion | C2O42- | oxalate ion |
| H3O+ | hydronium ion | CrO42- | chromate ion | MnO4– | permanganate ion |
| CN– | cyanide ion | O22- | peroxide ion |
| Cr2O72- | dichromate ion | S2O32- | thiosulfate ion |
| OH– | hydroxide ion |
The good news, is there are dozens of other polyatomic ions that are relatively easy to figure just by knowing or observing a few patterns.
The -ate Ions
Many of the nonmetals combine with oxygen to make anions (negatively charged ions) that we are sometimes referred to as oxyanions. If you refer to the periodic table below, you will that a several nonmetals have been highlighted that commonly form such oxyanions.
These nonmetals for polyatomic ions that end in the suffix -ate. However, the number of oxygens and the charge on the ions varies as we can see in the following table.
| | | |
---
| BO33- borate ion | CO32- carbonate ion | NO3– nitrate ion |
| PO43- phosphate ion | SO42- sulfate ion | ClO3– chlorate ion |
| AsO43- arsenate ion | SeO42- selenate ion | BrO3– bromate ion |
| IO3– iodate ion |
Patterns for Polyatomic Oxyanions
Notice how the four polyatomic-ate ions in the center square (phosphate, arsenate, sulfate, and selenate) all have four oxygen atoms, while the polyatomic -ate ions on the outside all have three oxygen atoms.
As you start from right-hand side, the first column of polyatomic -ate ions (chlorate, bromate, iodate) all have a 1- charge. The second column (sulfate, selenate) all have a 2- charge, while the third column (phosphate, arsenate) all have a 3- charge. On the top row, the farthest right polyatomic -ate ion is nitrate which is 1-, followed by carbonate which is 2-, followed by borate which is 3-.
Within a column, the pattern also repeats. For example, if you know that chlorate is ClO3–, then you can predict that all of the other oxyanions made from nonmetals in group 17/VIIA follow the same pattern. Cl combines with three oxygens and has a 1- charge; Br combines with three oxygens and has a 1- charge; I combines with three oxygens and has a 1- charge. In group 16/VIA, both S and Se have the same pattern. In group 15/VA, P and As have the same pattern, but notice N is different, combining with only three oxygens and having only a 1- charge.
Changing the Number of Oxygens
Once you know the polyatomic ions that end in the suffix -ate there are only a few more patterns to know and you’ll be on your way to naming dozens and dozens of polyatomic ions!
| | | | |
--- --- |
| Increasing # Of Oxygen Atoms | Prefix | Root | Suffix |
| +1 O atom | per- | root | -ate |
| – | root | -ate |
| -1 O atom | root | -ite |
| -2 O atom | hypo- | root | -ite |
For example, let’s use chlorate as an example. We know that chlorate is ClO3–. This is the anion that ends in -ate. Let’s look at the chart above and see if we can figure out what perchlorate would be. Since it has the prefix per- and the suffix -ate, it must have one O atom more than chlorate, meaning the formula is ClO4–. Notice only the number of oxygen atoms changed, the charge did not.
What would chlorite be? The suffix -ite means it has one O atom less than chlorate, so it must be ClO2–.
How about hypochlorite? The prefix hypo- and the suffix -ite mean it has two less O atoms than chlorate, so it must be ClO–.
We can similarly predict sulfite and hyposulfite once we know the formula and charge of sulfate. (Ignore persulfate as it’s an exception to the general rule.)
| | | | |
--- --- |
| Name | Formula | Name | Formula |
| perchlorate ion | ClO4– |
| chlorate ion | ClO3– | sulfate ion | SO42- |
| chlorite ion | ClO2– | sulfite ion | SO32- |
| hypochlorite ion | ClO– | hyposulfite ion | SO22- |
Notice that the prefixes and suffixes don’t denote the total number of O atoms, just that it has more or less than the -ate polyatomic ion.
Example Problems
Predict the formula and charge for:
sulfite ion – We determine by sulfite is by first figuring out what sulfate. Sulfate comes from sulfur which is in the center box of the nonmetals so it have four oxygens. It’s two columns away from the right-hand side so its charge is 2-. Sulfate is SO42-. The suffix -ite tells me it lost one oxygen atom, so sulfite must be SO32-.
hypophosphite ion – Let’s figure out what phosphate is. The root comes from phosphrous which is in the center box of nonmetals so it must have four oxygen atoms. Phosphrous is three columns away from the right-hand side so its charge is 3-. Therefore, phosphate is PO43-. The prefix hypo- and the suffix -ite tell me to subtract two oxygen atoms, so hypophosphite must be PO23-.
perbromate ion – Bromate comes from bromine which is outside the center box of nonmetals, so it combines with three oxygens. It’s in the first column from the right so its charge is 1-. Bromate must be BrO3–. The prefix per- tells me to add an oxygen, so perbromate must be BrO4–.
nitrite ion – Nitrate is NO3– and the -ite suffix means I must remove an oxygen atom, so nitrite must be NO2–.
How would you name the following:
IO2– – Let’s determine what the -ate version of iodine’s polyatomic is. Iodine is outside the center box, so it must have three oxygen atoms. Iodine is in the first column from the right-hand side so the charge on the polyatomic must be 1- (or just -). Therefore, iodate must be IO3–. However, this polyatomic has one less oxygen atom which means the suffix changes to -ite, so we would name it iodite ion.
AsO23- – Arsenate would be AsO43- and this has two fewer oxygen atoms. According to our rules, two less oxygens means we add the prefix hypo- and change the suffix to -ite, so this would be the hypoarsenite ion.
SeO32– – Selenate is SeO42-. The given ion has one less oxygen atom, so we must change the suffix from -ate to -ite. Therefore, we would name this the selenite ion.
“Hydrogen” Polyatomic Ions
Another way to form polyatomic ions is by combining them with one or more hydrogen ions, H+. For example, we could combine H+ with carbonate, CO32- to form hydrogen carbonate, HCO3–. Notice the overall charge is 1- because the 1+ on H+ combine with the 2- on CO32-.
What would hydrogen sulfate be? “Hydrogen” tells me to add H+ to sulfate, SO42-. So hydrogen sulfate would be HSO4–.
When we add hydrogen ions to a polyatomic with a 3- charge, such as phosphate (PO43-), we could add either one or two H+ ions and it would still remain a polyatomic ion. (Adding three H+ would completely cancel out the 3- charge making it a neutral compound and no longer a polyatomic ion.) We have two options then, HPO42- and H2PO4–. The former has only one hydrogen ion so call it either hydrogen phosphate ion, or sometimes monohydrogen phosphate ion. The latter has two hydrogen ions so we refer to it as dihydrogen phosphate ion.
Example Problems
Predict the formula and charge for:
hydrogen hyposulfite – Sulfate is SO42- so hyposulfite tells me to reduce the number of oxyen atoms by two, leaving SO22-. Adding a hydrogen ion, H+, it becomes HSO2–.
9. hydrogen selenide – The suffix -ide on selenide tells me the ion probably is a monoatomic anion formed from a nonmetal, in this case from selenium. Selenium is two columns away from the noble gases so it needs two electrons, meaning selenide is Se2-. If I add H+ to this, I get HS–.
10. dihydrogen borate – Borate is BO33-. Dihydrogen tells me to add two H+ ions, which produces H2BO3–.
How would you name:
11. HSO4– – If I remove an H+ from the ion, I am left with SO42-. (The charge becomes 2- because I removed a positive hydrogen ion.) SO42- is named sulfate so this ion combined with hydrogen would be hydrogen sulfate ion.
H2AsO3– – Removing two H+ ions leaves me with AsO33-. I know that arsenate is AsO43- and this has one less oxygen atom, so it must be aresnite. I’ve added two hydrogens to the polyatomic ion, so the result is dihydrogen arsenite ion.
Related Posts:
Chemistry Practice Problems: Names & Formulas of Acids
[View the accompanying Lesson on Names & Formulas of Acids here.] [Download ..
Chemistry Practice Problems: Names & Formulas Molecular Compounds
[View the accompanying Lesson on Names & Formulas of Molecular ..
Chemistry Lesson: Enthalpy
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Sulfur Dioxide
Also known as: SO2, sulfurous anhydride, sulfuroxide, sulfurous oxide, sulfurous acid anhydride
Sulfur dioxide, SO2, is a colorless gas or liquid with a strong, choking odor. It is produced from the burning of fossil fuels (coal and oil) and the smelting of mineral ores (aluminum, copper, zinc, lead, and iron) that contain sulfur.
Sulfur dioxide dissolves easily in water to form sulfuric acid. Sulfuric acid is a major component of acid rain. Acid rain can damage forests and crops, change the acidity of soils, and make lakes and streams acidic and unsuitable for fish. Sulfur dioxide also contributes to the decay of building materials and paints, including monuments and statues.
Most of the sulfur dioxide released into the environment comes from electric utilities, especially those that burn coal. Some other sources of sulfur dioxide include petroleum refineries, cement manufacturing, paper pulp manufacturing, and metal smelting and processing facilities. Locomotives, large ships, and some non-road diesel equipment currently burn high-sulfur fuel and release sulfur dioxide into the air. In nature, volcanic eruptions can release sulfur dioxide into the air.
Some dried fruits are preserved using SO2 to prevent discoloration of the fruit. SO2 is also used in bleaching materials and as a fumigant. In homes, sulfur dioxide gas can result from tobacco smoke, improperly or inadequately vented gas appliances (such as stoves, ranges, furnaces, or clothes dryers), gas or kerosene heaters, wood or coal stoves, automobile exhaust from attached garages, and malfunctioning chimneys.
Close all Open all
Exposure Information
You can be exposed to SO2 by breathing it in the air or getting it on your skin. People who live near industrial sources of sulfur dioxide may be exposed to it in the air. You are most likely to be exposed if you work in industries where SO2 is produced, such as copper smelting or power plants, or where it is used in the production of sulfuric acid, paper, food preservatives, or fertilizers. People with malfunctioning appliances or chimneys in their homes may also be exposed to sulfur dioxide.
Most SO2 exposures are caused by people breathing contaminated outdoor air. Therefore, limit your activities outdoors when you know that air pollution levels are high. The EPA and the Wisconsin Department of Natural Resources (DNR) issue air quality alerts for high pollution days. People with existing respiratory difficulties, such as asthma, should pay special attention to these air advisories. Special care should be taken with child asthmatics to limit their outdoor activities during high pollution days.
To reduce the possibility of exposure to sulfur dioxide caused by a source in your home, you can:
Use gas appliances with electronic (pilotless) ignition. This will eliminate the continuous low-level pollutants from pilot lights.
Use exhaust fans over gas stoves that are vented to the outdoors instead of fans that recirculate the air indoors. Keep the metal mesh filters on your exhaust fans clean (most can be run through the dishwasher).
Choose vented appliances whenever possible, and make sure they are vented to the outdoors. Have a trained professional inspect your appliances annually.
Never heat your home with a gas range or stove.
Do not idle your car in the garage.
Do not smoke indoors.
Health Effects
Short-term exposure to high levels of sulfur dioxide can be life-threatening. Generally, exposures can cause a burning sensation in the nose and throat. Also, exposure can cause difficulty breathing, including changes in the body's ability to take a breath or breathe deeply, or take in as much air per breath. Long-term exposure to sulfur dioxide can cause changes in lung function and aggravate existing heart disease.
People with asthma may be sensitive to changes in respiratory effects due to SO2 exposure at even low concentrations. Sulfur dioxide is not classified as a human carcinogen (it has not been shown to cause cancer in humans).
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Introductory Statistics 2e
1.5 Data Collection Experiment
Introductory Statistics 2e1.5 Data Collection Experiment
Search for key terms or text.
Stats Lab
Data Collection Experiment
Class Time:
Names:
Student Learning Outcomes
The student will demonstrate the systematic sampling technique.
The student will construct relative frequency tables.
The student will interpret results and their differences from different data groupings.
Movie SurveyAsk five classmates from a different class how many movies they saw at the theater last month. Do not include rented movies.
Record the data.
In class, randomly pick one person. On the class list, mark that person’s name. Move down four names on the class list. Mark that person’s name. Continue doing this until you have marked 12 names. You may need to go back to the start of the list. For each marked name record the five data values. You now have a total of 60 data values.
For each name marked, record the data.
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| | | | | | | | | | | | |
| | | | | | | | | | | | |
| | | | | | | | | | | | |
| | | | | | | | | | | | |
| | | | | | | | | | | | |
Table 1.18
Order the DataComplete the two relative frequency tables below using your class data.
| Number of Movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
--- --- |
| 0 | | | |
| 1 | | | |
| 2 | | | |
| 3 | | | |
| 4 | | | |
| 5 | | | |
| 6 | | | |
| 7+ | | | |
Table 1.19 Frequency of Number of Movies Viewed
| Number of Movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
--- --- |
| 0–1 | | | |
| 2–3 | | | |
| 4–5 | | | |
| 6–7+ | | | |
Table 1.20 Frequency of Number of Movies Viewed
Using the tables, find the percent of data that is at most two. Which table did you use and why?
Using the tables, find the percent of data that is at most three. Which table did you use and why?
Using the tables, find the percent of data that is more than two. Which table did you use and why?
Using the tables, find the percent of data that is more than three. Which table did you use and why?
Discussion Questions
Is one of the tables “more correct” than the other? Why or why not?
In general, how could you group the data differently? Are there any advantages to either way of grouping the data?
Why did you switch between tables, if you did, when answering the question above?
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Interior and Exterior Angles of a Polygon
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Interior&Exterior Angles of a Polygon Task Cards - Math Centers
Created by
Adventures in Inclusion
This resource includes 24 task cards on finding the measures of interior and exterior angles of polygons. These are great for repeated practice, math intervention, math centers activities, math remediation or test prep! Includes work pages and answer key. For more middle school math resources check out: Middle School Math Task Cards BundleParallel Lines, Perpendicular Lines, Parallel lines cut by a transversal Cheat SheetsSolving Equations Task CardsFor more math, ELA, and Special Education Res
6 th - 10 th
Geometry, Math, Math Test Prep
Also included in:Middle School Math Task Cards Bundle
$2.00
Original Price $2.00
Rated 4.91 out of 5, based on 11 reviews
4.9(11)
Add to cart
Wish List
Math Pong: Measures of Interior and Exterior Angles of a Polygon
Created by
Swinny's Math Store
In this activity students will practice finding the sum of the measures of the interior angles of a polygon, the measure of each interior angle in a regular polygon, and the measure of each exterior angle of a regular polygon. They will do so by playing Math Pong. Each question bounces the student from one side of the worksheet to the other until they finally arrive at the final answer. Students can work individually or in small groups as they have fun learning math. For more activities for th
7 th - 12 th, Higher Education
Geometry, Math Test Prep, Measurement
$3.00
Original Price $3.00
Rated 5 out of 5, based on 3 reviews
5.0(3)
Add to cart
Wish List
Interior and Exterior Angles of a Polygon- Modified Geometry- Angles in Polygons
Created by
Modified Treasures
Do you need a set of practice sheets for students to showcase their skills for interior and exterior angles in polygons? They will find the measure of the interior angle sum of interior angles, the measure of one interior angle, and the measures of one exterior angle of polygons. This set of print and go practice sheets are just what you need. There are three worksheets, each with a modified version for students who need more guidance. They are excellent for differentiation. Included:Practice sh
8 th - 12 th
Algebra, Geometry, Math
$3.00
Original Price $3.00
Add to cart
Wish List
Interior and Exterior Angles of a Polygon
Created by
Math is Timeless
Worksheet on polygons - sum of the interior angles, each interior angle, sum of the exterior angles and each exterior angle - including key. Could be used as notes to introduce the concept, a worksheet to practice, or as a review. Shapes are given so students can write the angle measures on each shape. Also includes big versions of each shape that are super helpful.
7 th - 10 th
Geometry, Math
$1.50
Original Price $1.50
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
Interior and Exterior Angles of a Polygon Reference Sheets - Math Intervention
Created by
Adventures in Inclusion
My math reference sheets cover a variety of topics and are to support students who may struggle with memorizing multi-step processes. Providing students with reference sheets is a common accommodation on IEP's that I see in my school and these reference sheets can help you meet that accommodation. These reference sheets show the formula and steps to find the interior and exterior angles of polygons. These are great for math remediation, special education, test prep, notes. Includes free sample
6 th - 10 th
Geometry, Math, Math Test Prep
Also included in:Math Reference Sheets Bundle – Cheat Sheets for Middle School Math
$2.00
Original Price $2.00
Rated 5 out of 5, based on 2 reviews
5.0(2)
Add to cart
Wish List
Guided Notes - Lesson 3.2 - Interior&Exterior Angles of a Polygon
Created by
Counting Corner
Objective: Students will be able to find the measure of the interior and exterior angles of any regular polygon - warm up - Introduce polygon sides, sum of interior angles - examples - Exterior angles - Solve for unknown angles - Angles in a regular polygon theorem - Partner work
8 th - 11 th
Geometry, Math
CCSS
HSG-CO.C.9
, HSG-CO.C.10
, HSG-SRT.B.4
+12
Also included in:Geometry - Full Year Course!
$1.00
Original Price $1.00
Add to cart
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Interior and Exterior Angles of a Polygon- Color By Number Activity- Geoemtry
Created by
Modified Treasures
Make angle calculations fun and colorful with this engaging Geometry Color by Number Activity focused on the Interior and Exterior Angles of Regular Polygons! Perfect for reinforcing concepts through independent practice, review, or as a sub plan, this resource allows students to solve 12 problems and color their way to mastery. Students will calculate interior or exterior angles of regular polygons, then use their answers to color a vibrant picture based on a key. It’s a no-prep, self-checki
7 th - 12 th
Algebra, Geometry
$2.00
Original Price $2.00
Add to cart
Wish List
Interior and Exterior Angles of a Polygon Maze Worksheet
Created by
Maeve's Method
This is a worksheet created to help students practice finding interior and exterior angles of polygons.
6 th - 8 th
Geometry, Math, Other (Math)
$1.00
Original Price $1.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
Geometry Quiz (Triangles) - Interior&Exterior Angles of a Polygon
Created by
FlaminisSuperstars
Short quiz, complete with multiple choice, open-ended, and a fun greeting for the student(s) name (my favorite personal touch for assessments). Please let me know if you have any questions! Enjoy!
9 th - 12 th
Geometry, Math
$3.00
Original Price $3.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
Identifying Interior and Exterior Angles of a Polygon
Created by
Jan Dayton
Students identify the classification of the regular polygon, identify the sum of interior angles, identify the sum of exterior angles, identify each interior angle, and identify each exterior angle. Each correct answer reveals the piece of the puzzle. Students will know right away if they are correct or need to look at it again.
7 th - 9 th
Geometry, Math
$3.00
Original Price $3.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
Add to cart
Wish List
QR Math Pong: Measures of Interior and Exterior Angles of a Polygon
Created by
Swinny's Math Store
In this activity students will practice finding the sum of the measures of the interior angles of a polygon, the measure of each interior angle in a regular polygon, and the measure of each exterior angle of a regular polygon. They will do so by playing QR Math Pong. Each question bounces the student from one side of the worksheet to the other until they finally arrive at the final answer. They will use a QR code scanner that can be downloaded on any smartphone or tablet. Students can work indiv
7 th - 12 th, Higher Education
Geometry, Math Test Prep, Measurement
$3.00
Original Price $3.00
Add to cart
Wish List
Homework - Lesson 3.2 - Interior&Exterior Angles of a Polygon
Created by
Counting Corner
Classwork/homework associated with Guided Notes - Lesson 3.2 Objective: Students will be able to find the measure of the interior and exterior angles of any regular polygon
8 th - 11 th
Geometry, Math
CCSS
HSG-CO.C.9
, HSG-CO.C.10
, HSG-SRT.B.4
+12
Also included in:Geometry - Full Year Course!
$1.00
Original Price $1.00
Add to cart
Wish List
Interior and Exterior Angles of a Polygon Thought Map Poster
Created by
Anna Koltunova
This poster is a nifty way to get students to start thinking through some higher level problems that involve interior and exterior angles.
8 th - 12 th
Geometry
FREE
Log in to Download
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Interior and Exterior Angles Guided of Polygons Notes and Practice Worksheet
Created by
Generally Geometry
About this resource:Looking for a fun or new way to deliver notes? Then, doodle guides are for you! These guides encourage creativity while delivering new concepts. Doodle guides keep students engaged and makes note-taking more fun! This doodle guide teaches the concept of Angles of Polygons. Also included, is a worksheet that practices the topic. See the preview for details! Completed sample keys included! Follow Me:Click here to Follow Me!Available in the following bundle(s):Geometry Curriculu
8 th - 9 th
Geometry, Math
Also included in:Polygons and Quadrilaterals Unit Bundle | Geometry
$3.50
Original Price $3.50
Rated 4.89 out of 5, based on 9 reviews
4.9(9)
Add to cart
Wish List
Interior and Exterior Angles of Polygons Mazes
Created by
Miss Crafty Math Teacher
These self-checking mazes consist of 15 problems to practice the following on interior and exterior angles of polygons: - Finding the sum of the interior/exterior angles of an indicated convex polygon. - Finding the number of the sides of a convex polygon given the sum of the measures of the interior angles. - Find the measure of an interior/exterior angle of an indicated polygon. - Find the value of x given the diagram of a polygon. This product includes TWO mazes, along with an answer key! Ma
8 th - 10 th
Geometry, Math
Also included in:Geometry Mazes Bundle
$1.50
Original Price $1.50
Rated 4.77 out of 5, based on 13 reviews
4.8(13)
Add to cart
Wish List
Interior&Exterior Angles of Polygons Cut, Paste, Solve, Match Puzzle Activity
Created by
Secondary Math Shop
Interior and Exterior Angles of Polygons Cut, Paste, Solve, Match Puzzle Activity In this activity, students cut out 48 puzzle pieces that represent the different angles in regular polygons These include: a polygon classification (i.e. regular hexagon) the interior angle sum the measure of one interior angle the measure of one exterior angle Students need to match the pieces together and use the given information to solve for each angle calculation. Each completed puzzle has 4
9 th - 11 th
Geometry, Math Test Prep, Other (Math)
Also included in:Geometry Puzzles Bundle: Cut, Paste, Solve, Match Puzzle Activities
$3.00
Original Price $3.00
Rated 4.81 out of 5, based on 38 reviews
4.8(38)
Add to cart
Wish List
Interior and Exterior Angles of Polygons Pixel Art Scramble
Created by
Knowlton Math
Great no prep, self-checking activity for angles of polygons. This activity has students solve 16 problems involving applying theorems involving finding the interior and exterior angles of polygons and regular polygons. If the student answers correctly, the box turns green and part of the picture is unscrambled. If the student answers incorrectly, the answer box turns red. The whole picture is revealed once all questions are answered correctly. Problem included are: 2 problems finding the sum of
9 th - 12 th
Geometry, Math
$2.00
Original Price $2.00
Rated 5 out of 5, based on 4 reviews
5.0(4)
Add to cart
Wish List
Interior and Exterior Angles of Polygons BINGO Game
Created by
Newton's Solutions
Do you want students to practice working with the interior and exterior angles of polygons? This BINGO game is perfect for students to gain confidence when determining the sum or one interior or exterior angle of a polygon by providing repeated exposure of this skill. Students will stay engaged and have fun while honing their math skills! Students will need to work with both regular and non-regular polygons, interior and exterior angles, and the sum or one angle. There are 30 unique BINGO Boar
9 th - 11 th
Geometry, Math
CCSS
HSG-CO.C.11
$3.50
Original Price $3.50
Rated 5 out of 5, based on 3 reviews
5.0(3)
Add to cart
Wish List
Interior and Exterior Angles of Polygons Guided Notes
Created by
Miss R Squared
These guided notes cover the Polygon Interior Sum and Exterior Sum Theorems. The 12 examples in these notes cover the following skills:Using the Polygon Interior Sum Theorem to find the total number of interior angles for a polygonUsing the Polygon Interior Sum Theorem to solve for variables Determine the number of sides a polygon has by the sum of the interior anglesFind the measure of each interior angle of a regular polygon.Using the Polygon Exterior Sum Theorem Find the measure of each ext
9 th - 11 th
Geometry, Math
Also included in:Quadrilaterals Guided Notes Bundle
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Polygons - Interior and Exterior Angles of Polygons Christmas Riddle Worksheet
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Secondary Math Shop
Geometry Interior and Exterior Angles of Polygons Christmas Riddle Worksheet This is an 21 question practice worksheet that centers around the concept of the Interior and Exterior Angles of Polygons. It requires students to find: the sum of the interior angles, the measure of one interior angle (given the number of sides), the number of sides (given the sum of the interior angles) the measure of one exterior angle (given the number of sides), Once they find the a
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Polygons - Interior&Exterior Angles Of Polygons Investigation Act. & Homework
Created by
Secondary Math Shop
Polygons Interior and Exterior Angles Of Polygons Investigation Activity And Assignment This is an activity designed to lead students to the formulas for: 1) one interior angle of a regular polygon 2)the interior angle sum of a regular polygon 3)one exterior angle of a regular polygon 4)the exterior angle sum of a regular polygon 5)the number of diagonals from one vertex of a regular polygon 6)the number of triangles formed by one diagonal The students do this by drawing and measuring an inte
8 th - 12 th
Applied Math, Geometry, Math Test Prep
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Interior and Exterior Angles of Polygons Mazes (GOOGLE Slides)
Created by
Miss Crafty Math Teacher
These self-checking mazes in Google Slides consist of 15 problems to practice the following on interior and exterior angles of polygons: - Finding the sum of the interior/exterior angles of an indicated convex polygon. - Finding the number of the sides of a convex polygon given the sum of the measures of the interior angles. - Find the measure of an interior/exterior angle of an indicated polygon. - Find the value of x given the diagram of a polygon. This product includes TWO mazes, along with
8 th - 10 th
Geometry, Math
Also included in:Geometry GOOGLE Digital Mazes Bundle
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Interior&Exterior Angles Of Regular Polygons Coloring Activities 8th Grade
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Math Rocks Eh
Engage your students with six differentiated activities on interior and exterior angles of regular polygons, designed for middle and early high school geometry classes. This resource covers everything from basic angle calculations with clear diagrams to algebraic word problems and “work backwards” challenges where students use the Exterior Angle Theorem to determine the number of sides of a polygon. Perfect for Grades 7–9, these activities align with Common Core geometry standards and the
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CCSS
7.G.B.5
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Snowman Math Interior and Exterior Angles of Polygons
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Volunteacher
Students will solve problems involving interior and exterior angles of polygons. They will have to find the sum of the interior angles of a convex polygon, sum of the exterior angles of a convex polygon, each interior angle in a regular polygon and each exterior angle in a regular polygon. They should choose the correct answer and select the snowman accessory that matches their answer. Great for Christmas, January, February, Winter activity.
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Geometry
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https://www.youtube.com/watch?v=b5_uyyXKUZg
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Matrices: Reduced row echelon form 3 | Vectors and spaces | Linear Algebra | Khan Academy
ailabhcmus
462 subscribers
Description
53 views
Posted: 21 Nov 2016
Transcript:
I have here three linear equations of four unknowns and like the first video where I talked about reduced row Echelon form and solving systems of linear equations using augmented matrices at least my gut feeling says look I have fewer equations than variables so I probably won't be able to constrain this enough or maybe I'll have an infinite number of solutions but let's see if I'm right so let's construct the augmented Matrix for this system of equations so my coefficients on the X1 terms are 1 1 and two coefficients on the X2 are 2 two and 4 2 2 and 4 coefficient on X3 are 1 2 and 0er 1 2 and zero there's of course no X3 term so we can view it as a zero coefficient coefficients the next four 1 - one and 6 1 - one and 6 and then on the right hand side of the equal sign I have 8 12 and 4 8 12 and and four there's my augmented Matrix now let's put this guy into reduced row Echelon form so the first thing I want to do is I want to zero out these two rows right here so what can we do so I'm going to keep my first row the same for now so it's 1 2 1 one8 that's that line essentially represents my equal sign and if I let's see what I can do is let me subtract me replace the second row with the second row minus the first row so 1 - 1 is 0 2 - 2 is 0 2 - 1 is 1 1 -1 -1 is -2 and then 12 - 8 is 4 there you go that looks good so far it looks like column or X2 which which is represented by column do looks like it might be a free variable but we're not 100% sure yet let's do all of our rows so let's take let's to get rid of this guy right here let's replace our third equation with our third equation minus 2 times our first equation so we get 2 - 2 1 is 0 4 - 2 2 well that's zero 0 - 2 1 that's -2 6 - 2 1 well that's 4 right 6 - 2 and then 4 - 2 8 - 2 8 is -6 4 - 16 is -12 -12 now what can we do well let's get let's see if we can get rid of this minus two term right there so let me rewrite my let me rewrite my augmented Matrix I'm going to keep row two the same this time so I get a zero 0 1 minus 2 and my essentially my equal sign or the augmented part of the Matrix and now let's see what I can do well actually let me get rid of this zero up here because I want to get in a reduced row Echelon form so any of my Pivot entries which are all always going to have the coefficient one or the entry one it should be the only nonzero term in my row so how do I get rid of this one here well I can subtract row I can replace Row one with Row one minus row two so 1 - 0 is 1 2 - 0 is 2 1 - 1 is 0 1 - -2 that's 1 + 2 which is 3 and then 8 - 4 is four there you go and now how can I get rid of this guy well let let me replace let me replace Row three with Row three + 2 Row one or sorry with Row 3 + 2 row two right because I you to have min-2 + 2 this and they'd cancel out so let's see the Zer 0 + 2 0 that's 0 0 + 2 0 that's 0 - 2 + 2 1 is 0 4 + 2 - 2 that's 4 - 4 that's 0 and then I have -12 + 2 4 that's -12 + 8 that's -4 -4 now this is interesting right now this is interesting I've essentially put this in reduced row Echelon form I have two pivot entries that's a pivot entry right there that's a pivot entry right there they're the only nonzero term in their respective columns this and this is just kind of a style issue but this pivot entry is in is in a lower row than that one so it's to the right it's it's in a column to the right of this one right there and when I just inspect it this looks like a you know this column two looks kind of like a free variable there's no pivot entry there no pivot entry there but let's see let's let's map this back to our to our system of equations these are just numbers to me and I just kind of mechanically almost like a computer put this in reduced row Echelon form actually almost exactly like a computer but let me put it back to my system of linear equations to see what our result is so we get 1 X1 let me write it in yellow so I get 1 X1 + 2 X2 + 0 X3 + 3 X4 is equal to 4 obviously I could ignore this term right there I didn't even have to write it then actually I'm not going to going to write that then I get 0 X1 + 0 X2 + 1 X3 so I could just write that I just write the one 1 x3 - 2 X4 is equal to 4 and then this last term what do I get I get 0 X1 plus 0 X2 plus 0 X3 plus 0 X4 well all of that's equal to zero and I've got to write something on the left hand side so let me just write a 0o and that's got to be equal to -4 well this doesn't make any sense whatsoever 0 equal -4 this is this is a nonsensical constraint this is impossible 0 can never equal Min -4 this is impossible this is impossible which means that it is essentially impossible to find an intersection of these three systems of equations or a solution set that satisfies all of them so when we looked at this initially at the beginning of the of the video we said oh you know there's only three equations have four unknowns maybe there's going to be an infinite set of solutions but it turns out that these three I guess you can call them these three surfaces don't intersect in R4 right these are all four-dimensional we're dealing in R4 right here because each because we have uh I guess each Vector has four components or we have four variables I guess is the way you could think about it and it's always hard to visualize things in R4 but if we were doing things in R3 we can imagine that situation where you know let's say we had two planes in R3 so that's one plane right there and then I had another completely parallel plane to that one so I had another completely parallel plane to that first one even though these would be two planes in R3 so let me give an example so let's say that you know this first plane was represented by the equation by the equation 3x + 6 y + 9 Z is equal to 5 and the second plane was represented by the equation 3x + 6 y + 9 Z is equal to two these two planes in R3 this is the case of R3 so this is R3 right here these two planes clearly they'll never intersect because obviously this one has my the the same coefficients adding up to five this one has the same coefficients adding up to two and when if we just looked at this initially if it wasn't so obvious said oh we have three only two equations with three unknowns maybe this has an infinite set of solutions but it won't be the case cuz you can actually just subtract this equation from the bottom equation from the top equation and what do you get you would get a very familiar so if you just subtract the bottom equation from the top equation you get 3x - 3x 6 y - 6 y 9 Z - 9 Z actually let me do it right here so if that minus that you get 0 is equal to 5 - 2 which is 3 which is a very similar result that we got up there so when you have two parallel planes in this case in R3 or really do any kind of two parallel equations or a set of parallel equations they won't intersect and you're going to get when you do when you either put in reduced row Echelon form or you just do basic elimination or you solve the system you're going to get a St statement that 0o is equal to something and that means that there are no Solutions no Solutions so the general takeaway if you have zero equals something no Solutions if you have the same same number of pivot variables the same number of pivot entries as you do columns so if you get the situation so let me write this down this is good to know if you have zero is equal to anything then that means no solution if you're dealing in R3 you're probably dealing with parallel planes in R2 you're dealing with parallel lines if you have if you have the situation where you have the same number of pivot entries as columns so it's just you know one one one one and it just you know and I'm just this is the case of R4 maybe right there I think you get the idea you know that equals a b c d then you have a unique solution then you have a unique solution now if you have any free variables so free variables look like this so let's say have 1 0 1 zero and then I have the entry one one let me be careful 0 let me do it like this 1 0 0 and then I have the entry 1 two and then I have a bunch of zeros over here and then this has to equal zero remember if this was a bunch of zeros equaling some variable then I would have no solution or equaling some constant but let's say this is equal to five is equal to two if this is our reduced row Echelon form that we eventually get to then we have a few free variables this is a free or I guess we could call this column a free column to some degree this one would be too because it has no pivot entries right these are the pivot entries so this is let's say variable X2 and that's variable X4 then these would be free we can set them equal to anything so then here we have unlimited Solutions or no no unique Solutions and that was actually the first example we saw and these are really the three cases that you're going to see every time and it's good to get familiar with them so you're never going to get stumped up when you have something like 0 = -4 or 0 = 3 or if you have just a bunch of zeros in a bunch of rows but I want to make that very clear sometimes you see a bunch of zeros here on the left hand side of kind of the augmented divide and you might say oh maybe I have no unique solution they have an infinite number of solutions but you got to look at this entry right here this if only if this whole thing is zero and you have free variables then you have an infinite number of solutions if you have a statement like 0 is equal to a if this is is equal to 7 right here then all of a sudden you would have no solution to this that you're dealing with parallel surfaces anyway hopefully you found that helpful
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Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|28 Act on Your Strategy Pressure ratio: P1 = 10 000 kPa P2 = 100 kPa pressure ratio > 1 is 10 000 kPa 100 kPa Substitution to solve for V2: 2 1 pressure ratio 20 mL × 10 000 kPa V V u 100 kPa 3 2 10 L u Number of balloons that can be filled: 3 total volume of helium number of balloons volume of 1 balloon 2 L 10 u 2 L 3 1 10 u 3 Therefore, 1 10 u , or 1000, balloons can be filled. Check Your Solution The initial pressure is greater than the final pressure and the initial volume is less than the final volume. This inverse relationship between volume and pressure is consistent with Boyle’s law. The units are correct and the answer correctly shows one significant digit. Section 11.3 Gases and Temperature Changes Solutions for Practice Problems Student Edition page 522 Note: Assume that the pressure and amount of gas are constant in all of the problems from question 11 to question 19. 11. Practice Problem (page 522) A gas has a volume of 6.0 L at a temperature of 250 K. What volume will the gas have at 450K? What Is Required? You need to find the final volume, V2, of the gas after it has been warmed to 450 K. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|29 What Is Given? You know the volume and temperature of the gas for the initial set of conditions and the temperature for the final set of conditions: V1 = 6.0 L T1 = 250 K T2 = 450 K Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Multiply each side of the equation by T2 to isolate the variable V2. Substitute the numbers and units for the known variables in the formula and solve for V2. Act on Your Strategy Isolation of the variable V2: 1 2 1 2 1 2 2 1 2 V V T T V V T T T 2 T 1 2 2 1 VT V T Substitution to solve for V2: 1 2 2 1 (6.0 L)(450 K VT V T ) 250 K 10.8 L 11 L The final volume of the gas is 11 L. Alternative Solution Plan Your Strategy According to Charles’s law, an increase in Kelvin temperature will cause a directly proportional increase in volume. Therefore, the final volume will be greater than the initial volume. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|30 Determine the ratio of the initial temperature and the final temperature that is greater than 1. Multiply the initial volume by the temperature ratio determined to obtain the final volume. Act on Your Strategy Temperature ratio: T1 = 250 K T2 = 450 K temperature ratio > 1 is 450 K 250 K Substitution to solve for V2: 2 1 temperature ratio 450 K 6.0 L × V V u 250 K 10.8 L 11 L The final volume of the gas is 11 L. Check Your Solution Charles’s law predicts a direct relationship between the temperature in Kelvin and the volume. The Kelvin temperature increased by slightly less than 2 times. The volume also increased by slightly less than 2 times. The answer is reasonable and correctly shows two significant digits. 12. Practice Problem (page 522) A syringe is filled with 30.0 mL of air at 298.15 K. If the temperature is raised to 353.25 K, what volume will the syringe indicate? What Is Required? You need to find the final volume, V2, of air in a syringe after it has been warmed. What Is Given? You know the volume and temperature of the air for the initial set of conditions and the temperature for the final set of conditions: V1 = 30.0 mL T1 = 298.15 K T2 = 353.25 K Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|32 temperature ratio > 1 is 353.25 K 298.15 K Substitution to solve for V2: 2 1 temperature ratio 353.25 K 30.0 mL V V u u 298.15 K 35.544 mL 35.5 mL The syringe will indicate a final volume of 35.5 mL. Check Your Solution Volume units remain when the temperature units are cancelled out. From Charles’s law, if the Kelvin temperature increases, the volume is expected to increase proportionally. The answer represents an increase in volume. The answer is reasonable and correctly shows three significant digits. 13. Practice Problem (page 522) The temperature of a 2.25 L sample of gas decreases from 35.0°C to 20.0°C. What is the new volume? What Is Required? You need to find the final volume, V2, of a sample of gas. What Is Given? You know the volume and temperature of the gas for the initial set of conditions and the temperature for the final set of conditions: V1 = 2.25 L T1 = 35.0°C T2 = 20.0°C Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Charles’s law expresses a direct relationship between Kelvin temperature and volume. Convert the initial and final temperatures from degrees Celsius to kelvin units: K = °C + 273.15 Multiply each side of the equation by T2 to isolate the variable V2. Substitute the numbers and units for the known variables in the formula and solve for V2. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|33 Act on Your Strategy Temperature conversion: 1 2 35 C 273.15 308.15 K 20 C 273.15 293.15 K T T q q Isolation of the variable V2: 1 2 1 2 1 2 2 1 2 V V T T V V T T T 2 T 1 2 2 1 VT V T Substitution to solve for V2: 1 2 2 1 2.25 L 293.15 K VT V T u 308.15 K 2.14047 L 2.14 L The new volume of the gas is 2.14 L. Alternative Solution Plan Your Strategy According to Charles’s law, a decrease in Kelvin temperature will cause a directly proportional decrease in volume. Therefore, the final volume will be less than the initial volume. Determine the ratio of the initial temperature and the final temperature that is less than 1. Multiply the initial volume by the temperature ratio determined to obtain the final volume. Act on Your Strategy Temperature ratio: T1 = 308.15 K T2 = 293.15 K Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|34 temperature ratio < 1 is 293.15 K 308.15 K Substitution to solve for V2: 2 1 temperature ratio 293.15 K 2.25 L V V u u 308.15 K 2.14047 L 2.14 L The new volume of the gas is 2.14 L. Check Your Solution Volume units remain when the temperature units are cancelled out. From Charles’s law, if the Kelvin temperature decreases, the volume is expected to decrease proportionally. The answer represents a smaller value for the volume. The answer is reasonable and correctly shows three significant digits. 14. Practice Problem (page 522) A balloon is inflated with air in a room in which the air temperature is 27°C. When the balloon is placed in a freezer at í20.0°C, the volume is 80.0 L. What was the original volume of the balloon? What Is Required? You need to find the initial volume, V1, of air in a balloon. What Is Given? You know the temperature of the air in the balloon for the initial set of conditions and the temperature and volume for the final set of conditions: T1 = 27°C T2 = –20.0°C V2 = 80.0 L Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Charles’s law expresses a direct relationship between Kelvin temperature and volume. Convert the initial and final temperatures from degrees Celsius to kelvin units: K = °C + 273.15 Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|36 Act on Your Strategy Temperature ratio: T1 = 300.15 K T2 = 253.15 K temperature ratio > 1 is 300.15 K 253.15 K Substitution to solve for V1: 1 2 temperature ratio 300.15 K 80.0 L V V u u 253.15 K 94.852 L 95 L The initial volume of air in the balloon was 95 L. Check Your Solution Volume units remain when the temperature units are cancelled out. From Charles’s law, if the Kelvin temperature decreases, the volume is expected to decrease proportionally. The initial volume should be greater than the final volume when the temperature is decreased. The answer is reasonable and correctly shows two significant digits. 15. Practice Problem (page 522) At a summer outdoor air temperature of 30.0°C, a particular size of bicycle tire has an interior volume of 685 cm3. The bicycle has been left outside in the winter and the outdoor air temperature drops to –25.0°C. Assuming the tire had been filled with air in the summer, to what volume would the tire have been reduced at the winter air temperature? What Is Required? You need to find the final volume, V2, of air in a bicycle tire. What Is Given? You know the volume and temperature of the air in the tire for the first set of conditions and the temperature for the second set of conditions: V1 = 685 cm3 T1 = 30.0°C T2 = –25.0°C Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|37 Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Charles’s law expresses a direct relationship between Kelvin temperature and volume. Convert the initial and final temperatures from degrees Celsius to kelvin units: K = °C + 273.15 Multiply each side of the equation by T2 to isolate the variable V2. Substitute the numbers and units for the known variables in the formula and solve for V2. Act on Your Strategy Temperature conversion: 1 30 C 273.15 303.15 K T q 2 –25.0 C 273.15 248.15 K T q Isolation of the variable V2: 1 2 1 2 1 2 2 1 2 V V T T V V T T T 2 T 1 2 2 1 VT V T Substitution to solve for V2: 1 2 3 2 1 (685 )( cm 248.15 K VT V T ) 303.15 K 3 3 560.721 cm 561 cm The volume of air in the tire would have been reduced to 561 cm3. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|38 Alternative Solution Plan Your Strategy According to Charles’s law, a decrease in Kelvin temperature will cause a directly proportional decrease in volume. Therefore, the final volume of air will be less than the initial volume. Determine the ratio of the initial temperature and the final temperature that is less than 1. Multiply the initial volume by the temperature ratio determined to obtain the final volume. Act on Your Strategy Temperature ratio: T1 = 303.15 K T2 = 248.15 K temperature ratio < 1 is 248.15 K 303.15 K Substitution to solve for V2: 2 1 3 temperature ratio 248.15 K 685 cm V V u u 303.15 K 3 3 560.721 cm 561 cm The volume of air in the tire would have been reduced to 561 cm3. Check Your Solution Volume units remain when the temperature units are cancelled out. From Charles’s law, if the Kelvin temperature decreases, the volume is expected to decrease proportionally. The answer represents a smaller value for the volume. The answer is reasonable and correctly shows three significant digits. 16. Practice Problem (page 522) At 275 K, a gas has a volume of 25.5 mL. What is its temperature if the volume increases to 50.0 mL? What Is Required? You need to find the final Kelvin temperature, T2, of a gas. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|40 Alternative Solution Plan Your Strategy According to Charles’s law, an increase in the Kelvin temperature will cause a directly proportional increase in the volume. Therefore, the final temperature will be greater than the initial temperature. Determine the ratio of the initial volume and the final volume that is greater than 1. Multiply the initial temperature by the volume ratio determined to obtain the final temperature. Act on Your Strategy Volume ratio: V1 = 25.5 mL V2 = 50.0 mL volume ratio > 1 is 50.0 mL 25.5 mL Substitution to solve for T2: 2 1 volume ratio 50.0 mL 275 K T T u u 25.5 mL 539.215 K 539 K The final temperature of the gas is 539 K. Check Your Solution From Charles’s law, if the volume increases the Kelvin temperature must increase proportionally. The volume approximately doubles and the Kelvin temperature approximately doubles. The answer is reasonable and correctly shows three significant digits. 17. Practice Problem (page 522) A sealed syringe contains 37.0 mL of trapped air. The temperature of the air in the syringe is 295 K. The sun shines on the syringe, causing the temperature of the air inside it to increase. If the volume increases to 38.6 mL, what is the new temperature of the air in the syringe? What Is Required? You need to find the final Kelvin temperature, T2, of the air in the syringe. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|41 What Is Given? You know the volume and temperature of the air in the syringe for the first set of conditions and the volume for the second set of conditions: V1 = 37.0 mL T1 = 295 K V2 = 38.6 mL Plan Your Strategy Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Multiply each side of the equation first by T2 and then by 1 1 T V to isolate the variable T2. Substitute the numbers and units for the known variables in the formula and solve for T2. Act on Your Strategy Isolation of the variable T2: 1 2 1 2 1 2 2 1 2 V V T T V V T T T § · ¨ ¸ © ¹ 2 T § · ¨ ¸ ¨ ¸ © ¹ 1 V 1 T 1 T § · ¨ ¸ ¨ ¸ © ¹ 1 V 1 2 2 1 2 1 2 1 T T V V V T T V § · § · ¨ ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ Substitution to solve for T2: 2 1 2 1 (38.6 mL V T T V )( 295 K) 37.0 mL 307.756 K 308 K The new temperature of the air in the syringe is 308 K. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|42 Alternative Solution Plan Your Strategy According to Charles’s law, an increase in volume will be caused by a directly proportional increase in the Kelvin temperature. Therefore, the final temperature will be greater than the initial temperature. Determine the ratio of the initial volume and the final volume that is greater than 1. Multiply the initial temperature by the volume ratio determined to obtain the final temperature. Act on Your Strategy Volume ratio: V1 = 37.0 mL V2 = 38.6 mL volume ratio > 1 is 38.6 mL 37.0 mL Substitution to solve for T2: 2 1 volume ratio 38.6 mL 295 K T T u u 37.0 mL 307.756 K 308 K The new temperature of the air in the syringe is 308 K. Check Your Solution From Charles’s law, if the volume increases then the Kelvin temperature must increase proportionally. The final temperature is greater than the initial temperature. The answer is reasonable for the small increase in volume and correctly shows three significant digits. 18. Practice Problem (page 522) A beach ball is inflated to a volume of 25 L of air in the cool of the morning at 15°C. During the afternoon, the volume changes to 26 L. What was the Celsius air temperature in the afternoon? What Is Required? You need to find the final Celsius temperature, T2, of the gas. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|45 Final temperature conversion: 2 K 273.15 299.676 K 273.15 26.526 C 27 C T q q The Celsius air temperature in the afternoon was 27°C. Check Your Solution From Charles’s law, if the volume increases then the Kelvin temperature must increase proportionally. The final temperature is greater than the initial temperature. The answer is reasonable for the small increase in volume and correctly shows two significant digits. 19. Practice Problem (page 522) The volume of a 1.50 L balloon at room temperature increases by 25.0 percent when placed in a hot-water bath. How does the temperature of the water bath compare with room temperature? What Is Required? You need to find the final temperature, T2, of a water bath and compare it with the original room temperature. What Is Given? You know the volume and temperature of the air in the balloon for the first set of conditions and the volume for the second set of conditions: V1 = 1.50 L T1 = T1 (room temperature) V2 = 1.50 L + 25% of 1.50 L Plan Your Strategy Determine the final volume, V2. Temperature and volume are changing at constant pressure and amount of gas. Use the equation for Charles’s law: 1 2 1 2 V V T T Multiply each side of the equation first by T2 and then by 1 1 T V to isolate the variable T2. Substitute the numbers and units for the known variables in the formula and solve for T2. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|46 Act on Your Strategy Calculation of final volume: 2 1.50 L (0.25)(1.50 L) 1.5 L + 0.375 L 1.875 L V Isolation of the variable T2: 1 2 1 2 1 2 2 1 2 V V T T V V T T T § · ¨ ¸ © ¹ 2 T § · ¨ ¸ ¨ ¸ © ¹ 1 V 1 T 1 T § · ¨ ¸ ¨ ¸ © ¹ 1 V 1 2 2 1 2 1 2 1 T T V V V T T V § · § · ¨ ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ Substitution to solve for T2: 2 1 2 1 1.875 L V T T V 1 1.50 L T u 1 1.25T The Kelvin temperature of the water bath is 1.25 times greater than the Kelvin temperature of the room. Alternative Solution Plan Your Strategy According to Charles’s law, an increase in volume will be caused by a directly proportional increase in the Kelvin temperature. Therefore, the final temperature (water bath) will be greater than the initial temperature (room). Determine the ratio of the initial volume and the final volume that is greater than 1. Multiply the initial temperature by the volume ratio determined to obtain the final temperature. Chemistry 11 Solutions 978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 11PropertiesofGases•MHR|47 Act on Your Strategy Volume ratio: V1 = 1.50 L V2 = 1.50 L + 25% of 1.50 L = 1.875 L volume ratio > 1 is 1.875 L 1.50 L Substitution to solve for T2: 2 1 1 1 volume ratio 1.875 L 1.50 L 1.25 T T T T u u The Kelvin temperature of the water bath is 1.25 times greater than the Kelvin temperature of the room. Check Your Solution From Charles’s law, if the volume increases then the Kelvin temperature must increase proportionally. The final temperature is greater than the initial temperature. The answer is reasonable and correctly shows three significant digits. 20. Practice Problem (page 522) Compressed gases can be condensed when they are cooled. A 5.00 × 102 mL sample of carbon dioxide gas at room temperature (assume 25.0°C) is compressed by a factor of four, and then is cooled so that its volume is reduced to 25.0 mL. What must the final temperature be (in °C)? (Hint: Use both Boyle’s law and Charles’s law to answer the question.) What Is Required? You must find the final temperature, in degrees Celsius, of a sample of gas that is first compressed and then cooled. What Is Given? You know the initial volume of the gas: V1 = 5.00 × 102 mL You know the initial temperature of the gas: T1 = 25.0°C You know the final volume of the gas: V2 = 25.0 mL You know that the final pressure of the gas is 4 times the initial pressure: P2 = 4P1
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https://www.cs.purdue.edu/homes/hmaji/teaching/Spring%202019/lectures/31.pdf
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Lecture 31: Discrete Fourier Analysis on the Boolean Hypercube (Basics) Fourier Analysis Recall I Objective: Study function f : {0, 1}n →R Interpret function {0, 1}n →R as vectors in RN, where N = 2n Fourier Basis: A basis for the space RN with appropriate properties Character Functions: For S ∈{0, 1}n, we define χS(x) := (−1)S1x1+ ···+Snxn, where x = x1x2 . . . xn and S = S1S2 . . . Sn.
We define the inner-product of two functions as ⟨f , g⟩= 1 N X x∈{0,1}n f (x)g(x) With respect to this inner-product the Fourier basis {χ0, χ1, . . . , χN−1} is orthonormal Fourier Analysis Recall II Now, every function f can be written as f = X S∈{0,1}n b f (S)χS The mapping f 7→b f is the Fourier transformation There exists an N × N matrix F such that f · F = b f , for all f This result proves that the Fourier transformation is linear, that is, \ (f + g) = b f + b g and d (cf ) = c b f We saw that F · F = 1 N · IN×N. This result implies that F is full rank and b f = b g if and only if f = g. So, for any function f , we have d b f = 1 N f Fourier Analysis Recall III We saw two identities 1 Plancherel’s Theorem: ⟨f , g⟩= P S∈{0,1}n b f (S)b g(S), and 2 Parseval’s Identity: ⟨f , f ⟩= P S∈{0,1}n b f (S)2.
Fourier Analysis Objective The objective of this lecture is to associate “properties of a function f ” to “properties of the function b f ” In the sequel we shall consider a few such properties Fourier Analysis Min-Entropy/Collision Probability I Let X be a random variable over the sample space {0, 1}n We shall use X to represent the corresponding function {0, 1}n →R defined as follows X(x) := P [X = x] Collision Probability. The probability that when we draw two independent samples according to the distribution X, the two samples turn out to be identical. Note that this probability is col(X) := P x∈{0,1}n X(x)2 = N⟨X, X⟩ Fourier Analysis Min-Entropy/Collision Probability II We can translate “collision probability” as a property of f into an alternate property of b f as follows Lemma col(X) = N X S∈{0,1}n b X(S)2 This lemma is a direct consequence of the Parseval’s identity Note that if we say that “X has low collision probability” then it is equivalent to saying that “P S∈{0,1}n b X(S)2 is small” So, we can use “P S∈{0,1}n b X(S)2 is small” as a proxy for the guarantee that “X has low collision probability” Min Entropy. We say that the min-entropy of X is ⩾k, if P [X = x] ⩽2−k = 1 K , for all x ∈{0, 1}n Fourier Analysis Min-Entropy/Collision Probability III We can similarly get a property of a high min-entropy distribution X Lemma If the min-entropy of X is ⩾k, then we have X S∈{0,1}n b X(S)2 ⩽ 1 NK The proof follows from the ovservation that if the min-entropy of X is ⩾k, then we have col(X) = X x∈{0,1}n X(x)2 ⩽ X x∈{0,1}n X(x) · 2−k = 1 K Fourier Analysis Min-Entropy/Collision Probability IV Intuitively, if a distribution X has “high min-entropy” then it has “low collision probability,” which, in turn, implies that “P S∈{0,1}n b X(S)2 is small” Fourier Analysis Vector Spaces over Finite Fields I We need to understand vector spaces over finite fields to understand the next result In this document, we shall restrict our attention to finite fields of size p, where p is a prime. In general, finite fields can have size q, where q is a prime-power A finite field is defined by three objects (Zp, +, ×) The set Zp = {0, 1, . . . , p −1} The addition operator +. This operator is integer addition mod p.
The multiplication operator ×. This operator is integer multiplication mod p.
For example, consider the finite field (Z5, +, ×). We have 3 + 4 = 2 and 2 × 4 = 3 Every element x ∈Zp has an additive inverse, represented by −x such that x + (−x) = 0. For example, −3 = 2 Fourier Analysis Vector Spaces over Finite Fields II Every element x ∈Z∗ p := Zp \ {0} has a multiplicative inverse, represented by 1/x, such that x × (1/x) = 1. For example, 1/3 = 2.
We can interpret Zn p as a vector space over the finite field (Zp, +, ×) We shall consider vector subspace V of Zn p that is spanned by the rows of the matrix G of the following form.
G = h Ik×k Pk×(n−k) i We consider the corresponding subspace V ⊥of Zn p that is spanned by the rows of the matrix H of the form H = h −P⊺ I(n−k)×(n−k) i Fourier Analysis Vector Spaces over Finite Fields III We define the dot-product of two vectors u, v ∈Zn p as u1v1 +· · · + unvn, where u = (u1, . . . , un) and v = (v1, . . . , vn) Note that the dot-product of any row of G with any row of H is 0. This result follows from the fact that G · H⊺= 0k×n−k.
This observation implies that the dot-product of any vector in V with any vector in V ⊥is 0 Note that V has dimension k and V ⊥has dimension (n −k) The vector space V ⊥is referred to as the dual vector space of V Note that the size of the vector space V is pk and the size of the vector space V ⊥is pn−k Fourier Analysis Vector Spaces over Finite Fields IV Let us consider an example. We shall work over the finite field (Z2, +, ×). Consider the following matrix G = 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 The corresponding matrix H is defined as follows H = " 1 1 0 1 0 0 1 1 0 1 # Note that the dot-product of any row of G with any row of H is 0. Consequently, the dot-product of any vector in the span of the rows-of-G with any vector in the span of the rows-of-H is always 0 Fourier Analysis Vector Spaces over Finite Fields V Actually, any vector space V ⊆Zn p has an associated V ⊥⊆Zn p such that the dot-product of their vectors is 0.
(Think how to prove this result) Fourier Analysis Fourier Transform of Vector Spaces I Let V be a vector sub-space of {0, 1}n of dimension k. Let V ⊥ be the dual vector sub-space of {0, 1}n of dimension (n −k).
Let f = 1 |V |1{V }. That is, the function f is the following probability distribution f (x) = ( 1 K , if x ∈V 0, if x ̸∈V Then, we have the following result.
Lemma b f (S) = ( 1 N , if S ∈V ⊥ 0, if S ̸∈V ⊥ Fourier Analysis Fourier Transform of Vector Spaces II Proof Outline. Suppose S ∈V ⊥.
b f (S) = ⟨f , χS⟩= 1 N X x∈{0,1}n f (x)χS(x) = 1 N X x∈V f (x)χS(x) = 1 NK X x∈V (−1)S·x = 1 NK X x∈V 1 = 1 NK · K = 1 N Fourier Analysis Fourier Transform of Vector Spaces III Now, note that ⟨f , f ⟩= 1 N X x∈{0,1}n f (x)2 = 1 N X x∈V 1 K 2 = 1 NK Next note that X S∈{0,1}n b f (S)2 = X S∈V ⊥ b f (S)2 + X S̸∈V ⊥ b f (S)2 = (N/K) 1 N2 + X S̸∈V ⊥ b f (S)2 = 1 NK + X S̸∈V ⊥ b f (S)2 Fourier Analysis Fourier Transform of Vector Spaces IV By Parseval’s identity, we have ⟨f , f ⟩= P S∈{0,1}n b f (S)2. So, we get that X S̸∈V ⊥ b f (S)2 = 0 That is, for every S ∈V ⊥, we have b f (S) = 0 We can write the entire result tersely as follows \ 1{V } |V | = 1 N 1{V ⊥} As a corollary of this result, we can conclude that b δ0 = 1 N 1{{0,1}n} Recall that δ0 is the delta function that is 1 only at x = 0; 0 elsewhere. Furthermore, the function 1{{0,1}n} is the constant function that evaluates to 1 at every x Fourier Analysis Fourier Transform of Vector Spaces V Recursively use this result and the fact that (V ⊥)⊥= V to verify that d b f = 1 N f Fourier Analysis
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logic - How can one intuit that P → Q ≡ ¬P ∨ (P ∧ Q)? - Philosophy Stack Exchange
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How can one intuit that P → Q ≡ ¬P ∨ (P ∧ Q)?
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I have not succeeded in intuiting P → Q ≡ ¬P ∨ Q in the sense of imagining how one would conjecture or divine the equivalence without any "foreknowledge" of ¬P ∨ Q to invoke formal proofs or truth tables.
Maybe the following (that I emended to ameliorate readability) can aid.
[ Source : ] "P implies Q" is logically equivalent to "(P and Q) or (not-P)", which is at least somewhat intuitive [:]
[1.] P is false[;] so it doesn't matter if Q is true or false.
[2.] or
[3.] either P is true and therefore Q is true as well.
But 1 ≡ ¬P, and 3 ≡ P ∧ Q.
So symbolising 1, 2, and 3 produces: ¬P ∨ (P ∧ Q). Ergo my title question.
logic
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edited Apr 13, 2017 at 12:42
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asked Nov 14, 2016 at 14:02
user8572 user8572
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7 You are running in a loop... If you want "intution", you have to leave with tautological equivalence. If you want tautological equivalence, using distributivity (that is quite intuitive) the above is simply : ¬P ∨ (P ∧ Q) ≡ (¬P ∨ P) ∧ (¬P ∨ Q) i.e. ¬P ∨ Q.Mauro ALLEGRANZA –Mauro ALLEGRANZA 2016-11-14 14:42:40 +00:00 Commented Nov 14, 2016 at 14:42
1 This is virtually equivalent to answers given to the original question. Please read them.user9166 –user9166 2016-11-14 17:27:11 +00:00 Commented Nov 14, 2016 at 17:27
You know thanks for asking this question. I don't know the answer but P->Q looks like glitch in the matrix to me. I understood one of my problems. To me P->Q looks like a standalone statement floating in air. I am not able to form a clear picture of "local universe". How it is Divided between True and false and how these two interact. I don't have a picture of world where these "logical" things exist. Thanks. I hope you get your answer.Ashish Shukla –Ashish Shukla 2025-08-10 03:38:02 +00:00 Commented Aug 10 at 3:38
But a sort of intuition can be: no P without Q, that is the gist of being a necessary condition.Mauro ALLEGRANZA –Mauro ALLEGRANZA 2025-08-10 08:32:42 +00:00 Commented Aug 10 at 8:32
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I am not sure why nobody pointed this out, but the intuition behind it is simply reasoning by cases. Given P → Q supposing P delivers Q, hence P ∧ Q. And supposing ¬P delivers, well, ¬P. And conversely, excluding ¬P we get both P and Q, so P must entail Q. This works for both expressions ¬P ∨ (P ∧ Q) and ¬P ∨ Q, in fact. Of course, it only works assuming that the P or ¬P dichotomy is exhaustive, i.e. assuming the law of excluded middle.
But this confirms the intuition. This is how it must be because in the intuitionistic logic, which rejects the law of excluded middle, the P → Q ≡ ¬P ∨ Q ≡ ¬P ∨ (P ∧ Q) equivalences fail along with it. Not only that, but intuitionistic implication is not compositional at all, truth values aren't enough to tell if it holds, P → Q roughly means that given a proof of P one can constructively convert it into a proof of Q, see Can one prove by contraposition in intuitionistic logic?
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edited Apr 13, 2017 at 12:19
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answered Nov 15, 2016 at 23:30
ConifoldConifold
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One way to view this intuitively is to create a graph of the relationship of P and Q when P → Q. Here is one such graph:
The domain is the part within the square. Each of the regions is represented by a conjunction of P and Q or their negations.
The question would be whether ¬P ∨ (P ∧ Q) covers the entire domain or not in this case where P implies Q. One can see that it does.
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answered Aug 2, 2018 at 5:48
Frank HubenyFrank Hubeny
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From your referenced answer:
However, one way of thinking about it is that "P implies Q" is logically equivalent to "(P and Q) or (not-P)", which is at least somewhat intuitive--either P is true and therefore Q is true as well, or P is false so it doesn't matter if Q is true or false.
...the author here is making note that the only case where P is true and the supposition "if P, then Q" is true is when Q is also true in the "if P, then Q" supposition. If P is false then "(P and Q) or (not-P)" will evaluate to "(false and Q) or (true)" and since (false and Q) will evaluate to (false), we are left with ((false) or (true)). The "or" here is an "inclusive or" ("exclusive or"s are usually explicitly denoted as "xor") and so long as one of the constituent propositions, or operands, in an evaluation proposition with an "inclusive or" operator is true, then the expression evaluates to true. Hence, "[when P is false] it does not matter if Q is true or false."
Apologies if this is already familiar to you, but note that "P is false" is not the same as "not P" - "not P" only evaluates to true when "P" is false. If "P" is true, then "not P" evaluates to false. If "P" is false, then "not P" evaluates to true.
To get an intuitive sense of this, it might help to note that when there are two propositions (for ease of discussion I will use the capital letters P, Q to represent non-identical, non-equivalent propositions) and each proposition has only one of two aspects (in this case, either "is true" or "is false" such that P is exclusively true or P is exclusively false) then the maximum count of possible arrangements is equal to the number of propositions to the power of the number of aspects. In other words, there are 2 propositions (P, Q). There are 2 possible aspects ("is true", "is false"). The total number of possible combinations is two propositions to the power two aspects, or 2 2 and 2 2 = 4. Note that when there are 3 propositions with only two aspects, there are 2 3 and two times two times two is eight.
Note: it is not necessary to only calculate for "is true" or "is false" - these are arbitrary considerations. You could arbitrarily do the same for the number of possible values when rolling two six sided dice. 2 dice, 6 aspects to each die, total number of potential outcomes when each roll results in exclusively one of the six aspect values amounts to each roll of two dice having 6 2 possible combinations, or, 36 "outcomes". Three dice have 6 3, or 216 possible combinations with every roll.
Back to our set of propositions [P, Q] and our set of aspects ["is true", "is false"]. For this next illustration I will shorten ["is true", "is false"] to [T, F]. We can then compose a compact visual shorthand for comparing the propositions and their aspects - a "table" if you will pardon the expression - like so:
P | Q
---|---
T | T
T | F
F | T
F | F
Note that there are several ways to arrange these elements which display the same relationship, for example:
P | Q
---|---
T | T
F | T
T | F
F | F
or
Q | P
---|---
T | T
T | F
F | T
F | F
or
Q | P
---|---
T | T
F | T
T | F
F | F
or
Q | P
---|---
T | T
F | F
F | T
T | F
...and so on.
It's worth noting that the familiar alphabetical order of the letters p and q have no bearing on the relationship of any propositions represented by and which have been shortened to P and Q. Also note that I am not examining the relationship of the content of propositions to each other, only the rendered values of exclusively either "is true" or "is false". For example, comparing the truth values of "Caesar had epileptic seizures" and "Henry VIII had gouty arthritis" is not a comparison of the rulers medical conditions, nor conjecture whether or not they did suffer these maladies. Comparing the truth values of the two statements is a comparison of the set of statements equivalent to ["that it is true that Caesar had epileptic seizures", "that it is true that Henry VIII had gouty arthritis"] and for the purposes of demonstrating logical relationships it is enough that this set of statements logically equivalent to [("is true" exclusively or "is false" exclusively), ("is true" exclusively or "is false" exclusively)].
P | Q |"not P"|"not Q"
----------|--------
T | T | F | F
T | F | F | T
F | T | T | F
F | F | T | T
Note here that we are not now discussing two propositions [P, Q] and four aspects ["is true", "is false", "is not true", "is not false"] or 2 4 total combinations due to the logical constant "not" resulting in an identical equivalency of terms, i.e. ("is true" <=> "is not false") and ("is false" <=> "is not true"). We are simply visualizing the counter-propositions, the "opposite" truth values, the propositional "negation" or "inversion".
Lastly, for a visual example, if we were explicitly articulating every combination of the truth or falsity (two aspects) of three propositions(2 3), note one way we could draw them all to coincide with our convention for the drawing of 2 2 propositional aspects:
X | B | R
----------
T | T | T
T | T | F
T | F | T
T | F | F
F | T | T
F | T | F
F | F | T
F | F | F
So, how then to compare P, Q, not P, not Q, P --> Q, P and Q, P or Q inclusively, P or Q exclusively and all the possible variations? Well, this is where truth tables come in handy to make a visual representation which gets at expressing an intuitive or explicitly articulated sense of the logical relationship of propositional truth values.
As discussed here, we can visually represent "If P, then Q" (i.e. P --> Q, P implies Q, P then Q, etc.) like so:
P | Q | if P, then Q
------------------
T | T | T
T | F | F
F | T | T
F | F | T
Before we can compare [if P, then Q] to ["not P" "inclusive or" (P and Q)], let us evaluate the constituent propositions in ["not P" "inclusive or" (P and Q)], specifically I will show "P and Q" in a truth table:
P | Q | [not]P | P and Q | P [inclusive]or Q ----------------
T | T | F | T | T |
T | F | F | F | T |
F | T | T | F | T |
F | F | T | F | F |
Side note: this is a good time to also display the differences of "inclusive or" and "exclusive or":
P | Q | P and Q | P [inclusive]or Q | P [exclusive]or Q ------------------------
T | T | T | T | T |
T | F | F | T | F |
F | T | F | T | F |
F | F | F | F | T |
...and I will comment on this later.
We can now evaluate ["not P" "inclusive or" (P and Q)]
P | Q | P and Q | [not]P | [not]P [incl]or (P and Q)
--------------------------------------------
T | T | T | F | T
T | F | F | F | F
F | T | F | T | T
F | F | F | T | T
...and here we have demonstrated the equivalence of "if P, then Q" and "not P inclusive_or (P and Q)"
P | Q | "not P" | (P and Q) | "not P" incl_or (P and Q) | "if P, then Q"
----------------------------------------|---------------
T | T | F | T | T | T
T | F | F | F | F | F
F | T | T | F | T | T
F | F | T | F | T | T
Lastly, as I suspect the difference between the logical constants "inclusive or" and "exclusive or" may be causing some confusion (and I may be completely incorrect on this account) I will also provide a visual representation of "if P, then Q" and "not P exclusive_or (P and Q)"
P | Q |"not P"| (P and Q) | "not P" Xor (P and Q)
-------------------------------
T | T | F | T | F
T | F | F | F | F
F | T | T | F | F
F | F | T | F | F
To sum up: "if P, then Q" is logically equivalent to "[not]P [inclusive]or (P and Q)".
Q.E.D.
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edited Aug 10 at 1:10
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answered Nov 14, 2016 at 15:45
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Yes. P -> Q means either, P is false, or if true then Q is implied to be so too. Ergo ~P v (P & Q). (via Law of Excluded Middle)
Also ~P v (P & Q) means that if P is true, then so must be Q. Ergo P -> Q.
Thus the statements are equivalent in Classical Logic.
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answered Aug 2, 2018 at 7:29
community wiki
Graham Kemp
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Intuitively, if P ∧ Q then P ↔ Q. This is intuitive because 'P and Q' is true if and only if P,Q are true simultaneously, and 'P iff Q' means P,Q have the same truth value. Now just throw in ¬ P. If ¬ P ∨ (P ∧ Q) then P → Q, because the possibility that P is false means it is possible if Q then P is false(Q could be true P could be false), so you can't assert P ↔ Q anymore, you can only assert P → Q. You can assert that because if P is false, then P → Q is vacuously true, and if P ∧ Q then they have the same truth value so P ↔ Q, which means (if P then Q) and (if Q then P) is true, so if P then Q is true.
So intuitively if ¬ P ∨ (P ∧ Q) then P → Q. Now you just have to intuit the converse.
Let P → Q be true. P is true or P is false, by the law of the excluded middle.
Case 1: P
If P and P → Q, then by modus ponens Q, so P ∧ Q.
Case 2: ¬ P
In this case by addition ¬ P ∨ (P ∧ Q).
Cases 1 and 2 are collectively exhaustive, so
(P ∧ Q) ∨ (¬ P ∨ (P ∧ Q)), which reduces to
¬ P ∨ (P ∧ Q).
Thus the converse is true, that is,
If P → Q then ¬ P ∨ (P ∧ Q).
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edited Nov 15, 2024 at 4:11
answered Nov 15, 2024 at 3:30
lee pappaslee pappas
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Either not p or p.
If not p then that's all we know (~p) but if p then definitely q follows (p & q).
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answered Aug 10 at 2:00
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From Stephen Abbott's - understanding analysis there is a section in the text which says:
"The finite set {0,1,2,3,4} is a field when addition and multiplication are computed modulo 5."
I wasn't so familiar with the term "modulo" so quick googling yielded this
Is the idea that an addition or multiplication which in theory yields a number greater than 4, is supposed to "wrap around" the set and begin from zero again?
so 4+1=0 ?
Otherwise, I do not see how this satisfies the properties of a field, namely having an additive identity and multiplicative inverse.
modular-arithmetic
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asked Feb 16, 2015 at 13:40
elbartoelbarto
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Yes, 4+1=0,3⋅3=4 etc.
– Daniel Fischer
Commented
Feb 16, 2015 at 13:42
The additive identity is 0.
– steedsnisps
Commented
Feb 16, 2015 at 13:45
@wowlolbrommer Sorry - meant additive inverse, which I am thinking is satisfied when the set loops around itself after traversing 4, is this correct?
– elbarto
Commented
Feb 16, 2015 at 13:50
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I don't understand what you mean. 4 and 1 are each others' additive inverses (4+1=5=0 modulo 5), and 2 and 3 also (2+3=5=0 modulo 5).
– steedsnisps
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Feb 16, 2015 at 13:52
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It turns out that the set of numbers {0,…,p−1} with everything computed modulo p is a field if and only if p is prime. On the other hand, consider Z/6Z (a notation for {0,…,5} modulo 6 — another common notation is Z6). This is not a field, because 2 has no multiplicative inverse. (If it did, then we would have 3=(2−1⋅2)⋅3=2−1⋅(2⋅3)=2−1⋅0=0. Remember that 2⋅3=0 modulo 6.)
– Akiva Weinberger
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Jul 22, 2016 at 14:17
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You can think of modular arithmetic either as the numbers wrapping around, or what's more typical is to think of the numbers as remainders when divided by 5 (in this case). You can write any integer m as m=5n+r then m≡r mod 5. 0 is the additive identity. 1 is the multiplicative identity. Both operations are commutative and associative. To see that every non-zero element has a multiplicative inverse you could just write out the multiplication table... The additive inverse is just the negative (as usual). So −2=−1∗5+3 and hence −2≡3 mod 5. So the additive inverse of 2 is 3 (because 2+3=5≡0).
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answered Feb 16, 2015 at 13:50
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Multiplicative inverses:
1⋅1≡1(mod5)2⋅3≡1(mod5)3⋅2≡1(mod5)4⋅4≡1(mod5)
Generally: addition mod m is a field if and only if m is prime.
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answered Feb 16, 2015 at 14:08
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This is because, modulo 5, the numbers 1, 6, and 16 are all the same.
– Akiva Weinberger
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https://cs.nyu.edu/shasha/papers/glife/algorithm.html
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Algorithm
Algorithms
Our goal is to choose one tree from each gene, then combine those trees into a direct acyclic graph with the fewest interbreeding events. The algorithm can be described in two stages: tree selecting and tree combining.
Tree Selecting -------------- ### Consensus Method
Our goal is to select one tree from each gene. Therefore we would like to select popular trees since they immediately "cover" may genes. If there is a single tree suggested by all genes, then it suffices to select only that tree. But usually we will see the following situation:
gene g1 suggests trees t1, t3, t5
gene g2 suggests trees t1, t2, t4
gene g3 suggests trees t1, t6, t4
gene g4 suggests trees t1, t6, t5
gene g5 suggests trees t2, t4, t7
gene g6 suggests trees t3, t5, t8
In this example we can select trees {t1,t2,t3} to cover all genes (at least one tree from each gene). These kind of problems are called hitting set problems. The fewer trees we select, the easier to combine them with a few interbreeding events (This statement is questionable since it would be harder to combine two totally different trees than to combine many similar trees. In practice, however, it turns out that popular trees usually share some common structures and will be similar after all). Therefore we would like to pick the smallest hitting set. So in the above example, {t4,t5} will be the right choice. Such problem is called minimum hitting set problem. In this software, we have developed a hitting set solver to solve it efficiently. See algorithms of the hitting set solver for more details.
Another thing we might want to do is to ignore trees that rarely appear in the input data since they are supported by very few evidence. Therefore we set a cutoff frequency F in the software to discard trees that appear less than F times in the data. If some genes are so eccentric such that every trees they suggest are discarded, then we ignore these genes as well. The developed algorithm is called Consensus Method:
U<-- the set of all gene trees (duplicates removed)
U<-- U - {t | tree t appear less than F times in the input}
S <-- empty set
for each gene gdo
R<-- {t | t belongs to U and is suggested by g}
Append R to S
for i in [1..MAX_ITERATION] do
C i<-- Hitting_Set_Solver(S, U)
return tree collection C x which requires the fewest interbreeding events to be combined
The results can be found in the software webpage.
Simulated Annealing Method
Consensus method selects trees based on their popularity. However, if all trees appear only once (or very few times) in the data, it is much meaningless to use this method. That is, in that case it works better to select similar trees from each gene instead of looking for identical trees. For this purpose we first define the similarity between trees:
Definition:
The distance between two trees:
distance(T 1, T 2) = the smallest number of interbreeding events needed to combine T 1 and T 2.
2. Similarity: two trees are said to be similar if the distance between them is small.
Once we define the similarity, we can easily come up with some greedy algorithms for tree selecting:
Similar to first heuristics:
Pick an arbitrary tree T 1 from the first gene
for the i th gene do
Pick the tree T i such that distance(T 1, T i) is minimized
return {T 1, T 2, ...}
Similar to last heuristics:
Pick an arbitrary tree T 1 from the first gene
for the i th gene do
Pick the tree T i such that distance(T i-1, T i) is minimized
return {T 1, T 2, ...}
We develop the following algorithm by combining the above two heuristics. We further randomize it such that it will not always run into the same dead end. The modified algorithm:
Randomized hybrid heuristics:
1. Pick an arbitrary tree T 1 from arbitrary gene
2. for each other gene (in a random order) do
3. Pick the tree T i such that _distance_(T 1, T i) + _distance_(T i-1, T i) is minimized
4. return {T 1, T 2, ...}
All these heuristics are nearly deterministic. Therefore they easily fall into some local minima. To get out of local minima solutions, a popular method is called simulated annealing. We first give some definitions before writing down the whole algorithm:
Definitions:
Evaluation of a tree collection:
Let C be a collection of gene trees (one tree from each gene).Evaluation(C) = (The smallest number of interbreeding events needed to combine C into a graph) K + (Total score/rank of trees in C) Here K is a large enough constant such that the first term will dominate the function (Remember the scores of trees are merely used for tie-breaking). A tree collection is better if the evaluation score is smaller.
Neighbor of a tree collection:
Let C be a collection of gene trees (one tree from each gene). A neighbor C' of C is obtained by the following random process: for each gene with probability 1-p we include the corresponding gene tree in C to C'. Otherwise (with probability p) randomly pick a gene tree from that gene and add it to C'. In our software p was set to the inverse of the number of genes. Therefore the expected number of changes from C to C' will be 1.
Algorithm:
Simulated_Annealing(T 0, T, alpha): 1. Pick a starting collection C using randomized hybrid heuristics
2. t <-- T 0
3. while t > T do
4. Choose a neighbor C' of C
5. d <-- Evaluation(C) - Evaluation(C')
6. if d > 0 then
7. C <-- C'
8. else with probability e d/t :
9. C <-- C'
10. t <-- t alpha
11. return the best tree collection seen through the whole process
Here T 0 and T stands for initial and final temperature, respectively, while the temperature decreases exponentially with a factor alpha every iteration. Note that even if the neighbor is worse than the current solution, there is still some chance for the algorithm to switch to it (especially when t is large since the probability of switching = e d/t). That means it is possible to get out of local minimum, which is an important feature. The spirit of this algorithm is that the initial high temperature will make the whole system unstable. Then the temperature cools down slowly, making it harder to switch without benefit. See an introduction to simulated annealing for more insight.
In this problem, a single bad tree could poison the solution seriously. So in actual experiments the original greedy solution was rarely improved much. Therefore instead of running a long annealing process, we also might want to restart the algorithm several times to explore the search space as much as we can. This is the final algorithm we developed. The experimental results can be found in the software webpage.
Tree Combining --------------
We use the following example (Trees are given in the Newick tree format):
geneA: ((s1 (s2 s3) s4 s5) (s6 s7 s8))
geneB: ((s6 (s1 s2) s7 (s3 s4)) (s5 s8))
Orthologs:
s1: {geneA_11, geneB_121}
s2: {geneA_121, geneB_122}
s3: {geneA_122, geneB_141}
s4: {geneA_13, geneB_142}
s5: {geneA_14, geneB_21}
s6: {geneA_21, geneB_11}
s7: {geneA_22, geneB_13}
s8: {geneA_23, geneB_22}
All trees here are unordered, that is, the order of siblings is unimportant. So the Newick tree representation is not unique. For example, the following two trees are equivalent to the above two:
geneA: ((s6 s7 s8) (s1 (s2 s3) s4 s5))
geneB: ((s6 s7 (s1 s2) (s3 s4)) (s5 s8))
Also, in Newick tree format, a pair of parentheses corresponds to an internal node of the tree which is the nearest common ancestor for all species enclosed by that parentheses. For a better illustration, see our introduction.
Definitions:
Group:
A subset of species S forms a group if each gene tree has a Newick tree representation such that we can insert a pair of parentheses into that representation to enclose only species in S without destroying the well-parenthesized structure in every tree.
Minimal Group:
A group which contains no smaller group. Note that if A and B are both groups, then the intersection of them will also be a group. Therefore for any two groups A and B,one of the following relationship holds: (1) A and B are disjoint. (2) A is a subset of B. (3) B is a subset of A. This proposition ensures the existence of the minimal group.
Relationship:
Species A and B have a closer relationship than do A and C if the nearest common ancestor of A and B is a descendent of the nearest common ancestor of A and C.
In parentheses representation, A and B have a closer relationship than do A and C if the innermost parentheses enclosing both A and B does not enclose C.
In subscript representation, A and B have a closer relationship than do A and C if the longest common prefix of the subscripts of A and B is longer than is the longest common prefix of subscripts of A and C.
Example:
In the above example, a pair of parentheses can be inserted in the first tree to enclose [s1 s2 s3] but not in the second tree. So [s1 s2 s3] is not a group. In fact, there are only three groups:
1. [s1 s2 s3 s4]
[s6 s7]
[s1 s2 s3 s4 s5 s6 s7 s8]
The third one is the trivial group which contains all species. Group [s1 s2 s3 s4] and [s6 s7] are both minimal while the trivial group is not.
In the first tree, s3 and s5 have a closer relationship than s3 and s7. In the second tree, however, they have a farther relationship than s3 and s7.
Theorem:
1. If s1 and s3 have a closer relationship than do s1 and s2 in some gene tree, then any group containing both s1 and s2 must also contains s3.
2. Assume G is the optimum combined graph, i.e., the one with the fewest interbreeding events. If a subset of species S forms a group, then G has a subtree T whose leaf set is exactly S. The root of T is the common ancestor of species in S. In other words, the descendant set of the root of T is S.
3. Let S be the descendant set of species A. In that case, the subscripts of orthologs in A will be the longest common prefixes of all corresponding orthologs in species of S.
Therefore, the algorithm can be described as the following:
Main algorithm
Grouping: Find all groups in the input data.
Find a minimal group whose species set is denoted by S.
For every gene tree extract the subtree containing only S. Combine these subtrees into a small graph G with the fewest interbreeding events.
In the original trees, replace S by their common ancestor which is the root of G.
Insert G to our final graph.
Repeat step 2~5 till there is only one species left. That species will be the root of the final graph.
Example:
Assume our program chooses [s1 s2 s3 s4] as the first minimal group to process, the corresponding subtrees will be:
(s1 (s2 s3) s4)
((s1 s2) (s3 s4)) -- Combining Problem 1
The next step is to combine these two trees and replace [s1 s2 s3 s4] by a hypothetical extinct ancestral species miss1 in original trees:
geneA: ((s1 (s2 s3) s4 s5) (s6 s7 s8)) --> ((miss1 s5) (s6 s7 s8))
geneB: ((s6 (s1 s2) s7 (s3 s4)) (s5 s8)) --> ((s6 miss1 s7) (s5 s8))
miss1 contains orthologs {geneA_1, geneB_1}
The next minimal group will be [s6 s7]. The corresponding subtrees:
(s6 s7)
(s6 s7) -- Combining Problem 2
Replace [s6 s7] by a hypothetical extinct ancestral species miss2 in original trees:
geneA: ((miss1 s5) (s6 s7 s8)) --> ((miss1 s5) (miss2 s8))
geneB: ((s6 miss1 s7) (s5 s8)) --> ((miss2 miss1) (s5 s8))
miss2 contains orthologs {geneA_2, geneB_1}
Now only the trivial group [s5 s8 miss1 miss2] is left and we get the final subtree combining problem:
((miss1 s5) (miss2 s8))
((miss2 miss1) (s5 s8)) -- Combining Problem 3
Replace [s5 s8 miss1 miss2] by a hypothetical extinct ancestral species miss3 in original trees:
geneA: ((miss1 s5) (miss2 s8)) --> miss3
geneB: ((miss2 miss1) (s5 s8)) --> miss3
miss3 contains orthologs {geneA_, geneB_}
miss3 will be the root of the combined graph.
Next step will be solving the tree combining problems in the previous stage. The basic strategy is to remove a minimum number of problematic species that cause conflicts, and then combine all gene trees into one big tree (That is, after removing those species all trees become mutually consistent). After that, insert some intermediate missing species to complete the gene deriving steps. Finally, pair up the removed species with proper parents and transform the big tree into a graph.
Definitions:
Conflict Triples: species A, B and C form a conflict triple if A and B have a closer relationship than do A and C in gene tree T 1, but have a farther relationship than do A and C in tree T 2
Theorem:
Several trees can be combined into a big tree if and only if there is no conflict triple.
Example:
2 conflict triples (s1,s2,s3) and (s2,s3,s4) in combining problem 1:
(s1 (s2 s3) s4)
((s1 s2) (s3 s4))
0 conflict triples in combining problem 2:
(s6 s7)
(s6 s7)
4 conflict triples (miss1,miss2,s5), (miss1,miss2,s8), (miss1,s5,s8) and (miss2,s5,s8) in combining problem 3:
((miss1 s5) (miss2 s8))
((miss2 miss1) (s5 s8))
Therefore our approach can be described as the following:
Choose a minimal set of species S and remove them such that there is no conflict triple left
Combine all trees into one tree
Insert the removed species back with proper parents
Here we assume the species removed came from interbreeding events. So at the last step we will actually associate each of them with multiple parents.
Clearly, if at the beginning we have a conflict triple (A,B,C), either A or B or C must be in S, otherwise the removal of S will not eliminate this triple. So again we get a hitting set problem:
Definition:
Hitting set: Let C be a collection of sets. H is a hitting set of C if the intersection of H and any set in C is non-empty.
A hitting set is minimal if the removal of any element will destroy the hitting set. A hitting set is minimum if it has the smallest size over all hitting sets.
Minimum hitting set problem H(C,K): C is a collection of subsets of K. Find the minimum hitting set H of C. In this case H will always be a subset of K.
Example:
Let K = {1,2,3,4,5,6,7}, C = {{1,4,5}, {1,2,3}, {2,6,7}, {1,3,7}}. {3,4,6} will be a minimal hitting set (yet not minimum). In this example, {1,2}, {1,6} and {1,7} are the only minimum hitting sets.
In combining problem 1, the minimum hitting set of { (s1,s2,s3), (s2,s3,s4)} is either {s2} or {s3}.
In combining problem 3, the minimum hitting set of { (miss1,miss2,s5), (miss1,miss2,s8), (miss1,s5,s8), (miss2,s5,s8)} has size 2. In fact, any size 2 subset is a minimum hitting set.
We can model each conflict triple as a set of size 3 and K the set of all species. Then the species removing problem in step 1 becomes a hitting set problem. A hitting set solver was developed to solve this NP-complete problem (The hitting set problem remains NP-complete even under the constraint that every set has size 3). Once we find the hitting set, we remove those species from all trees. Now we should be able to combine the rest into a big tree (See Theorem above). In fact this can be done by the same steps in the main algorithm (See the following example).
Example:
Assume we picked {s2} as the minimum hitting set and removed it from the combining problem 1:
geneA: (s1 (s2 s3) s4) --> (s1 (s3) s4)
geneB: ((s1 s2) (s3 s4)) --> ((s1) (s3 s4))
Now the [s3 s4] is a minimal group. Replace it by a hypothetical species miss4 :
geneA: (s1 (s3) s4) --> (s1 miss4)
geneB: ((s1) (s3 s4)) --> ((s1) miss4)
miss4 contains orthologs {geneA_1, geneB_14}
Then only the trivial group [s1 miss4] is left and we replace it by species miss5 :
geneA: (s1 miss4) --> miss5
geneB: ((s1) miss4)--> miss5
miss5 contains orthologs {geneA_1, geneB_1}
miss5 will be the root of the subgraph. Note that miss1 mentioned above has the same ortholog set with miss5, so they are actually the same species. Let's use miss1 to represent the root. The combined subgraph:
After combining the subtrees, we complete gene deriving steps by inserting intermediate hypothetical species.
Example:
From above we inferred that miss4 {geneA_1, geneB_14} is an ancestor of s3 {geneA_122, geneB_141}. But there are 3 gene mutation steps happening in between: geneA_1 --> geneA_12, geneA_12 --> geneA_122 and geneB_14 --> geneB_141. Therefore there are 3 possible evolutionary path:
1. {geneA_1, geneB_14} --> {geneA_12, geneB_14} --> {geneA_122, geneB_14} --> {geneA_122, geneB_141}
{geneA_1, geneB_14} --> {geneA_12, geneB_14} --> {geneA_12, geneB_141} --> {geneA_122, geneB_141}
{geneA_1, geneB_14} --> {geneA_1, geneB_141} --> {geneA_12, geneB_141} --> {geneA_122, geneB_141}
Deciding which path yields fewer interbreeding events in the final graph is an NP-hard problem. Therefore our software will randomly choose one path. Say we take the first one, the intermediate species inserted will be miss6 {geneA_12, geneB_14} and miss7 {geneA_122, geneB_14}. Similarly, assume we insert the following paths (edges) to this graph:
miss4 {geneA_1, geneB_14} --> miss6 {geneA_12, geneB_14} --> miss7 {geneA_122, geneB_14} --> s3 {geneA_122, geneB_141}
miss4 {geneA_1, geneB_14} --> miss8 {geneA_13, geneB_14} --> s4 {geneA_13, geneB_142}
miss1 {geneA_1, geneB_1} --> miss9 {geneA_11, geneB_14} --> miss10 {geneA_11, geneB_12} --> s1 {geneA_11, geneB_121}
miss1 {geneA_1, geneB_1} --> miss4 {geneA_1, geneB_14}
The combined graph becomes:
Finally, we infer the parents of removed species by their orthologs.
Example:
Here is only one removed species: s2 {geneA_121, geneB_122}. Recall that geneA_121 must be derived from geneA_12 and miss6 is the only species containing that ortholog in this subgraph. Therefore miss6 must be a parent of s2. Similarly, since miss10 is the only species containing ortholog geneB_12 for deriving geneB_122, it must also be a parent of s2. The final subgraph:
Following this algorithm, assume we remove s5 and s8 from combining problem 3, 2 more intermediate species will be inserted:
miss11 {geneA_, geneB_1}
miss12 {geneA_, geneB_2}
The complete combined graph:
### Grouping
We start with a set of two species and recursively bring in species which are close to some species in the set. The algorithm can be described as the following:
Arbitrarily pick species s1 and s2
s+<-- {s1, s2}
Until s+ does not change do
Consider a species t outside s+ :
for each gene g and any two species x, y in s+ :
if x and t have a closer relationship than do y and t then
Add t into s+
break
return s+
Repeat this procedure with all possible combinations of s1 and s2 such that we can obtain all possible groups (There are some minor things to be worried about. For more details, see the paper).
### Hitting Set Solver
Here we introduce three approximation/heuristic algorithms solving the general hitting set problem.
Algorithm 1 - Three approximation algorithm:
1. H <-- empty set
2. while H not yet hit all set in C do
3. Pick a set not yet hit, add every element in that set to H
4. return H
Note in our case that every set in C has size 3, this algorithm has a theoretical guarantee: if the minimum hitting set has size k, then the size of the returned hitting set will not exceed 3k. In computer science jargon, we say this algorithm yields an approximation factor of 3. However, practically it doesn't work very will. This algorithm might not even report a minimal hitting set! For the above example, the output would be {1,4,5,2,6,7} whose size is exactly 3 times the optimum size. In our program, redundant elements in the solution are removed such that the final output will be minimal. However, the performance is still not that great in general. We kept this implementation merely because it reaches the best known approximation ratio.
Algorithm 2 - Frequency heuristic (greedy) algorithm:
1. H <-- empty set
2. while H not yet hit all set in C do
3. Choose the element hitting the most un-hit sets and add it to H
4. return H
This algorithm also has a theoretical guarantee: an approximation factor of O (ln N), where N is the total number of sets. However, this algorithm is popular not because of the theoretical aspect, but because this seems like the most natural thing to do. For above example, the algorithm will pick 1 first. After that {2,6,7} becomes the only un-hit set and any one element of it will be picked in the second iteration. The final output will be the minimum solution. Unfortunately, this algorithm is almost deterministic therefore it easily falls into some local minima, which prevents further improvements of the solution.
Algorithm 3 - Random hitting algorithm:
H <-- K
P <-- a random permutation of K
for e in P do
if H-e is still a hitting set then
H <-- H-e
return H
This strategy is quite different. Instead of starting with an empty set, we initially takes all elements in K, which is guaranteed to be a hitting set. Then the algorithm tries to discard elements in a random order. The good news is it always has the opportunity to achieve the optimum solution. If we repeat the algorithm enough times, the probability that it returns a nearly optimum solution will be high. However, if the problem size is huge, we will need to repeat the algorithm many times for a good solution, which makes the software too inefficient.
Our hitting set solver combines all three strategies since each of them has its pros and cons. The final solver runs the first two deterministic algorithm once and the third randomized algorithm MAX_ITERATION times. Then it reports the best solution seen.
back to software homepage
Chien-I Liao
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In trigonometry, secant function (sec x) is the reciprocal of cosine (cos x) and forms an essential part of calculus. Because tan x = sin x / cos x, having a clear knowledge of these relationships facilitates effective differentiating of sec x. One can use a number of approaches to derive its derivative, some of which include the first principle, quotient rule, and chain rule. In this article, we’ll explore the derivative of sec x, its formula, graphical interpretations, proofs, different differentiation methods, and practical examples to deepen your understanding of this important function.
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Derivative of sec x
The ratio of hypotenuse to the adjacent side of an angle in a right triangle is defined as the trigonometric function secant of an angle.
sec(A)=cbsec(A)=cb
where A denotes the angle, c the hypotenuse, and b the adjacent side
One of the first transcendental functions introduced in Differential Calculus is the Secant Derivative. Secant times tangent, or \sec x .\tan x is the derivative of the secant function (x). where A denotes the angle, c the hypotenuse, and b the adjacent side
This derivative can be proven using limits and trigonometric identities.
\frac{d}{dx}\left ( \sec x \right )
\left ( \sec x \right )’ =\sec x .\tan x
Derivative of Sec X Using First Principle
We will use first principles (or the definition of the derivative to demonstrate that the derivative of sec x is sec x tan x. When attempting to find the derivative, avoid using the first principle unless specifically mentioned in the question. Using the first principle may take longer than the traditional method and may necessitate a good understanding of various forms of formulas related to algebra, trigonometry, and a bit of manipulation. Let’s start.
Assume that f(x)=secxf(x)=secx
f(x+h)sec(x+h)f(x+h)sec(x+h)
ddxf(x)=limh→0[f(x+h)−f(x)h]ddxf(x)=limh→0[f(x+h)−f(x)h]
ddx(secx)=limh→0[sec(x+h)−secxh]ddx(secx)=limh→0[sec(x+h)−secxh]
ddx(secx)=limh→0[1h(1cos(x+h)−1cosx)]ddx(secx)=limh→0[1h(1cos(x+h)−1cosx)]
ddx(secx)=limh→0[1h(cos(x)−cos(x+h)cos(x+h)cosx)]ddx(secx)=limh→0[1h(cos(x)−cos(x+h)cos(x+h)cosx)]
ddx(secx)=limh→0[1h2sin(2x+h2sin(x+h−x2))h.cos(x+h)cosx]
ddx(secx)=limh→0[sin(x+h2)cos(x+h)cosx].limh→0[sinh2h2]
ddx(secx)=sinxcosx.cosx×1
ddx(secx)=secx.tanx
Hence, proved.
Derivative of Sec X Using Quotient Rule
Using the quotient rule, we will demonstrate that the differentiation of sec x with respect to x yields sec x tan x.
We will assume that f\left ( x \right )= \sec x, which can be written as f\left ( x \right )= \frac{1}{\cos x}
f(x)=.cos−[cosx].1cos2x
=0.cosx−(−sinx).1cos2x
=0+sinxcos2x
=sinxcos2x
=sinxcosx.1cosx
=1cosx.sinxcosx
=secx.tanx
Hence proved
Derivative of Sec x by Chain Rule
It is , sin(x)cos(x2)
sec(x)=1cos(x)( \frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\cos \left ( x \right )} =\frac{\mathrm{d} }{\mathrm{d} x} \left ( \cos \left ( x \right )^{-1} \right ) )
This is equivalent to the chain rule.
ddx(cos(x)−1)=−cos(x)−2.
ddxcos(x)
=–1cos(x)2(−sin(x))
=sin(x)cos(x)2
=tan(x)sec(x)
Hence proved
Solved Examples of Derivative of sec x
Problem: 1What is (secx)2 derivative?
Solution:
f(x)=(secx)2
By using power rule and chain rule
f′(x)=2secxddx(secx)
=2secx.(secx.tanx)
=2sec2xtanx
The derivative of given function is =2sec2xtanx
Problem: 2Determine the derivative of sec−1x
Solution:
y=sec−1x
Then,
secy=x …..(1)
Differentiating both sides in terms of x,
secy.tany(dydx)=1
dydx=1secy.tany
According to one of the trigonometric identities,
tany=√sec2y−1=√x2−1
dydx=1(x√x2−1)
Derivative of sec−1x is 1(x√x2−1)
Grasping the derivative of sec x is essential for proficiency in calculus, particularly for the solution of problems relating to rates of change and integrals. Using the rules such as the quotient rule and chain rule, we deduced that d/dx (sec x) = sec x tan x. The result comes into play when performing advanced math involving physics and engineering applications. With a firm understanding of this differentiation process, you can comfortably challenge associated trigonometric derivatives and use them for practical problems, strengthening your total calculus abilities.
Derivative of Sec X FAQs
How to find the derivative of sec x?
By using below formula we can find the derivative of sec x(secx)=secx.tanx
What is the derivative of sec x by first principle?
We will use first principle to demonstrate that the derivative of secx isf(x)=secx
Is sec x differentiable at 0?
No, it is not.
How do you prove the integral of sec x?
Integration by parts cannot be used to calculate the integration of a secant function. As a result, finding its integral necessitates a unique procedure.∫secdx=ln|secx+tanx|+C
Is sec x continuous function?
Yes, it is. secx is defined as the ratio of two continuous functions.
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不规则多边形的面积
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不规则多边形的面积
简介
我想和你分享一个计算求一般多边形面积的巧妙方法
多边形可以是规则的（所有角和边都是相等的）或不规则的
规则 不规则
多边形例子
我们用这个多边形为例：
坐标
第一步是把所有顶点（角）转换成坐标，如图所示：
一条线段下面的面积
接下来，求每条线段下面与 x轴之间的面积。
我们怎样去求每个面积？
求两个高度的平均值，再乘以宽度
例子：在上面突显的图形中，我们求两个高度（"y" 坐标 2.28 和 4.71）的平均：
(2.28+4.71)/2 = 3.495
求宽度（"x" 坐标 2.66 和 0.72 的差）
2.66-0.72 = 1.94
面积是 宽度×高度：
1.94 × 3.495 = 6.7803
加起来
最后把全部面积加起来！
但窍门是前进时（正宽度)的面积要加上，而后退时（负宽度）的面积便要减去。
一直顺时针方向走，并且永远用第二个"x"坐标减去，计算就不会出错，像这样：
| 从 | 到 | | | |
:---: |
| x | y | x | y | 平均高度 | 宽度 (+/-) | 面积 (+/-) |
| 0.72 | 2.28 | 2.66 | 4.71 | 3.495 | 1.94 | 6.7803 |
| 2.66 | 4.71 | 5 | 3.5 | 4.105 | 2.34 | 9.6057 |
| 5 | 3.5 | 3.63 | 2.52 | 3.01 | -1.37 | -4.1237 |
| 3.63 | 2.52 | 4 | 1.6 | 2.06 | 0.37 | 0.7622 |
| 4 | 1.6 | 1.9 | 1 | 1.3 | -2.1 | -2.7300 |
| 1.9 | 1 | 0.72 | 2.28 | 1.64 | -1.18 | -1.9352 |
| | | | | | Total: | 8.3593 |
图是这样的：
做好了！面积是 8.3593
多边形面积工具
很感谢你看到这里！作为奖励，你可以用求多边形面积绘图工具来帮你做所有的计算。这工具也接受手动输入坐标。
问题1)问题2)问题3)问题4)问题5)问题6)问题7)问题8)问题9)问题10)
几何索引
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Molecular mechanisms of the microsomal mixed function oxidases and biological and pathological implications - PMC
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Redox Biol
. 2014 Nov 27;4:60–73. doi: 10.1016/j.redox.2014.11.008
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Molecular mechanisms of the microsomal mixed function oxidases and biological and pathological implications
Arthur I Cederbaum
Arthur I Cederbaum
1 Department of Pharmacology and Systems Therapeutics, Icahn School of Medicine at Mount Sinai, Box 1603, 1 Gustave L Levy Place, New York, NY 10029, USA
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1 Department of Pharmacology and Systems Therapeutics, Icahn School of Medicine at Mount Sinai, Box 1603, 1 Gustave L Levy Place, New York, NY 10029, USA
Received 2014 Oct 31; Revised 2014 Nov 13; Accepted 2014 Nov 16; Collection date 2015 Apr.
© 2014 The Authors
This is an open access article under the CC BY-NC-ND license (
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PMCID: PMC4309856 PMID: 25498968
Abstract
The cytochrome P450 mixed function oxidase enzymes play a major role in the metabolism of important endogenous substrates as well as in the biotransformation of xenobiotics. The liver P450 system is the most active in metabolism of exogenous substrates. This review briefly describes the liver P450 (CYP) mixed function oxidase system with respect to its enzymatic components and functions. Electron transfer by the NADPH-P450 oxidoreductase is required for reduction of the heme of P450, necessary for binding of molecular oxygen. Binding of substrates to P450 produce substrate binding spectra. The P450 catalytic cycle is complex and rate-limiting steps are not clear. Many types of chemical reactions can be catalyzed by P450 enzymes, making this family among the most diverse catalysts known. There are multiple forms of P450s arranged into families based on structural homology. The major drug metabolizing CYPs are discussed with respect to typical substrates, inducers and inhibitors and their polymorphic forms. The composition of CYPs in humans varies considerably among individuals because of sex and age differences, the influence of diet, liver disease, presence of potential inducers and/or inhibitors. Because of such factors and CYP polymorphisms, and overlapping drug specificity, there is a large variability in the content and composition of P450 enzymes among individuals. This can result in large variations in drug metabolism by humans and often can contribute to drug–drug interactions and adverse drug reactions. Because of many of the above factors, especially CYP polymorphisms, there has been much interest in personalized medicine especially with respect to which CYPs and which of their polymorphic forms are present in order to attempt to determine what drug therapy and what dosage would reflect the best therapeutic strategy in treating individual patients.
Keywords: Cytochrome P450, Liver microsomal drug metabolism system, NADPH-cytochrome P450 reductase, P450 induction, P450 polymorphisms, P450 catalytic cycle, P450 multiple forms, Personalized medicine
Graphical abstract
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Highlights
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The CYP P450 system is important in metabolism of endogenous substrates and drugs.
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About 150 forms of CYPs have been identified and they are grouped into families.
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CYPs catalyze a wide variety of reactions and are among the most diverse catalysts known.
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Electrons are passed to the CYP via NADPH+NADPH-cytochrome P450 reductase.
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Metabolism of certain compounds by CYPs generate reactive intermediates which are toxic.
Introduction
The cytochrome P450 mixed function oxidase system plays a major role in the metabolism of important endogenous substrates such as in cholesterol biosynthesis and cholesterol conversion to bile acids, formation of steroid hormones, androgens and estrogens, metabolism of vitamin D 3 to the active 1,25-dihydroxyvitamin D 3, omega hydroxylation of fatty acids, as well as biotransformation of exogenous xenobiotics. The biological effectiveness and the potential toxicity of many drugs are strongly influenced by their metabolism, much of which is accomplished by P450-dependent monoxygenase systems. The wide array of chemical reactions performed by P450 makes this enzyme one of the most versatile catalysts known. The liver, lung and skin microsomal P450s in particular are important in converting lipophilic xenobiotics including drugs, insecticides, carcinogens, food additives, and environmental pollutants to more polar compounds which are easier to excrete. Intestinal CYPs, especially CYP3A4, may be very important in promoting first pass metabolism of many drugs. Since many of these compounds, lose their activity or potency after being metabolized to polar and excretable metabolites, P450 was considered to be important as a cellular detoxification system. However, with certain compounds although the parent xenobiotic is not toxic, metabolism by the P450 system can generate reactive intermediates which are highly toxic e.g. CCL 4, nitrosamines and acetaminophen.
The term P450 designates a broad family of heme-containing proteins found in bacteria, yeast, plants, invertebrates and vertebrates. About 150 forms of P450 have been identified and a nomenclature based on structural homology, largely deduced from the corresponding cDNAs is used to classify these multiple forms of P450 . The nomenclature is based on evolutionary relationships between CYP450 enzymes and not on similarity in substrate profiles because of the overlapping substrate profiles of many CYP enzymes and the ability of multiple CYPs to modify a single substrate at the same or even at different sites . Many in vitro studies using isolated microsomal fractions or intact hepatocytes or cell lines have provided critical and basic information on drug metabolism by CYPs despite the relative short term limitations of such systems including lack of conjugation and other cytosolic enzymes, loss of CYPs in culture, limited number of liver functions expressed, lower levels of CYPs compared to in vivo. Studies to improve such in vitro systems in order to be more reflective of the in vivo state are an important research front.
Several of the contributions to the common theme series in Redox Biology entitled “Role of CYP2E1 and Oxidative/Nitrosative stress in the Hepatotoxic Actions of Alcohol” discuss the P450 enzyme CYP2E1. The goal of this brief overview is to summarize the molecular mechanisms of the cytochrome P450 microsomal drug oxidation system and perhaps be helpful as an educational tool analogous to the Graphical Review by Dr. B. Kalyanaraman on “Oxidants, Antioxidants and Disease Mechanisms” published in Redox Biology .
Cytochrome P450 General characteristics
The presence of a carbon monoxide-binding pigment in rat liver microsomes was initially reported in 1958 [4,5]. The oxidized and reduced spectrum of one member of the cytochrome P450 family of enzymes, CYP2E1, is shown in Fig.1. The pigment, when reduced, displayed a maximal absorbance at a wavelength of 450 nm when binding carbon monoxide and was called P (pigment)-450. Spectral evidence revealed that P450 was a heme-containing protein and P450 plus cytochrome b 5 accounted for most of the hemoproteins found in liver microsomes. The content of cytochrome P450 (nmol/mg microsomal protein) can be calculated from the ferrous carbon monoxide P450 versus ferrous P450 difference spectrum as described in detail in . Cytochrome P450 was subsequently shown to function as the oxygen-activating oxidase associated with microsomal oxygenation reactions such as steroid C-21 hydroxylation, xenobiotic hydroxylation and oxidative dealkylations [8,9]. Elevated activity of the microsomal mixed function oxidase system after in vivo administration of certain drugs was shown to be related to cytochrome P450 and its inducibility . CYPs are b-type cytochromes, containing protoporphyrin IX as the prosthetic group.
Fig.1.
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Spectral characteristics of cytochrome P450. A: Absolute spectra of CYP2E1 purified from pyrazole-treated rats . The sample cuvette contained 0.39 nmol of CYP2E1 in 0.1 M KPi, pH 7.4 buffer, 20% glycerol, 0.1 mM EDTA, 0.2% Emulgen 911 and 0.5% sodium cholate in a final volume of 1 ml. The reference cuvette was identical except for the omission of the CYP2E1. The scanned spectra were: oxidized CYP2E1 (solid line); dithionite reduced CYP2E1 (dot/dashed lines -.-.-.-.-.); carbon monoxide-bound reduced CYP2E1 (dashed line - - - -). B: The CO-CYP2E1 difference spectrum. Dithionite-reduced CYP2E1 (0.39 nmol/ml) was present in the sample and reference cuvettes. The sample cuvette was saturated with CO and spectra recorded over the indicated wavelengths.
Cytochrome P450 is present in various vertebrates, invertebrates and plants. In mammals, P450, while present at highest levels in microsomes from liver (where it plays a major role in detoxification reactions) is also present in microsomes from kidney, small intestine, lungs, adrenal cortex, skin, brain, testis, placenta and other tissues . Mitochondria, especially from liver and endocrine tissue, contain P450. The nuclear envelope and plasma membranes contain low amounts of P450 [11,12]. In plants, P450s are involved in the synthesis of lignins and alkaloids.
Most P450s are made up of about 400–500 amino acids with molecular weights of about 50,000 Da. About half of the amino acids are non-polar. While earlier studies proposed P450 to be buried in the membrane, it is now recognized that much of the P450 sticks out of the membrane and is attached at the amino terminus to the microsomal membrane (Fig.2). Similarly, the NADPH-cytochrome P450 reductase (discussed below) also largely sticks out of the membrane (Fig.2). The axial ligand for the heme iron is a cysteine residue which is well conserved through evolution near the carboxy terminus of the protein and attaches the heme to the protein. The sixth coordination position of the iron may be water or be empty in the oxidized state and oxygen or carbon monoxide in the reduced state. The addition of a substrate may displace the water with subsequent effects on the redox potential of the heme. The oxidation reduction potential as determined by titrations with redox dyes is very low, with a midpoint potential of about −0.34 to −0.40 V. Agents that react with sulfhydryl groups or disrupt hydrophobic interactions convert P450 to an inactive form called P420, based on the absorption maximum of the ferrous P420–carbon monoxide complex. Conversion of P450 to P420 usually results in loss of substrate-induced spectral changes discussed below.
Fig.2.
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Cartoon of the membrane topography of cytochrome P450 and of NADPH-cytochrome P450 reductase. Electrons for the reduction of molecular oxygen are provided by the reductant/cofactor NADPH which is generated by the pentose phosphate pathway or mitochondrial transhydrogenase, and transferred via NADPH-cytochrome P450 reductase to the heme of cytochrome P450. Both the NADPH-P450 reductase and P450 are microsomal membrane associated, large parts of these enzymes protrude into the cytosol and are attached to the membrane via their amino terminus.
Substrates may change the optical properties of P450, so called substrate binding spectra [13,14]. The substrate binding may promote the high spin state of the P450 with an absorption maximum at about 390 nm and a trough at about 420 nm (Type I substrate binding spectrum) or promote transition to the low spin state with an absorption maximum at about 417 nm and a trough at about 390 nm (Type II substrate binding spectrum) [13,14]. Fig.3 shows the binding spectra when increasing concentrations of DMSO or the chemical pyrazole, known ligands for CYP2E1, are added to purify CYP2E1. Type II binding spectra were produced. The inset of Fig.3 shows a Lineweaver–Burk plot of these data in order to calculate the spectral dissociation constants for DMSO (21 mM) or pyrazole (0.04 mM) binding to CYP2E1. The substrate binding spectra of ligands for cytochrome P450 can be intensified after induction of the specific CYP which interacts with the ligands. For example, CYP2E1 was induced by chronic ethanol feeding and microsomes from the ethanol-fed rats and the dextrose-pair fed controls were incubated with pyrazole or 4-methylpyrazole . The magnitude of the Type II binding spectrum was intensified with the microsomes from the ethanol-fed rats and the affinity for pyrazole and 4-methylpyrazole, as determined from Hanes–Wolf plots was elevated by ethanol treatment in association with a 3–5 folds increase in CYP2E1 content .
Fig.3.
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Substrate binding spectra to CYP2E1. CYP2E1 and various concentrations of the CYP2E1 ligands DMSO (14–364 mM) or pyrazole (6–51 µM) were added and the spectrum was recorded over the indicated wavelengths until no further changes were observed. A CO-CYP2E1 binding spectrum is shown by the solid lines with a peak at 451 nm. DMSO produces a type II substrate binding spectrum with CYP2E1 with a peak at 419 nm and trough at 386 nm. Pyrazole also produces a type II binding spectrum with a peak at 425 nm and trough at 390 nm. A Lineweaver–Burk plot of the concentration dependence was linear (insert) and the spectral dissociation constants were calculated to be 21 mM for DMSO and 0.04 mM for pyrazole. Results are from Ref. .
Electron transfer to P450
Reducing equivalents from either NADH or NADPH are required for P450 to activate molecular oxygen for subsequent mixed function oxidation. The first P450 system to be purified and reconstituted from its molecular components was the 11 β-hydroxylase system of adrenal cortex mitochondria . This system was separated into a P450-containing membrane fraction and a soluble NADPH-P450 reductase activity; the latter was further resolved into an iron–sulfur protein and a flavoprotein reductase and the sequence
NADPH→flavoprotein→iron sulfur protein→P450
was established . A similar electron transfer pathway that used NADH in place of NADPH was found for the camphor hydroxylase system of the bacterium Pseudomonas putida . Iron sulfur proteins, while required for mitochondrial P450 electron reduction are not involved in microsomal P450 electron reduction. The microsomal drug metabolism system contains two major components, the P450 and its reductant, the NADPH cytochrome P450 reductase (Fig.4). The mammalian microsomal cytochrome P450 reductase contains 2 mol of flavin, one FAD and one FMN, per mol of reductase enzyme. The activity of NADPH-cytochrome P450 reductase can be assayed via its ability to reduce other acceptors, usually ferric cytochrome c, following the increase in absorbance at a wavelength of 550 nm as described in . The reductase is also active in providing reducing equivalents for microsomal heme oxygenase activity. A recent review of the roles of the reductase in physiology, pharmacology and toxicology can be found in .
Fig.4.
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Electron transfer to cytochrome P450. NADPH is the preferred reductant for microsomal cytochrome P450. Electrons are passed from NADPH to the FAD cofactor of the reductase which passes electrons to the FMN cofactor which than reduces the heme of P450. NADH is less effective than NADPH in reacting with the P450 reductase and reducing P450. NADH effectively reduces the FAD component of the NADH-cytochrome b 5 reductase which then reduces the second major heme enzyme in microsomes, cytochrome b 5. The system is important in fatty acid desaturation. However, it is not very efficient, relative to the NADPH cytochrome P450 reductase in reducing the heme of cytochrome P450. For reduction of mitochondrial cytochrome P450, iron sulfur proteins are required. The scheme depicts the roles of the FAD adrenodoxin reductase in reducing the adrenodoxin iron–sulfur protein and subsequently, mitochondrial cytochrome P450.
Cytochrome P450s use molecular oxygen and reducing equivalents to catalyze the monooxygenation of a variety of substrates (RH) by the following general reaction:
RH+NAD(P)H+O 2→ROH+NAD(P)+H 2 O
This reaction is referred to as a mixed function oxidase reaction since one atom of oxygen is incorporated into the substrate (hydroxylation) while the other is reduced to water. P450s cannot directly accept electrons from NADH or NADPH, therefore a cytochrome P450 reductase containing FMN and FAD is necessary to reduce the heme of microsomal P450. For mitochondrial P450 reduction, a FAD containing reductase and an iron–sulfur protein are necessary (Fig.4). For microsomal P450 reduction, NADPH rather then NADH is the preferred reductant for transferring reducing equivalents to the NADPH-P450 reductase and then to cytochrome P450. NADH generally is about 10% as effective as NADPH in promoting microsomal P450 activity. NADH transfers electrons to the FAD containing flavoprotein NADH-cytochrome b 5 reductase which reduces cytochrome b 5 (Fig.4). Experiments with inhibitors, antibodies and reconstitution experiments with purified enzymes have provided clear evidence for the major role of the NADPH-P450 reductase in reducing P450. The reduced b 5 may to some extent reduce P450. In some cases, NADH may further increase NADPH-dependent P450 catalytic activity by providing the second electron required for the P450 catalytic cycle (described below). The NADH/ NADH cytochrome b 5 system is normally involved in transferring electrons to the cyanide sensitive factor required for fatty acid desaturation. A summary of the electron transfer pathways involved in P450 catalyzed reactions [9,20,21] is shown in Fig.4.
Rat liver microsomes contain about 0.05 nmol/mg protein of NADPH-cytochrome P450 reductase and 0.05 nmol/mg of NADH-cytochrome b 5 reductase, about 0.5 nmol/mg of cytochrome b 5 and about 0.5–1 nmol P450/mg protein. Therefore, the molar ratios of the two cytochromes are about an order of magnitude greater than that of the corresponding reductase. Molecular weights of cytochrome b 5, the NADH reductase and the NADPH reductase are about 17,000, 40,000 and 80,000 Da, respectively [9,22], while that for cytochrome P450s are about 46–60,000 Da depending on the molecular form. Fig.5 shows the absorption spectrum of oxidized cytochrome b 5 and after its reduction with dithionite. The two reductases and b 5 are typical integral membrane proteins that are amphipathic, attaching to the microsomal membrane via a short hydrophobic region and containing a large hydrophilic catalytic domain which protrudes into the cytosol of the cell and thus are accessible to reducing equivalents. As mentioned above, most of the P450 also protrudes out of the microsomal membrane, being attached to the membrane at its amino terminus. The amino terminus is believed to anchor P450 to the endoplasmic reticulum and to act as a microsomal membrane insertional signal [23,24]. For example, removal of the amino terminus of CYP2E1 promotes its translocation to the mitochondria .
Fig.5.
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Absorption spectrum of microsomal cytochrome b 5. Solid line is the spectrum for oxidized cytochrome b 5 with a peak at 412 nm. The dashed line is the absorption spectrum after reduction with dithionite, with a peak at 423 nm. Reduced cytochrome b 5 does not bind CO.
Phase 1 (I) and phase 2 (II) drug metabolism
The general reaction catalyzed by cytochrome P450s is to oxidize lipophilic substrates to more hydrophilic products e.g. RH→ROH. This is referred to as phase 1 drug metabolism and generally involves oxidation, reduction and hydrolysis reactions and is catalyzed by a number of enzymes, the most important being CYPs [26,27]. The phase 2 metabolic reactions catalyze conjugation reactions of lipophilic compounds with endogenous cofactors such as glucuronic acid, sulfate, glutathione or acetate (acetyl CoA). Phase 2 conjugation of a drug can occur in the absence of phase 1 metabolism. The hydroxylated product produced from phase 1 reactions can also be conjugated by conjugation enzymes to more hydrophilic products, easier to excrete. Often, phases 1 and 2 pathways work together to help in the removal of the xenobiotic. An example is shown in Fig.6 for the conversion of benzene to phenyl sulfate. Among the conjugation enzymes (and the conjugated product) are those promoting glucuronidation, the UDP-glucuronyl transferases (R-Glucuronide), sulfation by sulfotransferases (R-Sulfate), acetylation, by N-acetyl transferases (acetylated R), methylation by methyltransferases (R-CH 3) and glutathionylation by various glutathione transferases (R-SG) (Fig.6). The conjugation reactions are not only important for excretion of the drug but also help to minimize drug toxicity. For example, the widely used drug acetaminophen (N-acetyl-p-aminophenol) is normally metabolized largely by glucuronidation and sulfation to form nontoxic acetaminophen-conjugated products. A small amount of acetaminophen can be oxidized by CYPs, especially CYP2E1 to a reactive N-acetyl-p-quinone imine (NAPQI), which alkylates proteins, especially mitochondrial proteins. This reactive, toxic metabolite can be removed by conjugation with GSH. Acetaminophen toxicity is magnified when conjugation with glucuronic acid or sulfate is compromised or after intake of very high levels of acetaminophen which saturate the conjugation pathways thereby increasing formation of the reactive NAPQI. NAPQI toxicity is magnified especially when glutathione conjugation is impaired because of depletion of hepatic levels of GSH e.g. by alcohol, thereby causing accumulation of NAPQI [28,29]. These reactions are summarized in Fig.7.
Fig.6.
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Phase 2 conjugation enzymes.
Fig.7.
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Phases 1 and 2 acetaminophen metabolism. Acetaminophen consumed in “normal” amounts can be removed by direct conjugation by sulfotransferase or glucuronidation by glucuronyl transferases to produce acetaminophen sulfate or acetaminophen glucuronide, which are excreted. When present at high levels, the conjugation reactions become limiting and acetaminophen can be oxidized by several CYPS, especially CYP2E1 to form the reactive N-acetyl-p-benzoquinone imine. The quinone imine can be removed by conjugation with GSH. However, when formed in high amounts at high levels of acetaminophen or when GSH levels are low because of liver disease or alcohol intake, the reactive quinone imine forms covalent adducts with proteins, especially mitochondrial proteins. Impairment of mitochondrial bioenergetics leads to liver necrosis. Acetaminophen toxicity is one of the leading causes of liver damage and emergency room visits.
Catalytic activity of cytochrome P450
There is tremendous diversity in the reactions catalyzed by cytochrome P450, which may be one of the most versatile catalysts known [30–33]. The reactions catalyzed by P450 may be broadly categorized as: metabolism of endogenous constituents including steroids, cholesterol, bile acids, vitamins, fatty acids, and eicosanoids; conversion of lipophilic exogenous xenobiotics into more polar products which can be conjugated to more soluble products for detoxification and removal from the body and metabolism of certain xenobiotics such as CCL 4, acetaminophen, benzene, halothane and nitrosamines into more reactive products that are toxic or carcinogenic. Conversion of drugs into toxic products as a consequence of metabolism by the microsomal mixed-function oxidase system is a major reason for failure of many drugs in pre and post clinical trials [34–36].
Oxidative reactions that are catalyzed by P450 enzymes have been characterized and discussed by Guengerich and colleagues [34,37,38] and representative ones are shown in Fig.8 – aliphatic hydroxylations, aromatic hydroxylation, epoxidation, N-dealkylation, O-dealkylation, S-dealkylation, N-hydroxylation, oxidative deamination, oxidative denitrosation, oxidative dehalogenation, oxidative desulfuration, sulfoxidation. P450 can also catalyze reductive reactions under anaerobic or hypoxic conditions e.g. nitro reduction of nitropyrenes, azo reduction (aminoazobenzene), reductive dehalogenation of CCL 4 to the trichloromethyl radical. The latter plays a critical role in the use of CCL 4 in toxicological reactions to cause lipid peroxidation followed by liver damage, fibrosis and cirrhosis. In general, some of these reactions may be carried out by one form of P450 while others may be carried out by many forms. Some substrates may be oxidized by one form of P450 at a low concentration but another form at a high concentration e.g. low levels of the cancer-inducing N,N-nitrosodimetylamine are demethylated by CYP2E1 (low K m demethylase) whereas higher concentrations are demethylated by several P450s including CYPs 2B1 and 2B2 . The structural features necessary for a substrate to be oxidized by a particular P450 are still not clear despite much investigation.
Fig.8.
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General type of reactions catalyzed by cytochrome P450 enzymes. Other reactions, not shown, include aryl migration, ring contraction, ring formation, ring coupling, dimer formation, group migration (shown in Ref. ).
The cytochrome P450 catalytic cycle
A general mechanism for microsomal cytochrome P450 catalysis as described in Refs. [20,32–35] is shown in Fig.9. Substrates for P450s have K m values ranging from nM to mM. Binding of the substrate to the oxidized ferric P450 (RH, step 1) may or may not be rapid (depending on the P450 and the substrate) and may or may not change the oxidation/reduction potential of the heme. Binding of substrate to the heme of P450 leads to displacement of water as the 6th ligand to the heme iron, changing the spin state of the iron from low spin to high spin . A change to a more positive redox potential would accelerate acceptance of electrons from the NADPH reductase to reduce the heme from the ferric to the ferrous heme (step 2). Molecular oxygen binds rapidly to the substrate–ferrous P450 (step 3) to form the ternary P450-oxygen–substrate complex (oxy-P450). Carbon monoxide competes favorably with oxygen for binding to ferrous P450 thereby inhibiting P450 activity and providing the basis for detecting and quantifying P450 in biological systems at 450 nm as the ferrous P450– carbon monoxide complex . The ferrous O 2 complex can display a resonance structure in which an electron from the ferrous is transferred to the O 2 to produce the ferric–O 2− complex. Decay of this complex produces the ferric-P450–substrate complex with release of superoxide anion radical, the focus of production of normally small amounts of reactive oxygen intermediates during the P450 cycle [40–42]. The decay of the P450–substrate-superoxide complex is dependent on the nature of the substrate, the form of P450 and the efficacy of input of a second electron to reduce the oxy ferrous P450 to the peroxy P450 complex (substrate–ferrous P450-superoxide or substrate–ferric P450-peroxide) (step 4). Decay of the latter produces H 2 O 2 plus substrate–ferric P450. The second electron may come from the NADPH P450 reductase (like the first electron) or it may come from reduced cytochrome b 5[43,44]. The ability of cytochrome b 5 to provide the second electron to reduce oxy-P450 and continue the P450 cycle may explain the ability of b 5 to stimulate (or inhibit) certain P450-dependent reactions . Subsequent steps of the cycle (steps 5–8) are not very clear but it is believed that heterolytic cleavage of the O–O bond of peroxy P450 yields H 2 O and a ferryl type oxidant (step 5). The ferryl like oxidant abstracts a hydrogen from the substrate (RH) to produce a substrate radical (R•, step 6) and the bound equivalent of a hydroxyl radical (FeOH). Rapid oxygen rebound to the substrate radical (radical recombination) yields the hydroxylated substrate product (step 7) and regenerates the ferric P450 (step 8) [20,30,32–35].
Fig.9.
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The CYP catalytic cycle. Please see text for discussion. Oxygen surrogates such as hydroperoxides can replace NADPH, oxygen and the NADPH-cytochrome P450 reductase in promoting certain CYP-catalyzed reactions by forming active oxygenated–CYP complexes. For example:
RH+Fe 3+ (CYP)→RH-Fe 3+ (CYP)
RH-Fe 3+ (CYP)+XOOH→XOH+R• (Fe–OH)3+ (CYP)
R∙ (Fe–OH)3+ (CYP)→ROH-Fe 3+ (CYP)→ROH+Fe 3+ CYP.
With such a complex overall reaction pathway, it is difficult to ascertain what step(s) is rate limiting for P450-catalyzed reactions. It appears that depending on the P450 and the substrate, different steps may be rate-limiting. Input of the first and especially the second electron may at times be limiting. Substrate may facilitate entry of the first electron . Carbon–hydrogen bond breakage of the substrate to produce the substrate radical may be limiting based upon kinetic isotope effects with deuterated substrates . Cytochrome b 5 may stimulate, have no effect or inhibit. Decay of the oxygenated P450 complex may vary among different forms of P450 and thereby limit the monoxygenase reaction e.g. CYP2E1 appears to be a “loosely coupled” form of P450 associated with elevated rates of production of superoxide and H 2 O 2[48,49]. Depending on the form of P450, there may be a 1:1:1 stoichiometry between the amount of oxygen consumed, the amount of product formed and the amount of NADPH used. In some cases, the stoichiometry is poor unless other products of molecular oxygen reduction such as superoxide or peroxide or especially water are taken into account. For example, with CYP2E1, a considerable amount of the oxygen and NADPH consumed produces water . Excellent reviews on the P450 catalytic cycle can be found in Refs. [42,50,51].
So-called oxygen surrogates may replace NADPH, oxygen and the reductase in promoting certain P450-catalyzed reactions. Agents such as hydroperoxides (e.g. cumene hydroperoxide) can react directly with P450 to produce an active oxygenated P450 complex that can oxygenate certain substrates . Usually these reactions are much more rapid than the NADPH-dependent reactions indicating that electron transfer from the reductase to P450 is a limiting step in the overall P450 mechanism.
Multiple forms of cytochrome P450s
The very broad substrate specificity, the differential effect of inducers and inhibitors on P450 reactions, the multiple bands on SDS-gels, and multiphasic Lineweaver–Burk plots, all pointed to the presence of many forms of P450 enzymes. This was confirmed by purifications of different P450s and by molecular biology. It is now recognized that more than 150 separate forms of P450 exist. The human genome project has identified 57 human genes coding for the various P450 enzymes ( OR (
Numerous names were given to these different forms as they were purified or identified and the older literature can be confusing on nomenclature. A systematic nomenclature is now used which assigns proteins into families and subfamilies based on their amino acid similarities. Proteins within a family exhibit greater than 40% homology while those with at least 55% homology are in the same subfamily. The written notation is an Arabic number designating the gene family followed by a letter for the subfamily and an Arabic number for individual genes within a subfamily. Thus CYP1A1 and CYP1A2 are two individual P450s in family 1 subfamily A.
One way of categorizing the multiple forms of human CYPs is by the substrates they oxidized (Table 1, derived from Ref. ). About one half of the human CYPs metabolize endogenous substrates such as steroids, bile acids, fatty acids, eicosanoids and vitamins. The other 50% function to metabolize xenobiotics . In general, members of the CYPs 1, 2 and 3 families are the most important in catalyzing oxidation of exogenous drugs and are generally present in highest amounts in human liver (Table 2, derived from Ref 26,31,34,37 53). In the P450 reactions with drugs, 90–95% of clinical drugs are metabolized by 5 of the 57 human CYPs: 1A2, 2C9, 2C19, 2D6 and 3A4 . CYPs 3A4/5 oxidize about 40% of the drugs used clinically. CYPs 2C, 2D and 1A also metabolize many clinically-used drugs. Together, CYPs 2A6, 2B6 and 2E1 oxidize less than 10% of clinical drugs (Table 2). Recently, Rendic and Guengerich reported on the contributions of human P450 enzymes in carcinogen metabolism. Their Fig.2B of Ref 55 showed the fraction of P450 activation reactions for carcinogens attributed to individual human P450 enzymes as follows: CYP1A1, 20%; CYP1A2, 17%; CYP1B1, 11%; CYP2A6, 8%; CYP2E1, 11%; CYP3A4, 10%; others 23%. Members of the CYP4 family are active in metabolizing fatty acids, especially arachidonic acid while CYPs 7, 11, 17 and 19 are most important in steroid metabolism (Table 1). CYPs 24A1, 27B1 and 26B1 play a role in vitamin D and retinoic acid metabolism, respectively (Table 1). Approaches that are used to predict or estimate the contribution of an individual CYP to the metabolism of a drug include comparison of rates measured with recombinant P450s, correlation of the catalytic activity of the drug with marker activities of individual CYPs, use of specific selective inhibitors or antibodies against the CYP and evaluation of the drug metabolism remaining activity, and determining the effect of a specific CYP inducer on the drug oxidation activity .
Table 1.
Classification of CYPS based on the substrates they oxidize.
| Sterols | Xenobiotics | Fatty acids | Eicosanoids | Vitamins |
:--- :---
| 1B1 | 1A1 | 2J2 | 4F2 | 2R1 |
| 7A1 | 1A2 | 4A11 | 4F3 | 24A1 |
| 7B1 | 2A6 | 4B1 | 4F8 | 26A1 |
| 8B1 | 2A13 | 4F12 | 5A1 | 26B1 |
| 11A1 | 2B6 | | 8A1 | 26C1 |
| 11B1 | 2C8 | | | 27B1 |
| 11B2 | 2C9 | | | |
| 17A1 | 2C18 | | | |
| 19A1 | 2C19 | | | |
| 21A2 | 2D6 | | | |
| 27A1 | 2E1 | | | |
| 39A1 | 2F1 | | | |
| 46A1 | 3A4 | | | |
| 51A1 | 3A5 | | | |
| | 3A7 | | | |
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Modified from Refs. [33,34,37]. Other CYPs such as CYP2A7, 2S1, 2U1, 2W1, 3A43, 4A22, 4F11, 4F22 4V2, 4X1, 4Z1, 20A1, 27C1 are referred to as orphan CYPs as specific substrates have not been identified .
Table 2.
Human liver drug metabolizing CYPS-content and % drug metabolism.
| CYP | Percent drug metabolized | Percent of total human liver P450 |
:---
| 1A2 | 10–12 | 5 |
| 2A6 | 3–5 | 2 |
| 2B6 | 3–5 | 2–4 |
| 2C8/9 | 15 | 11 |
| 2C19 | 7–9 | 5 |
| 2D6 | 30 | 20–30 |
| 2E1 | 3–5 | 2–4 |
| 3A4/5 | 35–40 | 40–45 |
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Table 3 summarizes some representative substrates for the major human liver drug metabolizing CYPs. Table 4 summarizes representative inhibitors and inducers (discussed below) for the major human liver drug metabolizing CYPs.
Table 3.
Representative substrates for human liver drug metabolizing CYPS.
| CYP1A2- | tacrine, theophylline, propranol, imipramine, amitryptyline, phenacetin, caffeine, nifedipine. |
:--- |
| CYP2D6- | debrisoquine, sparteine, bufuralol, codeine, desipramine, nortriptyline, dextromethorphan. |
| CYP2E1- | ethanol, acetaminophen, chlorzoxazone, halothane, enflurane, nitrosamines, benzene, CCl 4. |
| CYP2C9- | tolbutamide, mephenytoin, S-warfarin, diclofenac. ibuprofen, phenytoin, sulfamethoxazole. |
| CYP2C19- | propranolol, clomipramine, diazepam, omeprazole, progesterone, cyclophosphamide. |
| CYP2A6- | coumarin, nitrosamines, aflatoxin, nicotine, cotinine, testosterone. |
| CYP3A4/5- | nifedipine, cyclosporine, simvastatin, quinidine, midazolam, lovastatin, dexamethasone. |
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Table 4.
Representative CYP inhibitors and inducers.
| Isozyme | Inhibitor | Inducer |
:---
| CYP1A2 | Cimetidine, ellipticine, benzo(a)pyrene, furafylline, α-naphthoflavone. | Phenobarbital, omeprazole, rifampin, ritonvair, cyclic aromatic hydrocarbon |
| CYP2D6 | Quinidine, yohimbine, haloperidol, fluoxetine, thioridazine, bupropion. | Dexamethasone, carbamazepine, phenytoin, rifampin, ritonavir |
| CYP2C9 | Amiodarone, chloramphenicol, ritonavir, ketoconazole, fluconazole | Rifampin, phenobarbital, phenytoin |
| CYP2C19 | Cimetidine, fluoxetine, omeprazole, ritonavir, ketoconazole, sertraline | Carbamazepine, phenytoin, rifampin, phenobarbital |
| CYP2E1 | Chlormethiazole, 4-methylpyrazole, disulfiram, diallyl disulfide | Ethanol, acetone, isoniazid pyrazole, diabetes |
| CYP3A4/5 | Cimetidine, ketoconazole, ritonavir, azole anti-fungals, grapefruit juice | Dexamethazone,barbiturates, carbamazepine, phenytoin, rifampin |
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Differences in expression of specific CYPs and active CYPs in humans can be due to differences in gene regulation and the presence of genetic polymorphisms. CYPs 2A6, 2C9, 2C19 and 2D6 in particular display polymorphisms [56,57]. As an example, a listing of many cytochrome P450 genes and their polymorphisms can be found in Table 1 of Ref. and in The effect of the polymorphism may include formation of an inactive CYP enzyme, CYP gene deletion, formation of an unstable CYP which rapidly degrades, formation of a CYP with lower affinity for the cytochrome P450 reductase, formation of a CYP with altered substrate specificity or altered substrate affinity, or even increased CYP activity due to a gene duplication . An example of such effects for polymorphisms of CYP2D6 is shown in Table 5 (derived from [58,59]).
Table 5.
Polymorphisms of CYP2D6.
| Major variant | Mutation | Consequence |
:---
| 2D62Xn | Gene duplication | Increased activity |
| 2D64 | Defective splicing | Inactive enzyme |
| 2D65 | Gene deletion | No enzyme |
| 2D610 | P34S, S486T | Unstable enzyme |
| 2D617 | T107L, R296C, S486T | Reduced affinity for substrate |
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Several of the major variants of CYP2D6, the nature of the mutation and the consequences of the polymorphism on content and/or activity of CYP2D6 are shown. Results modified from Refs. [58,59].
Induction of cytochrome P450s
The ability of certain xenobiotics to elevate their own metabolism has long been known and the mechanism proposed was that the agent increased the content of the P450 responsible for its metabolism. Table 4 lists classical inducers of liver microsomal P450s: for example, polycyclic aromatic hydrocarbons and barbiturates such as phenobarbital were among the first identified inducers of CYP1A1 and CYP2B1/2B2 respectively, while glucocorticoids such as dexamethasone increase CYP3A4 and ethanol elevates CYP2E1. Not all forms of P450 are inducible especially the steroid metabolizing P450s characteristic of the 11, 17, 19, 21 and 26 families. Not all P450s in a particular family are inducible e.g. CYPs 2B and 2E are inducible (by phenobarbital or ethanol respectively) but CYPs 2A, 2C and 2D are not. Induction of most (but not all) P450s involves activation of the respective gene and increased de novo protein synthesis. In some cases, specific cell receptors which interact with the inducing agent have been identified e.g. induction of CYP1A1 by polycyclic aromatic hydrocarbons such as 3-methylcholanthrene involves initial binding of the inducer to the Ah receptor, translocation into the nucleus and eventual activation of the CYP1A1 gene. Induction of CYP2E1 by ethanol is complex and involves both transcriptional and post-transcriptional mechanisms . The CYP2E1 gene is under transcriptional regulation during development and in rats is activated immediately after birth. Following fasting or induced diabetes, CYP2E1 mRNA is increased several fold due to post-transcriptional mRNA stability . The elevation of CYP2E1 by many low molecular weight chemicals including ethanol. acetone, pyrazole is largely due to protein stabilization and increases in protein half-life [60,62,63]. Ethanol, at very high levels can also increase CYP2E1 by a transcriptional mechanism and increased mRNA synthesis . Thus, multiple mechanisms can exist by which a cytochrome P450 such as CYP2E1 can be induced.
In many cases, induction of a specific P450 by a chemical inducer requires binding of the inducer to a nuclear receptor, followed by translocation of the receptor-inducer complex into the nucleus and subsequent interaction and activation of the gene for the P450 [65,66]. Nuclear receptors involved in activating P450 genes include the constitutive androstane receptor (CAR), the pregnane X receptor (PXR), the peroxisome proliferator-activated receptors (PPARs), the aromatic hydrocarbon receptor (AhR) and the farnesoid X receptor (FXR). The Ah receptor in cooperation with its binding partner ARNT is involved in induction of CYPs 1A1, 1A2 and 1B1, critical CYPs involved in carcinogenesis; CYPs 1A1 and 1B1 are mainly extrahepatic. PXR promotes the induction of CYP3A after forming a complex with its binding partner retinoid X receptor (RXR); CYP3A4 is responsible for metabolism of many drugs in humans. PPARs are activated by fatty acids and lipid lowering drugs and induce fatty acid oxidation enzymes such as peroxisomal acyl CoA oxidase to promote fat oxidation and removal. Genes induced by CAR include CYPs 2B6,2C9 and 3A4 as well as several phase 2 conjugation enzymes. Examples of drugs and substrates for these nuclear receptors include omeprazole for AhR, phenobarbital for CAR, rifampin for PXR, bile acids for FXR, fibrates for PPAR.
Table 4 lists representative inducers and inhibitors of the major human liver drug metabolizing CYPs .
Sex, age and species differences
Levels of the hepatic microsomal mixed function oxidase system are very low during fetal development but increase rapidly soon after birth. There are varying patterns of development depending on the form of P450. In general, drug metabolism activities stabilize by early adulthood. Studies in rats of varying ages have suggested that aging is often accompanied by a decline in metabolism of certain drugs (e.g. ). Sex differences, especially in rats in metabolism of certain drugs have been reported and certain forms of P450 are present in males e.g. constitutive CYPs 2C11 and 2C13 whereas other forms are present in females e.g. CYP2C12. Male rats can metabolize certain drugs faster than females whereas the opposite occurs with other drugs .
In mammals, there are wide differences in drug metabolism by different species. For example, the half life of the barbiturate hexobarbital has been reported to be 19, 60, 140, and 260 min and estimated to be 360 min for mice, rabbits, rats, dogs and humans, respectively , results in general agreement with the in vitro rates of P450-catalyzed metabolism. Imidazole can elevate CYP2E1 in rabbits but not in rats . Strain differences in drug oxidation are frequently observed in mice. Qualitative and quantitative differences in the amount of and form of cytochrome P450s present account for many of these aging, sex and species differences found. Human genetic polymorphisms, as discussed above, not only in drug metabolism P450 enzymes but also in phase 2 conjugation enzymes such as individuals who are rapid acetylators versus slow acetylators are especially important in drug development, efficacy and toxicity. For example, the drugs sparteine and debrisoquine are metabolized by CYP2D1 . Poor metabolizers of these drugs are deficient in CYP2D1 and represent a high risk group with respect to adverse drug effects .
Diet can have major effects on drug metabolism by modulating CYP activity [72,73]. Poor nutrition can depress P450-catalyzed drug oxidations by lowering synthesis and induction of CYPs, by effects on heme biosynthesis thus altering holoCYP formation and by modulation of cofactor availability. Various components in the diet can affect hepatic drug oxidation for example, St. John's wort can increase hepatic levels of CYP3A4 . CYP inhibitors and inducers are present in foods . This is especially notable for components in grapefruit juice such as naringin which is a potent inhibitor of CYP3A4. Consumption of grapefruit juice is known to alter the half life and bioavailability of many important clinical drugs e.g. statins. Isothiocyanates present in cruciferous vegetables and diallyl disulfide present in garlic inhibit CYP2E1 catalytic activity. Alcohol increases levels of CYP2E1. Charring of food or smoke can induce CYP1A2 which can activate many carcinogens. Levels of CYPs can vary in the fasted versus the fed nutritional state; for example, levels of CYP2E1 are increased after 24 h of starvation .
Liver disease can alter the elimination and half life of xenobiotics by altering blood flow to the liver due to hepatic vascular resistance or intrahepatic shunting or extrahepatic shunting of drugs given orally. Impaired synthesis of plasma proteins such as albumin may interfere with drug transport of highly protein-bound drugs in the blood. A decrease in the number of viable hepatocytes or lower activity of residual hepatocytes will lower P450-catalyzed drug oxidation. Cofactor availability, not always considered, may also be compromised by glycogen depletion or pentose cycle impairment and under certain conditions, NADPH may be a limiting factor in overall drug oxidation by P450 . There is a zonal distribution of P450 in the liver acinus as the centrilobular region of the liver generally contains a higher amount of P450 than the periportal zone. For example, CYP2E1 is present, especially after induction by ethanol in the centrilobular zone of the liver . Interestingly, this may, along with other factors (lower oxygen tension, lower levels of GSH), contribute to the initiation and increased toxicity of agents oxidized by CYP2E1 to reactive products including CCl 4, acetaminophen, benzene and ethanol itself in the centrilobular zone. A review of the effects of diseases and environmental factors on human CYPs can be found in .
Reactive oxygen formation (ROS)
As mentioned above, small amounts of ROS such as superoxide radical anion and H 2 O 2 are produced during the P450 catalytic cycle and cytochrome P450 enzymes are a significant source of ROS in biological systems, especially tissues like the liver where P450 is present in high amounts. Several factors determine the generation of ROS by P450s including the specific form of P450, entry of the second electron into the P450 cycle, the presence of substrate and nature of the substrate [40–43]. The toxicity of many reagents is due, in part, to increased production of ROS when they are metabolized by cytochrome P450s e.g. CCL 4, halogenated hydrocarbons, benzene, acetaminophen, anesthetics, nitrosamines etc. CYP2E1 appears to be significant generator of ROS and this may play a role in alcohol-induced liver toxicity . Using CYP2E1 as the representative P450 enzyme, some examples of ROS production by microsomes are briefly presented below. Fig.10 shows the generation of superoxide from liver microsomes isolated from chronic ethanol-fed rats (in which CYP2E1 is elevated) and their pair-fed dextrose controls [80,81]. Superoxide was assayed by electron resonance spectroscopy from the interaction of superoxide with a chemical probe which produces a stable nitroxyl radical with defined splitting constants. Superoxide was produced when either NADPH or NADH was the microsomal reductant, however, rates of superoxide production were about 3–4 times greater with NADPH (Fig.10 spectra b and c) than with NADH (Fig.10 spectra d and e) with either microsomes from the ethanol-fed or the control rats, in agreement with NADPH being the preferred reductant for the liver microsomal mixed function oxidase system. The rates of superoxide production by microsomes from the ethanol-fed rats were greater with NADH or especially NADPH as the reductants as compared to the microsomes from the control rats. All ESR signals were decreased more than 80% by superoxide dismutase validating the role of superoxide in producing the ESR signal. The increase in the ESR signal by the microsomes from the ethanol-fed rats was blunted by an antibody against CYP2E1, suggesting that induction of CYP2E1 by ethanol played a role in the increase in microsomal superoxide generation.
Fig.10.
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Microsomal production of superoxide anion radical. Microsomes were isolated from chronic ethanol-fed rats and dextrose pair-fed controls. Generation of superoxide was assayed by determining the superoxide dismutase-sensitive production of the nitroxyl radical formed from the interaction of superoxide with 1-hydroxy-2,2,6,6-tetramethyl-4-oxo-piperidine. Either NADPH (spectra b and c) or NADH (spectra d and e) were used as cofactors. ESR measurements were carried out at room temperature in a flat quartz cuvette using a Bruker E-300 spectrometer. Spectrum a is that of 0.04 mM of a standard nitroxyl radical. Splitting constants for the resulting triplet were A N=16.0 G and g=2.005. Results are from Refs. [80,81].
Fig.11 shows that chronic ethanol feeding for 4 weeks increased liver lipid peroxidation as compared to the pair-fed controls as determined by a TBARs assay. Immunohistochemistry showed that the ethanol feeding increased levels of 3-nitrotyrosine protein adducts and 4-hydroxynonenal protein adducts. Increases in these adducts occurred mostly in the pericentral zone of the liver acinus, the area where alcohol injury in the liver is initiated. Chlormethiazole, a potent inhibitor of CYP2E1, blocked these increases in TBARs and 3-NT and 4-HNE adducts (Fig.11) . Another approach to evaluate the role of a CYP in ROS generation is to use knockout (KO) mice [83,84]. Fig.12 shows that chronic ethanol treatment increased levels of TBARs in wild type (WT) mice about 2-fold. No such increases were observed in CYP2E1 KO mice, whereas even greater increases in TBARs were seen in CYP2E1 KI mice in which high levels of CYP2E1 were restored in the KO mice . Similar results were found with respect to hepatic GSH levels; GSH was lowered by ethanol feeding in WT and KI mice but not in KO mice (Fig.12). Thus, CYPs such as CYP2E1 can produce ROS and increases in ROS production can occur due to induction of CYP2E1. Use of specific inhibitors or appropriate knockout mice are tools to allow evaluation of the role of a specific CYP in ROS production or the elevation in ROS production by a specific treatment.
Fig.11.
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Chlormethiazole (CMZ) lowers alcohol-induced oxidant stress. SV129 female mice were fed the Lieber-DeCarli ethanol or dextrose liquid diets for 4 weeks. At 2 weeks, some of the ethanol-fed mice were also treated with the CYP2E1 inhibitor, CMZ, at a dose of 50 mg/kg body wt, IP, every other day and fed ethanol for the remaining 2 weeks. Lipid peroxidation in liver homogenates was determined by a thiobarbituric acid (TBARs) assay. Immunohistochemical staining for 3-nitrotyrosine (3-NT) protein adducts or 4-hydroxynonenal (4-HNE) protein adducts was carried out using anti-3-NT or 4-HNE antibodies, followed by a rabbit ABC staining system (Santa Cruz). Results are from Ref. .
Fig.12.
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Alcohol-induced oxidant stress is lower in CYP2E1 knockout (KO) mice. CYP2E1 KO mice and CYP2E1 knockin (KI) mice in which the human 2E1 transgene was introduced into the corresponding mouse null mice to produce a humanized CYP2E1 in the mouse background in the absence of the mouse CYP2E1, were kindly provided by Dr. Frank Gonzalez, NCI, NIH [83,84]. SV129 female wild type (WT) and the KO and the KI mice were fed the ethanol or dextrose diets for 4 weeks. Liver homogenates (1:10) were prepared in 150 mM KCl and assayed for TBARs and for total GSH levels. Results are from Ref. .
Personalized medicine
Personalized medicine refers to the customization of healthcare using molecular analysis and tailoring an individual's medical approach and treatment according to his/her genetic information [86,87]. Selection of an individual's appropriate and optimal therapy is based, in part, on the patient's genetic content. This has been successfully applied to pharmacogenomics of the applied therapy which is largely based on evaluation of the presence or absence of specific CYPs and of CYP polymorphisms which are present. Personalized medicine can be used to predict a person's risk for a particular disease based on the presence or absence of a specific gene. With respect to drug therapy, the drug chosen and the dosage used may be due, in part, to the content of CYPs and the presence of certain CYP polymorphisms. Knowledge of the CYPs present and their polymorphic forms can be useful for inclusion or exclusion of certain individual’s in clinical trials of new drugs and may increase safety and lower adverse outcomes caused by the drug being tested [88,89]. This information may also lower trial and error approaches to find the most appropriate therapy effective for the patient [88,89]. According to the Personalized Medical Coalition, 137 FDA-approved drugs have pharmacogenomic information in their labeling and 155 pharmacogenomic biomarkers are included in FDA-approved drug labeling. Many of these labels include information on the role of CYPs and effect of polymorphic CYPs on metabolism of the drug [88,89].
The major factor for interindividual differences in drug response is the variable pharmacokinetics, which are due mainly to differences in the content and activity of the CYPs which metabolize the drug. About 50% of human drug metabolism is mainly carried out by the highly polymorphic CYP enzymes, CYP2C9, 2C19 and 2D6 (Table 2). Resulting phenotypes in response to a drug may depend on the specific polymorphism e.g single nucleotide polymorphism, insertion, deletion, copy number variation (Table 5 for CYP2D6) which can give rise to ultrarapid metabolizers (UMs), extensive metabolizers (EMs), intermediate metabolizers (IMs) and poor metabolizers (PMs) [53,56,59]. Problems associated with UMs include lack of a response to the drug when administered at the normal drug maintenance concentration and perhaps formation of toxic metabolites. PMs have a risk of no therapeutic response and increased risk of adverse drug reactions. For example, for the CYP2D61/2XN which displays gene duplication, there is poor or no response to antiemetics or antidepressants; the latter was associated with increased suicide risk . For the CYP2D64 or 5 polymorphisms, which are PMs, there is reduced response to antianalgesics . Besides detoxifying and eliminating drugs, the liver microsomal CYPs are often required to activate certain prodrugs to the active form. Many opiod analgesics are activated by CYP2D6 and in view of the many polymorphisms of CYP2D6, many individuals are relatively insensitive to the analgesic actions of e.g. codeine and there is much interindividual variation in the efficacy of pain relief when “normal” doses of codeine or ethylmorphine are administered. A good example on how CYP2D6 polymorphisms affect the necessary maintenance dosage of a drug is the study by Zanger et al. . The doses of nortriptyline required to achieve therapeutics were about 50 mg in PMs, about 50–100 mg in IMs, 100–150 mg in EMs and 500 mg in UMs. Such differences in required maintenance dosages were in accordance with rates of bufuralol hydroxylation, a marker substrate for CYP2D6 activity.
Warfarin is a widely used oral anticoagulant but has a narrow therapeutic window and there is difficulty in management of effective maintenance doses. Warfarin is metabolized by CYP2C9 and ethnic differences in warfarin maintenance doses requirements are related to CYP2C9 polymorphisms . African-Americans require a higher warfarin dose (about 6 mg/day) while Asians require a lower warfarin maintenance dose (about 3.5 mg/day) compared to Caucasians (about 5 mg/day) . There are many polymorphic forms of CYP2C9; Caucasians possess mainly the CYP2C92 and 3 forms and Asians contain mainly the CYP2C93 form. These polymorphisms reduce CYP2C9 metabolism of warfarin leading to requirement of lower maintenance doses than in African-Americans in which these two polymorphic forms of CYP2C9 are lower and the authors concluded that additional studies on polymorphic forms of CYP2C9 are needed . Allele frequencies also can differ between Caucasians and Orientals e.g. the CYP2D64 allele shows a 12–21% frequency is Caucasians but only a 1% frequency is Orientals while the CYP2D610 allele has a frequency of 1–2% in Caucasians but a 50% frequency in Orientals , likely contributing to differences in drug responses between these two ethnic groups. Table 3 of Ref. shows the ethnic distribution of the most common variant alleles of CYP2D6 among Caucasians, Asians, Black Africans and Ethiopians/ Saudi Arabians e.g. the latter have the highest frequency of the CYP2D62Xn allele which displays gene duplication and increased enzyme activity; such considerations play critical roles in personalized medicine.
Besides the importance of phase I CYPs in personalized medicine, there are also polymorphisms in phase 2 conjugation enzymes which play significant roles in individual differences to drugs e.g. the N-acetyl transferases which can cause the “slow acetylator” phenotype with respect to certain drugs such as isoniazid or hydralazine or sulfamethoxazole .
In view of the importance of the liver microsomal CYPs in personalized medicine, identifying which CYPs and which polymorphic forms are present is of major value in choosing the most appropriate drug and drug dosage. The concept of CYP on a chip has been developed as a major tool for this. For example, the AmpliChip CYP450 (Roche Molecular Diagnostic) test evaluates the CYP2D6 genotype and makes phenotype predictions . This chip allows simultaneous analysis of 33 CYP2D6 alleles . Rabsamen et al. verified the genotyping accuracy of the chip for the 5 major CYP2D6 alleles and confirmed these results with real time PCR. An updated version of the AmpliChip contains 3 CYP19 alleles in addition to the 33 CYP2D6 alleles with the goal “to help clinicians determine therapeutic strategy and treatment doses for therapeutic metabolism by the CYP2D6 and CYP2C19” .
Conclusions
Because of many of the above described factors modulating levels of CYPs as well as polymorphisms of many CYPs, there is large variability in content of the xenobiotic metabolizing CYPs in families 1, 2 and 3 in human livers, which likely accounts for the large variability in drug oxidation by humans. Guengerich evaluated the levels of CYPs 1A2, 2E1 and 3A4 in 18 human liver samples . There were 5–10 folds differences in levels of each of the 3 CYPs among the 18 samples . Because of this variability in levels of a specific P450, the presence of multiple CYPs with overlapping substrate specificity and the ability of a CYP to metabolize many structurally distinct substrates, there is extensive overlapping substrate specificities which likely contributes to drug–drug interactions and adverse drug reactions . Attempts to minimize such adverse effects will require further studies on structure–function relationships of CYP enzymes, further information on links between CYP polymorphisms and disease/toxicity, identification of what are the substrates for orphan CYPs , and developing bioinformatic models to predict whether certain drugs and chemicals may be toxic or carcinogenic or act to induce a CYP. Development of designer CYPs which may be useful in removal of toxins, pollutants, oil etc. and engineering CYPs to extend their catalytic capabilities as discussed in Refs. [99,100] are likely to further extend the activities of this versatile family of enzymes.
Acknowledgments
Original data were supported by grants from the National Institute on Alcohol Abuse and Alcoholism, USPHS AA-017425, AA-018790 and AA-021362.
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Abstract
Graphical abstract
Highlights
Introduction
Cytochrome P450 General characteristics
Electron transfer to P450
Phase 1 (I) and phase 2 (II) drug metabolism
Catalytic activity of cytochrome P450
The cytochrome P450 catalytic cycle
Multiple forms of cytochrome P450s
Induction of cytochrome P450s
Sex, age and species differences
Reactive oxygen formation (ROS)
Personalized medicine
Conclusions
Acknowledgments
References
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SOCIETY FOR ENDOCRINOLOGY ENDOCRINE EMERGENCY GUIDANCE: Emergency management of acute adrenal insufficiency (adrenal crisis) in adult patients - PMC
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. 2016 Oct 5;5(5):G1–G3. doi: 10.1530/EC-16-0054
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SOCIETY FOR ENDOCRINOLOGY ENDOCRINE EMERGENCY GUIDANCE: Emergency management of acute adrenal insufficiency (adrenal crisis) in adult patients
Wiebke Arlt
Wiebke Arlt
1 Institute of Metabolism and Systems Research, University of Birmingham, Birmingham, UK
2 Centre for Endocrinology, Diabetes and Metabolism, Birmingham Health Partners, Birmingham, UK
Find articles by Wiebke Arlt
1,2,✉; the Society for Endocrinology Clinical Committee 3
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1 Institute of Metabolism and Systems Research, University of Birmingham, Birmingham, UK
2 Centre for Endocrinology, Diabetes and Metabolism, Birmingham Health Partners, Birmingham, UK
3 The Society for Endocrinology, 22 Apex Court, Woodlands, Bradley Stoke, Bristol, UK
✉
Correspondence should be addressed to W Arlt; Email: w.arlt@bham.ac.uk
Received 2016 Aug 3; Accepted 2016 Aug 3; Collection date 2016 Sep.
© 2016 The authors
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
PMC Copyright notice
PMCID: PMC5314805 PMID: 27935813
Introduction
Acute adrenal insufficiency, also termed adrenal crisis, is a life-threatening endocrine emergency brought about by a lack of production of the adrenal hormone cortisol, the major glucocorticoid. Identifying patients at risk and prompt management can save lives. This guideline aims to take the non-specialist through the initial phase of assessment and management.
Underlying conditions
Primary adrenal insufficiency is caused by loss of function of the adrenal gland itself, for example due to autoimmune-mediated destruction of adrenocortical tissue or surgical removal of the adrenal glands or due to inborn disruption of adrenal cortisol production in congenital adrenal hyperplasia.
Secondary adrenal insufficiency is caused if the regulation of adrenal cortisol production by the pituitary is compromised, this can be the consequence of tumours in the hypothalamic–pituitary area. However, pituitary regulation of cortisol production is also switched off in patients who receive chronic exogenous glucocorticoid treatment with doses ≥5 mg prednisolone equivalent for more than 4 weeks. This may also be caused by long-lasting glucocorticoid injections into joints or chronic application of glucocorticoid cream or inhalers.
In primary adrenal insufficiency cortisol deficiency is aggravated by a lack of adrenal aldosterone production, a hormone important for blood pressure and electrolyte regulation. This puts primary adrenal insufficiency patients at a somewhat higher risk of adrenal crisis.
Clinical presentation
Clinical signs and symptoms:
Fatigue, lack of energy, weight loss
Low blood pressure, postural dizziness and hypotension (≥20 mmHg drop in BP from supine to standing position), dizziness, collapse, in severe cases hypovolaemic shock
Abdominal pain, tenderness and guarding, nausea, vomiting (in particular in primary adrenal insufficiency), history of weight loss
Fever
Confusion, somnolence, in severe cases delirium or coma
Back and leg cramps/spasms are commonly reported and can be distracting if not recognised for what they are (electrolyte derangement in large muscles?)
In primary adrenal insufficiency: generalised skin hyperpigmentation, in particular in areas exposed to mechanical shear stress (palmar creases, nipples, scars, inside of oral mucosa)
In secondary adrenal insufficiency: alabaster-like, pale skin; dependent on underlying conditions also signs and symptoms of other pituitary axis deficiencies
Lab findings:
Hyponatraemia (in primary and secondary adrenal insufficiency)
Hyperkalaemia (in primary adrenal insufficiency)
Pre-renal failure (increased serum creatinine due to hypovolaemia)
Normochromic anaemia, sometimes also lymphocytosis and eosinophilia
Hypoglycaemia (primarily in affected children; can cause long-term neurological deficits, if not promptly treated)
Investigations for suspected adrenal crisis in patients not already known to have adrenal failure
Adrenal insufficiency should be ruled out in any acutely ill patient with signs or symptoms potentially suggestive of acute adrenal insufficiency
Assess blood pressure and fluid balance status; if clinically feasible, measure blood pressure from supine to standing to check for postural drop
Take drug history (glucocorticoids?)
Bloods:
Sodium, potassium, urea, creatinine
Full blood counts
TSH, fT 4 (hyperthyroidism can trigger adrenal crisis; acute adrenal insufficiency can increase TSH due to loss of inhibitory control of TRH release, do not replace with thyroxine if TSH ≤ 10 mU/L)
Paired serum cortisol and plasma ACTH
Diagnostic measures should never delay prompt treatment of a suspected adrenal crisis! There are no adverse consequences of initiating life-saving hydrocortisone treatment and diagnosis can be safely and formally established once the patient has clinically recovered
If the patient is haemodynamically stable, consider performing a short Synacthen test (serum cortisol at baseline and 30 min after i.v. injection of 250 micrograms ACTH 1–24); however, if the patient is severely ill, confirmation of diagnosis can be safely left until after clinical recovery following implementation of emergency dose hydrocortisone treatment
Serum/plasma aldosterone and plasma renin (aldosterone will be low and renin high in primary adrenal insufficiency; observe special sample collection and transport conditions; can be left to confirmation of diagnosis after clinical recovery)
Management of adrenal crisis
Hydrocortisone (immediate bolus injection of 100 mg hydrocortisone i.v. or i.m. followed by continuous intravenous infusion of 200 mg hydrocortisone per 24 h (alternatively 50 mg hydrocortisone per i.v. or i.m. Injection every 6 h)
Rehydration with rapid intravenous infusion of 1000 mL of isotonic saline infusion within the first hour, followed by further intravenous rehydration as required (usually 4–6 L in 24 h; monitor for fluid overload in case of renal impairment and in elderly patients)
Contact an endocrinologist for urgent review of the patient, advice on further tapering of hydrocortisone, investigation of the underlying cause of disease including diagnosis of primary vs secondary adrenal insufficiency
Tapering of hydrocortisone can be started after clinical recovery guided by an endocrinologist. In patients with primary adrenal insufficiency, mineralocorticoid replacement needs to be initiated (starting dose 100 micrograms fludrocortisone once daily) as soon as the daily glucocorticoid dose is below 50 mg hydrocortisone/24 h
Precipitating factors
In more than half of patients with adrenal insufficiency the diagnosis of adrenal failure is only established after presentation with an acute adrenal crisis
However, patients with established adrenal insufficiency and those receiving chronic exogenous supraphysiological glucocorticoid treatment (e.g. for asthma or autoimmune disease) are at permanent risk of adrenal crisis. Most frequent causes are:
Chronic glucocorticoid intake is suddenly stopped
Failure to observe Sick Day Rule 1: the need to double daily oral glucocorticoid dose during intercurrent illness with fever that requires bed rest and/or antibiotics
Failure to observe Sick Day Rule 2: the need to administer glucocorticoids per i.v. or i.m. injection or iv infusion during prolonged vomiting or diarrhoea, during preparation for colonoscopy or in case of acute trauma or surgery requiring general anaesthesia
After emergency care: how to prevent an adrenal crisis
Regular review of the patient by an endocrinologist, initially monthly, in the long-term every 6–12 months
Education of patients and partner/parents regarding symptom awareness and the correct adjustment of glucocorticoid replacement dose:
Sick Day Rule 1: the need to double daily oral glucocorticoid dose during illness with fever that requires bed rest and/or antibiotics
Ensure they have an additional supply of hydrocortisone tablets so that they can double their dose for at least 7 days
Sick Day Rule 2: the need to administer glucocorticoids per i.v. or i.m. injection during prolonged vomiting or diarrhoea, during preparation for colonoscopy or in case of acute trauma or surgery
Teach the patient and partner/parents how to self-administer and inject hydrocortisone and provide them with a Hydrocortisone Emergency Injection kit (100 mg hydrocortisone sodium succinate for injection; hyperlink to ADSHG and Pit foundation where there are picture tutorials on using this); check regularly that their kit is up to date
Provide the patient with a Steroid Emergency Cardwww.endocrinology.org/adrenal-crisis and encourage them to wear medical alert bracelets, in addition to keeping the steroid emergency card with them at all times and showing it to any health care professional they are dealing with
Provide them with emergency phone numbers and contact details for the patient self-help groups
Further information
For further information and to request a steroid card, please go to the Society for Endocrinology’s website www.endocrinology.org/adrenal-crisis.
Disclaimer
The document should be considered as a guideline only; it is not intended to determine an absolute standard of medical care. The doctors concerned must make the management plan for an individual patient.
Sources
1.Wass JA, Arlt W. How to avoid precipitating an acute adrenal crisis. British Medical Journal 2012. 345 e6333 ( 10.1136/bmj.e6333) [DOI] [PubMed] [Google Scholar]
2.Husebye ES, Allolio B, Arlt W, Badenhoop K, Bensing S, Betterle C, Falorni A, Gan EH, Hulting AL, Kasperlik-Zaluska A, et al. Consensus statement on the diagnosis, treatment and follow-up of patients with primary adrenal insufficiency. Journal of Internal Medicine 2014. 275 104–115. ( 10.1111/joim.12162) [DOI] [PubMed] [Google Scholar]
3.Hahner S, Spinnler C, Fassnacht M, Burger-Stritt S, Lang K, Milovanovic D, Beuschlein F, Willenberg HS, Quinkler M, Allolio B. High incidence of adrenal crisis in educated patients with chronic adrenal insufficiency: a prospective study. Journal of Clinical Endocrinology and Metabolism 2015. 100 407–416. ( 10.1210/jc.2014-3191) [DOI] [PubMed] [Google Scholar]
4.Bancos I, Hahner S, Tomlinson JW, Arlt W. Diagnosis and management of adrenal insufficiency. Lancet Diabetes & Endocrinology 2015. 3 216–226. [DOI] [PubMed] [Google Scholar]
5.Allolio B. Extensive expertise in endocrinology: adrenal crisis. European Journal of Endocrinology 2015. 172 R115–R124. ( 10.1530/eje-14-0824) [DOI] [PubMed] [Google Scholar]
6.Bornstein S, Allolio B, Arlt W, Barthel A, Don-Wauchope A, Hammer GD, Husebye E, Merke DP, Murad MH, Stratakis CA, et al. Management of primary adrenal insufficiency: an Endocrine Society clinical practice guideline. Journal of Clinical Endocrinology and Metabolism 2016. 101 364–389. ( 10.1210/jc.2015-1710) [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from Endocrine Connections are provided here courtesy of Bioscientifica Ltd.
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Table of Contents
What is “Build an Error Free Model”?
Key Learning Points
Build an Error Free Model – Model Integrity and Error Checking
Best Practices for Modeling to Minimize Error Creation
Error Checking Techniques – An Example
Recommended Product
Home / Free Resources / Financial Modeling / Build an Error Free Model
Build an Error Free Model
By Victoria Collin |
November 26, 2021
What is “Build an Error Free Model”?
A fully-functioning financial model needs to be accurate and error-free. To build an error-free model, construction best practices should be adhered to and integrity checks should be applied throughout the process. These include:
1.Sense checking – is the output reasonable?
2.Structure checking – is formula construction consistent and as expected?
3.Stress testing – does the model react to changes as predicted?
These three checking techniques can be used to audit the constructed model, to ensure that it is accurate and error-free. It can be applied to the balance sheet, income statement, and cash flow statement. Further, there are certain best practices that modelers must adopt in order to effectively tackle the errors in model construction so that the model remains consistent and in good working order.
Key Learning Points
It is highly important that a financial model is rigorously checked for errors
There are a set of checking techniques i.e. integrity checks – sense, structure, and stress testing – that are used to audit a constructed model
In addition to integrity checking, the modeler should adopt certain best practices for modeling to minimize error creation
One of the most time-consuming parts of building a 3 statement model is finding mistakes that cause the balance sheet to be out of balance
Build an Error Free Model – Model Integrity and Error Checking
The three components required to build an error-free model, termed error checking techniques, are sense, structure, and stress testing. This involves asking three questions – does the output seem reasonable, is the formula structure consistent and expected, and does the model react predictably to changes?
Sense checking: this focuses on the model’s output. In order to check the reasonableness of the model results, they must usually be benchmarked. This can be done by trend analysis over time i.e. are values within the expected range? Peer comparison is another way to sense check the outputs – are the results consistent with benchmarks? A good sense check would also be analysis of interrelated metrics – do related values move as expected relative to one another?
Structure checking: a logical and consistent model layout allows timely and efficient error diagnosis. A well-built model will use a consistent column structure that allows for copyable formula construction. This also means the column reference used in formulas will be consistent thereby allowing the user to identify any “off column” references with ease.
Stress testing: this process involves making a deliberate change to the model (typically an assumption such as a growth rate) and checking to establish that it has behaved as predicted. Thought should be given to the change that will give the maximum information to the model builder. It is best to make an extreme change so the output effect will be easily seen. The stress testing process involves changing an assumption and predicting the output change as a result of a change in assumption. If the model behaves as predicted, then the analyst must remember to undo the stress test so the model reverts to its correct status!
Best Practices for Modeling to Minimize Error Creation
To minimize error creation, it is a good idea to take on best practices for modeling. The following should be applied where possible:
Consistency – the more consistent the model, the easier it is to find errors
Use similar formulas and similar construction in the model – e.g. if your formula is ‘Sales + growth rate assumption = forecast sales’ then it is a good idea to keep that formula-style consistent down the P&L
Use consistent data sourcing
Build formulas once only and link thereafter – this can avoid typing errors
Build copyable formulas – this will facilitate a consistent and easy to error check model
Calculate all subtotals once and then copy them
Input historic and hard data once only– avoid duplication and potential error
Check little and oftenas you work through the model – you can often correct an error early on
Build, check, copy– in that order to keep the model consistent
Consistency is the key to minimizing model construction errors. It applies to both data layout and formula construction. The column structure should be consistent throughout. For example forecast year 1 should be in the same column (e.g. Column F) on each sheet used in the workbook. Also, where possible the order of data row by row should be similar. For example, operating current assets should be listed in the same order on the balance sheet as in the operating working capital calculation. The structure of the data in rows and columns is often referred to as the ‘model matrix’.
Another application of the consistency principle is that similar formulas should be constructed in a similar way. The final application of the consistency principles is to link data to the same source where possible. For example, if there is a backup calculation for PP&E, then depreciation on the balance sheet, depreciation, and CAPEX in the cash flow statement can all be sourced from the PP&E BASE analysis.
A fast and efficient modeler will often build formulas only once and then copy them. Rebuilding formulas creates an extra opportunity for a construction error. Further, subtotals can be a notorious source of errors. To minimize this, they should always be calculated and copied. They should never be input or downloaded or linked from another version of the subtotal.
Next, it is important to input historic or hard-coded data only once, and finally, it is helpful to narrow the universe of what is being checked. It is significantly easier to check an income statement rather than an entire model. So best practice is to check little and often, and always check before copying across.
It may be pertinent to be aware that one of the most time-consuming parts of building a 3 statement model is finding mistakes that cause the balance sheet to be out of balance. There are some techniques that should be used to help resolve these issues.
Error Checking Techniques – An Example
In the following example, CAPEX as a percentage (%) of sales has been changed from 3.7% to 50%. PP&E will increase, the business will run out of cash and a revolver drawdown will be necessary. This is a good error check as the increase is significant and should instigate movement across the model. We can see that the model is working as expected.
However, in the next example, given below, something is wrong with cash and revolver. PP&E has increased as expected but cash is negative and the revolver is still zero. The error should be found and fixed.
Finally, any output that is zero or that has been left blank (as an interim measure) should always be stress-tested since sense and structure checking may not work as normal. The dividends payout assumption is changed from 0.0% to 32.0% and the balance sheet is out of balance. The error should be investigated and fixed.
In conclusion, one of the most time-consuming parts of building a 3-statement model is finding mistakes that cause the balance sheet to be out of balance. A combination of sense checks, structure checks, and mathematical checks should be used to ensure that the balance sheet balances.
The balance sheet usually ‘balances’ the entire model should be checked in detail using these three checking techniques. Once this has been completed, stress testing the entire model can be very powerful and will often find subtle errors or those that are not causing an obvious problem.
A final tip is to always stress-test assumptions that are likely to change or which are currently producing a zero amount in the income or cash flow statements, or an amount equal to the historical data in the balance sheet.
Download the accompanying Excel exercise sheets to learn financial modeling and for a full explanation of the correct answer.
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Which term of the AP: @$\begin{align}53,48,43, \ldots\end{align}@$ is the first negative term?
Which term of the AP: @$\begin{align}53,48,43, \ldots\end{align}@$ is the first negative term?
To find the first negative term of the arithmetic progression (AP) @$\begin{align}53, 48, 43, \ldots\end{align}@$, we need to determine the term number @$\begin{align}n\end{align}@$ for which the term is negative. The general formula for the @$\begin{align}n\end{align}@$-th term of an AP is: @$$\begin{align} a_n = a_1 + (n-1) \cdot d \end{align}@$$ where @$\begin{align}a_1\end{align}@$ is the first term and @$\begin{align}d\end{align}@$ is the common difference. 1. Identify the first term (@$\begin{align}a_1\end{align}@$) and the common difference (@$\begin{align}d\end{align}@$): @$\begin{align}a_1 = 53\end{align}@$ @$\begin{align}d = 48 - 53 = -5\end{align}@$ 2. Set up the inequality for the first negative term: @$$\begin{align} a_n < 0 \end{align}@$$ Substitute the formula for @$\begin{align}a_n\end{align}@$: @$$\begin{align} 53 + (n-1)(-5) < 0 \end{align}@$$ 3. Solve the inequality: @$$\begin{align} 53 - 5(n-1) < 0 \end{align}@$$ @$$\begin{align} 53 - 5n + 5 < 0 \end{align}@$$ @$$\begin{align} 58 - 5n < 0 \end{align}@$$ @$$\begin{align} 58 < 5n \end{align}@$$ @$$\begin{align} n > \frac{58}{5} \end{align}@$$ @$$\begin{align} n > 11.6 \end{align}@$$ Since @$\begin{align}n\end{align}@$ must be a whole number, the smallest integer greater than 11.6 is 12. Thus, the first negative term is the 12th term.
To find the first negative term of the arithmetic progression (AP) @$\begin{align}53, 48, 43, \ldots\end{align}@$, we need to determine the term number @$\begin{align}n\end{align}@$ for which the term is negative.
The general formula for the @$\begin{align}n\end{align}@$-th term of an AP is:
@$$\begin{align}
a_n = a_1 + (n-1) \cdot d
\end{align}@$$
where @$\begin{align}a_1\end{align}@$ is the first term and @$\begin{align}d\end{align}@$ is the common difference.
Identify the first term (@$\begin{align}a_1\end{align}@$) and the common difference (@$\begin{align}d\end{align}@$):
Set up the inequality for the first negative term:
@$$\begin{align}
a_n < 0
\end{align}@$$
Substitute the formula for @$\begin{align}a_n\end{align}@$:
@$$\begin{align}
53 + (n-1)(-5) < 0
\end{align}@$$
@$$\begin{align}
53 - 5(n-1) < 0
\end{align}@$$
@$$\begin{align}
53 - 5n + 5 < 0
\end{align}@$$
@$$\begin{align}
58 - 5n < 0
\end{align}@$$
@$$\begin{align}
58 < 5n
\end{align}@$$
@$$\begin{align}
n > \frac{58}{5}
\end{align}@$$
@$$\begin{align}
n > 11.6
\end{align}@$$
Since @$\begin{align}n\end{align}@$ must be a whole number, the smallest integer greater than 11.6 is 12.
Thus, the first negative term is the 12th term.
Try Asking:
Find the sum:
1+(-2)+(-5)+(-8)+\ldots+(-236)
Verify that each of the following is an AP, and then write its next three terms.
a+b,(a+1)+b,(a+1)+(b+1), \ldots
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ffirs.qxd 2/13/12 5:47 PM Page ii E n g i n e e r i n g M e c h a n i c s Dynamics ffirs.qxd 2/13/12 5:47 PM Page i ffirs.qxd 2/13/12 5:47 PM Page ii E n g i n e e r i n g M e c h a n i c s V o l u m e 2 Dynamics Seventh Edition J. L. Meriam L. G. Kraige Virginia Polytechnic Institute and State University John Wiley & Sons, Inc.
ffirs.qxd 2/13/12 5:47 PM Page iii On the Cover: NASA and the European Space Agency are collaborating on the design of future missions which will gather samples of Martian surface material and return them to the earth. This artist’s view shows a spacecraft carrying a sample-retrieving rover and an ascent vehicle as it approaches Mars. The rover would collect previously gath-ered materials and deliver them to the ascent vehicle, which would then rendezvous with another spacecraft already in orbit about Mars. This orbiting spacecraft would then travel to the earth. Such missions are planned for the 2020’s.
Vice President & Executive Publisher Don Fowley Associate Publisher Dan Sayre Executive Editor Linda Ratts Editorial Assistant Christopher Teja Content Manager Kevin Holm Production Editor Jill Spikereit, Production Management Services provided by Camelot Editorial Services, LLC Marketing Manager Christopher Ruel Senior Designer Maureen Eide Cover Design Kristine Carney Cover Photo Courtesy of NASA/JPL-Caltech Electronic Illustrations Precision Graphics Senior Photo Editor Lisa Gee Product Designer Jennifer Welter Content Editor Wendy Ashenberg Media Editor Jennifer Mullin This book was set in 10.5/12 ITC Century Schoolbook by PreMediaGlobal, and printed and bound by RR Donnelley. The cover was printed by RR Donnelley.
This book is printed on acid-free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2012 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com.
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9780470614815 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ffirs.qxd 2/14/12 2:47 PM Page iv This series of textbooks was begun in 1951 by the late Dr. James L. Meriam. At that time, the books represented a revolutionary transformation in undergraduate mechanics education.
They became the definitive textbooks for the decades that followed as well as models for other engineering mechanics texts that have subsequently appeared. Published under slightly differ-ent titles prior to the 1978 First Editions, this textbook series has always been characterized by logical organization, clear and rigorous presentation of the theory, instructive sample prob-lems, and a rich collection of real-life problems, all with a high standard of illustration. In addi-tion to the U.S. versions, the books have appeared in SI versions and have been translated into many foreign languages. These texts collectively represent an international standard for un-dergraduate texts in mechanics.
The innovations and contributions of Dr. Meriam (1917–2000) to the field of engineer-ing mechanics cannot be overstated. He was one of the premier engineering educators of the second half of the twentieth century. Dr. Meriam earned his B.E., M. Eng., and Ph.D.
degrees from Yale University. He had early industrial experience with Pratt and Whitney Aircraft and the General Electric Company. During the Second World War he served in the U.S. Coast Guard. He was a member of the faculty of the University of California–Berkeley, Dean of Engineering at Duke University, a faculty member at the California Polytechnic State University–San Luis Obispo, and visiting professor at the University of California– Santa Barbara, finally retiring in 1990. Professor Meriam always placed great emphasis on teaching, and this trait was recognized by his students wherever he taught. At Berkeley in 1963, he was the first recipient of the Outstanding Faculty Award of Tau Beta Pi, given pri-marily for excellence in teaching. In 1978, he received the Distinguished Educator Award for Outstanding Service to Engineering Mechanics Education from the American Society for Engineering Education, and in 1992 was the Society’s recipient of the Benjamin Garver Lamme Award, which is ASEE’s highest annual national award.
Dr. L. Glenn Kraige, coauthor of the Engineering Mechanics series since the early 1980s, has also made significant contributions to mechanics education. Dr. Kraige earned his B.S., M.S., and Ph.D. degrees at the University of Virginia, principally in aerospace v Foreword ffirs.qxd 2/13/12 5:47 PM Page v engineering, and he currently serves as Professor of Engineering Science and Mechanics at Virginia Polytechnic Institute and State University. During the mid 1970s, I had the singular pleasure of chairing Professor Kraige’s graduate committee and take particular pride in the fact that he was the first of my forty-five Ph.D. graduates. Professor Kraige was invited by Professor Meriam to team with him and thereby ensure that the Meriam legacy of textbook authorship excellence was carried forward to future generations. For the past three decades, this highly successful team of authors has made an enormous and global impact on the educa-tion of several generations of engineers.
In addition to his widely recognized research and publications in the field of spacecraft dynamics, Professor Kraige has devoted his attention to the teaching of mechanics at both introductory and advanced levels. His outstanding teaching has been widely recognized and has earned him teaching awards at the departmental, college, university, state, regional, and national levels. These include the Francis J. Maher Award for excellence in education in the Department of Engineering Science and Mechanics, the Wine Award for excellence in uni-versity teaching, and the Outstanding Educator Award from the State Council of Higher Ed-ucation for the Commonwealth of Virginia. In 1996, the Mechanics Division of ASEE bestowed upon him the Archie Higdon Distinguished Educator Award. The Carnegie Foun-dation for the Advancement of Teaching and the Council for Advancement and Support of Education awarded him the distinction of Virginia Professor of the Year for 1997. During 2004–2006, he held the W. S. “Pete” White Chair for Innovation in Engineering Education, and in 2006 he teamed with Professors Scott L. Hendricks and Don H. Morris as recipients of the XCaliber Award for Teaching with Technology. In his teaching, Professor Kraige stresses the development of analytical capabilities along with the strengthening of physical insight and engineering judgment. Since the early 1980s, he has worked on personal-computer software designed to enhance the teaching/learning process in statics, dynamics, strength of materials, and higher-level areas of dynamics and vibrations.
The Seventh Edition of Engineering Mechanics continues the same high standards set by previous editions and adds new features of help and interest to students. It contains a vast collection of interesting and instructive problems. The faculty and students privileged to teach or study from Professors Meriam and Kraige’s Engineering Mechanics will benefit from the several decades of investment by two highly accomplished educators. Following the pattern of the previous editions, this textbook stresses the application of theory to ac-tual engineering situations, and at this important task it remains the best.
John L. Junkins Distinguished Professor of Aerospace Engineering Holder of the George J. Eppright Chair Professorship in Engineering Texas A&M University College Station, Texas vi Foreword ffirs.qxd 2/13/12 5:47 PM Page vi vii Preface Engineering mechanics is both a foundation and a framework for most of the branches of engineering. Many of the topics in such areas as civil, mechanical, aerospace, and agricul-tural engineering, and of course engineering mechanics itself, are based upon the subjects of statics and dynamics. Even in a discipline such as electrical engineering, practitioners, in the course of considering the electrical components of a robotic device or a manufacturing process, may find themselves first having to deal with the mechanics involved.
Thus, the engineering mechanics sequence is critical to the engineering curriculum.
Not only is this sequence needed in itself, but courses in engineering mechanics also serve to solidify the student’s understanding of other important subjects, including applied math-ematics, physics, and graphics. In addition, these courses serve as excellent settings in which to strengthen problem-solving abilities.
Philosophy The primary purpose of the study of engineering mechanics is to develop the capacity to predict the effects of force and motion while carrying out the creative design functions of engineering. This capacity requires more than a mere knowledge of the physical and mathematical principles of mechanics; also required is the ability to visualize physical config-urations in terms of real materials, actual constraints, and the practical limitations which govern the behavior of machines and structures. One of the primary objectives in a mechan-ics course is to help the student develop this ability to visualize, which is so vital to problem formulation. Indeed, the construction of a meaningful mathematical model is often a more important experience than its solution. Maximum progress is made when the principles and their limitations are learned together within the context of engineering application.
There is a frequent tendency in the presentation of mechanics to use problems mainly as a vehicle to illustrate theory rather than to develop theory for the purpose of solving prob-lems. When the first view is allowed to predominate, problems tend to become overly idealized and unrelated to engineering with the result that the exercise becomes dull, academic, and uninteresting. This approach deprives the student of valuable experience in formulating problems and thus of discovering the need for and meaning of theory. The second view pro-vides by far the stronger motive for learning theory and leads to a better balance between theory and application. The crucial role played by interest and purpose in providing the strongest possible motive for learning cannot be overemphasized.
Furthermore, as mechanics educators, we should stress the understanding that, at best, theory can only approximate the real world of mechanics rather than the view that the real world approximates the theory. This difference in philosophy is indeed basic and distinguishes the engineering of mechanics from the science of mechanics.
Over the past several decades, several unfortunate tendencies have occurred in engineer-ing education. First, emphasis on the geometric and physical meanings of prerequisite mathe-matics appears to have diminished. Second, there has been a significant reduction and even elimination of instruction in graphics, which in the past enhanced the visualization and repre-sentation of mechanics problems. Third, in advancing the mathematical level of our treat-ment of mechanics, there has been a tendency to allow the notational manipulation of vector operations to mask or replace geometric visualization. Mechanics is inherently a subject which depends on geometric and physical perception, and we should increase our efforts to develop this ability.
A special note on the use of computers is in order. The experience of formulating problems, where reason and judgment are developed, is vastly more important for the student than is the manipulative exercise in carrying out the solution. For this reason, computer usage must be carefully controlled. At present, constructing free-body diagrams and formulating governing equations are best done with pencil and paper. On the other hand, there are instances in which the solution to the governing equations can best be carried out and displayed using the com-puter. Computer-oriented problems should be genuine in the sense that there is a condition of design or criticality to be found, rather than “makework” problems in which some parameter is varied for no apparent reason other than to force artificial use of the computer. These thoughts have been kept in mind during the design of the computer-oriented problems in the Seventh Edition. To conserve adequate time for problem formulation, it is suggested that the student be assigned only a limited number of the computer-oriented problems.
As with previous editions, this Seventh Edition of Engineering Mechanics is written with the foregoing philosophy in mind. It is intended primarily for the first engineering course in mechanics, generally taught in the second year of study. Engineering Mechanics is written in a style which is both concise and friendly. The major emphasis is on basic principles and methods rather than on a multitude of special cases. Strong effort has been made to show both the cohesiveness of the relatively few fundamental ideas and the great variety of prob-lems which these few ideas will solve.
Pedagogical Features The basic structure of this textbook consists of an article which rigorously treats the par-ticular subject matter at hand, followed by one or more Sample Problems, followed by a group of Problems. There is a Chapter Review at the end of each chapter which summarizes the main points in that chapter, followed by a Review Problem set.
Problems The 124 sample problems appear on specially colored pages by themselves. The solu-tions to typical dynamics problems are presented in detail. In addition, explanatory and cautionary notes (Helpful Hints) in blue type are number-keyed to the main presentation.
viii Preface There are 1541 homework exercises, of which approximately 45 percent are new to the Seventh Edition. The problem sets are divided into Introductory Problems and Representa-tive Problems. The first section consists of simple, uncomplicated problems designed to help students gain confidence with the new topic, while most of the problems in the second sec-tion are of average difficulty and length. The problems are generally arranged in order of increasing difficulty. More difficult exercises appear near the end of the Representative Problems and are marked with the symbol . Computer-Oriented Problems, marked with an asterisk, appear in a special section at the conclusion of the Review Problems at the end of each chapter. The answers to all problems have been provided in a special section at the end of the textbook.
In recognition of the need for emphasis on SI units, there are approximately two prob-lems in SI units for every one in U.S. customary units. This apportionment between the two sets of units permits anywhere from a 50–50 emphasis to a 100-percent SI treatment.
A notable feature of the Seventh Edition, as with all previous editions, is the wealth of interesting and important problems which apply to engineering design. Whether directly identified as such or not, virtually all of the problems deal with principles and procedures inherent in the design and analysis of engineering structures and mechanical systems.
Illustrations In order to bring the greatest possible degree of realism and clarity to the illustrations, this textbook series continues to be produced in full color. It is important to note that color is used consistently for the identification of certain quantities: • red for forces and moments • green for velocity and acceleration arrows • orange dashes for selected trajectories of moving points Subdued colors are used for those parts of an illustration which are not central to the problem at hand. Whenever possible, mechanisms or objects which commonly have a cer-tain color will be portrayed in that color. All of the fundamental elements of technical illus-tration which have been an essential part of this Engineering Mechanics series of textbooks have been retained. The author wishes to restate the conviction that a high standard of il-lustration is critical to any written work in the field of mechanics.
Special Features While retaining the hallmark features of all previous editions, we have incorporated these improvements: • The main emphasis on the work-energy and impulse-momentum equations is now on the time-order form, both for particles in Chapter 3 and rigid bodies in Chapter 6.
• New emphasis has been placed on three-part impulse-momentum diagrams, both for particles and rigid bodies. These diagrams are well integrated with the time-order form of the impulse-momentum equations.
• Within-the-chapter photographs have been added in order to provide additional connection to actual situations in which dynamics has played a major role.
• Approximately 45 percent of the homework problems are new to this Seventh Edition.
All new problems have been independently solved in order to ensure a high degree of accuracy.
Preface ix • New Sample Problems have been added, including ones with computer-oriented solutions.
• All Sample Problems are printed on specially colored pages for quick identification.
• All theory portions have been reexamined in order to maximize rigor, clarity, readability, and level of friendliness.
• Key Concepts areas within the theory presentation have been specially marked and highlighted.
• The Chapter Reviews are highlighted and feature itemized summaries.
Organization The logical division between particle dynamics (Part I) and rigid-body dynamics (Part II) has been preserved, with each part treating the kinematics prior to the kinetics. This arrangement promotes thorough and rapid progress in rigid-body dynamics with the prior benefit of a comprehensive introduction to particle dynamics.
In Chapter 1, the fundamental concepts necessary for the study of dynamics are established.
Chapter 2 treats the kinematics of particle motion in various coordinate systems, as well as the subjects of relative and constrained motion.
Chapter 3 on particle kinetics focuses on the three basic methods: force-mass-acceleration (Section A), work-energy (Section B), and impulse-momentum (Section C). The special topics of impact, central-force motion, and relative motion are grouped together in a special applications section (Section D) and serve as optional material to be assigned according to instructor preference and available time. With this arrangement, the attention of the stu-dent is focused more strongly on the three basic approaches to kinetics.
Chapter 4 on systems of particles is an extension of the principles of motion for a single particle and develops the general relationships which are so basic to the modern compre-hension of dynamics. This chapter also includes the topics of steady mass flow and variable mass, which may be considered as optional material.
In Chapter 5 on the kinematics of rigid bodies in plane motion, where the equations of relative velocity and relative acceleration are encountered, emphasis is placed jointly on solution by vector geometry and solution by vector algebra. This dual approach serves to reinforce the meaning of vector mathematics.
In Chapter 6 on the kinetics of rigid bodies, we place great emphasis on the basic equations which govern all categories of plane motion. Special emphasis is also placed on forming the direct equivalence between the actual applied forces and couples and their and resultants. In this way the versatility of the moment principle is em-phasized, and the student is encouraged to think directly in terms of resultant dynamics effects.
Chapter 7, which may be treated as optional, provides a basic introduction to three-dimensional dynamics which is sufficient to solve many of the more common space-motion problems. For students who later pursue more advanced work in dynamics, Chapter 7 will provide a solid foundation. Gyroscopic motion with steady precession is treated in two ways.
The first approach makes use of the analogy between the relation of force and linear-momentum vectors and the relation of moment and angular-momentum vectors. With this treatment, the student can understand the gyroscopic phenomenon of steady precession and can handle most of the engineering problems on gyroscopes without a detailed study of three-dimensional dynamics. The second approach employs the more general momentum equations for three-dimensional rotation where all components of momentum are ac-counted for.
I ma x Preface Chapter 8 is devoted to the topic of vibrations. This full-chapter coverage will be espe-cially useful for engineering students whose only exposure to vibrations is acquired in the basic dynamics course.
Moments and products of inertia of mass are presented in Appendix B. Appendix C con-tains a summary review of selected topics of elementary mathematics as well as several nu-merical techniques which the student should be prepared to use in computer-solved problems. Useful tables of physical constants, centroids, and moments of inertia are con-tained in Appendix D.
Supplements The following items have been prepared to complement this textbook: Instructor’s Manual Prepared by the authors and independently checked, fully worked solutions to all odd-numbered problems in the text are available to faculty by contacting their local Wiley representative.
Instructor Lecture Resources The following resources are available online at www.wiley.com/college/meriam. There may be additional resources not listed.
WileyPlus: A complete online learning system to help prepare and present lectures, assign and manage homework, keep track of student progress, and customize your course content and delivery. See the description in front of the book for more information, and the website for a demonstration. Talk to your Wiley representative for details on setting up your Wiley-Plus course.
Lecture software specifically designed to aid the lecturer, especially in larger classrooms.
Written by the author and incorporating figures from the textbooks, this software is based on the Macromedia Flash® platform. Major use of animation, concise review of the theory, and numerous sample problems make this tool extremely useful for student self-review of the material.
All figures in the text are available in electronic format for use in creating lecture presen-tations.
All Sample Problems are available as electronic files for display and discussion in the classroom.
Acknowledgments Special recognition is due Dr. A. L. Hale, formerly of Bell Telephone Laboratories, for his continuing contribution in the form of invaluable suggestions and accurate checking of the manuscript. Dr. Hale has rendered similar service for all previous versions of this entire series of mechanics books, dating back to the 1950s. He reviews all aspects of the books, in-cluding all old and new text and figures. Dr. Hale carries out an independent solution to each new homework exercise and provides the author with suggestions and needed correc-tions to the solutions which appear in the Instructor’s Manual. Dr. Hale is well known for being extremely accurate in his work, and his fine knowledge of the English language is a great asset which aids every user of this textbook. Preface xi I would like to thank the faculty members of the Department of Engineering Science and Mechanics at VPI&SU who regularly offer constructive suggestions. These include Scott L. Hendricks, Saad A. Ragab, Norman E. Dowling, Michael W. Hyer, Michael L. Madi-gan, and J. Wallace Grant. Jeffrey N. Bolton of Bluefield State College is recognized for his significant contributions to this textbook series.
The following individuals (listed in alphabetical order) provided feedback on recent editions, reviewed samples of the Seventh Edition, or otherwise contributed to the Seventh Edition: Michael Ales, U.S. Merchant Marine Academy Joseph Arumala, University of Maryland Eastern Shore Eric Austin, Clemson University Stephen Bechtel, Ohio State University Peter Birkemoe, University of Toronto Achala Chatterjee, San Bernardino Valley College Jim Shih-Jiun Chen, Temple University Yi-chao Chen, University of Houston Mary Cooper, Cal Poly San Luis Obispo Mukaddes Darwish, Texas Tech University Kurt DeGoede, Elizabethtown College John DesJardins, Clemson University Larry DeVries, University of Utah Craig Downing, Southeast Missouri State University William Drake, Missouri State University Raghu Echempati, Kettering University Amelito Enriquez, Canada College Sven Esche, Stevens Institute of Technology Wallace Franklin, U.S. Merchant Marine Academy Christine Goble, University of Kentucky Barry Goodno, Georgia Institute of Technology Robert Harder, George Fox University Javier Hasbun, University of West Georgia Javad Hashemi, Texas Tech University Robert Hyers, University of Massachusetts, Amherst Matthew Ikle, Adams State College Duane Jardine, University of New Orleans Mariappan Jawaharlal, California Polytechnic State University, Pomona Qing Jiang, University of California, Riverside Jennifer Kadlowec, Rowan University Robert Kern, Milwaukee School of Engineering John Krohn, Arkansas Tech University Keith Lindler, United States Naval Academy Francisco Manzo-Robledo, Washington State University Geraldine Milano, New Jersey Institute of Technology Saeed Niku, Cal Poly San Luis Obispo Wilfrid Nixon, University of Iowa Karim Nohra, University of South Florida Vassilis Panoskaltsis, Case Western Reserve University Chandra Putcha, California State University, Fullerton Blayne Roeder, Purdue University Eileen Rossman, Cal Poly San Luis Obispo xii Preface Nestor Sanchez, University of Texas, San Antonio Joseph Schaefer, Iowa State University Scott Schiff, Clemson University Sergey Smirnov, Texas Tech University Ertugrul Taciroglu, UCLA Constantine Tarawneh, University of Texas John Turner, University of Wyoming Chris Venters, Virginia Tech Sarah Vigmostad, University of Iowa T. W. Wu, University of Kentucky Mohammed Zikry, North Carolina State University The contributions by the staff of John Wiley & Sons, Inc., including Executive Editor Linda Ratts, Production Editor Jill Spikereit, Senior Designer Maureen Eide, and Photo-graph Editor Lisa Gee, reflect a high degree of professional competence and are duly recog-nized. I wish to especially acknowledge the critical production efforts of Christine Cervoni of Camelot Editorial Services, LLC. The talented illustrators of Precision Graphics continue to maintain a high standard of illustration excellence.
Finally, I wish to state the extremely significant contribution of my family. In addition to providing patience and support for this project, my wife Dale has managed the prepara-tion of the manuscript for the Seventh Edition and has been a key individual in checking all stages of the proof. In addition, both my daughter Stephanie Kokan and my son David Kraige have contributed problem ideas, illustrations, and solutions to a number of the prob-lems over the past several editions.
I am extremely pleased to participate in extending the time duration of this textbook series well past the sixty-year mark. In the interest of providing you with the best possible educational materials over future years, I encourage and welcome all comments and sugges-tions. Please address your comments to kraige@vt.edu.
Blacksburg, Virginia Preface xiii xiv PART I DYNAMICS OF PARTICLES 1 CHAPTER 1 INTRODUCTION TO DYNAMICS 3 1/1 History and Modern Applications 3 1/2 Basic Concepts 4 1/3 Newton’s Laws 6 1/4 Units 6 1/5 Gravitation 8 1/6 Dimensions 11 1/7 Solving Problems in Dynamics 12 1/8 Chapter Review 15 CHAPTER 2 KINEMATICS OF PARTICLES 21 2/1 Introduction 21 2/2 Rectilinear Motion 22 Contents 2/3 Plane Curvilinear Motion 40 2/4 Rectangular Coordinates (x-y) 43 2/5 Normal and Tangential Coordinates (n-t ) 54 2/6 Polar Coordinates (r-u) 66 2/7 Space Curvilinear Motion 79 2/8 Relative Motion (Translating Axes) 88 2/9 Constrained Motion of Connected Particles 98 2/10 Chapter Review 106 CHAPTER 3 KINETICS OF PARTICLES 117 3/1 Introduction 117 SECTION A FORCE, MASS, AND ACCELERATION 118 3/2 Newton’s Second Law 118 3/3 Equation of Motion and Solution of Problems 122 3/4 Rectilinear Motion 124 3/5 Curvilinear Motion 138 SECTION B WORK AND ENERGY 154 3/6 Work and Kinetic Energy 154 3/7 Potential Energy 175 SECTION C IMPULSE AND MOMENTUM 191 3/8 Introduction 191 3/9 Linear Impulse and Linear Momentum 191 3/10 Angular Impulse and Angular Momentum 205 SECTION D SPECIAL APPLICATIONS 217 3/11 Introduction 217 3/12 Impact 217 3/13 Central-Force Motion 230 3/14 Relative Motion 244 3/15 Chapter Review 255 CHAPTER 4 KINETICS OF SYSTEMS OF PARTICLES 267 4/1 Introduction 267 4/2 Generalized Newton’s Second Law 268 Contents xv 4/3 Work-Energy 269 4/4 Impulse-Momentum 271 4/5 Conservation of Energy and Momentum 275 4/6 Steady Mass Flow 288 4/7 Variable Mass 303 4/8 Chapter Review 315 PART II DYNAMICS OF RIGID BODIES 323 CHAPTER 5 PLANE KINEMATICS OF RIGID BODIES 325 5/1 Introduction 325 5/2 Rotation 327 5/3 Absolute Motion 338 5/4 Relative Velocity 348 5/5 Instantaneous Center of Zero Velocity 362 5/6 Relative Acceleration 372 5/7 Motion Relative to Rotating Axes 385 5/8 Chapter Review 402 CHAPTER 6 PLANE KINETICS OF RIGID BODIES 411 6/1 Introduction 411 SECTION A FORCE, MASS, AND ACCELERATION 413 6/2 General Equations of Motion 413 6/3 Translation 420 6/4 Fixed-Axis Rotation 431 6/5 General Plane Motion 443 SECTION B WORK AND ENERGY 459 6/6 Work-Energy Relations 459 6/7 Acceleration from Work-Energy; Virtual Work 477 SECTION C IMPULSE AND MOMENTUM 486 6/8 Impulse-Momentum Equations 486 6/9 Chapter Review 503 xvi Contents CHAPTER 7 INTRODUCTION TO THREE-DIMENSIONAL DYNAMICS OF RIGID BODIES 513 7/1 Introduction 513 SECTION A KINEMATICS 514 7/2 Translation 514 7/3 Fixed-Axis Rotation 514 7/4 Parallel-Plane Motion 515 7/5 Rotation about a Fixed Point 515 7/6 General Motion 527 SECTION B KINETICS 539 7/7 Angular Momentum 539 7/8 Kinetic Energy 542 7/9 Momentum and Energy Equations of Motion 550 7/10 Parallel-Plane Motion 552 7/11 Gyroscopic Motion: Steady Precession 558 7/12 Chapter Review 576 CHAPTER 8 VIBRATION AND TIME RESPONSE 583 8/1 Introduction 583 8/2 Free Vibration of Particles 584 8/3 Forced Vibration of Particles 600 8/4 Vibration of Rigid Bodies 614 8/5 Energy Methods 624 8/6 Chapter Review 632 APPENDICES APPENDIX A AREA MOMENTS OF INERTIA 639 APPENDIX B MASS MOMENTS OF INERTIA 641 B/1 Mass Moments of Inertia about an Axis 641 B/2 Products of Inertia 660 APPENDIX C SELECTED TOPICS OF MATHEMATICS 671 C/1 Introduction 671 C/2 Plane Geometry 671 Contents xvii C/3 Solid Geometry 672 C/4 Algebra 672 C/5 Analytic Geometry 673 C/6 Trigonometry 673 C/7 Vector Operations 674 C/8 Series 677 C/9 Derivatives 677 C/10 Integrals 678 C/11 Newton’s Method for Solving Intractable Equations 681 C/12 Selected Techniques for Numerical Integration 683 APPENDIX D USEFUL TABLES 687 Table D/1 Physical Properties 687 Table D/2 Solar System Constants 688 Table D/3 Properties of Plane Figures 689 Table D/4 Properties of Homogeneous Solids 691 INDEX 695 PROBLEM ANSWERS 701 xviii Contents PART I Dynamics of Particles c01.qxd 2/8/12 7:02 PM Page 1 This astronaut is anchored to a foot restraint on the International Space Station’s Canadarm2.
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c01.qxd 2/8/12 7:02 PM Page 2 3 1/1 History and Modern Applications Dynamics is that branch of mechanics which deals with the motion of bodies under the action of forces. The study of dynamics in engineer-ing usually follows the study of statics, which deals with the effects of forces on bodies at rest. Dynamics has two distinct parts: kinematics, which is the study of motion without reference to the forces which cause motion, and kinetics, which relates the action of forces on bodies to their resulting motions. A thorough comprehension of dynamics will provide one of the most useful and powerful tools for analysis in engineering.
History of Dynamics Dynamics is a relatively recent subject compared with statics. The beginning of a rational understanding of dynamics is credited to Galileo (1564–1642), who made careful observations concerning bodies in free fall, motion on an inclined plane, and motion of the pendulum. He was largely responsible for bringing a scientific approach to the investigation of physical problems. Galileo was continually under severe criticism for refusing to accept the established beliefs of his day, such as the philoso-phies of Aristotle which held, for example, that heavy bodies fall more rapidly than light bodies. The lack of accurate means for the measure-ment of time was a severe handicap to Galileo, and further significant development in dynamics awaited the invention of the pendulum clock by Huygens in 1657.
Newton (1642–1727), guided by Galileo’s work, was able to make an accurate formulation of the laws of motion and, thus, to place dynamics Galileo Galilei Portrait of Galileo Galilei (1564–1642) (oil on canvas), Sustermans, Justus (1597–1681) (school of)/Galleria Palatina, Florence, Italy/Bridgeman Art Library 1/1 History and Modern Applications 1/2 Basic Concepts 1/3 Newton’s Laws 1/4 Units 1/5 Gravitation 1/6 Dimensions 1/7 Solving Problems in Dynamics 1/8 Chapter Review CHAPTER OUTLINE 1 Introduction to Dynamics © Fine Art Images/SuperStock c01.qxd 2/8/12 7:02 PM Page 3 on a sound basis. Newton’s famous work was published in the first edi-tion of his Principia, which is generally recognized as one of the great-est of all recorded contributions to knowledge. In addition to stating the laws governing the motion of a particle, Newton was the first to cor-rectly formulate the law of universal gravitation. Although his mathe-matical description was accurate, he felt that the concept of remote transmission of gravitational force without a supporting medium was an absurd notion. Following Newton’s time, important contributions to mechanics were made by Euler, D’Alembert, Lagrange, Laplace, Poinsot, Coriolis, Einstein, and others.
Applications of Dynamics Only since machines and structures have operated with high speeds and appreciable accelerations has it been necessary to make calculations based on the principles of dynamics rather than on the principles of statics. The rapid technological developments of the present day require increasing application of the principles of mechanics, particularly dy-namics. These principles are basic to the analysis and design of moving structures, to fixed structures subject to shock loads, to robotic devices, to automatic control systems, to rockets, missiles, and spacecraft, to ground and air transportation vehicles, to electron ballistics of electrical devices, and to machinery of all types such as turbines, pumps, recipro-cating engines, hoists, machine tools, etc.
Students with interests in one or more of these and many other activities will constantly need to apply the fundamental principles of dynamics.
1/2 Basic Concepts The concepts basic to mechanics were set forth in Art. 1/2 of Vol. 1 Statics. They are summarized here along with additional comments of special relevance to the study of dynamics.
Space is the geometric region occupied by bodies. Position in space is determined relative to some geometric reference system by means of linear and angular measurements. The basic frame of reference for the laws of Newtonian mechanics is the primary inertial system or astro-nomical frame of reference, which is an imaginary set of rectangular axes assumed to have no translation or rotation in space. Measurements show that the laws of Newtonian mechanics are valid for this reference system as long as any velocities involved are negligible compared with the speed of light, which is 300 000 km/s or 186,000 mi/sec. Measure-ments made with respect to this reference are said to be absolute, and this reference system may be considered “fixed” in space.
A reference frame attached to the surface of the earth has a some-what complicated motion in the primary system, and a correction to the basic equations of mechanics must be applied for measurements made 4 Chapter 1 Introduction to Dynamics Artificial hand The original formulations of Sir Isaac Newton may be found in the translation of his Principia (1687), revised by F. Cajori, University of California Press, 1934.
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c01.qxd 2/8/12 7:02 PM Page 4 relative to the reference frame of the earth. In the calculation of rocket and space-flight trajectories, for example, the absolute motion of the earth becomes an important parameter. For most engineering problems involving machines and structures which remain on the surface of the earth, the corrections are extremely small and may be neglected. For these problems the laws of mechanics may be applied directly with mea-surements made relative to the earth, and in a practical sense such mea-surements will be considered absolute.
Time is a measure of the succession of events and is considered an absolute quantity in Newtonian mechanics.
Mass is the quantitative measure of the inertia or resistance to change in motion of a body. Mass may also be considered as the quantity of matter in a body as well as the property which gives rise to gravita-tional attraction.
Force is the vector action of one body on another. The properties of forces have been thoroughly treated in Vol. 1 Statics.
A particle is a body of negligible dimensions. When the dimensions of a body are irrelevant to the description of its motion or the action of forces on it, the body may be treated as a particle. An airplane, for ex-ample, may be treated as a particle for the description of its flight path.
A rigid body is a body whose changes in shape are negligible com-pared with the overall dimensions of the body or with the changes in po-sition of the body as a whole. As an example of the assumption of rigidity, the small flexural movement of the wing tip of an airplane fly-ing through turbulent air is clearly of no consequence to the description of the motion of the airplane as a whole along its flight path. For this purpose, then, the treatment of the airplane as a rigid body is an accept-able approximation. On the other hand, if we need to examine the inter-nal stresses in the wing structure due to changing dynamic loads, then the deformation characteristics of the structure would have to be exam-ined, and for this purpose the airplane could no longer be considered a rigid body.
Vector and scalar quantities have been treated extensively in Vol.
1 Statics, and their distinction should be perfectly clear by now. Scalar quantities are printed in lightface italic type, and vectors are shown in boldface type. Thus, V denotes the scalar magnitude of the vector V. It is important that we use an identifying mark, such as an underline V, for all handwritten vectors to take the place of the boldface designation in print. For two nonparallel vectors recall, for example, that V1 V2 and V1 V2 have two entirely different meanings.
We assume that you are familiar with the geometry and algebra of vectors through previous study of statics and mathematics. Students who need to review these topics will find a brief summary of them in Ap-pendix C along with other mathematical relations which find frequent use in mechanics. Experience has shown that the geometry of mechan-ics is often a source of difficulty for students. Mechanics by its very na-ture is geometrical, and students should bear this in mind as they review their mathematics. In addition to vector algebra, dynamics re-quires the use of vector calculus, and the essentials of this topic will be developed in the text as they are needed.
Article 1/2 Basic Concepts 5 c01.qxd 2/8/12 7:02 PM Page 5 Dynamics involves the frequent use of time derivatives of both vec-tors and scalars. As a notational shorthand, a dot over a symbol will fre-quently be used to indicate a derivative with respect to time. Thus, means dx/dt and stands for d2x/dt2.
1/3 Newton’s Laws Newton’s three laws of motion, stated in Art. 1/4 of Vol. 1 Statics, are restated here because of their special significance to dynamics. In modern terminology they are: Law I. A particle remains at rest or continues to move with uniform velocity (in a straight line with a constant speed) if there is no unbal-anced force acting on it.
Law II. The acceleration of a particle is proportional to the resul-tant force acting on it and is in the direction of this force. Law III. The forces of action and reaction between interacting bod-ies are equal in magnitude, opposite in direction, and collinear.
These laws have been verified by countless physical measurements.
The first two laws hold for measurements made in an absolute frame of reference, but are subject to some correction when the motion is mea-sured relative to a reference system having acceleration, such as one at-tached to the surface of the earth.
Newton’s second law forms the basis for most of the analysis in dy-namics. For a particle of mass m subjected to a resultant force F, the law may be stated as (1/1) where a is the resulting acceleration measured in a nonaccelerating frame of reference. Newton’s first law is a consequence of the second law since there is no acceleration when the force is zero, and so the par-ticle is either at rest or is moving with constant velocity. The third law constitutes the principle of action and reaction with which you should be thoroughly familiar from your work in statics.
1/4 Units Both the International System of metric units (SI) and the U.S. cus-tomary system of units are defined and used in Vol. 2 Dynamics, al-though a stronger emphasis is placed on the metric system because it is replacing the U.S. customary system. However, numerical conversion from one system to the other will often be needed in U.S. engineering F ma x ¨ x ˙ 6 Chapter 1 Introduction to Dynamics To some it is preferable to interpret Newton’s second law as meaning that the resultant force acting on a particle is proportional to the time rate of change of momentum of the particle and that this change is in the direction of the force. Both formulations are equally correct when applied to a particle of constant mass.
c01.qxd 2/8/12 7:02 PM Page 6 Article 1/4 Units 7 practice for some years to come. To become familiar with each system, it is necessary to think directly in that system. Familiarity with the new system cannot be achieved simply by the conversion of numerical re-sults from the old system.
Tables defining the SI units and giving numerical conversions be-tween U.S. customary and SI units are included inside the front cover of the book. Charts comparing selected quantities in SI and U.S. custom-ary units are included inside the back cover of the book to facilitate con-version and to help establish a feel for the relative size of units in both systems.
The four fundamental quantities of mechanics, and their units and symbols for the two systems, are summarized in the following table: The U.S. standard kilogram at the National Bureau of Standards Also spelled metre.
As shown in the table, in SI the units for mass, length, and time are taken as base units, and the units for force are derived from Newton’s second law of motion, Eq. 1/1. In the U.S. customary system the units for force, length, and time are base units and the units for mass are de-rived from the second law.
The SI system is termed an absolute system because the standard for the base unit kilogram (a platinum-iridium cylinder kept at the In-ternational Bureau of Standards near Paris, France) is independent of the gravitational attraction of the earth. On the other hand, the U.S.
customary system is termed a gravitational system because the stan-dard for the base unit pound (the weight of a standard mass located at sea level and at a latitude of 45) requires the presence of the gravita-tional field of the earth. This distinction is a fundamental difference be-tween the two systems of units.
In SI units, by definition, one newton is that force which will give a one-kilogram mass an acceleration of one meter per second squared. In the U.S. customary system a 32.1740-pound mass (1 slug) will have an acceleration of one foot per second squared when acted on by a force of one pound. Thus, for each system we have from Eq. 1/1 SI UNITS N kgm/s2 (1 N) (1 kg)(1 m/s2) U.S. CUSTOMARY UNITS slug lb sec2/ft (1 lb) (1 slug)(1 ft/sec2) Omikron/PhotoResearchers, Inc.
c01.qxd 2/8/12 7:02 PM Page 7 In SI units, the kilogram should be used exclusively as a unit of mass and never force. Unfortunately, in the MKS (meter, kilogram, sec-ond) gravitational system, which has been used in some countries for many years, the kilogram has been commonly used both as a unit of force and as a unit of mass.
In U.S. customary units, the pound is unfortunately used both as a unit of force (lbf) and as a unit of mass (lbm). The use of the unit lbm is especially prevalent in the specification of the thermal properties of liq-uids and gases. The lbm is the amount of mass which weighs 1 lbf under standard conditions (at a latitude of 45 and at sea level). In order to avoid the confusion which would be caused by the use of two units for mass (slug and lbm), in this textbook we use almost exclusively the unit slug for mass. This practice makes dynamics much simpler than if the lbm were used. In addition, this approach allows us to use the symbol lb to always mean pound force.
Additional quantities used in mechanics and their equivalent base units will be defined as they are introduced in the chapters which follow.
However, for convenient reference these quantities are listed in one place in the first table inside the front cover of the book.
Professional organizations have established detailed guidelines for the consistent use of SI units, and these guidelines have been followed throughout this book. The most essential ones are summarized inside the front cover, and you should observe these rules carefully.
1/5 Gravitation Newton’s law of gravitation, which governs the mutual attraction between bodies, is (1/2) where F the mutual force of attraction between two particles G a universal constant called the constant of gravitation m1, m2 the masses of the two particles r the distance between the centers of the particles The value of the gravitational constant obtained from experimental data is . Except for some spacecraft applications, the only gravitational force of appreciable magnitude in engineering is the force due to the attraction of the earth. It was shown in Vol. 1 Stat-ics, for example, that each of two iron spheres 100 mm in diameter is at-tracted to the earth with a gravitational force of 37.1 N, which is called its weight, but the force of mutual attraction between them if they are just touching is only 0.000 000 095 1 N.
Because the gravitational attraction or weight of a body is a force, it should always be expressed in force units, newtons (N) in SI units and pounds force (lb) in U.S. customary units. To avoid confusion, the word “weight” in this book will be restricted to mean the force of gravita-tional attraction.
G 6.673(1011) m3/(kg s2) F G m1m2 r2 8 Chapter 1 Introduction to Dynamics c01.qxd 2/8/12 7:02 PM Page 8 Effect of Altitude The force of gravitational attraction of the earth on a body depends on the position of the body relative to the earth. If the earth were a perfect homogeneous sphere, a body with a mass of exactly 1 kg would be attracted to the earth by a force of 9.825 N on the surface of the earth, 9.822 N at an altitude of 1 km, 9.523 N at an altitude of 100 km, 7.340 N at an altitude of 1000 km, and 2.456 N at an altitude equal to the mean radius of the earth, 6371 km. Thus the variation in gravita-tional attraction of high-altitude rockets and spacecraft becomes a major consideration.
Every object which falls in a vacuum at a given height near the sur-face of the earth will have the same acceleration g, regardless of its mass. This result can be obtained by combining Eqs. 1/1 and 1/2 and canceling the term representing the mass of the falling object. This com-bination gives where me is the mass of the earth and R is the radius of the earth. The mass me and the mean radius R of the earth have been found through experimental measurements to be 5.976(1024) kg and 6.371(106) m, re-spectively. These values, together with the value of G already cited, when substituted into the expression for g, give a mean value of g 9.825 m/s2.
The variation of g with altitude is easily determined from the gravi-tational law. If g0 represents the absolute acceleration due to gravity at sea level, the absolute value at an altitude h is where R is the radius of the earth.
Effect of a Rotating Earth The acceleration due to gravity as determined from the gravita-tional law is the acceleration which would be measured from a set of axes whose origin is at the center of the earth but which does not ro-tate with the earth. With respect to these “fixed” axes, then, this value may be termed the absolute value of g. Because the earth rotates, the acceleration of a freely falling body as measured from a position at-tached to the surface of the earth is slightly less than the absolute value.
Accurate values of the gravitational acceleration as measured rela-tive to the surface of the earth account for the fact that the earth is a rotating oblate spheroid with flattening at the poles. These values may g g0 R2 (R h)2 g Gme R2 Article 1/5 Gravitation 9 It can be proved that the earth, when taken as a sphere with a symmetrical distribution of mass about its center, may be considered a particle with its entire mass concentrated at its center.
c01.qxd 2/8/12 7:02 PM Page 9 10 Chapter 1 Introduction to Dynamics You will be able to derive these relations for a spherical earth after studying relative mo-tion in Chapter 3.
Standard Value of g The standard value which has been adopted internationally for the gravitational acceleration relative to the rotating earth at sea level and at a latitude of 45 is 9.806 65 m/s2 or 32.1740 ft/sec2. This value differs very slightly from that obtained by evaluating the International Gravity Formula for 45. The reason for the small difference is that the earth is not exactly ellipsoidal, as assumed in the formulation of the In-ternational Gravity Formula.
The proximity of large land masses and the variations in the density of the crust of the earth also influence the local value of g by a small but detectable amount. In almost all engineering applications near the sur-face of the earth, we can neglect the difference between the absolute and relative values of the gravitational acceleration, and the effect of local Figure 1/1 be calculated to a high degree of accuracy from the 1980 International Gravity Formula, which is where is the latitude and g is expressed in meters per second squared.
The formula is based on an ellipsoidal model of the earth and also ac-counts for the effect of the rotation of the earth.
The absolute acceleration due to gravity as determined for a nonro-tating earth may be computed from the relative values to a close approxi-mation by adding 3.382(102) cos2 m/s2, which removes the effect of the rotation of the earth. The variation of both the absolute and the relative values of g with latitude is shown in Fig. 1/1 for sea-level conditions. g 9.780 327(1 0.005 279 sin2 0.000 023 sin4 …) c01.qxd 2/8/12 7:02 PM Page 10 variations. The values of 9.81 m/s2 in SI units and 32.2 ft/sec2 in U.S.
customary units are used for the sea-level value of g.
Apparent Weight The gravitational attraction of the earth on a body of mass m may be calculated from the results of a simple gravitational experiment. The body is allowed to fall freely in a vacuum, and its absolute acceleration is measured. If the gravitational force of attraction or true weight of the body is W, then, because the body falls with an absolute acceleration g, Eq. 1/1 gives (1/3) The apparent weight of a body as determined by a spring balance, calibrated to read the correct force and attached to the surface of the earth, will be slightly less than its true weight. The difference is due to the rotation of the earth. The ratio of the apparent weight to the appar-ent or relative acceleration due to gravity still gives the correct value of mass. The apparent weight and the relative acceleration due to gravity are, of course, the quantities which are measured in experiments con-ducted on the surface of the earth.
1/6 Dimensions A given dimension such as length can be expressed in a number of different units such as meters, millimeters, or kilometers. Thus, a di-mension is different from a unit. The principle of dimensional homogene-ity states that all physical relations must be dimensionally homogeneous; that is, the dimensions of all terms in an equation must be the same. It is customary to use the symbols L, M, T, and F to stand for length, mass, time, and force, respectively. In SI units force is a derived quantity and from Eq. 1/1 has the dimensions of mass times acceleration or One important use of the dimensional homogeneity principle is to check the dimensional correctness of some derived physical relation. We can derive the following expression for the velocity v of a body of mass m which is moved from rest a horizontal distance x by a force F: where the is a dimensionless coefficient resulting from integration.
This equation is dimensionally correct because substitution of L, M, and T gives Dimensional homogeneity is a necessary condition for correctness of a physical relation, but it is not sufficient, since it is possible to construct [MLT2][L] [M][LT1]2 1 2 Fx 1 2mv2 F ML/T2 W mg Article 1/6 Dimensions 11 c01.qxd 2/8/12 7:02 PM Page 11 an equation which is dimensionally correct but does not represent a cor-rect relation. You should perform a dimensional check on the answer to every problem whose solution is carried out in symbolic form.
1/7 Solving Problems in Dynamics The study of dynamics concerns the understanding and description of the motions of bodies. This description, which is largely mathematical, en-ables predictions of dynamical behavior to be made. A dual thought process is necessary in formulating this description. It is necessary to think in terms of both the physical situation and the corresponding mathematical description. This repeated transition of thought between the physical and the mathematical is required in the analysis of every problem.
One of the greatest difficulties encountered by students is the in-ability to make this transition freely. You should recognize that the mathematical formulation of a physical problem represents an ideal and limiting description, or model, which approximates but never quite matches the actual physical situation.
In Art. 1/8 of Vol. 1 Statics we extensively discussed the approach to solving problems in statics. We assume therefore, that you are familiar with this approach, which we summarize here as applied to dynamics.
Approximation in Mathematical Models Construction of an idealized mathematical model for a given engi-neering problem always requires approximations to be made. Some of these approximations may be mathematical, whereas others will be physical. For instance, it is often necessary to neglect small distances, angles, or forces compared with large distances, angles, or forces. If the change in velocity of a body with time is nearly uniform, then an as-sumption of constant acceleration may be justified. An interval of mo-tion which cannot be easily described in its entirety is often divided into small increments, each of which can be approximated.
As another example, the retarding effect of bearing friction on the motion of a machine may often be neglected if the friction forces are small compared with the other applied forces. However, these same fric-tion forces cannot be neglected if the purpose of the inquiry is to deter-mine the decrease in efficiency of the machine due to the friction process. Thus, the type of assumptions you make depends on what infor-mation is desired and on the accuracy required.
You should be constantly alert to the various assumptions called for in the formulation of real problems. The ability to understand and make use of the appropriate assumptions when formulating and solving engi-neering problems is certainly one of the most important characteristics of a successful engineer.
Along with the development of the principles and analytical tools needed for modern dynamics, one of the major aims of this book is to provide many opportunities to develop the ability to formulate good mathematical models. Strong emphasis is placed on a wide range of practical problems which not only require you to apply theory but also force you to make relevant assumptions.
12 Chapter 1 Introduction to Dynamics c01.qxd 2/8/12 7:02 PM Page 12 Application of Basic Principles The subject of dynamics is based on a surprisingly few fundamental concepts and principles which, however, can be extended and applied over a wide range of conditions. The study of dynamics is valuable partly be-cause it provides experience in reasoning from fundamentals. This experi-ence cannot be obtained merely by memorizing the kinematic and dynamic equations which describe various motions. It must be obtained through ex-posure to a wide variety of problem situations which require the choice, use, and extension of basic principles to meet the given conditions.
In describing the relations between forces and the motions they pro-duce, it is essential to define clearly the system to which a principle is to be applied. At times a single particle or a rigid body is the system to be isolated, whereas at other times two or more bodies taken together con-stitute the system.
Article 1/7 Solving Problems in Dynamics 13 Method of Attack An effective method of attack is essential in the solution of dynam-ics problems, as for all engineering problems. Development of good habits in formulating problems and in representing their solutions will be an invaluable asset. Each solution should proceed with a logical se-quence of steps from hypothesis to conclusion. The following sequence of steps is useful in the construction of problem solutions.
1. Formulate the problem: (a) State the given data.
(b) State the desired result.
(c) State your assumptions and approximations.
2. Develop the solution: (a) Draw any needed diagrams, and include coordinates which are appropriate for the problem at hand.
(b) State the governing principles to be applied to your solution.
(c) Make your calculations.
(d) Ensure that your calculations are consistent with the accuracy justified by the data.
(e) Be sure that you have used consistent units throughout your calculations.
(f) Ensure that your answers are reasonable in terms of magni-tudes, directions, common sense, etc.
(g) Draw conclusions.
The arrangement of your work should be neat and orderly. This will help your thought process and enable others to understand your work.
The discipline of doing orderly work will help you to develop skill in prob-lem formulation and analysis. Problems which seem complicated at first often become clear when you approach them with logic and discipline.
KEY CONCEPTS c01.qxd 2/8/12 7:02 PM Page 13 The definition of the system to be analyzed is made clear by con-structing its free-body diagram. This diagram consists of a closed out-line of the external boundary of the system. All bodies which contact and exert forces on the system but are not a part of it are removed and replaced by vectors representing the forces they exert on the isolated system. In this way, we make a clear distinction between the action and reaction of each force, and all forces on and external to the system are accounted for. We assume that you are familiar with the technique of drawing free-body diagrams from your prior work in statics.
Numerical versus Symbolic Solutions In applying the laws of dynamics, we may use numerical values of the involved quantities, or we may use algebraic symbols and leave the answer as a formula. When numerical values are used, the magnitudes of all quantities expressed in their particular units are evident at each stage of the calculation. This approach is useful when we need to know the magnitude of each term.
The symbolic solution, however, has several advantages over the numerical solution: 1. The use of symbols helps to focus attention on the connection between the physical situation and its related mathematical description.
2. A symbolic solution enables you to make a dimensional check at every step, whereas dimensional homogeneity cannot be checked when only numerical values are used.
3. We can use a symbolic solution repeatedly for obtaining answers to the same problem with different units or different numerical values.
Thus, facility with both forms of solution is essential, and you should practice each in the problem work.
In the case of numerical solutions, we repeat from Vol. 1 Statics our convention for the display of results. All given data are taken to be exact, and results are generally displayed to three significant figures, unless the leading digit is a one, in which case four significant figures are displayed.
Solution Methods Solutions to the various equations of dynamics can be obtained in one of three ways.
1. Obtain a direct mathematical solution by hand calculation, using ei-ther algebraic symbols or numerical values. We can solve the large majority of the problems this way.
2. Obtain graphical solutions for certain problems, such as the deter-mination of velocities and accelerations of rigid bodies in two-dimensional relative motion.
3. Solve the problem by computer. A number of problems in Vol. 2 Dy-namics are designated as Computer-Oriented Problems. They ap-pear at the end of the Review Problem sets and were selected to illustrate the type of problem for which solution by computer offers a distinct advantage.
14 Chapter 1 Introduction to Dynamics c01.qxd 2/8/12 7:02 PM Page 14 The choice of the most expedient method of solution is an important aspect of the experience to be gained from the problem work. We em-phasize, however, that the most important experience in learning me-chanics lies in the formulation of problems, as distinct from their solution per se.
Article 1/8 Chapter Review 15 Virgin Galactic SpaceShip2 in gliding flight after release from its mother-ship WhiteKnight2.
© Mark Greenberg/VirginGalactic/Zuma Press 1/8 CHAPTER REVIEW This chapter has introduced the concepts, definitions, and units used in dynamics, and has given an overview of the approach used to formulate and solve problems in dynamics. Now that you have finished this chapter, you should be able to do the following: 1. State Newton’s laws of motion.
2. Perform calculations using SI and U.S. customary units.
3. Express the law of gravitation and calculate the weight of an object.
4. Discuss the effects of altitude and the rotation of the earth on the acceleration due to gravity.
5. Apply the principle of dimensional homogeneity to a given physical relation.
6. Describe the methodology used to formulate and solve dynamics problems.
c01.qxd 2/8/12 7:02 PM Page 15 16 Chapter 1 Introduction to Dynamics SAMPLE PROBLEM 1/1 A space-shuttle payload module weighs 100 lb when resting on the surface of the earth at a latitude of 45 north.
(a) Determine the mass of the module in both slugs and kilograms, and its surface-level weight in newtons.
(b) Now suppose the module is taken to an altitude of 200 miles above the surface of the earth and released there with no velocity relative to the center of the earth.
Determine its weight under these conditions in both pounds and newtons.
(c) Finally, suppose the module is fixed inside the cargo bay of a space shuttle.
The shuttle is in a circular orbit at an altitude of 200 miles above the surface of the earth. Determine the weight of the module in both pounds and newtons under these conditions.
For the surface-level value of the acceleration of gravity relative to a rotat-ing earth, use g 32.1740 ft/sec2 (9.80665 m/s2). For the absolute value relative to a nonrotating earth, use g 32.234 ft/sec2 (9.825 m/s2). Round off all answers using the rules of this textbook.
Solution.
(a) From relationship 1/3, we have Ans.
Here we have used the acceleration of gravity relative to the rotating earth, be-cause that is the condition of the module in part (a). Note that we are using more significant figures in the acceleration of gravity than will normally be required in this textbook (32.2 ft/sec2 and 9.81 m/s2 will normally suffice).
From the table of conversion factors inside the front cover of the textbook, we see that 1 pound is equal to 4.4482 newtons. Thus, the weight of the module in newtons is Ans.
Finally, its mass in kilograms is Ans.
As another route to the last result, we may convert from pounds mass to kilograms. Again using the table inside the front cover, we have We recall that 1 lbm is the amount of mass which under standard conditions has a weight of 1 lb of force. We rarely refer to the U.S. mass unit lbm in this text-book series, but rather use the slug for mass. The sole use of slug, rather than the unnecessary use of two units for mass, will prove to be powerful and simple.
m 100 lbm 0.45359 kg 1 lbm 45.4 kg [W mg] m W g 445 N 9.80665 m/s2 45.4 kg W 100 lb 4.4482 N 1 lb 445 N [W mg] m W g 100 lb 32.1740 ft/sec2 3.11 slugs Helpful Hints Our calculator indicates a result of 3.108099 slugs. Using the rules of significant figure display used in this textbook, we round the written result to three significant figures, or 3.11 slugs. Had the numerical result begun with the digit 1, we would have rounded the displayed answer to four significant figures.
A good practice with unit conversion is to multiply by a factor such as , which has a value of 1, because the numerator and the de-nominator are equivalent. Be sure that cancellation of the units leaves the units desired—here the units of lb cancel, leaving the desired units of N.
Note that we are using a previously calculated result (445 N). We must be sure that when a calculated num-ber is needed in subsequent calcula-tions, it is obtained in the calculator to its full accuracy (444.82 ). If necessary, numbers must be stored in a calculator storage register and then brought out of the register when needed. We must not merely punch 445 into our calculator and proceed to divide by 9.80665—this practice will result in loss of numeri-cal accuracy. Some individuals like to place a small indication of the storage register used in the right margin of the work paper, directly beside the number stored.
4.4482 N 1 lb c01.qxd 2/8/12 7:02 PM Page 16 Article 1/8 Chapter Review 17 SAMPLE PROBLEM 1/1 (CONTINUED) (b) We begin by calculating the absolute acceleration of gravity (relative to the nonrotating earth) at an altitude of 200 miles.
The weight at an altitude of 200 miles is then Ans.
We now convert Wh to units of newtons.
Ans.
As an alternative solution to part (b), we may use Newton’s universal law of gravitation. In U.S. units, which agrees with our earlier result. We note that the weight of the module when at an altitude of 200 mi is about 90% of its surface-level weight—it is not weightless. We will study the effects of this weight on the motion of the module in Chapter 3.
(c) The weight of an object (the force of gravitational attraction) does not depend on the motion of the object. Thus the answers for part (c) are the same as those in part (b).
Ans.
This Sample Problem has served to eliminate certain commonly held and persistent misconceptions. First, just because a body is raised to a typical shuttle altitude, it does not become weightless. This is true whether the body is released with no velocity relative to the center of the earth, is inside the orbiting shuttle, or is in its own arbitrary trajectory. And second, the acceleration of gravity is not zero at such altitudes. The only way to reduce both the acceleration of gravity and the corresponding weight of a body to zero is to take the body to an infinite distance from the earth.
Wh 90.8 lb or 404 N 90.8 lb Wh Gmem (R h)2 [3.439(108)][4.095(1023)][3.11] [(3959 200)(5280)]2 F Gm1m2 r2 Wh 90.8 lb 4.4482 N 1 lb 404 N Wh mgh 3.11(29.2) 90.8 lb g g0 R2 (R h)2 gh 32.234 39592 (3959 200)2 29.2 ft/sec2 c01.qxd 2/8/12 7:02 PM Page 17 18 Chapter 1 Introduction to Dynamics 1/6 Two uniform aluminum spheres are positioned as shown. Determine the gravitational force which sphere A exerts on sphere B. The value of R is 50 mm.
Problem 1/6 1/7 At what altitude h above the north pole is the weight of an object reduced to one-half of its earth-surface value? Assume a spherical earth of radius R and ex-press h in terms of R.
1/8 Determine the absolute weight and the weight rela-tive to the rotating earth of a 90-kg man if he is stand-ing on the surface of the earth at a latitude of 40°.
1/9 A space shuttle is in a circular orbit at an altitude of 150 mi. Calculate the absolute value of g at this alti-tude and determine the corresponding weight of a shuttle passenger who weighs 200 lb when standing on the surface of the earth at a latitude of 45°. Are the terms “zero-g” and “weightless,” which are sometimes used to describe conditions aboard orbit-ing spacecraft, correct in the absolute sense?
1/10 Determine the angle at which a particle in Jupiter’s circular orbit experiences equal attractions from the sun and from Jupiter. Use Table of Appendix D as needed.
Problem 1/10 Not to scale Sun Jupiter m θ D/2 8R B R 2R A x y 30° PROBLEMS (Refer to Table D/2 in Appendix D for relevant solar-system values.) 1/1 Determine your mass in slugs. Convert your weight to newtons and calculate the corresponding mass in kilograms.
1/2 Determine the weight in newtons of a car which has a mass of 1500 kg. Convert the given mass of the car to slugs and calculate the corresponding weight in pounds.
Problem 1/2 1/3 The weight of one dozen apples is 5 lb. Determine the average mass of one apple in both SI and U.S. units and the average weight of one apple in SI units. In the present case, how applicable is the “rule of thumb” that an average apple weighs 1 N?
1/4 For the given vectors and , determine , , , , and . Consider the vectors to be nondimensional.
Problem 1/4 1/5 The two 100-mm-diameter spheres constructed of dif-ferent metals are located in deep space. Determine the gravitational force F which the copper sphere exerts on the titanium sphere if (a) , and (b) Problem 1/5 d x Copper Titanium d 4 m.
d 2 m 30° V2 = 15 V1 = 12 y x 4 3 V1 V2 V1 V2 V1 V2 V1 V2 V1 V2 V2 V1 m = 1500 kg c01.qxd 2/8/12 7:02 PM Page 18 Article 1/8 Problems 19 1/11 Calculate the distance d from the center of the earth at which a particle experiences equal attractions from the earth and from the moon. The particle is restricted to the line through the centers of the earth and the moon. Justify the two solutions physically.
Refer to Table D/2 of Appendix D as needed.
Problem 1/11 1/12 Consider a woman standing on the earth with the sun directly overhead. Determine the ratio of the force which the earth exerts on the woman to the force which the sun exerts on her. Neglect the effects of the rotation and oblateness of the earth.
1/13 Consider a woman standing on the surface of the earth when the moon is directly overhead. Deter-mine the ratio of the force which the earth ex-erts on the woman to the force which the moon exerts on her. Neglect the effects of the rotation and oblateness of the earth. Find the same ratio if we now move the woman to a corresponding position on the moon.
Rem Res d Moon Not to scale Earth 1/14 Determine the ratio of the force exerted by the sun on the moon to that exerted by the earth on the moon for position A of the moon. Repeat for moon position B.
Problem 1/14 1/15 Check the following equation for dimensional homo-geneity: where m is mass, v is velocity, F is force, is an angle, and t is time.
mv t2 t1 (F cos ) dt RA c01.qxd 2/8/12 7:02 PM Page 19 Even if this car maintains a constant speed along the winding road, it accelerates laterally, and this acceleration must be considered in the design of the car, its tires, and the roadway itself.
© Daniel DempsterPhotography/Alamy c02.qxd 2/8/12 7:10 PM Page 20 21 2/1 Introduction Kinematics is the branch of dynamics which describes the motion of bodies without reference to the forces which either cause the motion or are generated as a result of the motion. Kinematics is often described as the “geometry of motion.” Some engineering applications of kinematics include the design of cams, gears, linkages, and other machine elements to control or produce certain desired motions, and the calculation of flight trajectories for aircraft, rockets, and spacecraft. A thorough work-ing knowledge of kinematics is a prerequisite to kinetics, which is the study of the relationships between motion and the corresponding forces which cause or accompany the motion.
Particle Motion We begin our study of kinematics by first discussing in this chapter the motions of points or particles. A particle is a body whose physical di-mensions are so small compared with the radius of curvature of its path that we may treat the motion of the particle as that of a point. For ex-ample, the wingspan of a jet transport flying between Los Angeles and New York is of no consequence compared with the radius of curvature of 2/1 Introduction 2/2 Rectilinear Motion 2/3 Plane Curvilinear Motion 2/4 Rectangular Coordinates (x-y) 2/5 Normal and Tangential Coordinates (n-t) 2/6 Polar Coordinates (r-) 2/7 Space Curvilinear Motion 2/8 Relative Motion (Translating Axes) 2/9 Constrained Motion of Connected Particles 2/10 Chapter Review CHAPTER OUTLINE 2 Kinematics of Particles c02.qxd 2/8/12 7:10 PM Page 21 its flight path, and thus the treatment of the airplane as a particle or point is an acceptable approximation.
We can describe the motion of a particle in a number of ways, and the choice of the most convenient or appropriate way depends a great deal on experience and on how the data are given. Let us obtain an overview of the several methods developed in this chapter by referring to Fig. 2/1, which shows a particle P moving along some general path in space. If the particle is confined to a specified path, as with a bead sliding along a fixed wire, its motion is said to be constrained. If there are no physical guides, the motion is said to be unconstrained. A small rock tied to the end of a string and whirled in a circle undergoes con-strained motion until the string breaks, after which instant its motion is unconstrained.
Choice of Coordinates The position of particle P at any time t can be described by specify-ing its rectangular coordinates x, y, z, its cylindrical coordinates r, , z, or its spherical coordinates R, , . The motion of P can also be de-scribed by measurements along the tangent t and normal n to the curve.
The direction of n lies in the local plane of the curve.† These last two measurements are called path variables.
The motion of particles (or rigid bodies) can be described by using co-ordinates measured from fixed reference axes (absolute-motion analysis) or by using coordinates measured from moving reference axes (relative-motion analysis). Both descriptions will be developed and applied in the articles which follow.
With this conceptual picture of the description of particle motion in mind, we restrict our attention in the first part of this chapter to the case of plane motion where all movement occurs in or can be repre-sented as occurring in a single plane. A large proportion of the motions of machines and structures in engineering can be represented as plane motion. Later, in Chapter 7, an introduction to three-dimensional mo-tion is presented. We begin our discussion of plane motion with recti-linear motion, which is motion along a straight line, and follow it with a description of motion along a plane curve.
2/2 Rectilinear Motion Consider a particle P moving along a straight line, Fig. 2/2. The po-sition of P at any instant of time t can be specified by its distance s mea-sured from some convenient reference point O fixed on the line. At time t t the particle has moved to P and its coordinate becomes s s.
The change in the position coordinate during the interval t is called the displacement s of the particle. The displacement would be negative if the particle moved in the negative s-direction.
22 Chapter 2 Kinematics of Particles B P R Path x y r y x z n t z φ θ A Figure 2/1 O s – s + s P s P′ Δ Figure 2/2 Often called Cartesian coordinates, named after René Descartes (1596–1650), a French mathematician who was one of the inventors of analytic geometry.
†This plane is called the osculating plane, which comes from the Latin word osculari mean-ing “to kiss.” The plane which contains P and the two points A and B, one on either side of P, becomes the osculating plane as the distances between the points approach zero.
c02.qxd 2/8/12 7:10 PM Page 22 Velocity and Acceleration The average velocity of the particle during the interval t is the dis-placement divided by the time interval or vav s/t. As t becomes smaller and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the particle, which is v or (2/1) Thus, the velocity is the time rate of change of the position coordinate s.
The velocity is positive or negative depending on whether the corre-sponding displacement is positive or negative.
The average acceleration of the particle during the interval t is the change in its velocity divided by the time interval or aav v/t. As t becomes smaller and approaches zero in the limit, the average accelera-tion approaches the instantaneous acceleration of the particle, which is a or (2/2) The acceleration is positive or negative depending on whether the ve-locity is increasing or decreasing. Note that the acceleration would be positive if the particle had a negative velocity which was becoming less negative. If the particle is slowing down, the particle is said to be decelerating.
Velocity and acceleration are actually vector quantities, as we will see for curvilinear motion beginning with Art. 2/3. For rectilinear mo-tion in the present article, where the direction of the motion is that of the given straight-line path, the sense of the vector along the path is de-scribed by a plus or minus sign. In our treatment of curvilinear motion, we will account for the changes in direction of the velocity and accelera-tion vectors as well as their changes in magnitude.
By eliminating the time dt between Eq. 2/1 and the first of Eqs. 2/2, we obtain a differential equation relating displacement, velocity, and ac-celeration. This equation is (2/3) Equations 2/1, 2/2, and 2/3 are the differential equations for the rec-tilinear motion of a particle. Problems in rectilinear motion involving fi-nite changes in the motion variables are solved by integration of these basic differential relations. The position coordinate s, the velocity v, and the acceleration a are algebraic quantities, so that their signs, positive or negative, must be carefully observed. Note that the positive direc-tions for v and a are the same as the positive direction for s.
v dv a ds or s ˙ ds ˙ s ¨ ds a dv dt v ˙ or a d2s dt2 s ¨ lim tl0 v t v ds dt s ˙ lim tl0 s t This sprinter will undergo rectilinear acceleration until he reaches his ter-minal speed.
Article 2/2 Rectilinear Motion 23 Differential quantities can be multiplied and divided in exactly the same way as other algebraic quantities.
© Datacraft/Age FotostockAmerica, Inc.
c02.qxd 2/8/12 7:10 PM Page 23 Graphical Interpretations Interpretation of the differential equations governing rectilinear motion is considerably clarified by representing the relationships among s, v, a, and t graphically. Figure 2/3a is a schematic plot of the variation of s with t from time t1 to time t2 for some given rectilinear motion. By constructing the tangent to the curve at any time t, we obtain the slope, which is the velocity v ds/dt. Thus, the velocity can be determined at all points on the curve and plotted against the corresponding time as shown in Fig. 2/3b. Similarly, the slope dv/dt of the v-t curve at any in-stant gives the acceleration at that instant, and the a-t curve can there-fore be plotted as in Fig. 2/3c.
We now see from Fig. 2/3b that the area under the v-t curve during time dt is v dt, which from Eq. 2/1 is the displacement ds. Consequently, the net displacement of the particle during the interval from t1 to t2 is the corresponding area under the curve, which is Similarly, from Fig. 2/3c we see that the area under the a-t curve during time dt is a dt, which, from the first of Eqs. 2/2, is dv. Thus, the net change in velocity between t1 and t2 is the corresponding area under the curve, which is Note two additional graphical relations. When the acceleration a is plotted as a function of the position coordinate s, Fig. 2/4a, the area under the curve during a displacement ds is a ds, which, from Eq. 2/3, is v dv d(v2/2). Thus, the net area under the curve between position co-ordinates s1 and s2 is When the velocity v is plotted as a function of the position coordinate s, Fig. 2/4b, the slope of the curve at any point A is dv/ds. By constructing the normal AB to the curve at this point, we see from the similar trian-gles that dv/ds. Thus, from Eq. 2/3, v(dv/ds) a, the accel-eration. It is necessary that the velocity and position coordinate axes have the same numerical scales so that the acceleration read on the po-sition coordinate scale in meters (or feet), say, will represent the actual acceleration in meters (or feet) per second squared.
The graphical representations described are useful not only in visu-alizing the relationships among the several motion quantities but also in obtaining approximate results by graphical integration or differentia-tion. The latter case occurs when a lack of knowledge of the mathemati-cal relationship prevents its expression as an explicit mathematical function which can be integrated or differentiated. Experimental data and motions which involve discontinuous relationships between the variables are frequently analyzed graphically.
CB CB/v v2 v1 v dv s2 s1 a ds or 1 2 (v2 2 v1 2) (area under a-s curve) v2 v1 dv t2 t1 a dt or v2 v1 (area under a-t curve) s2 s1 ds t2 t1 v dt or s2 s1 (area under v-t curve) 24 Chapter 2 Kinematics of Particles v = = s · ds — dt a = = v · dv — dt 1 1 t t1 dt t2 t1 t2 t1 t2 t t v v a s a (c) (b) (a) t dt Figure 2/3 dv — ds 1 B C s1 s2 s1 s2 v a a A a s s v (b) (a) ds Figure 2/4 c02.qxd 2/8/12 7:10 PM Page 24 Article 2/2 Rectilinear Motion 25 Analytical Integration If the position coordinate s is known for all values of the time t, then successive mathematical or graphical differentiation with respect to t gives the velocity v and acceleration a. In many problems, however, the functional relationship between position coordinate and time is un-known, and we must determine it by successive integration from the ac-celeration. Acceleration is determined by the forces which act on moving bodies and is computed from the equations of kinetics discussed in subse-quent chapters. Depending on the nature of the forces, the acceleration may be specified as a function of time, velocity, or position coordinate, or as a combined function of these quantities. The procedure for integrating the differential equation in each case is indicated as follows.
(a) Constant Acceleration. When a is constant, the first of Eqs. 2/2 and 2/3 can be integrated directly. For simplicity with s s0, v v0, and t 0 designated at the beginning of the interval, then for a time inter-val t the integrated equations become Substitution of the integrated expression for v into Eq. 2/1 and integra-tion with respect to t give These relations are necessarily restricted to the special case where the acceleration is constant. The integration limits depend on the initial and final conditions, which for a given problem may be different from those used here. It may be more convenient, for instance, to begin the integra-tion at some specified time t1 rather than at time t 0.
(b) Acceleration Given as a Function of Time, a ƒ(t). Substitu-tion of the function into the first of Eqs. 2/2 gives ƒ(t) dv/dt. Multiply-ing by dt separates the variables and permits integration. Thus, v v0 dv t 0 ƒ(t) dt or v v0 t 0 ƒ(t) dt Caution: The foregoing equations have been integrated for constant acceleration only. A common mistake is to use these equations for problems involving variable ac-celeration, where they do not apply.
s s0 ds t 0 (v0 at) dt or s s0 v0 t 1 2 at2 v v0 v dv a s s0 ds or v2 v0 2 2a(s s0) v v0 dv a t 0 dt or v v0 at KEY CONCEPTS c02.qxd 2/8/12 7:10 PM Page 25 From this integrated expression for v as a function of t, the position co-ordinate s is obtained by integrating Eq. 2/1, which, in form, would be If the indefinite integral is employed, the end conditions are used to es-tablish the constants of integration. The results are identical with those obtained by using the definite integral.
If desired, the displacement s can be obtained by a direct solution of the second-order differential equation ƒ(t) obtained by substitution of ƒ(t) into the second of Eqs. 2/2.
(c) Acceleration Given as a Function of Velocity, a ƒ(v). Substi-tution of the function into the first of Eqs. 2/2 gives ƒ(v) dv/dt, which permits separating the variables and integrating. Thus, This result gives t as a function of v. Then it would be necessary to solve for v as a function of t so that Eq. 2/1 can be integrated to obtain the po-sition coordinate s as a function of t.
Another approach is to substitute the function a ƒ(v) into the first of Eqs. 2/3, giving v dv ƒ(v) ds. The variables can now be separated and the equation integrated in the form Note that this equation gives s in terms of v without explicit reference to t.
(d) Acceleration Given as a Function of Displacement, a ƒ(s).
Substituting the function into Eq. 2/3 and integrating give the form Next we solve for v to give v g(s), a function of s. Now we can substi-tute ds/dt for v, separate variables, and integrate in the form which gives t as a function of s. Finally, we can rearrange to obtain s as a function of t.
In each of the foregoing cases when the acceleration varies according to some functional relationship, the possibility of solving the equations by direct mathematical integration will depend on the form of the function.
In cases where the integration is excessively awkward or difficult, integra-tion by graphical, numerical, or computer methods can be utilized.
s s0 ds g(s) t 0 dt or t s s0 ds g(s) v v0 v dv s s0 ƒ(s) ds or v2 v0 2 2 s s0 ƒ(s) ds v v0 v dv ƒ(v) s s0 ds or s s0 v v0 v dv ƒ(v) t t 0 dt v v0 dv ƒ(v) s ¨ s s0 ds t 0 v dt or s s0 t 0 v dt 26 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:10 PM Page 26 SAMPLE PROBLEM 2/1 The position coordinate of a particle which is confined to move along a straight line is given by s 2t3 24t 6, where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t 0, (b) the ac-celeration of the particle when v 30 m/s, and (c) the net displacement of the particle during the interval from t 1 s to t 4 s.
Solution.
The velocity and acceleration are obtained by successive differentia-tion of s with respect to the time. Thus, (a) Substituting v 72 m/s into the expression for v gives us 72 6t2 24, from which t 4 s. The negative root describes a mathematical solution for t before the initiation of motion, so this root is of no physical interest. Thus, the desired result is Ans.
(b) Substituting v 30 m/s into the expression for v gives 30 6t2 24, from which the positive root is t 3 s, and the corresponding acceleration is Ans.
(c) The net displacement during the specified interval is Ans.
which represents the net advancement of the particle along the s-axis from the position it occupied at t 1 s to its position at t 4 s.
To help visualize the motion, the values of s, v, and a are plotted against the time t as shown. Because the area under the v-t curve represents displacement, we see that the net displacement from t 1 s to t 4 s is the positive area s24 less the negative area s12. 54 m s [2(43) 24(4) 6] [2(13) 24(1) 6] s s4 s1 or a 12(3) 36 m/s2 t 4 s a 12t m/s2 [a v ˙] v 6t2 24 m/s [v s ˙] Article 2/2 Rectilinear Motion 27 Helpful Hints Be alert to the proper choice of sign when taking a square root. When the situation calls for only one an-swer, the positive root is not always the one you may need.
Note carefully the distinction be-tween italic s for the position coordi-nate and the vertical s for seconds.
Note from the graphs that the val-ues for v are the slopes of the s-t curve and that the values for a are the slopes of the v-t curve. Sug-gestion: Integrate v dt for each of the two intervals and check the answer for s. Show that the total distance traveled during the interval t 1 s to t 4 s is 74 m.
(v ˙) (s ˙) s, m t, s t, s t, s a, m/s2 v, m/s 48 30 38 s2 – 4 72 36 3 4 2 1 0 0 3 4 2 1 0 0 6 –26 3 4 2 1 0 0 –24 Δ s1 – 2 Δ c02.qxd 2/8/12 7:10 PM Page 27 SAMPLE PROBLEM 2/2 A particle moves along the x-axis with an initial velocity vx 50 ft/sec at the origin when t 0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force which gives it a constant acceleration ax 10 ft/sec2. Calculate the velocity and the x-coordinate of the particle for the condi-tions of t 8 sec and t 12 sec and find the maximum positive x-coordinate reached by the particle.
Solution.
The velocity of the particle after t 4 sec is computed from and is plotted as shown. At the specified times, the velocities are Ans.
The x-coordinate of the particle at any time greater than 4 seconds is the dis-tance traveled during the first 4 seconds plus the distance traveled after the dis-continuity in acceleration occurred. Thus, For the two specified times, Ans.
The x-coordinate for t 12 sec is less than that for t 8 sec since the motion is in the negative x-direction after t 9 sec. The maximum positive x-coordinate is, then, the value of x for t 9 sec which is Ans.
These displacements are seen to be the net positive areas under the v-t graph up to the values of t in question.
xmax 5(92) 90(9) 80 325 ft t 12 sec, x 5(122) 90(12) 80 280 ft t 8 sec, x 5(82) 90(8) 80 320 ft ds v dt x 50(4) t 4 (90 10t) dt 5t2 90t 80 ft t 12 sec, vx 90 10(12) 30 ft/sec t 8 sec, vx 90 10(8) 10 ft/sec dv a dt vx 50 dvx 10 t 4 dt vx 90 10t ft/sec 28 Chapter 2 Kinematics of Particles Helpful Hints Learn to be flexible with symbols.
The position coordinate x is just as valid as s.
Note that we integrate to a general time t and then substitute specific values.
Show that the total distance traveled by the particle in the 12 sec is 370 ft.
1 vx, ft/sec t, sec –10 50 00 4 8 12 –30 c02.qxd 2/8/12 7:10 PM Page 28 SAMPLE PROBLEM 2/3 The spring-mounted slider moves in the horizontal guide with negligible friction and has a velocity v0 in the s-direction as it crosses the mid-position where s 0 and t 0. The two springs together exert a retarding force to the motion of the slider, which gives it an acceleration proportional to the displace-ment but oppositely directed and equal to a k2s, where k is constant. (The constant is arbitrarily squared for later convenience in the form of the expres-sions.) Determine the expressions for the displacement s and velocity v as func-tions of the time t.
Solution I.
Since the acceleration is specified in terms of the displacement, the differential relation v dv a ds may be integrated. Thus, When s 0, v v0, so that C1 and the velocity becomes The plus sign of the radical is taken when v is positive (in the plus s-direction).
This last expression may be integrated by substituting v ds/dt. Thus, With the requirement of t 0 when s 0, the constant of integration becomes C2 0, and we may solve the equation for s so that Ans.
The velocity is v which gives Ans.
Solution II.
Since a the given relation may be written at once as This is an ordinary linear differential equation of second order for which the so-lution is well known and is where A, B, and K are constants. Substitution of this expression into the differ-ential equation shows that it satisfies the equation, provided that K k. The ve-locity is v which becomes The initial condition v v0 when t 0 requires that A v0/k, and the condition s 0 when t 0 gives B 0. Thus, the solution is Ans.
s v0 k sin kt and v v0 cos kt v Ak cos kt Bk sin kt s ˙, s A sin Kt B cos Kt s ¨ k2s 0 s ¨, v v0 cos kt s ˙, s v0 k sin kt ds v0 2 k2s2 dt C2 a constant, or 1 k sin1 ks v0 t C2 v v0 2 k2s2 v0 2/2, v dv k2s ds C1 a constant, or v2 2 k2s2 2 C1 Article 2/2 Rectilinear Motion 29 s This motion is called simple har-monic motion and is characteristic of all oscillations where the restoring force, and hence the acceleration, is proportional to the displacement but opposite in sign.
Again try the definite integral here as above.
Helpful Hints We have used an indefinite integral here and evaluated the constant of integration. For practice, obtain the same results by using the definite integral with the appropriate limits.
c02.qxd 2/8/12 7:10 PM Page 29 SAMPLE PROBLEM 2/4 A freighter is moving at a speed of 8 knots when its engines are suddenly stopped. If it takes 10 minutes for the freighter to reduce its speed to 4 knots, de-termine and plot the distance s in nautical miles moved by the ship and its speed v in knots as functions of the time t during this interval. The deceleration of the ship is proportional to the square of its speed, so that a kv2.
Solution.
The speeds and the time are given, so we may substitute the expres-sion for acceleration directly into the basic definition a dv/dt and integrate.
Thus, Now we substitute the end limits of v 4 knots and t hour and get Ans.
The speed is plotted against the time as shown.
The distance is obtained by substituting the expression for v into the defi-nition v ds/dt and integrating. Thus, Ans.
The distance s is also plotted against the time as shown, and we see that the ship has moved through a distance s ln ln 2 0.924 mi (nautical) dur-ing the 10 minutes.
4 3 (1 6 6) 4 3 8 1 6t ds dt t 0 8 dt 1 6t s 0 ds s 4 3 ln (1 6t) 4 8 1 8k(1/6) k 3 4 mi1 v 8 1 6t 1 6 10 60 1 v 1 8 kt v 8 1 8kt kv2 dv dt dv v2 k dt v 8 dv v2 k t 0 dt 30 Chapter 2 Kinematics of Particles 0 0 2 4 6 8 2 t, min v, knots 4 6 8 10 0 1.0 0.8 0.6 0.4 0.2 0 2 t, min s, mi (nautical) 4 6 8 10 Helpful Hints Recall that one knot is the speed of one nautical mile (6076 ft) per hour.
Work directly in the units of nauti-cal miles and hours.
We choose to integrate to a general value of v and its corresponding time t so that we may obtain the variation of v with t.
c02.qxd 2/8/12 7:10 PM Page 30 PROBLEMS Introductory Problems Problems 2/1 through 2/6 treat the motion of a particle which moves along the s-axis shown in the figure.
Problems 2/1–2/6 2/1 The velocity of a particle is given by 50, where v is in meters per second and t is in seconds.
Plot the velocity v and acceleration a versus time for the first 6 seconds of motion and evaluate the velocity when a is zero.
2/2 The displacement of a particle is given by where s is in feet and t is in seconds.
Plot the displacement, velocity, and acceleration as functions of time for the first 12 seconds of motion.
Determine the time at which the velocity is zero.
2/3 The velocity of a particle which moves along the s-axis is given by where t is in seconds and v is in meters per second. Evaluate the displacement s, velocity v, and acceleration a when The parti-cle is at the origin when 2/4 The velocity of a particle along the s-axis is given by where s is in millimeters and v is in millime-ters per second. Determine the acceleration when s is 2 millimeters.
2/5 The position of a particle in millimeters is given by where t is in seconds. Plot the s-t and v-t relationships for the first 9 seconds. Deter-mine the net displacement during that interval and the total distance D traveled. By inspection of the s-t relationship, what conclusion can you reach re-garding the acceleration?
2/6 The velocity of a particle which moves along the s-axis is given by where t is in seconds.
Calculate the displacement of the particle during the interval from to 2/7 Calculate the constant acceleration a in g’s which the catapult of an aircraft carrier must provide to produce a launch velocity of 180 mi/hr in a distance of 300 ft. Assume that the carrier is at anchor.
2/8 A particle moves along a straight line with a velocity in millimeters per second given by , where t is in seconds. Calculate the net displacement and total distance D traveled during the first 6 seconds of motion.
s v 400 16t2 t 4 s.
t 2 s s 40 3t2 m/s, s ˙ s s 27 12t t2, v 5s3/2, t 0.
s 0 t 4 s.
v 2 5t3/2, 30t2 100t 50, s 2t3 v 20t2 100t –1 0 1 2 3 + s, ft or m Article 2/2 Problems 31 2/9 The acceleration of a particle is given by where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at is and the initial velocity is 2/10 During a braking test, a car is brought to rest begin-ning from an initial speed of 60 mi/hr in a distance of 120 ft. With the same constant deceleration, what would be the stopping distance s from an initial speed of 80 mi/hr?
2/11 Ball 1 is launched with an initial vertical velocity Three seconds later, ball 2 is launched with an initial vertical velocity Deter-mine if the balls are to collide at an altitude of 300 ft. At the instant of collision, is ball 1 ascending or descending?
Problem 2/11 2/12 A projectile is fired vertically with an initial velocity of 200 m/s. Calculate the maximum altitude h reached by the projectile and the time t after firing for it to return to the ground. Neglect air resistance and take the gravitational acceleration to be con-stant at 2/13 A ball is thrown vertically upward with an initial speed of 80 ft/sec from the base A of a 50-ft cliff.
Determine the distance h by which the ball clears the top of the cliff and the time t after release for the ball to land at B. Also, calculate the impact velocity Neglect air resistance and the small horizontal motion of the ball.
Problem 2/13 B h 50′ A v0 vB.
9.81 m/s2.
v1, v2 1 2 v2 v2.
v1 160 ft/sec.
v0 3 m/s.
s0 5 m, t 0 a 4t 30, c02.qxd 2/8/12 7:10 PM Page 31 2/17 The car is traveling at a constant speed km/h on the level portion of the road. When the 6-percent ( ) incline is encountered, the driver does not change the throttle setting and con-sequently the car decelerates at the constant rate Determine the speed of the car (a) 10 sec-onds after passing point A and (b) when Problem 2/16 Representative Problems 2/18 In traveling a distance of 3 km between points A and D, a car is driven at 100 km/h from A to B for t sec-onds and 60 km/h from C to D also for t seconds. If the brakes are applied for 4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Problem 2/18 2/19 During an 8-second interval, the velocity of a particle moving in a straight line varies with time as shown.
Within reasonable limits of accuracy, determine the amount by which the acceleration at exceeds the average acceleration during the interval. What is the displacement during the interval?
Problem 2/19 14 12 10 8 6 4 2 0 0 2 4 t, s v, m/s 6 8 s t 4 s a A B C D s 100 km/h 60 km/h 3 km θ A s v0 s 100 m.
g sin .
tan 6/100 v0 100 32 Chapter 2 Kinematics of Particles 2/14 In the pinewood-derby event shown, the car is re-leased from rest at the starting position A and then rolls down the incline and on to the finish line C.
If the constant acceleration down the incline is and the speed from B to C is essentially con-stant, determine the time duration for the race.
The effects of the small transition area at B can be neglected.
Problem 2/14 2/15 Starting from rest at home plate, a baseball player runs to first base (90 ft away). He uniformly acceler-ates over the first 10 ft to his maximum speed, which is then maintained until he crosses first base. If the overall run is completed in 4 seconds, determine his maximum speed, the acceleration over the first 10 feet, and the time duration of the acceleration.
Problem 2/15 2/16 The graph shows the displacement-time history for the rectilinear motion of a particle during an 8-second interval. Determine the average velocity during the interval and, to within reasonable limits of accu-racy, find the instantaneous velocity v when Problem 2/16 10 8 6 4 2 0 0 2 4 t, s s, m 6 8 t 4 s.
vav 10′ 80′ t = 0 t = 4 sec 20° A B C 12′ 10′ tAC 9 ft/sec2 c02.qxd 2/8/12 7:10 PM Page 32 2/20 A particle moves along the positive x-axis with an acceleration in meters per second squared which increases linearly with x expressed in millimeters, as shown on the graph for an interval of its motion. If the velocity of the particle at is determine the velocity at Problem 2/20 2/21 A girl rolls a ball up an incline and allows it to re-turn to her. For the angle and ball involved, the acceleration of the ball along the incline is con-stant at 0.25g, directed down the incline. If the ball is released with a speed of 4 m/s, determine the distance s it moves up the incline before re-versing its direction and the total time t required for the ball to return to the child’s hand.
Problem 2/21 θ s ax, m/s2 0 4 2 x, mm 40 120 x 120 mm.
0.4 m/s, x 40 mm ax Article 2/2 Problems 33 2/22 A train which is traveling at 80 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is ob-served to be 60 mi/hr as it passes a point 1/2 mi be-yond A. A car moving at 50 mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver “steps on the gas.” Calculate the constant ac-celeration a that the car must have in order to beat the train to the crossing by 4 sec and find the veloc-ity v of the car as it reaches the crossing.
Problem 2/22 2/23 Car A is traveling at a constant speed at a location where the speed limit is 100 km/h. The police officer in car P observes this speed via radar.
At the moment when A passes P, the police car be-gins to accelerate at the constant rate of until a speed of 160 km/h is achieved, and that speed is then maintained. Determine the distance required for the police officer to overtake car A. Neglect any nonrectilinear motion of P.
Problem 2/23 2/24 Repeat the previous problem, only now the driver of car A is traveling at as it passes P, but over the next 5 seconds, the car uniformly decel-erates to the speed limit of 100 km/h, and after that the speed limit is maintained. If the motion of the police car P remains as described in the previous problem, determine the distance required for the police officer to overtake car A.
vA 130 km/h A vA P 6 m/s2 vA 130 km/h B A 80 mi/hr 1 mi Train Car 50 mi/hr 1.3 mi c02.qxd 2/8/12 7:10 PM Page 33 Problem 2/28 2/29 A particle starts from rest at and moves along the x-axis with the velocity history shown. Plot the corresponding acceleration and the displacement histories for the 2 seconds. Find the time t when the particle crosses the origin.
Problem 2/29 2/30 A retarding force is applied to a body moving in a straight line so that, during an interval of its mo-tion, its speed v decreases with increased position co-ordinate s according to the relation , where k is a constant. If the body has a forward speed of 2 in./sec and its position coordinate is 9 in. at time , determine the speed v at sec.
t 3 t 0 v2 k/s v, m/s 0 3 t, s 0.5 1.0 1.5 –1 2.0 0 x 2 m 0 20 40 60 80 100 120 140 160 0 2 4 6 8 10 t, sec v, ft/sec 34 Chapter 2 Kinematics of Particles 2/25 Repeat Prob. 2/23, only now the driver of car A sees and reacts very unwisely to the police car P. Car A is traveling at as it passes P, but over the next 5 seconds, the car uniformly accelerates to 150 km/h, after which that speed is maintained. If the motion of the police car P remains as described in Prob. 2/23, determine the distance required for the police officer to overtake car A.
2/26 The 14-in. spring is compressed to an 8-in. length, where it is released from rest and accelerates block A.
The acceleration has an initial value of and then decreases linearly with the x-movement of the block, reaching zero when the spring regains its original 14-in. length. Calculate the time t for the block to go (a) 3 in. and (b) 6 in.
Problem 2/26 2/27 A single-stage rocket is launched vertically from rest, and its thrust is programmed to give the rocket a constant upward acceleration of . If the fuel is exhausted 20 s after launch, calculate the maxi-mum velocity and the subsequent maximum alti-tude h reached by the rocket.
2/28 An electric car is subjected to acceleration tests along a straight and level test track. The resulting v-t data are closely modeled over the first 10 seconds by the function , where t is the time in seconds and v is the velocity in feet per second.
Determine the displacement s as a function of time over the interval sec and specify its value at time sec.
t 10 0 t 10 v 24t t2 5t vm 6 m/s2 A x 8″ 14″ 400 ft/sec2 vA 130 km/h c02.qxd 2/8/12 7:10 PM Page 34 2/31 The deceleration of the mass center G of a car dur-ing a crash test is measured by an accelerometer with the results shown, where the distance x moved by G after impact is 0.8 m. Obtain a close approximation to the impact velocity v from the data given.
Problem 2/31 2/32 A sprinter reaches his maximum speed in 2.5 seconds from rest with constant acceleration. He then maintains that speed and finishes the 100 yards in the overall time of 9.60 seconds. Determine his maximum speed Problem 2/32 100 yd t = 0 t = 2.5 sec t = 9.60 sec vmax.
vmax 0 10g 8g 6g 4g 2g 0.2 0.4 x, m Deceleration 0.6 0.8 x G Article 2/2 Problems 35 2/33 If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from 20 m/s to a value approaching zero at de-termine the acceleration a of the particle when and show that the particle never reaches the 30-m displacement.
Problem 2/33 2/34 A car starts from rest with an acceleration of which decreases linearly with time to zero in 10 sec-onds, after which the car continues at a constant speed. Determine the time t required for the car to travel 400 m from the start.
2/35 Packages enter the 10-ft chute at A with a speed of 4 ft/sec and have a 0.3g acceleration from A to B. If the packages come to rest at C, calculate the constant acceleration a of the packages from B to C. Also find the time required for the packages to go from A to C.
Problem 2/35 10′ 12′ B C A 6 m/s2 0 0 20 30 v , m/s s, m s 15 m s 30 m, c02.qxd 2/8/12 7:10 PM Page 35 2/39 The body falling with speed strikes and maintains contact with the platform supported by a nest of springs. The acceleration of the body after impact is where c is a positive constant and y is measured from the original platform position. If the maximum compression of the springs is observed to be , determine the constant c.
Problem 2/39 2/40 Particle 1 is subjected to an acceleration particle 2 is subjected to and particle 3 is subjected to . All three particles start at the origin with an initial velocity at time and the magnitude of k is 0.1 for all three particles (note that the units of k vary from case to case). Plot the position, velocity, and acceleration ver-sus time for each particle over the range 2/41 The steel ball A of diameter D slides freely on the horizontal rod which leads to the pole face of the electromagnet. The force of attraction obeys an inverse-square law, and the resulting acceleration of the ball is , where K is a measure of the strength of the magnetic field. If the ball is released from rest at determine the velocity v with which it strikes the pole face.
Problem 2/41 2/42 A certain lake is proposed as a landing area for large jet aircraft. The touchdown speed of 100 mi/hr upon contact with the water is to be reduced to 20 mi/hr in a distance of 1500 ft. If the deceleration is propor-tional to the square of the velocity of the aircraft through the water, , find the value of the design parameter K, which would be a measure of the size and shape of the landing gear vanes that plow through the water. Also find the time t elapsed during the specified interval.
a Kv2 L A D B x x 0, a K/(L x)2 0 t 10 s.
t 0, v0 10 m/s s 0 a ks a kt, a kv, v0 y ym a g cy, v0 36 Chapter 2 Kinematics of Particles 2/36 In an archery test, the acceleration of the arrow de-creases linearly with distance s from its initial value of at A upon release to zero at B after a travel of 24 in. Calculate the maximum velocity v of the arrow.
Problem 2/36 2/37 The 230,000-lb space-shuttle orbiter touches down at about 220 mi/hr. At 200 mi/hr its drag parachute deploys. At 35 mi/hr, the chute is jettisoned from the orbiter. If the deceleration in feet per second squared during the time that the chute is deployed is (speed v in feet per second), determine the corresponding distance traveled by the orbiter. Assume no braking from its wheel brakes.
Problem 2/37 2/38 Reconsider the rollout of the space-shuttle orbiter of the previous problem. The drag chute is deployed at 200 mi/hr, the wheel brakes are applied at 100 mi/hr until wheelstop, and the drag chute is jettisoned at 35 mi/hr. If the drag chute results in a deceleration of (in feet per second squared when the speed v is in feet per second) and the wheel brakes cause a constant deceleration of , determine the distance traveled from 200 mi/hr to wheelstop.
5 ft/sec2 0.0003v2 0.0003v2 A B s 24″ 16,000 ft/sec2 c02.qxd 2/8/12 7:10 PM Page 36 2/43 The electronic throttle control of a model train is pro-grammed so that the train speed varies with position as shown in the plot. Determine the time t required for the train to complete one lap.
Problem 2/43 2/44 A particle moving along the s-axis has a velocity given by where t is in seconds.
When the position of the particle is given by For the first 5 seconds of motion, deter-mine the total distance D traveled, the net displace-ment , and the value of s at the end of the interval.
2/45 The cone falling with a speed strikes and pene-trates the block of packing material. The accelera-tion of the cone after impact is where c is a positive constant and y is the penetration dis-tance. If the maximum penetration depth is ob-served to be determine the constant c.
Problem 2/45 y v0 ym, a g cy2, v0 s s0 3 ft.
t 0, v 18 2t2 ft/sec, 1 m 2 m 1 m s 2 + π _ 2 2 +3π __ 2 2 + π 4 + π 4 + 2π 2 0 0 0.125 Speed v, m/s 0.250 Distance s, m Article 2/2 Problems 37 2/46 The acceleration of the piston in a small recipro-cating engine is given in the following table in terms of the position x of the piston measured from the top of its stroke. From a plot of the data, determine to within two-significant-figure accuracy the maximum velocity reached by the piston.
x, m x, m 0 4950 0.075 0.0075 4340 0.090 0.015 3740 0.105 0.030 2580 0.120 0.045 1490 0.135 0.060 476 0.150 Problem 2/46 2/47 The aerodynamic resistance to motion of a car is nearly proportional to the square of its velocity. Ad-ditional frictional resistance is constant, so that the acceleration of the car when coasting may be written where and are constants which depend on the mechanical configuration of the car. If the car has an initial velocity when the engine is disengaged, derive an expression for the distance D required for the car to coast to a stop.
Problem 2/47 v v0 C2 C1 a C1 C2v2, x 3150 2910 2510 1960 1265 450 ax, m/s2 ax, m/s2 vmax ax c02.qxd 2/8/12 7:10 PM Page 37 2/51 A projectile is fired horizontally into a resisting medium with a velocity and the resulting decel-eration is equal to , where c and n are constants and v is the velocity within the medium. Find the expression for the velocity v of the projectile in terms of the time t of penetration.
2/52 The horizontal motion of the plunger and shaft is ar-rested by the resistance of the attached disk which moves through the oil bath. If the velocity of the plunger is in the position A where and , and if the deceleration is proportional to v so that , derive expressions for the velocity v and position coordinate x in terms of the time t. Also ex-press v in terms of x.
Problem 2/52 2/53 On its takeoff roll, the airplane starts from rest and accelerates according to where is the constant acceleration resulting from the engine thrust and is the acceleration due to aerody-namic drag. If , and v is in meters per second, determine the design length of runway required for the airplane to reach the takeoff speed of 250 km/h if the drag term is (a) ex-cluded and (b) included.
Problem 2/53 s v0 = 0 v = 250 km/h a0 2 m/s2, k 0.00004 m1 kv2 a0 a a0 kv2, Oil v x A a kv t 0 x 0 v0 cvn v0, 38 Chapter 2 Kinematics of Particles 2/48 A subway train travels between two of its station stops with the acceleration schedule shown. Deter-mine the time interval during which the train brakes to a stop with a deceleration of and find the distance s between stations.
Problem 2/48 2/49 Compute the impact speed of a body released from rest at an altitude (a) Assume a constant gravitational acceleration and (b) account for the variation of g with altitude (refer to Art. 1/5). Neglect the effects of atmospheric drag.
Problem 2/49 2/50 Compute the impact speed of body A which is re-leased from rest at an altitude mi above the surface of the moon. (a) First assume a constant gravitational acceleration and (b) then account for the variation of with altitude (refer to Art. 1/5).
Problem 2/50 2160 mi h A gm gm0 5.32 ft/sec2 h 750 R h gm0 32.2 ft/sec2 h 500 mi.
2 –2 0 1 t, s a, m/s2 8 6 10 Δt 2 m/s2 t c02.qxd 2/8/12 7:10 PM Page 38 2/54 A test projectile is fired horizontally into a viscous liquid with a velocity . The retarding force is pro-portional to the square of the velocity, so that the acceleration becomes . Derive expressions for the distance D traveled in the liquid and the cor-responding time t required to reduce the velocity to Neglect any vertical motion.
Problem 2/54 2/55 A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 40 ft/sec as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the com-pression exceeds 6 in. as shown on the graph. Deter-mine the maximum compression x of the outer springs.
Problem 2/55 2/56 When the effect of aerodynamic drag is included, the y-acceleration of a baseball moving vertically upward is , while the acceleration when the ball is moving downward is , where k is a positive constant and v is the speed in feet per sec-ond. If the ball is thrown upward at 100 ft/sec from essentially ground level, compute its maximum height h and its speed upon impact with the ground. Take k to be and assume that g is constant.
0.002 ft1 v ƒ ad g kv2 au g kv2 x, in.
Deceleration ft/sec2 40 ft/sec 0 6 12 3000 2000 1000 0 x v v0 v0/2.
a kv2 v0 Article 2/2 Problems 39 Problem 2/56 2/57 The vertical acceleration of a certain solid-fuel rocket is given by where k, b, and c are constants, v is the vertical velocity acquired, and g is the gravitational acceleration, essentially constant for atmospheric flight. The exponential term represents the effect of a decaying thrust as fuel is burned, and the term approximates the retardation due to atmospheric resistance. Determine the expression for the vertical velocity of the rocket t seconds after firing.
2/58 The preliminary design for a rapid-transit system calls for the train velocity to vary with time as shown in the plot as the train runs the two miles be-tween stations A and B. The slopes of the cubic tran-sition curves (which are of form ) are zero at the end points. Determine the total run time t between the stations and the maximum accel-eration.
a bt ct2 dt3 cv a kebt cv g, h ad = –g + kv2 au = –g – kv2 y 100 ft/sec Problem 2/58 2 mi A A B B v, mi/hr 80 0 15 15 Δt Cubic functions t, sec c02.qxd 2/8/12 7:11 PM Page 39 2/3 Plane Curvilinear Motion We now treat the motion of a particle along a curved path which lies in a single plane. This motion is a special case of the more general three-dimensional motion introduced in Art. 2/1 and illustrated in Fig. 2/1. If we let the plane of motion be the x-y plane, for instance, then the coordinates z and of Fig. 2/1 are both zero, and R becomes the same as r. As men-tioned previously, the vast majority of the motions of points or particles encountered in engineering practice can be represented as plane motion.
Before pursuing the description of plane curvilinear motion in any specific set of coordinates, we will first use vector analysis to describe the motion, since the results will be independent of any particular coor-dinate system. What follows in this article constitutes one of the most basic concepts in dynamics, namely, the time derivative of a vector.
Much analysis in dynamics utilizes the time rates of change of vector quantities. You are therefore well advised to master this topic at the outset because you will have frequent occasion to use it.
Consider now the continuous motion of a particle along a plane curve as represented in Fig. 2/5. At time t the particle is at position A, which is located by the position vector r measured from some convenient fixed ori-gin O. If both the magnitude and direction of r are known at time t, then the position of the particle is completely specified. At time t t, the particle is at A, located by the position vector r r. We note, of course, that this combination is vector addition and not scalar addition. The dis-placement of the particle during time t is the vector r which represents the vector change of position and is clearly independent of the choice of origin. If an origin were chosen at some different location, the position vector r would be changed, but r would be unchanged. The distance actually traveled by the particle as it moves along the path from A to A is the scalar length s measured along the path. Thus, we distinguish between the vector displacement r and the scalar distance s.
Velocity The average velocity of the particle between A and A is defined as vav r/t, which is a vector whose direction is that of r and whose magnitude is the magnitude of r divided by t. The average speed of 40 Chapter 2 Kinematics of Particles r + Δr Δr Δs Δv v v a r O A A Path of particle A v′ v′ A′ A′ Figure 2/5 c02.qxd 2/8/12 7:11 PM Page 40 the particle between A and A is the scalar quotient s/t. Clearly, the magnitude of the average velocity and the speed approach one another as the interval t decreases and A and A become closer together.
The instantaneous velocity v of the particle is defined as the limiting value of the average velocity as the time interval approaches zero. Thus, We observe that the direction of r approaches that of the tangent to the path as t approaches zero and, thus, the velocity v is always a vec-tor tangent to the path.
We now extend the basic definition of the derivative of a scalar quantity to include a vector quantity and write (2/4) The derivative of a vector is itself a vector having both a magnitude and a direction. The magnitude of v is called the speed and is the scalar At this point we make a careful distinction between the magnitude of the derivative and the derivative of the magnitude. The magnitude of the derivative can be written in any one of the several ways dr/dt v v and represents the magnitude of the velocity, or the speed, of the particle. On the other hand, the derivative of the magni-tude is written dr/dt dr/dt , and represents the rate at which the length of the position vector r is changing. Thus, these two derivatives have two entirely different meanings, and we must be extremely careful to distinguish between them in our thinking and in our notation. For this and other reasons, you are urged to adopt a consistent notation for handwritten work for all vector quantities to distinguish them from scalar quantities. For simplicity the underline v is recommended. Other handwritten symbols such as , , and are sometimes used.
With the concept of velocity as a vector established, we return to Fig.
2/5 and denote the velocity of the particle at A by the tangent vector v and the velocity at A by the tangent v. Clearly, there is a vector change in the velocity during the time t. The velocity v at A plus (vectorially) the change v must equal the velocity at A, so we can write v v v. In-spection of the vector diagram shows that v depends both on the change in magnitude (length) of v and on the change in direction of v. These two changes are fundamental characteristics of the derivative of a vector.
Acceleration The average acceleration of the particle between A and A is defined as v/t, which is a vector whose direction is that of v. The magnitude of this average acceleration is the magnitude of v divided by t.
v ˆ v ~ v 9 r ˙ s ˙ r ˙ v v ds dt s ˙ v dr dt r ˙ v lim tl0 r t Article 2/3 Plane Curvilinear Motion 41 c02.qxd 2/8/12 7:11 PM Page 41 The instantaneous acceleration a of the particle is defined as the limiting value of the average acceleration as the time interval ap-proaches zero. Thus, By definition of the derivative, then, we write (2/5) As the interval t becomes smaller and approaches zero, the direction of the change v approaches that of the differential change dv and, thus, of a. The acceleration a, then, includes the effects of both the change in magnitude of v and the change of direction of v. It is apparent, in gen-eral, that the direction of the acceleration of a particle in curvilinear motion is neither tangent to the path nor normal to the path. We do ob-serve, however, that the acceleration component which is normal to the path points toward the center of curvature of the path.
Visualization of Motion A further approach to the visualization of acceleration is shown in Fig. 2/6, where the position vectors to three arbitrary positions on the path of the particle are shown for illustrative purpose. There is a velocity vector tangent to the path corresponding to each position vector, and the relation is v . If these velocity vectors are now plotted from some ar-bitrary point C, a curve, called the hodograph, is formed. The derivatives of these velocity vectors will be the acceleration vectors a which are tangent to the hodograph. We see that the acceleration has the same re-lation to the velocity as the velocity has to the position vector.
The geometric portrayal of the derivatives of the position vector r and velocity vector v in Fig. 2/5 can be used to describe the derivative of any vector quantity with respect to t or with respect to any other scalar variable. Now that we have used the definitions of velocity and accelera-tion to introduce the concept of the derivative of a vector, it is important to establish the rules for differentiating vector quantities. These rules v ˙ r ˙ a dv dt v ˙ a lim tl0 v t 42 Chapter 2 Kinematics of Particles r3 v3 v2 v1 v2 = r · 2 v3 = r · 3 a3 = v · 3 a2 = v · 2 a1 = v · 1 v1 = r · 1 r2 r1 O C Path Hodograph Figure 2/6 c02.qxd 2/8/12 7:11 PM Page 42 are the same as for the differentiation of scalar quantities, except for the case of the cross product where the order of the terms must be pre-served. These rules are covered in Art. C/7 of Appendix C and should be reviewed at this point.
Three different coordinate systems are commonly used for describing the vector relationships for curvilinear motion of a particle in a plane: rec-tangular coordinates, normal and tangential coordinates, and polar coor-dinates. An important lesson to be learned from the study of these coordinate systems is the proper choice of a reference system for a given problem. This choice is usually revealed by the manner in which the mo-tion is generated or by the form in which the data are specified. Each of the three coordinate systems will now be developed and illustrated.
2/4 Rectangular Coordinates (x-y) This system of coordinates is particularly useful for describing mo-tions where the x- and y-components of acceleration are independently generated or determined. The resulting curvilinear motion is then ob-tained by a vector combination of the x- and y-components of the posi-tion vector, the velocity, and the acceleration.
Vector Representation The particle path of Fig. 2/5 is shown again in Fig. 2/7 along with x- and y-axes. The position vector r, the velocity v, and the acceleration a of the particle as developed in Art. 2/3 are represented in Fig. 2/7 to-gether with their x- and y-components. With the aid of the unit vectors i and j, we can write the vectors r, v, and a in terms of their x- and y-components. Thus, (2/6) As we differentiate with respect to time, we observe that the time deriv-atives of the unit vectors are zero because their magnitudes and direc-tions remain constant. The scalar values of the components of v and a are merely vx , vy and ax , ay . (As drawn in Fig. 2/7, ax is in the negative x-direction, so that would be a negative number.) As observed previously, the direction of the velocity is always tan-gent to the path, and from the figure it is clear that If the angle is measured counterclockwise from the x-axis to v for the configuration of axes shown, then we can also observe that dy/dx tan vy/vx.
a2 ax 2 ay 2 a ax 2 ay 2 v2 vx 2 vy 2 v vx 2 vy 2 tan vy vx x ¨ v ˙y y ¨ x ¨ v ˙x y ˙ x ˙ a v ˙ r ¨ x ¨i y ¨j v r ˙ x ˙i y ˙j r xi yj Article 2/4 Rectangular Coordinates (x-y) 43 Figure 2/7 Path j i xi yj r A x y A θ v a vy vx ax ay c02.qxd 2/8/12 7:11 PM Page 43 If the coordinates x and y are known independently as functions of time, x ƒ1(t) and y ƒ2(t), then for any value of the time we can com-bine them to obtain r. Similarly, we combine their first derivatives and to obtain v and their second derivatives and to obtain a. On the other hand, if the acceleration components ax and ay are given as functions of the time, we can integrate each one separately with re-spect to time, once to obtain vx and vy and again to obtain x ƒ1(t) and y ƒ2(t). Elimination of the time t between these last two parametric equations gives the equation of the curved path y ƒ(x).
From the foregoing discussion we can see that the rectangular-coordinate representation of curvilinear motion is merely the superposi-tion of the components of two simultaneous rectilinear motions in the x- and y-directions. Therefore, everything covered in Art. 2/2 on rectilin-ear motion can be applied separately to the x-motion and to the y-motion.
Projectile Motion An important application of two-dimensional kinematic theory is the problem of projectile motion. For a first treatment of the subject, we neglect aerodynamic drag and the curvature and rotation of the earth, and we assume that the altitude change is small enough so that the acceleration due to gravity can be considered constant. With these assumptions, rectangular coordinates are useful for the trajectory analysis.
For the axes shown in Fig. 2/8, the acceleration components are Integration of these accelerations follows the results obtained previ-ously in Art. 2/2a for constant acceleration and yields In all these expressions, the subscript zero denotes initial conditions, frequently taken as those at launch where, for the case illustrated, vy 2 (vy)0 2 2g(y y0) x x0 (vx)0 t y y0 (vy)0 t 1 2 gt2 vx (vx)0 vy (vy)0 gt ax 0 ay g y ¨ x ¨ y ˙ x ˙ 44 Chapter 2 Kinematics of Particles θ v0 vy vx v vy vx v g x y (vx)0 = v0 cos θ (vy)0 = v0 sin θ Figure 2/8 c02.qxd 2/8/12 7:11 PM Page 44 x0 y0 0. Note that the quantity g is taken to be positive throughout this text.
We can see that the x- and y-motions are independent for the simple projectile conditions under consideration. Elimination of the time t be-tween the x- and y-displacement equations shows the path to be parabolic (see Sample Problem 2/6). If we were to introduce a drag force which de-pends on the speed squared (for example), then the x- and y-motions would be coupled (interdependent), and the trajectory would be nonparabolic.
When the projectile motion involves large velocities and high alti-tudes, to obtain accurate results we must account for the shape of the projectile, the variation of g with altitude, the variation of the air den-sity with altitude, and the rotation of the earth. These factors introduce considerable complexity into the motion equations, and numerical inte-gration of the acceleration equations is usually necessary.
Article 2/4 Rectangular Coordinates (x-y) 45 This stroboscopic photograph of a bouncing ping-pong ball suggests not only the parabolic nature of the path, but also the fact that the speed is lower near the apex.
Andrew Davidhazy c02.qxd 2/8/12 7:11 PM Page 45 SAMPLE PROBLEM 2/5 The curvilinear motion of a particle is defined by vx 50 16t and y 100 4t2, where vx is in meters per second, y is in meters, and t is in seconds.
It is also known that x 0 when t 0. Plot the path of the particle and deter-mine its velocity and acceleration when the position y 0 is reached.
Solution.
The x-coordinate is obtained by integrating the expression for vx, and the x-component of the acceleration is obtained by differentiating vx. Thus, The y-components of velocity and acceleration are We now calculate corresponding values of x and y for various values of t and plot x against y to obtain the path as shown.
When y 0, 0 100 4t2, so t 5 s. For this value of the time, we have The velocity and acceleration components and their resultants are shown on the separate diagrams for point A, where y 0. Thus, for this condition we may write Ans.
Ans. a 16i 8j m/s2 v 30i 40j m/s a (16)2 (8)2 17.89 m/s2 v (30)2 (40)2 50 m/s vy 8(5) 40 m/s vx 50 16(5) 30 m/s ay d dt (8t) ay 8 m/s2 [ay v ˙y] vy d dt (100 4t2) vy 8t m/s [vy y ˙] ax d dt (50 16t) ax 16 m/s2 [ax v ˙x] x 0 dx t 0 (50 16t) dt x 50t 8t2 m dx vx dt 46 Chapter 2 Kinematics of Particles Helpful Hint We observe that the velocity vector lies along the tangent to the path as it should, but that the acceleration vector is not tangent to the path. Note espe-cially that the acceleration vector has a component that points toward the in-side of the curved path. We concluded from our diagram in Fig. 2/5 that it is impossible for the acceleration to have a component that points toward the out-side of the curve.
100 80 60 40 20 00 20 40 t = 5 s 1 2 3 4 A t = 0 y, m x, m 60 80 = 53.1° θ vx = –30 m/s vy = –40 m/s v = 50 m/s a = 17.89 m/s2 ay = –8 m/s2 ax = –16 m/s2 Path Path A A – – c02.qxd 2/8/12 7:11 PM Page 46 SAMPLE PROBLEM 2/6 A team of engineering students designs a medium-size catapult which launches 8-lb steel spheres. The launch speed is the launch angle is above the horizontal, and the launch posi-tion is 6 ft above ground level. The students use an athletic field with an adjoining slope topped by an 8-ft fence as shown. Determine: (a) the x-y coordinates of the point of first impact (b) the time duration of the flight (c) the maximum height h above the horizontal field attained by the ball (d) the velocity (expressed as a vector) with which the projectile strikes the ground Repeat part (a) for a launch speed of Solution.
We make the assumptions of constant gravitational acceleration and no aerodynamic drag. With the latter assumption, the 8-lb weight of the pro-jectile is irrelevant. Using the given x-y coordinate system, we begin by checking the y-displacement at the horizontal position of the fence.
(a) Because the y-coordinate of the top of the fence is the projectile clears the fence. We now find the flight time by setting Ans.
(b) Thus the point of first impact is .
Ans.
(c) For the maximum height: Ans.
(d) For the impact velocity: So the impact velocity is .
Ans.
If the time from launch to the fence is found by and the corresponding value of y is For this launch speed, we see that the projectile hits the fence, and the point of impact is Ans.
For lower launch speeds, the projectile could land on the slope or even on the level portion of the athletic field.
(x, y) (130, 24.9) ft [y y0 (vy)0 t 1 2 gt2] y 6 80 sin 35(2.12) 1 2 (32.2)(2.12)2 24.9 ft [x x0 (vx)0t] 100 30 (75 cos 35)t t 2.12 sec v0 75 ft/sec, v 65.5i 34.7j ft/sec [vy (vy)0 gt] vy 80 sin 35 32.2(2.50) 34.7 ft/sec [vx (vx)0] vx 80 cos 35 65.5 ft/sec [vy 2 (vy)0 2 2g(y y0)] 02 (80 sin 35)2 2(32.2)(h 6) h 38.7 ft (x, y) (164.0, 20) ft [x x0 (vx)0t] x 0 80 cos 35(2.50) 164.0 ft [y y0 (vy)0t 1 2 gt2] 20 6 80 sin 35(tƒ) 1 2 (32.2)tƒ 2 tƒ 2.50 s y 20 ft: 20 8 28 feet, [y y0 (vy)0t 1 2 gt2] y 6 80 sin 35(1.984) 1 2 (32.2)(1.984)2 33.7 ft [x x0 (vx)0t] 100 30 0 (80 cos 35)t t 1.984 sec v0 75 ft/sec.
tƒ 35 v0 80 ft/sec, Article 2/4 Rectangular Coordinates (x-y) 47 = 35° x y v0 = 80 ft/sec 100′ fence 30′ 6′ 8′ 20′ θ Helpful Hints Neglecting aerodynamic drag is a poor assumption for projectiles with relatively high initial velocities, large sizes, and low weights. In a vacuum, a baseball thrown with an initial speed of 100 ft/sec at above the horizontal will travel about 311 feet over a horizontal range. In sea-level air, the baseball range is about 200 ft, while a typical beachball under the same conditions will travel about 10 ft.
45 As an alternative approach, we could find the time at apex where then use that time in the y-displacement equation. Verify that the trajectory apex occurs over the 100-ft horizon-tal portion of the athletic field.
vy 0, c02.qxd 2/9/12 12:04 PM Page 47 2/65 A rocket runs out of fuel in the position shown and continues in unpowered flight above the atmosphere.
If its velocity in this position was 600 mi/hr, calcu-late the maximum additional altitude h acquired and the corresponding time t to reach it. The gravita-tional acceleration during this phase of its flight is Problem 2/65 2/66 A particle moves in the x-y plane with a y-component of velocity in feet per second given by with t in seconds. The acceleration of the particle in the x-direction in feet per second squared is given by with t in seconds. When and Find the equation of the path of the particle and calculate the magnitude of the velocity v of the particle for the instant when its x-coordinate reaches 18 ft.
2/67 A roofer tosses a small tool to the ground. What min-imum magnitude of horizontal velocity is required to just miss the roof corner B? Also determine the distance d.
Problem 2/67 v0 A B C 1.2 m 2.4 m 0.9 m 3 m d v0 vx 0.
t 0, y 2 ft, x 0, ax 4t vy 8t 30° v = 600 mi/hr Vertical 30.8 ft/sec2.
48 Chapter 2 Kinematics of Particles PROBLEMS (In the following problems where motion as a projectile in air is involved, neglect air resistance unless otherwise stated and use g 9.81 m/s2 or g 32.2 ft/sec2.) Introductory Problems 2/59 At time the position vector of a particle mov-ing in the x-y plane is By time its position vector has become Deter-mine the magnitude of its average velocity during this interval and the angle made by the average velocity with the positive x-axis.
2/60 A particle moving in the x-y plane has a velocity at time given by and at its velocity has become Calculate the magnitude aav of its average acceleration during the 0.1-s interval and the angle it makes with the x-axis.
2/61 The velocity of a particle moving in the x-y plane is given by at time Its aver-age acceleration during the next 0.02 s is Determine the velocity v of the particle at and the angle between the average-acceleration vector and the velocity vector at 2/62 A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to and Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when 2/63 The x-coordinate of a particle in curvilinear motion is given by where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by If the particle has y-components and when find the magnitudes of the velocity v and acceleration a when Sketch the path for the first 2 seconds of motion, and show the velocity and accel-eration vectors for 2/64 The y-coordinate of a particle in curvilinear motion is given by where y is in inches and t is in seconds. Also, the particle has an acceleration in the x-direction given by If the ve-locity of the particle in the x-direction is when calculate the magnitudes of the velocity v and acceleration a of the particle when Construct v and a in your solution.
t 1 sec.
t 0, 4 in./sec ax 12t in./sec2.
y 4t3 3t, t 2 sec.
t 2 sec.
t 0, 4 ft/sec y ˙ y 0 ay 4t.
x 2t3 3t, t 2 s.
y 3t2 1 3t3.
x 2t2 4t t 3.67 s.
t 3.67 s 4i 6j m/s2.
t 3.65 s.
6.12i 3.24j m/s 4.3i 5.4j m/s.
t 6.1 s 4i 5j m/s, t 6 s vav 5.1i 0.4j m.
t 0.02 s, r 5i m.
t 0, c02.qxd 2/8/12 7:11 PM Page 48 2/68 Prove the well-known result that, for a given launch speed the launch angle yields the maxi-mum horizontal range R. Determine the maximum range. (Note that this result does not hold when aerodynamic drag is included in the analysis.) 2/69 Calculate the minimum possible magnitude u of the muzzle velocity which a projectile must have when fired from point A to reach a target B on the same horizontal plane 12 km away.
Problem 2/69 2/70 The center of mass G of a high jumper follows the trajectory shown. Determine the component , mea-sured in the vertical plane of the figure, of his take-off velocity and angle if the apex of the trajectory just clears the bar at A. (In general, must the mass center G of the jumper clear the bar during a suc-cessful jump?) Problem 2/70 3.5′ 3′ v0 A G 3.5′ θ v0 A u B 12 km 45 v0, Article 2/4 Problems 49 Representative Problems 2/71 The quarterback Q throws the football when the receiver R is in the position shown. The receiver’s velocity is constant at 10 yd/sec, and he catches passes when the ball is 6 ft above the ground. If the quarterback desires the receiver to catch the ball 2.5 sec after the launch instant shown, determine the initial speed and angle required.
Problem 2/71 2/72 The water nozzle ejects water at a speed at the angle Determine where, relative to the wall base point B, the water lands. Neglect the effects of the thickness of the wall.
Problem 2/72 2/73 Water is ejected from the water nozzle of Prob. 2/72 with a speed For what value of the angle will the water land closest to the wall after clearing the top? Neglect the effects of wall thick-ness and air resistance. Where does the water land?
2/74 A football player attempts a 30-yd field goal. If he is able to impart a velocity u of 100 ft/sec to the ball, compute the minimum angle for which the ball will clear the crossbar of the goal. (Hint: Let .) Problem 2/74 u 30 yd θ 10′ m tan v0 45 ft/sec.
60′ Not to scale 1′ A 3′ θ v0 B 40.
ft/sec v0 45 v0 vR R Q θ 7′ 30 yd v0 c02.qxd 2/8/12 7:11 PM Page 49 2/78 The basketball player likes to release his foul shots with an initial speed What value(s) of the initial angle will cause the ball to pass through the center of the rim? Neglect clearance considerations as the ball passes over the front por-tion of the rim.
Problem 2/78 2/79 A projectile is launched with an initial speed of 200 m/s at an angle of with respect to the horizontal.
Compute the range R as measured up the incline.
Problem 2/79 2/80 A rock is thrown horizontally from a tower at A and hits the ground 3.5 s later at B. The line of sight from A to B makes an angle of with the horizon-tal. Compute the magnitude of the initial velocity u of the rock.
Problem 2/80 A 50° u B 50 R 60° 20° A B 200 m/s 60 7′ θ 10′ 13′ 9′′ v0 v0 23.5 ft/sec.
50 Chapter 2 Kinematics of Particles 2/75 The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location A.
What angle with the horizontal should the pilot’s line of sight to the target make at the instant of re-lease? The airplane is flying horizontally at an alti-tude of 100 m with a velocity of 200 km/h.
Problem 2/75 2/76 During a baseball practice session, the cutoff man A executes a throw to the third baseman B. If the ini-tial speed of the baseball is what angle is best if the ball is to arrive at third base at essentially ground level?
Problem 2/76 2/77 If the tennis player serves the ball horizontally calculate its velocity v if the center of the ball clears the 36-in. net by 6 in. Also find the dis-tance s from the net to the point where the ball hits the court surface. Neglect air resistance and the effect of ball spin.
Problem 2/77 39′ s 36″ 8.5′ A θ v ( 0), v0 7′ θ B A 150′ v0 130 ft/sec, θ 200 km/h 100 m A c02.qxd 2/8/12 7:11 PM Page 50 2/81 The muzzle velocity of a long-range rifle at A is Determine the two angles of elevation which will permit the projectile to hit the mountain target B.
Problem 2/81 2/82 A projectile is launched with a speed from the floor of a 5-m-high tunnel as shown. Deter-mine the maximum horizontal range R of the projec-tile and the corresponding launch angle .
Problem 2/82 2/83 A projectile is launched from point A with the initial conditions shown in the figure. Determine the slant distance s which locates the point B of impact. Cal-culate the time of flight t.
Problem 2/83 v0 = 120 m/s = 40° θ 20° B s A 800 m 5 m A θ v0 = 25 m/s v0 25 m/s B 5 km 1.5 km 2 θ 1 θ u u A 400 m/s.
u Article 2/4 Problems 51 2/84 A team of engineering students is designing a cata-pult to launch a small ball at A so that it lands in the box. If it is known that the initial velocity vector makes a angle with the horizontal, determine the range of launch speeds for which the ball will land inside the box.
Problem 2/84 2/85 Ball bearings leave the horizontal trough with a ve-locity of magnitude u and fall through the 70-mm-diameter hole as shown. Calculate the permissible range of u which will enable the balls to enter the hole. Take the dashed positions to represent the lim-iting conditions.
Problem 2/85 2/86 A horseshoe player releases the horseshoe at A with an initial speed Determine the range for the launch angle for which the shoe will strike the 14-in. vertical stake.
Problem 2/86 v0 = 36 ft/sec 40′ A B θ 3 14″ v0 36 ft/sec.
120 mm 20 mm 80 mm 70 mm u v0 12″ 30° A 8″ 12′ 2′ v0 30 c02.qxd 2/8/12 7:11 PM Page 51 2/90 The pilot of an airplane pulls into a steep 45° climb at 300 km/h and releases a package at position A.
Calculate the horizontal distance s and the time t from the point of release to the point at which the package strikes the ground.
Problem 2/90 2/91 Compare the slant range and flight time for the depicted projectile with the range R and flight time t for a projectile (launched with speed and inclina-tion angle ) which flies over a horizontal surface.
Evaluate your four results for Problem 2/91 2/92 A projectile is launched from point A and lands on the same level at D. Its maximum altitude is h. De-termine and plot the fraction of the total flight time that the projectile is above the level where is a fraction which can vary from zero to 1. State the value of for Problem 2/92 A D C ƒ1h v0 α h B ƒ 1 3 4.
ƒ 2 ƒ 1 ƒ 1h, ƒ 2 A B Ri v0 α α 30.
v0 ti Ri s A 300 km/h 500 m 45° 52 Chapter 2 Kinematics of Particles 2/87 A fireworks shell is launched vertically from point A with speed sufficient to reach a maximum altitude of 500 ft. A steady horizontal wind causes a constant horizontal acceleration of but does not af-fect the vertical motion. Determine the deviation at the top of the trajectory caused by the wind.
Problem 2/87 2/88 Consider the fireworks shell of the previous problem.
What angle compensates for the wind in that the shell peaks directly over the launch point A? All other information remains as stated in the previous problem, including the fact that the initial launch velocity if vertical would result in a maximum altitude of 500 ft. What is the maximum height h possible in this problem?
Problem 2/88 2/89 Determine the location h of the spot toward which the pitcher must throw if the ball is to hit the catcher’s mitt. The ball is released with a speed of 40 m/s.
Problem 2/89 2.2 m 20 m v0 θ 0.6 m 1 m h Wind A h v0 α v0 v0 δ Wind A 500′ 0.5 ft/sec2, c02.qxd 2/8/12 7:11 PM Page 52 2/93 A projectile is ejected into an experimental fluid at time The initial speed is and the angle to the horizontal is . The drag on the projectile results in an acceleration term where k is a con-stant and v is the velocity of the projectile. Deter-mine the x- and y-components of both the velocity and displacement as functions of time. What is the terminal velocity? Include the effects of gravitational acceleration.
Problem 2/93 2/94 An experimental fireworks shell is launched verti-cally from point A with an initial velocity of magni-tude In addition to the acceleration due to gravity, an internal thrusting mechanism causes a constant acceleration component of 2g in the direction shown for the first 2 seconds of flight, after which the thruster ceases to function.
Determine the maximum height h achieved, the total flight time, the net horizontal displacement from point A, and plot the entire trajectory. Neglect any acceleration due to aerodynamics.
Problem 2/94 v0 2g 60° A 60 v0 100 ft/sec.
y x θ v0 aD kv, v0 t 0.
Article 2/4 Problems 53 2/95 A projectile is launched with speed from point A.
Determine the launch angle which results in the maximum range R up the incline of angle (where ). Evaluate your results for and Problem 2/95 2/96 A projectile is launched from point A with the initial conditions shown in the figure. Determine the x- and y-coordinates of the point of impact.
Problem 2/96 v0 = 225 ft/sec 1000′ 30° C B x y A 500′ v0 R A B α θ 45.
0, 30, 0 90 v0 c02.qxd 2/8/12 7:11 PM Page 53 2/5 Normal and Tangential Coordinates (n-t ) As we mentioned in Art. 2/1, one of the common descriptions of curvilinear motion uses path variables, which are measurements made along the tangent t and normal n to the path of the particle. These coor-dinates provide a very natural description for curvilinear motion and are frequently the most direct and convenient coordinates to use. The n- and t-coordinates are considered to move along the path with the par-ticle, as seen in Fig. 2/9 where the particle advances from A to B to C.
The positive direction for n at any position is always taken toward the center of curvature of the path. As seen from Fig. 2/9, the positive n-direction will shift from one side of the curve to the other side if the curvature changes direction.
Velocity and Acceleration We now use the coordinates n and t to describe the velocity v and acceleration a which were introduced in Art. 2/3 for the curvilinear mo-tion of a particle. For this purpose, we introduce unit vectors en in the n-direction and et in the t-direction, as shown in Fig. 2/10a for the posi-tion of the particle at point A on its path. During a differential incre-ment of time dt, the particle moves a differential distance ds along the curve from A to A. With the radius of curvature of the path at this posi-tion designated by , we see that ds d, where is in radians. It is unnecessary to consider the differential change in between A and A because a higher-order term would be introduced which disappears in the limit. Thus, the magnitude of the velocity can be written v ds/dt d/dt, and we can write the velocity as the vector (2/7) The acceleration a of the particle was defined in Art. 2/3 as a dv/dt, and we observed from Fig. 2/5 that the acceleration is a vector which reflects both the change in magnitude and the change in direc-tion of v. We now differentiate v in Eq. 2/7 by applying the ordinary rule for the differentiation of the product of a scalar and a vector and get (2/8) where the unit vector et now has a nonzero derivative because its direc-tion changes.
To find we analyze the change in et during a differential incre-ment of motion as the particle moves from A to A in Fig. 2/10a. The unit vector et correspondingly changes to , and the vector difference det is shown in part b of the figure. The vector det in the limit has a magnitude equal to the length of the arc et d d obtained by swinging the unit vector et through the angle d expressed in radians.
e t e ˙t a dv dt d(vet) dt ve ˙t v ˙et v vet ˙et Figure 2/10 54 Chapter 2 Kinematics of Particles A B C n n n t t t Figure 2/9 v′ v v A′ A n C Path t dvn dv dvt det et ρ ds = d ρ β e′ t et en at an a (a) (c) (b) v′ e′ t β d d β d β See Art. C/7 of Appendix C.
c02.qxd 2/8/12 7:11 PM Page 54 Article 2/5 Normal and Tangential Coordinates (n-t) 55 The direction of det is given by en. Thus, we can write det en d.
Dividing by d gives Dividing by dt gives det/dt (d/dt)en, which can be written (2/9) With the substitution of Eq. 2/9 and from the relation v Eq.
2/8 for the acceleration becomes (2/10) where an at a We stress that is the time rate of change of the speed v. Finally, we note that at This relation, however, finds little use because we seldom have reason to compute Geometric Interpretation Full understanding of Eq. 2/10 comes only when we clearly see the geometry of the physical changes it describes. Figure 2/10c shows the ve-locity vector v when the particle is at A and v when it is at A. The vector change in the velocity is dv, which establishes the direction of the accelera-tion a. The n-component of dv is labeled dvn, and in the limit its magni-tude equals the length of the arc generated by swinging the vector v as a radius through the angle d. Thus, dvn v d and the n-component of acceleration is an dvn/dt v(d/dt) as before. The t-component of dv is labeled dvt, and its magnitude is simply the change dv in the magni-tude or length of the velocity vector. Therefore, the t-component of accel-eration is at dv/dt as before. The acceleration vectors resulting from the corresponding vector changes in velocity are shown in Fig. 2/10c.
It is especially important to observe that the normal component of acceleration an is always directed toward the center of curvature C. The tangential component of acceleration, on the other hand, will be in the positive t-direction of motion if the speed v is increasing and in the nega-tive t-direction if the speed is decreasing. In Fig. 2/11 are shown schematic representations of the variation in the acceleration vector for a particle moving from A to B with (a) increasing speed and (b) decreas-ing speed. At an inflection point on the curve, the normal acceleration v2/ goes to zero because becomes infinite.
s ¨ v ˙ v ˙ ˙.
˙ ˙.
¨ d( ˙)/dt v ˙ v ˙ at an 2 at 2 v ˙ s ¨ v2 ˙2 v ˙ a v2 en v ˙et ˙, ˙ e ˙t ˙en det d en c02.qxd 2/8/12 7:11 PM Page 55 Circular Motion Circular motion is an important special case of plane curvilinear motion where the radius of curvature becomes the constant radius r of the circle and the angle is replaced by the angle measured from any convenient radial reference to OP, Fig. 2/12. The velocity and the accel-eration components for the circular motion of the particle P become (2/11) We find repeated use for Eqs. 2/10 and 2/11 in dynamics, so these relations and the principles behind them should be mastered.
at v ˙ r ¨ an v2/r r ˙2 v ˙ v r ˙ Figure 2/12 56 Chapter 2 Kinematics of Particles A Speed increasing (a) Speed decreasing (b) Acceleration vectors for particle moving from A to B A B B Figure 2/11 θ r n an P t O at v An example of uniform circular motion is this car moving with constant speed around a wet skidpad (a circular roadway with a diameter of about 200 feet).
GaryTramontina/Bloomberg viaGetty Images c02.qxd 2/8/12 7:11 PM Page 56 Article 2/5 Normal and Tangential Coordinates (n-t) 57 SAMPLE PROBLEM 2/7 To anticipate the dip and hump in the road, the driver of a car applies her brakes to produce a uniform deceleration. Her speed is 100 km/h at the bottom A of the dip and 50 km/h at the top C of the hump, which is 120 m along the road from A. If the passengers experience a total acceleration of 3 m/s2 at A and if the radius of curvature of the hump at C is 150 m, calculate (a) the radius of curvature at A, (b) the acceleration at the inflection point B, and (c) the total acceleration at C.
Solution.
The dimensions of the car are small compared with those of the path, so we will treat the car as a particle. The velocities are We find the constant deceleration along the path from (a) Condition at A.
With the total acceleration given and at determined, we can easily compute an and hence from Ans.
(b) Condition at B.
Since the radius of curvature is infinite at the inflection point, an 0 and Ans.
(c) Condition at C.
The normal acceleration becomes With unit vectors en and et in the n- and t-directions, the acceleration may be written where the magnitude of a is Ans.
The acceleration vectors representing the conditions at each of the three points are shown for clarification.
[a an 2 at 2] a (1.286)2 (2.41)2 2.73 m/s2 a 1.286en 2.41et m/s2 [an v2/] an (13.89)2/150 1.286 m/s2 a at 2.41 m/s2 [an v2/] v2/an (27.8)2/1.785 432 m [a2 an 2 at 2] an 2 32 (2.41)2 3.19 an 1.785 m/s2 at 1 2s (vC 2 vA 2) (13.89)2 (27.8)2 2(120) 2.41 m/s2 v dv at ds vC vA v dv at s 0 ds vC 50 1000 3600 13.89 m/s vA 100 km h 1 h 3600 s1000 m km 27.8 m/s 150 m 60 m 60 m B A C A B at = –2.41 m/s2 C an = 1.785 m/s2 at = –2.41 m/s2 a = at = –2.41 m/s2 an = 1.286 m/s2 an = 2.73 m/s2 a = 3 m/s2 +n +t +t +t +n Helpful Hint Actually, the radius of curvature to the road differs by about 1 m from that to the path followed by the cen-ter of mass of the passengers, but we have neglected this relatively small difference.
c02.qxd 2/8/12 7:11 PM Page 57 58 Chapter 2 Kinematics of Particles SAMPLE PROBLEM 2/8 A certain rocket maintains a horizontal attitude of its axis during the pow-ered phase of its flight at high altitude. The thrust imparts a horizontal compo-nent of acceleration of 20 ft/sec2, and the downward acceleration component is the acceleration due to gravity at that altitude, which is g 30 ft/sec2. At the in-stant represented, the velocity of the mass center G of the rocket along the 15 direction of its trajectory is 12,000 mi/hr. For this position determine (a) the ra-dius of curvature of the flight trajectory, (b) the rate at which the speed v is in-creasing, (c) the angular rate of the radial line from G to the center of curvature C, and (d) the vector expression for the total acceleration a of the rocket.
Solution.
We observe that the radius of curvature appears in the expression for the normal component of acceleration, so we use n- and t-coordinates to de-scribe the motion of G. The n- and t-components of the total acceleration are ob-tained by resolving the given horizontal and vertical accelerations into their n-and t-components and then combining. From the figure we get (a) We may now compute the radius of curvature from Ans.
(b) The rate at which v is increasing is simply the t-component of acceleration.
Ans.
(c) The angular rate of line GC depends on v and and is given by Ans.
(d) With unit vectors en and et for the n- and t-directions, respectively, the total acceleration becomes Ans.
a 23.8en 27.1et ft/sec2 ˙ v/ 12,000(44/30) 13.01(106) 13.53(104) rad/sec [v ˙] ˙ v ˙ 27.1 ft/sec2 [v ˙ at] [an v2/] v2 an [(12,000)(44/30)]2 23.8 13.01(106) ft at 30 sin 15 20 cos 15 27.1 ft/sec2 an 30 cos 15 20 sin 15 23.8 ft/sec2 ˙ n G v = 12,000 mi/hr g = 30 ft/sec2 20 ft/sec2 Horiz.
t 15° C ρ 15° x ax = 20 ft/sec2 a g = 30 ft/sec2 · v at = et en v2 — – an = ρ Helpful Hints Alternatively, we could find the re-sultant acceleration and then re-solve it into n- and t-components.
To convert from mi/hr to ft/sec, multi-ply by which is easily remembered, as 30 mi/hr is the same as 44 ft/sec.
44 ft/sec 30 mi/hr 5280 ft/mi 3600 sec/hr c02.qxd 2/8/12 7:11 PM Page 58 Article 2/5 Problems 59 2/100 The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a parti-cle) is 2 m above the road. Calculate the speed v of the truck.
Problem 2/100 2/101 A bicycle is placed on a service rack with its wheels hanging free. As part of a bearing test, the front wheel is spun at the rate Assume that this rate is constant and determine the speed v and magnitude a of the acceleration of point A.
Problem 2/101 2/102 A ship which moves at a steady 20-knot speed (1 executes a turn to port by changing its compass heading at a constant coun-terclockwise rate. If it requires 60 s to alter course calculate the magnitude of the acceleration a of the ship during the turn.
2/103 A train enters a curved horizontal section of track at a speed of 100 km/h and slows down with con-stant deceleration to 50 km/h in 12 seconds. An ac-celerometer mounted inside the train records a horizontal acceleration of 2 when the train is 6 seconds into the curve. Calculate the radius of curvature of the track for this instant.
m/s2 90, knot 1.852 km/h) N A O 30° 27″ N 45 rev/min.
2 m A G PROBLEMS Introductory Problems 2/97 Determine the maximum speed for each car if the normal acceleration is limited to 0.88g. The roadway is unbanked and level.
Problem 2/97 2/98 A car is traveling around a circular track of 800-ft radius. If the magnitude of its total acceleration is at the instant when its speed is 45 mi/hr, determine the rate at which the car is changing its speed.
2/99 Six acceleration vectors are shown for the car whose velocity vector is directed forward. For each acceler-ation vector describe in words the instantaneous mo-tion of the car.
Problem 2/99 v a1 a2 a3 a4 a5 a6 10 ft/sec2 21 m 16 m B A c02.qxd 2/8/12 7:11 PM Page 59 2/104 The two cars A and B enter an unbanked and level turn. They cross line C-C simultaneously, and each car has the speed corresponding to a maximum normal acceleration of 0.9g in the turn. Determine the elapsed time for each car between its two cross-ings of line C-C. What is the relative position of the two cars as the second car exits the turn? Assume no speed changes throughout.
Problem 2/104 2/105 Revisit the two cars of the previous problem, only now the track has variable banking—a concept shown in the figure. Car A is on the unbanked por-tion of the track and its normal acceleration re-mains at 0.9g. Car B is on the banked portion of the track and its normal acceleration is limited to 1.12g. If the cars approach line C-C with speeds equal to the respective maxima in the turn, deter-mine the time for each car to negotiate the turn as delimited by line C-C. What is the relative position of the two cars as the second car exits the turn?
Assume no speed changes throughout.
Problem 2/105 A B θ 72 m C b a b a C A B 88 m 2/106 A particle moves along the curved path shown. If the particle has a speed of 40 ft/sec at A at time and a speed of 44 ft/sec at B at time , determine the average values of the acceleration of the parti-cle between A and B, both normal and tangent to the path.
Problem 2/106 2/107 The speed of a car increases uniformly with time from 50 km/h at A to 100 km/h at B during 10 sec-onds. The radius of curvature of the hump at A is 40 m. If the magnitude of the total acceleration of the mass center of the car is the same at B as at A, compute the radius of curvature of the dip in the road at B. The mass center of the car is 0.6 m from the road.
Problem 2/107 0.6 m A B 40 m B ρ B tB = 3.84 sec tA = 3.64 sec 36° 26° B A tB tA 60 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 60 Article 2/5 Problems 61 2/110 A satellite travels with constant speed v in a circu-lar orbit 320 km above the earth’s surface. Calcu-late v knowing that the acceleration of the satellite is the gravitational acceleration at its altitude.
(Note: Review Art. 1/5 as necessary and use the mean value of g and the mean value of the earth’s radius. Also recognize that v is the magnitude of the velocity of the satellite with respect to the cen-ter of the earth.) 2/111 The car is traveling at a speed of 60 mi/hr as it ap-proaches point A. Beginning at A, the car deceler-ates at a constant until it gets to point B, after which its constant rate of decrease of speed is as it rounds the interchange ramp. Deter-mine the magnitude of the total car acceleration (a) just before it gets to B, (b) just after it passes B, and (c) at point C.
Problem 2/111 2/112 Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when if 2 rad/sec and 4.025 rad/sec2.
Problem 2/112 x y n t 4′ θ G ¨ ˙ 60 200′ A B C 300′ 3 ft/sec2 7 ft/sec2 Representative Problems 2/108 The figure shows two possible paths for negotiating an unbanked turn on a horizontal portion of a race course. Path A-A follows the centerline of the road and has a radius of curvature , while path B-B uses the width of the road to good advan-tage in increasing the radius of curvature to . If the drivers limit their speeds in their curves so that the lateral acceleration does not exceed 0.8g, determine the maximum speed for each path.
Problem 2/108 2/109 Consider the polar axis of the earth to be fixed in space and compute the magnitudes of the velocity and acceleration of a point P on the earth’s sur-face at latitude 40° north. The mean diameter of the earth is 12 742 km and its angular velocity is .
Problem 2/109 N S P 40° rad/s 0.7292(104) ρB = 200 m ρA = 85 m A B A B B 200 m A 85 m c02.qxd 2/8/12 7:11 PM Page 61 2/113 The preliminary design for a “small” space station to orbit the earth in a circular path consists of a ring (torus) with a circular cross section as shown.
The living space within the torus is shown in sec-tion A, where the “ground level” is 20 ft from the center of the section. Calculate the angular speed N in revolutions per minute required to simulate standard gravity at the surface of the earth . Recall that you would be unaware of a gravitational field if you were in a nonrotating spacecraft in a circular orbit around the earth.
Problem 2/113 2/114 Magnetic tape is being transferred from reel A to reel B and passes around idler pulleys C and D. At a certain instant, point on the tape is in contact with pulley C and point is in contact with pulley D. If the normal component of acceleration of is and the tangential component of accelera-tion of is at this instant, compute the corresponding speed v of the tape, the magnitude of the total acceleration of , and the magnitude of the total acceleration of .
Problem 2/114 P1 100 mm 50 mm P2 A C D B P2 P1 30 m/s2 P2 40 m/s2 P1 P2 P1 Section A "Ground level" r r 240′ 20′ N A (32.17 ft/sec2) 2/115 The car C increases its speed at the constant rate of as it rounds the curve shown. If the mag-nitude of the total acceleration of the car is at the point A where the radius of curva-ture is 200 m, compute the speed v of the car at this point.
Problem 2/115 2/116 A football player releases a ball with the initial con-ditions shown in the figure. Determine the radius of curvature of the trajectory (a) just after release and (b) at the apex. For each case, compute the time rate of change of the speed.
Problem 2/116 2/117 For the football of the previous problem, determine the radius of curvature of the path and the time rate of change of the speed at times sec and sec, where is the time of release from the quarterback’s hand.
2/118 A particle moving in the x-y plane has a position vector given by , where r is in inches and t is in seconds. Calculate the radius of curva-ture of the path for the position of the particle when sec. Sketch the velocity v and the cur-vature of the path for this particular instant.
t 2 r 3 2t2i 2 3t3j t 0 t 2 t 1 v ˙ v0 = 80 ft/sec = 35° θ A C 2.5 m/s2 1.5 m/s2 62 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 62 Article 2/5 Problems 63 2/121 At a certain point in the reentry of the space shut-tle into the earth’s atmosphere, the total accelera-tion of the shuttle may be represented by two components. One component is the gravitational acceleration at this altitude. The sec-ond component equals due to atmos-pheric resistance and is directed opposite to the velocity. The shuttle is at an altitude of 48.2 km and has reduced its orbital velocity of 28 300 km/h to 15 450 km/h in the direction . For this instant, calculate the radius of curvature of the path and the rate at which the speed is changing.
Problem 2/121 2/122 The particle P starts from rest at point A at time and changes its speed thereafter at a con-stant rate of 2g as it follows the horizontal path shown. Determine the magnitude and direction of its total acceleration (a) just before point B, (b) just after point B, and (c) as it passes point C.
State your directions relative to the x-axis shown (CCW positive).
Problem 2/122 x A B P C 3 m 3.5 m t 0 v θ v ˙ 1.50 12.90 m/s2 g 9.66 m/s2 2/119 The design of a camshaft-drive system of a four-cylinder automobile engine is shown. As the engine is revved up, the belt speed v changes uniformly from 3 m/s to 6 m/s over a two-second interval. Cal-culate the magnitudes of the accelerations of points and halfway through this time interval.
Problem 2/119 2/120 A small particle P starts from point O with a negli-gible speed and increases its speed to a value , where y is the vertical drop from O.
When , determine the n-component of ac-celeration of the particle. (See Art. C/10 for the ra-dius of curvature.) Problem 2/120 v Horizontal P O x y = ( ) 2 ft y Vertical x –– 20 x 50 ft v 2gy v P2 P1 60 mm Camshaft sprocket Crankshaft sprocket Intermediate sprocket Drive belt tensioner P2 P1 c02.qxd 2/8/12 7:11 PM Page 63 2/123 For the conditions of the previous problem, deter-mine the magnitude and direction of the total ac-celeration of the particle P at times and .
2/124 Race car A follows path a-a while race car B follows path b-b on the unbanked track. If each car has a constant speed limited to that corresponding to a lateral (normal) acceleration of 0.8g, determine the times and for both cars to negotiate the turn as delimited by the line C-C.
Problem 2/124 2/125 The mine skip is being hauled to the surface over the curved track by the cable wound around the 30-in. drum, which turns at the constant clockwise speed of 120 rev/min. The shape of the track is de-signed so that , where x and y are in feet.
Calculate the magnitude of the total acceleration of the skip as it reaches a level of 2 ft below the top.
Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curva-ture is given by Problem 2/125 x y 30″ 1 dy dx 2 3/2 d2y dx2 y x2/40 72 m C C A B a b 88 m a b tB tA t 1.2 s t 0.8 s 2/126 An earth satellite which moves in the elliptical equatorial orbit shown has a velocity v in space of 17 970 km/h when it passes the end of the semi-minor axis at A. The earth has an absolute surface value of g of 9.821 m/s2 and has a radius of 6371 km.
Determine the radius of curvature of the orbit at A.
Problem 2/126 2/127 A particle which moves in two-dimensional curvi-linear motion has coordinates in millimeters which vary with time t in seconds according to and . For time , determine the ra-dius of curvature of the particle path and the mag-nitudes of the normal and tangential accelerations.
2/128 In a handling test, a car is driven through the slalom course shown. It is assumed that the car path is sinusoidal and that the maximum lateral acceleration is 0.7g. If the testers wish to design a slalom through which the maximum speed is 80 km/h, what cone spacing L should be used?
Problem 2/128 Sinusoidal v L 3 m 3 m t 3 s y 2t3 6 x 5t2 4 16 000 km 13 860 km 8000 km A r v 64 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 64 Article 2/5 Problems 65 2/129 The pin P is constrained to move in the slotted guides which move at right angles to one another.
At the instant represented, A has a velocity to the right of 0.2 m/s which is decreasing at the rate of 0.75 m/s each second. At the same time, B is mov-ing down with a velocity of 0.15 m/s which is de-creasing at the rate of 0.5 m/s each second. For this instant determine the radius of curvature of the path followed by P. Is it possible to also determine the time rate of change of ?
Problem 2/129 P B A 2/130 A particle which moves with curvilinear motion has coordinates in meters which vary with time t in sec-onds according to and .
Determine the coordinates of the center of curva-ture C at time .
t 1 s y 5t 2 x 2t2 3t 1 c02.qxd 2/8/12 7:11 PM Page 65 2/6 Polar Coordinates (r-) We now consider the third description of plane curvilinear motion, namely, polar coordinates where the particle is located by the radial dis-tance r from a fixed point and by an angular measurement to the ra-dial line. Polar coordinates are particularly useful when a motion is constrained through the control of a radial distance and an angular po-sition or when an unconstrained motion is observed by measurements of a radial distance and an angular position.
Figure 2/13a shows the polar coordinates r and which locate a particle traveling on a curved path. An arbitrary fixed line, such as the x-axis, is used as a reference for the measurement of . Unit vectors er and e are established in the positive r- and -directions, respectively.
The position vector r to the particle at A has a magnitude equal to the radial distance r and a direction specified by the unit vector er. Thus, we express the location of the particle at A by the vector Time Derivatives of the Unit Vectors To differentiate this relation with respect to time to obtain v and a we need expressions for the time derivatives of both unit vectors er and e. We obtain and in exactly the same way we derived in the preceding article. During time dt the coordinate directions rotate through the angle d, and the unit vectors also rotate through the same angle from er and e to and , as shown in Fig. 2/13b. We note that the vec-tor change der is in the plus -direction and that de is in the minus r-direction. Because their magnitudes in the limit are equal to the unit vector as radius times the angle d in radians, we can write them as der e d and de er d. If we divide these equations by d, we have If, on the other hand, we divide them by dt, we have der/dt (d/dt)e and de/dt (d/dt)er, or simply (2/12) Velocity We are now ready to differentiate r rer with respect to time. Using the rule for differentiating the product of a scalar and a vector gives With the substitution of from Eq. 2/12, the vector expression for the velocity becomes (2/13) v r ˙er r ˙e e ˙r v r ˙ r ˙er re ˙r e ˙r ˙e and e ˙ ˙er der d e and de d er e e r e ˙t e ˙ e ˙r v ˙, r ˙ r rer 66 Chapter 2 Kinematics of Particles Figure 2/13 e θ θ θ er e θ de θ e′ θ er e′ r der r r x y (a) (b) A O – r Path θ + θ d θ d c02.qxd 2/8/12 7:11 PM Page 66 where vr v v The r-component of v is merely the rate at which the vector r stretches. The -component of v is due to the rotation of r.
Acceleration We now differentiate the expression for v to obtain the acceleration a Note that the derivative of will produce three terms, since all three factors are variable. Thus, Substitution of and from Eq. 2/12 and collecting terms give (2/14) where ar a a We can write the -component alternatively as which can be verified easily by carrying out the differentiation. This form for a will be useful when we treat the angular momentum of par-ticles in the next chapter.
Geometric Interpretation The terms in Eq. 2/14 can be best understood when the geometry of the physical changes can be clearly seen. For this purpose, Fig. 2/14a is developed to show the velocity vectors and their r- and -components at position A and at position A after an infinitesimal movement. Each of these components undergoes a change in magnitude and direction as shown in Fig. 2/14b. In this figure we see the following changes: (a) Magnitude Change of vr. This change is simply the increase in length of vr or dvr and the corresponding acceleration term is in the positive r-direction.
(b) Direction Change of vr. The magnitude of this change is seen from the figure to be vr d and its contribution to the accelera-tion becomes which is in the positive -direction.
(c) Magnitude Change of v. This term is the change in length of v or and its contribution to the acceleration is and is in the positive -direction.
r ˙ ˙ r ¨ d(r ˙)/dt d(r ˙), r ˙ ˙ r ˙ d/dt r ˙ d, r ¨ dr ˙/dt dr ˙, a 1 r d dt (r2 ˙) ar 2 a 2 r ¨ 2r ˙ ˙ r ¨ r ˙2 a (r ¨ r ˙2)er (r ¨ 2r ˙ ˙)e e ˙ e ˙r a v ˙ (r ¨er r ˙e ˙r) (r ˙ ˙e r ¨e r ˙e ˙) r ˙e v ˙.
vr 2 v 2 r ˙ r ˙ Figure 2/14 Article 2/6 Polar Coordinates (r-) 67 θ d θ d θ d θ r · v θ v θ vr dvr vr v′ v A Path A v′ r vr = r · dr · r · d θ θ = θ r · (a) (b) θ d ( ) r · v′ θ v ′ ′ ′ c02.qxd 2/13/12 5:16 PM Page 67 (d) Direction Change of v. The magnitude of this change is v d and the corresponding acceleration term is observed to be in the negative r-direction.
Collecting terms gives ar and a as obtained previously. We see that the term is the acceleration which the particle would have along the radius in the absence of a change in . The term is the normal component of acceleration if r were constant, as in circular motion. The term is the tangential acceleration which the particle would have if r were constant, but is only a part of the accelera-tion due to the change in magnitude of v when r is variable. Finally, the term is composed of two effects. The first effect comes from that portion of the change in magnitude of v due to the change in r, and the second effect comes from the change in direction of vr. The term represents, therefore, a combination of changes and is not so easily perceived as are the other acceleration terms.
Note the difference between the vector change dvr in vr and the change dvr in the magnitude of vr. Similarly, the vector change dv is not the same as the change dv in the magnitude of v. When we divide these changes by dt to obtain expressions for the derivatives, we see clearly that the magnitude of the derivative dvr/dt and the derivative of the magnitude dvr/dt are not the same. Note also that ar is not and that a is not The total acceleration a and its components are represented in Fig.
2/15. If a has a component normal to the path, we know from our analy-sis of n- and t-components in Art. 2/5 that the sense of the n-component must be toward the center of curvature.
Circular Motion For motion in a circular path with r constant, the components of Eqs. 2/13 and 2/14 become simply This description is the same as that obtained with n- and t-components, where the - and t-directions coincide but the positive r-direction is in the negative n-direction. Thus, ar an for circular motion centered at the origin of the polar coordinates.
The expressions for ar and a in scalar form can also be obtained by direct differentiation of the coordinate relations x r cos and y r sin to obtain ax and ay Each of these rectangular components of ac-celeration can then be resolved into r- and -components which, when combined, will yield the expressions of Eq. 2/14.
y ¨.
x ¨ ar r ˙2 a r ¨ vr 0 v r ˙ v ˙.
v ˙r 2r ˙ ˙ d(r ˙) 2r ˙ ˙ r ¨ r ˙2 r ¨ 2r ˙ ˙ r ¨ r ˙2 r ¨ r ˙2 r ˙(d/dt) r ˙ d, 68 Chapter 2 Kinematics of Particles Figure 2/15 O Path r A a θ ar = r ·· – r 2 θ · a = r + 2r ·θ θ · θ ·· c02.qxd 2/8/12 7:11 PM Page 68 Article 2/6 Polar Coordinates (r-) 69 B O θ A r = 65.3° θ r = 0.56 m O v = 0.479 m/s vr = 0.24 m/s v = 0.414 m/s θ B = 65.3° θ O a = 0.601 m/s2 ar = –0.227 m/s2 a = 0.557 m/s2 θ B SAMPLE PROBLEM 2/9 Rotation of the radially slotted arm is governed by 0.2t 0.02t3, where is in radians and t is in seconds. Simultaneously, the power screw in the arm engages the slider B and controls its distance from O according to r 0.2 0.04t2, where r is in meters and t is in seconds. Calculate the magnitudes of the velocity and acceleration of the slider for the instant when t 3 s.
Solution.
The coordinates and their time derivatives which appear in the ex-pressions for velocity and acceleration in polar coordinates are obtained first and evaluated for t 3 s.
The velocity components are obtained from Eq. 2/13 and for t 3 s are Ans.
The velocity and its components are shown for the specified position of the arm.
The acceleration components are obtained from Eq. 2/14 and for t 3 s are Ans.
The acceleration and its components are also shown for the 65.3 po-sition of the arm.
Plotted in the final figure is the path of the slider B over the time interval 0 t 5 s. This plot is generated by varying t in the given ex-pressions for r and . Conversion from polar to rectangular coordinates is given by Helpful Hint We see that this problem is an example of constrained motion where the cen-ter B of the slider is mechanically constrained by the rotation of the slotted arm and by engagement with the turning screw.
x r cos y r sin a (0.227)2 (0.557)2 0.601 m/s2 [a ar 2 a 2] a 0.56(0.36) 2(0.24)(0.74) 0.557 m/s2 [a r ¨ 2r ˙ ˙] ar 0.08 0.56(0.74)2 0.227 m/s2 [ar r ¨ r ˙2] v (0.24)2 (0.414)2 0.479 m/s [v vr 2 v 2] v 0.56(0.74) 0.414 m/s [v r ˙] vr 0.24 m/s [vr r ˙] ¨ 0.12t ¨3 0.12(3) 0.36 rad/s2 ˙ 0.2 0.06t2 ˙3 0.2 0.06(32) 0.74 rad/s or 3 1.14(180/) 65.3 0.2t 0.02t3 3 0.2(3) 0.02(33) 1.14 rad r ¨ 0.08 r ¨3 0.08 m/s2 r ˙ 0.08t r ˙3 0.08(3) 0.24 m/s r 0.2 0.04t2 r3 0.2 0.04(32) 0.56 m –1.5 –1 –0.5 0 t = 3 s t = 0 t = 5 s r3 = 0.56 m r3 0.5 1 0.5 0 – 0.5 x, m y, m 3 = 65.3° θ 3 θ c02.qxd 2/8/12 7:11 PM Page 69 SAMPLE PROBLEM 2/10 A tracking radar lies in the vertical plane of the path of a rocket which is coasting in unpowered flight above the atmosphere. For the instant when 30, the tracking data give r 25(104) ft, 4000 ft/sec, and 0.80 deg/sec.
The acceleration of the rocket is due only to gravitational attraction and for its particular altitude is 31.4 ft/sec2 vertically down. For these conditions determine the velocity v of the rocket and the values of and .
Solution.
The components of velocity from Eq. 2/13 are Ans.
Since the total acceleration of the rocket is g 31.4 ft/sec2 down, we can easily find its r- and -components for the given position. As shown in the figure, they are We now equate these values to the polar-coordinate expressions for ar and a which contain the unknowns and Thus, from Eq. 2/14 Ans.
Ans. ¨ 3.84(104) rad/sec2 15.70 25(104) ¨ 2(4000)0.80 180 [a r ¨ 2r ˙ ˙] r ¨ 21.5 ft/sec2 27.2 r ¨ 25(104)0.80 180 2 [ar r ¨ r ˙2] ¨.
r ¨ a 31.4 sin 30 15.70 ft/sec2 ar 31.4 cos 30 27.2 ft/sec2 v (4000)2 (3490)2 5310 ft/sec [v vr 2 v 2] v 25(104)(0.80) 180 3490 ft/sec [v r ˙] vr 4000 ft/sec [vr r ˙] ¨ r ¨ ˙ r ˙ 70 Chapter 2 Kinematics of Particles θ θ r +r + = 30° θ v = 5310 ft /sec vr = 4000 ft /sec v = 3490 ft /sec θ = 30° θ a = g = 31.4 ft /sec2 ar = –27.2 ft /sec2 a = 15.70 ft /sec2 θ Helpful Hints We observe that the angle in polar coordinates need not always be taken positive in a counterclockwise sense.
Note that the r-component of accel-eration is in the negative r-direction, so it carries a minus sign.
We must be careful to convert from deg/sec to rad/sec.
˙ c02.qxd 2/8/12 8:26 PM Page 70 PROBLEMS Introductory Problems 2/131 The position of the slider P in the rotating slotted arm OA is controlled by a power screw as shown. At the instant represented, 8 rad/s and 20 rad/s2. Also at this same instant, , , and . For this instant deter-mine the r- and -components of the acceleration of P.
Problem 2/131 2/132 A model airplane flies over an observer O with con-stant speed in a straight line as shown. Determine the signs (plus, minus, or zero) for r, , , , , and for each of the positions A, B, and C.
Problem 2/132 x y B O r A v C θ ¨ ˙ r ¨ r ˙ θ θ r A P r O r ¨ 0 r ˙ 300 mm/s r 200 mm ¨ ˙ Article 2/6 Problems 71 2/133 A car P travels along a straight road with a con-stant speed . At the instant when the angle , determine the values of in ft/sec and in deg/sec.
Problem 2/133 2/134 The sphere P travels in a straight line with speed . For the instant depicted, determine the corresponding values of and as measured rela-tive to the fixed Oxy coordinate system.
Problem 2/134 2/135 If the 10-m/s speed of the previous problem is con-stant, determine the values of and at the in-stant shown.
¨ r ¨ x y P r O 4 m 5 m 30° v θ ˙ r ˙ v 10 m/s r O 100′ v x y P θ ˙ r ˙ 60 v 65 mi/hr c02.qxd 2/8/12 7:11 PM Page 71 2/139 An internal mechanism is used to maintain a con-stant angular rate about the z-axis of the spacecraft as the telescopic booms are ex-tended at a constant rate. The length l is varied from essentially zero to 3 m. The maximum accel-eration to which the sensitive experiment modules P may be subjected is . Determine the maximum allowable boom extension rate .
Problem 2/139 2/140 The radial position of a fluid particle P in a certain centrifugal pump with radial vanes is approxi-mated by cosh Kt, where t is time and is the constant angular rate at which the impeller turns. Determine the expression for the magnitude of the total acceleration of the particle just prior to leaving the vane in terms of , R, and K.
Problem 2/140 Fixed reference axis R r r0 θ P r0 ˙ K r r0 l l z P P 1.2 m 1.2 m Ω l ˙ 0.011 m/s2 0.05 rad/s 72 Chapter 2 Kinematics of Particles 2/136 As the hydraulic cylinder rotates around O, the ex-posed length l of the piston rod P is controlled by the action of oil pressure in the cylinder. If the cylinder rotates at the constant rate and l is decreasing at the constant rate of 150 mm/s, calculate the magnitudes of the velocity v and acceleration a of end B when .
Problem 2/136 2/137 The drag racer P starts from rest at the start line S and then accelerates along the track. When it has traveled 100 m, its speed is 45 m/s. For that in-stant, determine the values of and relative to axes fixed to an observer O in the grandstand G as shown.
Problem 2/137 2/138 In addition to the information supplied in the pre-vious problem, it is known that the drag racer is ac-celerating forward at when it has traveled 100 m from the start line S. Determine the corre-sponding values of and .
¨ r ¨ 10 m/s2 G r S P 35 m O θ ˙ r ˙ 375 mm θ O l B l 125 mm 60 deg/s ˙ c02.qxd 2/8/12 7:11 PM Page 72 2/141 The slider P can be moved inward by means of the string S, while the slotted arm rotates about point O. The angular position of the arm is given by , where is in radians and t is in seconds. The slider is at when and thereafter is drawn inward at the constant rate of 0.2 m/s.
Determine the magnitude and direction (expressed by the angle relative to the x-axis) of the velocity and acceleration of the slider when .
Problem 2/141 2/142 The piston of the hydraulic cylinder gives pin A a constant velocity in the direction shown for an interval of its motion. For the instant when , determine , , , and where .
Problem 2/142 6 ″ θ O A r v r OA ¨ ˙ r ¨ r ˙ 60 v 3 ft/sec r P x S O y θ t 4 s t 0 r 1.6 m 0.8t t2 20 Article 2/6 Problems 73 2/143 The rocket is fired vertically and tracked by the radar station shown. When reaches 60°, other cor-responding measurements give the values , , and .
Calculate the magnitudes of the velocity and accel-eration of the rocket at this position.
Problem 2/143 2/144 A hiker pauses to watch a squirrel P run up a par-tially downed tree trunk. If the squirrel’s speed is when the position , determine the corresponding values of and .
Problem 2/144 r s P O A v θ 20 m 60° ˙ r ˙ s 10 m v 2 m/s r θ a v ˙ 0.02 rad/sec r ¨ 70 ft/sec2 30,000 ft r c02.qxd 2/8/12 7:11 PM Page 73 Representative Problems 2/147 Instruments located at O are part of the ground-traffic control system for a major airport. At a cer-tain instant during the takeoff roll of the aircraft P, the sensors indicate the angle and the range rate . Determine the corre-sponding speed v of the aircraft and the value of .
Problem 2/147 2/148 In addition to the information supplied in the pre-vious problem, the sensors at O indicate that . Determine the corresponding accel-eration a of the aircraft and the value of .
2/149 The cam is designed so that the center of the roller A which follows the contour moves on a limaçon de-fined by , where . If the cam does not rotate, determine the magnitude of the total acceleration of A in terms of if the slotted arm revolves with a constant counterclockwise angular rate .
Problem 2/149 r A O θ ˙ b c r b c cos ¨ r ¨ 14 ft/sec2 20° x 500′ O s r P S θ v ˙ r ˙ 140 ft/sec 50 74 Chapter 2 Kinematics of Particles 2/145 A jet plane flying at a constant speed v at an alti-tude is being tracked by radar located at O directly below the line of flight. If the angle is decreasing at the rate of 0.020 rad/s when , determine the value of at this instant and the magnitude of the velocity v of the plane.
Problem 2/145 2/146 A projectile is launched from point A with the ini-tial conditions shown. With the conventional defini-tions of r- and -coordinates relative to the Oxy coordinate system, determine r, , , , , and at the instant just alter launch. Neglect aerodynamic drag.
Problem 2/146 x y v0 O A d α ¨ r ¨ ˙ r ˙ O h r θ v r ¨ 60 h 10 km c02.qxd 2/8/12 7:11 PM Page 74 2/150 The slotted arm OA forces the small pin to move in the fixed spiral guide defined by . Arm OA starts from rest at and has a constant counterclockwise angular acceleration . De-termine the magnitude of the acceleration of the pin P when .
Problem 2/150 2/151 A rocket is tracked by radar from its launching point A. When it is 10 seconds into its flight, the following radar measurements are recorded: , , , , , and . For this instant determine the angle between the horizon-tal and the direction of the trajectory of the rocket and find the magnitudes of its velocity v and accel-eration a.
Problem 2/151 θ A r r θ β 0.0341 rad/s2 ¨ 0.0788 rad/s ˙ 22 r ¨ 4.66 m/s2 r ˙ 500 m/s r 2200 m r A P O θ 3/4 ¨ /4 r K Article 2/6 Problems 75 2/152 For an interval of motion the drum of radius b turns clockwise at a constant rate in radians per second and causes the carriage P to move to the right as the unwound length of the connecting cable is shortened. Use polar coordinates r and and derive expressions for the velocity v and accel-eration a of P in the horizontal guide in terms of the angle . Check your solution by a direct differ-entiation with time of the relation .
Problem 2/152 2/153 Car A is moving with constant speed v on the straight and level highway. The police officer in the stationary car P attempts to measure the speed v with radar. If the radar measures “line-of sight” ve-locity, what velocity will the officer observe?
Evaluate your general expression for the values , , and , and draw any appropriate conclusions.
Problem 2/153 L A v D P D 20 ft L 500 ft v 70 mi/hr v θ ω r P h x b x2 h2 r2 c02.qxd 2/8/12 7:11 PM Page 75 2/156 The member OA of the industrial robot telescopes and pivots about the fixed axis at point O. At the instant shown, , , , , , a n d . Determine the magnitudes of the velocity and acceleration of joint A of the robot. Also, sketch the velocity and acceleration of A and determine the angles which these vectors make with the posi-tive x-axis. The base of the robot does not revolve about a vertical axis.
Problem 2/156 2/157 The robot arm is elevating and extending simulta-neously. At a given instant, , constant, , , and .
Compute the magnitudes of the velocity v and acceleration a of the gripped part P. In addition, express v and a in terms of the unit vectors i and j.
Problem 2/157 l 0.75 m y P θ O x l ¨ 0.3 m/s2 l ˙ 0.2 m/s l 0.5 m ˙ 10 deg/s 30 P A x O y 15° 1.1 m 0.9 m θ 6 m/s2 OA ¨ OA ˙ 0.5 m/s OA 0.9 m rad/s2 0.8 ¨ 1.2 rad/s ˙ 60 76 Chapter 2 Kinematics of Particles 2/154 The hydraulic cylinder gives pin A a constant veloc-ity along its axis for an interval of mo-tion and, in turn, causes the slotted arm to rotate about O. Determine the values of , , and for the instant when . (Hint: Recognize that all acceleration components are zero when the velocity is constant.) Problem 2/154 2/155 The particle P moves along the parabolic surface shown. When , the particle speed is . For this instant determine the corre-sponding values of r, , , and . Both x and y are in meters.
Problem 2/155 x y y = 4x2 P r O θ ˙ r ˙ v 5 m/s x 0.2 m r v A 300 mm 30° O θ 30 ¨ r ¨ r ˙ v 2 m/s c02.qxd 2/8/12 7:11 PM Page 76 2/158 During a portion of a vertical loop, an airplane flies in an arc of radius with a constant speed . When the airplane is at A, the angle made by v with the horizontal is , and radar tracking gives and . Calculate , , , and for this instant.
Problem 2/158 2/159 The particle P starts from rest at point O at time , and then undergoes a constant tangential ac-celeration as it negotiates the circular slot in the counterclockwise direction. Determine r, , , and as functions of time over the first revolution.
Problem 2/159 x y r O P b θ ˙ r ˙ at t 0 r r A B v θ θ β ¨ ar v vr 30 r 800 m 30 v 400 km/h 600 m Article 2/6 Problems 77 2/160 The low-flying aircraft P is traveling at a constant speed of 360 km/h in the holding circle of radius 3 km. For the instant shown, determine the quanti-ties r, , , , , and relative to the fixed x-y coor-dinate system, which has its origin on a mountaintop at O. Treat the system as two-dimensional.
Problem 2/160 2/161 Pin A moves in a circle of 90-mm radius as crank AC revolves at the constant rate . The slotted link rotates about point O as the rod at-tached to A moves in and out of the slot. For the position , determine , , , and .
Problem 2/161 300 mm r r θ θ β A 90 mm C O ¨ ˙ r ¨ r ˙ 30 ˙ 60 rad/s θ 16 km 12 km 3 km r P C O v y x ¨ ˙ r ¨ r ˙ c02.qxd 2/8/12 7:11 PM Page 77 2/164 At time , the baseball player releases a ball with the initial conditions shown in the figure.
Determine the quantities r, , , , , and , all relative to the coordinate system shown, at time .
Problem 2/164 6′ = 30° v0 = 100 ft/sec y x α t 0.5 sec x-y ¨ ˙ r ¨ r ˙ t 0 78 Chapter 2 Kinematics of Particles 2/162 A fireworks shell P fired in a vertical trajectory has a y-acceleration given by , where the latter term is due to aerodynamic drag. If the speed of the shell is 15 m/s at the instant shown, deter-mine the corresponding values of r, , , , , and .
The drag parameter k has a constant value of .
Problem 2/162 2/163 An earth satellite traveling in the elliptical orbit shown has a velocity as it passes the end of the semiminor axis at A. The accelera-tion of the satellite at A is due to gravitational attraction and is directed from A to O. For position A calculate the values of , , , and .
Problem 2/163 8400 mi 7275 mi 4200 mi A P O r v θ ¨ ˙ r ¨ r ˙ 32.23[3959/8400]2 7.159 ft/sec2 v 12,149 mi/hr 200 m 100 m y x r P θ O v 0.01 m1 ¨ ˙ r ¨ r ˙ ay g kv2 c02.qxd 2/8/12 7:11 PM Page 78 2/7 Space Curvilinear Motion The general case of three-dimensional motion of a particle along a space curve was introduced in Art. 2/1 and illustrated in Fig. 2/1. Three coordinate systems, rectangular (x-y-z), cylindrical (r--z), and spherical (R--), are commonly used to describe this motion. These systems are indicated in Fig. 2/16, which also shows the unit vectors for the three co-ordinate systems. Before describing the use of these coordinate systems, we note that a path-variable description, using n- and t-coordinates, which we devel-oped in Art. 2/5, can be applied in the osculating plane shown in Fig.
2/1. We defined this plane as the plane which contains the curve at the location in question. We see that the velocity v, which is along the tan-gent t to the curve, lies in the osculating plane. The acceleration a also lies in the osculating plane. As in the case of plane motion, it has a com-ponent at tangent to the path due to the change in magnitude of the velocity and a component an v2/ normal to the curve due to the change in direction of the velocity. As before, is the radius of curva-ture of the path at the point in question and is measured in the osculat-ing plane. This description of motion, which is natural and direct for many plane-motion problems, is awkward to use for space motion be-cause the osculating plane continually shifts its orientation. We will confine our attention, therefore, to the three fixed coordinate systems shown in Fig. 2/16.
Rectangular Coordinates (x-y-z) The extension from two to three dimensions offers no particular dif-ficulty. We merely add the z-coordinate and its two time derivatives to the two-dimensional expressions of Eqs. 2/6 so that the position vector R, the velocity v, and the acceleration a become (2/15) Note that in three dimensions we are using R in place of r for the posi-tion vector.
Cylindrical Coordinates (r--z) If we understand the polar-coordinate description of plane motion, then there should be no difficulty with cylindrical coordinates because all that is required is the addition of the z-coordinate and its two time derivatives. The position vector R to the particle for cylindrical coordi-nates is simply R rer zk a v ˙ R ¨ x ¨i y ¨j z ¨k v R ˙ x ˙i y ˙j z ˙k R xi yj zk v ˙ Article 2/7 Space Curvilinear Motion 79 Figure 2/16 In a variation of spherical coordinates commonly used, angle is replaced by its complement.
z z R P k i j er eR O r x y R θ θ φ eφ eθ eθ φ c02.qxd 2/8/12 7:11 PM Page 79 In place of Eq. 2/13 for plane motion, we can write the velocity as (2/16) where vr v vz v Similarly, the acceleration is written by adding the z-component to Eq.
2/14, which gives us (2/17) where ar a az a Whereas the unit vectors er and e have nonzero time derivatives due to the changes in their directions, we note that the unit vector k in the z-direction remains fixed in direction and therefore has a zero time derivative.
Spherical Coordinates (R--) Spherical coordinates R, , are utilized when a radial distance and two angles are utilized to specify the position of a particle, as in the case of radar measurements, for example. Derivation of the expression for the velocity v is easily obtained, but the expression for the acceleration a is more complex because of the added geometry. Consequently, only the results will be cited here. First we designate unit vectors eR, e, e as shown in Fig. 2/16. Note that the unit vector eR is in the direction in which the particle P would move if R increases but and are held con-stant. The unit vector e is in the direction in which P would move if increases while R and are held constant. Finally, the unit vector e is in the direction in which P would move if increases while R and are held constant. The resulting expressions for v and a are (2/18) where vR v v R ˙ R ˙ cos R ˙ v vReR ve ve ar 2 a 2 az 2 z ¨ r ¨ 2r ˙ ˙ 1 r d dt (r2 ˙) r ¨ r ˙2 a (r ¨ r ˙2)er (r ¨ 2r ˙ ˙)e z ¨k vr 2 v 2 vz 2 z ˙ r ˙ r ˙ v r ˙er r ˙e z ˙k 80 Chapter 2 Kinematics of Particles For a complete derivation of v and a in spherical coordinates, see the first author’s book Dynamics, 2nd edition, 1971, or SI Version, 1975 (John Wiley & Sons, Inc.).
c02.qxd 2/13/12 2:18 PM Page 80 and (2/19) where aR a a Linear algebraic transformations between any two of the three coordinate-system expressions for velocity or acceleration can be devel-oped. These transformations make it possible to express the motion component in rectangular coordinates, for example, if the components are known in spherical coordinates, or vice versa. These transforma-tions are easily handled with the aid of matrix algebra and a simple computer program.
1 R d dt (R2 ˙) R ˙2 sin cos cos R d dt (R2 ˙) 2R ˙ ˙ sin R ¨ R ˙2 R ˙2 cos2 a aReR ae ae Article 2/7 Space Curvilinear Motion 81 These coordinate transformations are developed and illustrated in the first author’s book Dynamics, 2nd edition, 1971, or SI Version, 1975 (John Wiley & Sons, Inc.).
A portion of the track of this amusement-park ride is in the shape of a helix whose axis is horizontal.
© Howard Sayer/Alamy c02.qxd 2/8/12 7:11 PM Page 81 SAMPLE PROBLEM 2/11 The power screw starts from rest and is given a rotational speed which in-creases uniformly with time t according to kt, where k is a constant. Deter-mine the expressions for the velocity v and acceleration a of the center of ball A when the screw has turned through one complete revolution from rest. The lead of the screw (advancement per revolution) is L.
Solution.
The center of ball A moves in a helix on the cylindrical surface of ra-dius b, and the cylindrical coordinates r, , z are clearly indicated.
Integrating the given relation for gives For one revolution from rest we have giving Thus, the angular rate at one revolution is The helix angle of the path followed by the center of the ball governs the relation between the - and z-components of velocity and is given by tan L/(2b). Now from the figure we see that v v cos . Substituting v from Eq. 2/16 gives v v/cos With cos obtained from tan and with we have for the one-revolution position Ans.
The acceleration components from Eq. 2/17 become Now we combine the components to give the magnitude of the total acceler-ation, which becomes Ans. (b tan ) ¨ b L 2b k kL 2 az d dt (vz) d dt (v tan ) d dt (b ˙ tan ) [az z ¨ v ˙z] a bk 2(0)(2k) bk [a r ¨ 2r ˙ ˙] ar 0 b(2k)2 4bk [ar r ¨ r ˙2] 2k, ˙ b ˙/cos .
b ˙ r ˙ ˙ kt k(2/k) 2k t 2/k 2 1 2 kt2 ˙ dt 1 2 kt2.
˙ ˙ ˙ 82 Chapter 2 Kinematics of Particles z b r A 2r0 θ · r A ar az θ aθ γ z v Helpful Hints We must be careful to divide the lead L by the circumference 2b and not the diameter 2b to obtain tan .
If in doubt, unwrap one turn of the helix traced by the center of the ball.
Sketch a right triangle and recall that for tan a/b the cosine of becomes The negative sign for ar is consistent with our previous knowledge that the normal component of accelera-tion is directed toward the center of curvature.
b/a2 b2.
c02.qxd 2/8/12 7:11 PM Page 82 SAMPLE PROBLEM 2/12 An aircraft P takes off at A with a velocity v0 of 250 km/h and climbs in the vertical y-z plane at the constant 15 angle with an acceleration along its flight path of 0.8 m/s2. Flight progress is monitored by radar at point O. (a) Resolve the velocity of P into cylindrical-coordinate components 60 seconds after takeoff and find and for that instant. (b) Resolve the velocity of the aircraft P into spherical-coordinate components 60 seconds after takeoff and find and for that instant.
Solution. (a) The accompanying figure shows the velocity and ac-celeration vectors in the y-z plane. The takeoff speed is and the speed after 60 seconds is The distance s traveled after takeoff is The y-coordinate and associated angle are From the figure (b) of x-y projections, we have Ans.
So Ans.
Finally Ans.
(b) Refer to the accompanying figure (c), which shows the x-y plane and various velocity components projected into the vertical plane con-taining r and R. Note that From the figure, Ans.
Ans.
Ans.
˙ 6.95 6360 1.093(103) rad/s v R ˙ 30.4 cos 13.19 99.2 sin 13.19 6.95 m/s ˙ 8.88(103) rad/s, as in part (a) vR R ˙ 99.2 cos 13.19 30.4 sin 13.19 103.6 m/s R r2 z2 61902 14512 6360 m tan1 z r tan1 1451 6190 13.19 z y tan 15 5420 tan 15 1451 m z ˙ vz v sin 15 117.4 sin 15 30.4 m/s ˙ 55.0 6190 8.88(103) rad/s v r ˙ vxy cos 113.4 cos 61.0 55.0 m/s vr r ˙ vxy sin 113.4 sin 61.0 99.2 m/s vxy v cos 15 117.4 cos 15 113.4 m/s r 30002 54202 6190 m tan1 5420 3000 61.0 y 5610 cos 15 5420 m s s0 v0 t 1 2 at2 0 69.4(60) 1 2 (0.8)(60)2 5610 m v v0 at 69.4 0.8(60) 117.4 m/s v0 250 3.6 69.4 m/s ˙ ˙, R ˙, z ˙ ˙, r ˙, Article 2/7 Space Curvilinear Motion 83 P O v0 x y y′ z z r R A 3 km 15° z′ P a O v0 v vxy vr x y y y O y x z z z′ A y′ r r R s 3000 m 3000 m 15° = 61.0° = 13.19° v (a) (b) (c) y = 5420 m R z y O x r = 6190 m r = 6190 m 3000 m z =1451 m 99.2 m/s 99.2 m/s 55.0 m/s = 61.0° vxy = 113.4 m/s vR = R · vz = R · v c02.qxd 2/8/12 7:11 PM Page 83 2/168 The radar antenna at P tracks the jet aircraft A, which is flying horizontally at a speed u and an alti-tude h above the level of P. Determine the expres-sions for the components of the velocity in the spherical coordinates of the antenna motion.
Problem 2/168 2/169 The rotating element in a mixing chamber is given a periodic axial movement while it is rotating at the constant angular velocity .
Determine the expression for the maximum magni-tude of the acceleration of a point A on the rim of radius r. The frequency n of vertical oscillation is constant.
Problem 2/169 ω A r z z = z0 sin 2 nt π ˙ z z0 sin 2nt z h y x b u A P θ φ 84 Chapter 2 Kinematics of Particles PROBLEMS Introductory Problems 2/165 The velocity and acceleration of a particle are given for a certain instant by and . Determine the angle be-tween v and a, , and the radius of curvature in the osculating plane.
2/166 A projectile is launched from point O with an initial speed directed as shown in the fig-ure. Compute the x-, y-, and z-components of posi-tion, velocity, and acceleration 20 seconds after launch. Neglect aerodynamic drag.
Problem 2/166 2/167 An amusement ride called the “corkscrew” takes the passengers through the upside-down curve of a horizontal cylindrical helix. The velocity of the cars as they pass position A is 15 m/s, and the compo-nent of their acceleration measured along the tan-gent to the path is g cos at this point. The effective radius of the cylindrical helix is 5 m, and the helix angle is . Compute the magnitude of the ac-celeration of the passengers as they pass position A.
Problem 2/167 = 40° 5 m Vert.
A Horiz.
Horiz.
γ 40 60° 20° v0 z y x O v0 500 ft/sec v ˙ a 3i j 5k m/s2 v 6i 3j 2k m/s c02.qxd 2/8/12 7:11 PM Page 84 Representative Problems 2/170 The vertical shaft of the industrial robot rotates at the constant rate . The length h of the vertical shaft has a known time history, and this is true of its time derivatives and as well. Likewise, the values of l, , and are known. Determine the magnitudes of the velocity and acceleration of point P. The lengths and are fixed.
Problem 2/170 2/171 The car A is ascending a parking-garage ramp in the form of a cylindrical helix of 24-ft radius rising 10 ft for each half turn. At the position shown the car has a speed of 15 mi/hr, which is decreasing at the rate of 2 mi/hr per second. Determine the r-, -, and z-components of the acceleration of the car.
Problem 2/171 z A 10′ 24′ 24′ r θ ω h z h0 P l0 l l0 h0 l ¨ l ˙ h ¨ h ˙ Article 2/7 Problems 85 2/172 An aircraft takes off at A and climbs at a steady angle with a slope of 1 to 2 in the vertical y-z plane at a constant speed . The aircraft is tracked by radar at O. For the position B, deter-mine the values of , , and .
Problem 2/172 2/173 For the conditions of Prob. 2/172, find the values of , , and for the radar tracking coordinates as the aircraft passes point B. Use the results cited for Prob. 2/172.
2/174 The rotating nozzle sprays a large circular area and turns with the constant angular rate . Parti-cles of water move along the tube at the constant rate relative to the tube. Write expressions for the magnitudes of the velocity and acceleration of a water particle P for a given position l in the rotating tube.
Problem 2/174 z l P β θ = K θ · l ˙ c ˙ K ¨ ¨ R ¨ 300 m 1 2 500 m z A R v B y x O φ · θ ˙ ˙ R ˙ v 400 km/h c02.qxd 2/8/12 7:11 PM Page 85 2/177 The base structure of the firetruck ladder rotates about a vertical axis through O with a constant an-gular velocity . At the same time, the ladder unit OB elevates at a constant rate 7 , and section AB of the ladder extends from within section OA at the constant rate of 0.5 m/s.
At the instant under consideration, , , and . Determine the magni-tudes of the velocity and acceleration of the end B of the ladder.
Problem 2/177 2/178 The member OA of the industrial robot telescopes.
At the instant represented, , , , , , and . The base of the robot is revolving at the constant rate . Calculate the mag-nitudes of the velocity and acceleration of joint A.
Problem 2/178 P A x O z 15° 1.1 m 0.9 m φ ω 1.4 rad/s OA ¨ 6 m/s2 OA ˙ 0.5 m/s OA 0.9 m ¨ 0.8 rad/s2 1.2 rad/s ˙ 60 C O C Ω φ A B θ AB 6 m OA 9 m 30 deg/s ˙ 10 deg/s 86 Chapter 2 Kinematics of Particles 2/175 The small block P travels with constant speed v in the circular path of radius r on the inclined surface.
If at time , determine the x-, y-, and z-components of velocity and acceleration as func-tions of time.
Problem 2/175 2/176 An aircraft is flying in a horizontal circle of radius b with a constant speed u at an altitude h. A radar tracking unit is located at C. Write expressions for the components of the velocity of the aircraft in the spherical coordinates of the radar station for a given position .
Problem 2/176 C h z y x u O R b r φ θ β z x n r P 30° v y t 0 0 c02.qxd 2/8/12 7:11 PM Page 86 2/179 Consider the industrial robot of the previous prob-lem. The telescoping member OA is now fixed in length at 0.9 m. The other conditions remain at , , , , , and angle OAP is locked at 105°. Determine the magnitudes of the velocity and acceleration of the end point P.
2/180 In a design test of the actuating mechanism for a telescoping antenna on a spacecraft, the supporting shaft rotates about the fixed z-axis with an angular rate . Determine the R-, -, and -components of the acceleration a of the end of the antenna at the instant when and if the rates rad/s, , and are con-stant during the motion.
Problem 2/180 L R y x z θ θ β φ θ · L ˙ 0.9 m/s ˙ 3 2 rad/s ˙ 2 45 L 1.2 m ˙ ˙ 0 1.4 rad/s ¨ 0.8 rad/s2 ˙ 1.2 rad/s 60 Article 2/7 Problems 87 2/181 In the design of an amusement-park ride, the cars are attached to arms of length R which are hinged to a central rotating collar which drives the assem-bly about the vertical axis with a constant angular rate . The cars rise and fall with the track ac-cording to the relation . Find the R-, -, and -components of the velocity v of each car as it passes the position .
Problem 2/181 2/182 The particle P moves down the spiral path which is wrapped around the surface of a right circular cone of base radius b and altitude h. The angle be-tween the tangent to the curve at any point and a horizontal tangent to the cone at this point is con-stant. Also the motion of the particle is controlled so that is constant. Determine the expression for the radial acceleration of the particle for any value of .
Problem 2/182 x y z r P h b ar ˙ /4 rad z (h/2)(1 cos 2) ˙ y v ω h R z φ θ x h c02.qxd 2/8/12 7:11 PM Page 87 2/8 Relative Motion (Translating Axes) In the previous articles of this chapter, we have described particle motion using coordinates referred to fixed reference axes. The dis-placements, velocities, and accelerations so determined are termed ab-solute. It is not always possible or convenient, however, to use a fixed set of axes to describe or to measure motion. In addition, there are many engineering problems for which the analysis of motion is simpli-fied by using measurements made with respect to a moving reference system. These measurements, when combined with the absolute mo-tion of the moving coordinate system, enable us to determine the ab-solute motion in question. This approach is called a relative-motion analysis.
Choice of Coordinate System The motion of the moving coordinate system is specified with re-spect to a fixed coordinate system. Strictly speaking, in Newtonian me-chanics, this fixed system is the primary inertial system, which is assumed to have no motion in space. For engineering purposes, the fixed system may be taken as any system whose absolute motion is negligible for the problem at hand. For most earthbound engineering problems, it is sufficiently precise to take for the fixed reference system a set of axes attached to the earth, in which case we neglect the motion of the earth.
For the motion of satellites around the earth, a nonrotating coordinate system is chosen with its origin on the axis of rotation of the earth. For interplanetary travel, a nonrotating coordinate system fixed to the sun would be used. Thus, the choice of the fixed system depends on the type of problem involved.
We will confine our attention in this article to moving reference systems which translate but do not rotate. Motion measured in rotating systems will be discussed in Art. 5/7 of Chapter 5 on rigid-body kine-matics, where this approach finds special but important application. We will also confine our attention here to relative-motion analysis for plane motion.
Vector Representation Now consider two particles A and B which may have separate curvilinear motions in a given plane or in parallel planes, Fig. 2/17. We will arbitrarily attach the origin of a set of translating (nonrotating) axes x-y to particle B and observe the motion of A from our moving po-sition on B. The position vector of A as measured relative to the frame x-y is rA/B xi yj, where the subscript notation “A/B” means “A rel-ative to B” or “A with respect to B.” The unit vectors along the x- and y-axes are i and j, and x and y are the coordinates of A measured in the x-y frame. The absolute position of B is defined by the vector rB mea-sured from the origin of the fixed axes X-Y. The absolute position of A is seen, therefore, to be determined by the vector rA rB rA/B Figure 2/17 88 Chapter 2 Kinematics of Particles j i y A B O rA rB rA/B Y X x Relative motion is a critical issue in the midair refueling of aircraft.
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c02.qxd 2/8/12 7:11 PM Page 88 Article 2/8 Relative Motion (Translating Axes) 89 We now differentiate this vector equation once with respect to time to obtain velocities and twice to obtain accelerations. Thus, (2/20) (2/21) In Eq. 2/20 the velocity which we observe A to have from our position at B attached to the moving axes x-y is vA/B This term is the velocity of A with respect to B. Similarly, in Eq. 2/21 the acceleration which we observe A to have from our nonrotating posi-tion on B is This term is the acceleration of A with respect to B. We note that the unit vectors i and j have zero de-rivatives because their directions as well as their magnitudes remain unchanged. (Later when we discuss rotating reference axes, we must account for the derivatives of the unit vectors when they change di-rection.) Equation 2/20 (or 2/21) states that the absolute velocity (or acceler-ation) of A equals the absolute velocity (or acceleration) of B plus, vecto-rially, the velocity (or acceleration) of A relative to B. The relative term is the velocity (or acceleration) measurement which an observer at-tached to the moving coordinate system x-y would make. We can express the relative-motion terms in whatever coordinate system is convenient— rectangular, normal and tangential, or polar—and the formulations in the preceding articles can be used for this purpose. The appropriate fixed system of the previous articles becomes the moving system in the present article.
Additional Considerations The selection of the moving point B for attachment of the reference coordinate system is arbitrary. As shown in Fig. 2/18, point A could be used just as well for the attachment of the moving system, in which case the three corresponding relative-motion equations for position, velocity, and acceleration are It is seen, therefore, that rB/A rA/B, vB/A vA/B, and aB/A aA/B.
In relative-motion analysis, it is important to realize that the ac-celeration of a particle as observed in a translating system x-y is the same as that observed in a fixed system X-Y if the moving system has a constant velocity. This conclusion broadens the application of New-ton’s second law of motion (Chapter 3). We conclude, consequently, that a set of axes which has a constant absolute velocity may be used in place of a “fixed” system for the determination of accelerations. A translating reference system which has no acceleration is called an in-ertial system.
rB rA rB/A vB vA vB/A aB aA aB/A y ¨j.
x ¨i v ˙A/B r ¨A/B y ˙j.
x ˙i r ˙A/B r ¨A r ¨B r ¨A/B or aA aB aA/B r ˙A r ˙B r ˙A/B or vA vB vA/B y A B O rA rB rB/A Y X x Figure 2/18 c02.qxd 2/8/12 7:11 PM Page 89 SAMPLE PROBLEM 2/13 Passengers in the jet transport A flying east at a speed of 800 km/h observe a second jet plane B that passes under the transport in horizontal flight. Al-though the nose of B is pointed in the 45 northeast direction, plane B appears to the passengers in A to be moving away from the transport at the 60 angle as shown. Determine the true velocity of B.
Solution.
The moving reference axes x-y are attached to A, from which the relative observations are made. We write, therefore, Next we identify the knowns and unknowns. The velocity vA is given in both mag-nitude and direction. The 60 direction of vB/A, the velocity which B appears to have to the moving observers in A, is known, and the true velocity of B is in the 45 direction in which it is heading. The two remaining unknowns are the magni-tudes of vB and vB/A. We may solve the vector equation in any one of three ways.
(I) Graphical.
We start the vector sum at some point P by drawing vA to a convenient scale and then construct a line through the tip of vA with the known direction of vB/A. The known direction of vB is then drawn through P, and the in-tersection C yields the unique solution enabling us to complete the vector trian-gle and scale off the unknown magnitudes, which are found to be Ans.
(II) Trigonometric.
A sketch of the vector triangle is made to reveal the trigonometry, which gives Ans.
(III) Vector Algebra.
Using unit vectors i and j, we express the velocities in vector form as Substituting these relations into the relative-velocity equation and solving sepa-rately for the i and j terms give Solving simultaneously yields the unknown velocity magnitudes Ans.
It is worth noting the solution of this problem from the viewpoint of an observer in B. With reference axes attached to B, we would write vA vB vA/B. The ap-parent velocity of A as observed by B is then vA/B, which is the negative of vB/A.
vB/A 586 km/h and vB 717 km/h (j-terms) vB sin 45 vB/A sin 60 (i-terms) vB cos 45 800 vB/A cos 60 vB/A (vB/A cos 60)(i) (vB/A sin 60)j vA 800i km/h vB (vB cos 45)i (vB sin 45)j vB sin 60 vA sin 75 vB 800 sin 60 sin 75 717 km/h vB/A 586 km/h and vB 717 km/h vB vA vB/A 90 Chapter 2 Kinematics of Particles B A x y 60° 45° 60° 60° 45° 45° 75° 60° P P C Dir. of vB/A Dir. of vB vB vB/A vA vA = 800 km/h vA Helpful Hints We treat each airplane as a particle.
We assume no side slip due to cross wind.
Students should become familiar with all three solutions.
We must be prepared to recognize the appropriate trigonometric rela-tion, which here is the law of sines.
We can see that the graphical or trigonometric solution is shorter than the vector algebra solution in this particular problem.
c02.qxd 2/8/12 7:11 PM Page 90 Article 2/8 Relative Motion (Translating Axes) 91 SAMPLE PROBLEM 2/14 Car A is accelerating in the direction of its motion at the rate of 3 ft/sec2.
Car B is rounding a curve of 440-ft radius at a constant speed of 30 mi/hr. Deter-mine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi/hr for the positions represented.
Solution.
We choose nonrotating reference axes attached to car A since the motion of B with respect to A is desired.
Velocity.
The relative-velocity equation is and the velocities of A and B for the position considered have the magnitudes The triangle of velocity vectors is drawn in the sequence required by the equa-tion, and application of the law of cosines and the law of sines gives Ans.
Acceleration.
The relative-acceleration equation is The acceleration of A is given, and the acceleration of B is normal to the curve in the n-direction and has the magnitude The triangle of acceleration vectors is drawn in the sequence required by the equation as illustrated. Solving for the x- and y-components of aB/A gives us Ans.
The direction of aB/A may be specified by the angle which, by the law of sines, becomes Ans.
4.4 sin 2.34 sin 30 sin1 4.4 2.34 0.5 110.2 from which aB/A (0.810)2 (2.2)2 2.34 ft/sec2 (aB/A)y 4.4 sin 30 2.2 ft/sec2 (aB/A)x 4.4 cos 30 3 0.810 ft/sec2 aB (44)2/440 4.4 ft/sec2 [an v2/] aB aA aB/A vB/A 58.2 ft/sec 40.9 vA 45 5280 602 45 44 30 66 ft/sec vB 30 44 30 44 ft/sec vB vA vB/A y A B n 30° 440′ x 60° 30° vB/A aB/A vB = 44 ft/sec aB = 4.4 ft/sec2 aA = 3 ft/sec2 vA = 66 ft/sec θ β Helpful Hints Alternatively, we could use either a graphical or a vector algebraic solution.
Be careful to choose between the two values 69.8 and 180 69.8 110.2.
Suggestion: To gain familiarity with the manipulation of vector equations, it is suggested that the student rewrite the relative-motion equations in the form vB/A vB vA and aB/A aB aA and redraw the vector polygons to conform with these alternative relations.
Caution: So far we are only prepared to handle motion relative to nonrotating axes. If we had attached the reference axes rigidly to car B, they would rotate with the car, and we would find that the velocity and acceleration terms relative to the rotating axes are not the negative of those measured from the nonrotating axes moving with A. Rotating axes are treated in Art. 5/7.
c02.qxd 2/8/12 7:11 PM Page 91 PROBLEMS Introductory Problems 2/183 Car A rounds a curve of 150-m radius at a constant speed of 54 km/h. At the instant represented, car B is moving at 81 km/h but is slowing down at the rate of . Determine the velocity and accelera-tion of car A as observed from car B.
Problem 2/183 2/184 For the instant represented, car A is rounding the circular curve at a constant speed of 30 mi/hr, while car B is slowing down at the rate of 5 mi/hr per sec-ond. Determine the magnitude of the acceleration that car A appears to have to an observer in car B.
Problem 2/184 B A 500′ 30° x y x B A y 150 m 3 m/s2 2/185 The passenger aircraft B is flying east with a veloc-ity . A military jet traveling south with a velocity passes under B at a slightly lower altitude. What velocity does A appear to have to a passenger in B, and what is the direc-tion of that apparent velocity?
Problem 2/185 2/186 A marathon participant R is running north at a speed . A wind is blowing in the di-rection shown at a speed . (a) Deter-mine the velocity of the wind relative to the runner. (b) Repeat for the case when the runner is moving directly to the south at the same speed. Ex-press all answers both in terms of the unit vectors i and j and as magnitudes and compass directions.
Problem 2/186 N 35° x R y vR vW vW 15 mi/hr vR 10 mi/hr x y A B vA vB N vA 1200 km/h vB 800 km/h 92 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 92 Article 2/8 Problems 93 Representative Problems 2/189 A small ship capable of making a speed of 6 knots through still water maintains a heading due east while being set to the south by an ocean current.
The actual course of the boat is from A to B, a dis-tance of 10 nautical miles that requires exactly 2 hours. Determine the speed of the current and its direction measured clockwise from the north.
Problem 2/189 2/190 Hockey player A carries the puck on his stick and moves in the direction shown with a speed . In passing the puck to his stationary teammate B, by what angle should the direction of his shot trail the line of sight if he launches the puck with a speed of 7 m/s relative to himself?
Problem 2/190 45° vA B A α vA 4 m/s N(0°) A E(90°) 10° B vC 2/187 A small aircraft A is about to land with an airspeed of 80 mi/hr. If the aircraft is encountering a steady side wind of speed as shown, at what angle should the pilot direct the aircraft so that the absolute velocity is parallel to the runway?
What is the speed at touchdown?
Problem 2/187 2/188 The car A has a forward speed of 18 km/h and is ac-celerating at . Determine the velocity and ac-celeration of the car relative to observer B, who rides in a nonrotating chair on the Ferris wheel.
The angular rate of the Ferris wheel is constant.
Problem 2/188 Ω = 3 rev/min R = 9 m B A 45° y x 3 rev/min 3 m/s2 N α A vW vW 10 mi/hr c02.qxd 2/8/12 7:11 PM Page 93 2/191 A ferry is moving due east and encounters a south-west wind of speed as shown. The ex-perienced ferry captain wishes to minimize the effects of the wind on the passengers who are on the outdoor decks. At what speed should he proceed?
Problem 2/191 2/192 A drop of water falls with no initial speed from point A of a highway overpass. After dropping 6 m, it strikes the windshield at point B of a car which is traveling at a speed of 100 km/h on the horizontal road. If the windshield is inclined 50° from the ver-tical as shown, determine the angle relative to the normal n to the windshield at which the water drop strikes.
Problem 2/192 100 km / h 50° 6 m n A B t vW vB B N 40° vB vW 10 m/s 2/193 While scrambling directly toward the sideline at a speed 20 ft/sec, the football quarterback Q throws a pass toward the stationary receiver R. At what angle should the quarterback release the ball?
The speed of the ball relative to the quarterback is 60 ft/sec. Treat the problem as two-dimensional.
Problem 2/193 2/194 The speedboat B is cruising to the north at 75 mi/hr when it encounters an eastward current of speed but does not change its heading (relative to the water). Determine the subsequent velocity of the boat relative to the wind and express your result as a magnitude and compass direction.
The current affects the motion of the boat; the southwesterly wind of speed does not.
Problem 2/194 vC vW N B 30° vW 20 mi/hr vC 10 mi/hr 20 yd 5 yd Q R vQ v α vQ 94 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 94 Article 2/8 Problems 95 2/198 The spacecraft S approaches the planet Mars along a trajectory b-b in the orbital plane of Mars with an absolute velocity of 19 km/s. Mars has a velocity of 24.1 km/s along its trajectory a-a. Determine the angle between the line of sight S-M and the tra-jectory b-b when Mars appears from the spacecraft to be approaching it head on.
Problem 2/198 2/199 Two ships A and B are moving with constant speeds and , respectively, along straight inter-secting courses. The navigator of ship B notes the time rates of change of the separation distance r between the ships and the bearing angle . Show that and .
Problem 2/199 θ r B A vA vB r ¨ r ˙ 2 ¨ 2r ˙ ˙/r vB vA a b 15° a b M S 19 km /s 24.1 km /s β 2/195 Starting from the relative position shown, aircraft B is to rendezvous with the refueling tanker A. If B is to arrive in close proximity to A in a two-minute time interval, what absolute velocity vector should B acquire and maintain? The velocity of the tanker A is 300 mi/hr along the constant-altitude path shown.
Problem 2/195 2/196 Airplane A is flying horizontally with a constant speed of 200 km/h and is towing the glider B, which is gaining altitude. If the tow cable has a length and is increasing at the constant rate of 5 degrees per second, determine the magnitudes of the velocity v and acceleration a of the glider for the instant when .
Problem 2/196 2/197 If the airplane in Prob. 2/196 is increasing its speed in level flight at the rate of 5 km/h each second and is unreeling the glider tow cable at the constant rate while remains constant, determine the magnitude of the acceleration of the glider B.
r ˙ 2 m/s B A r vA θ 15 r 60 m y x 10,000′ 2000′ A B c02.qxd 2/8/12 7:11 PM Page 95 2/200 Airplane A is flying north with a constant horizon-tal velocity of 500 km/h. Airplane B is flying south-west at the same altitude with a velocity of 500 km/h.
From the frame of reference of A, determine the magnitude of the apparent or relative velocity of B. Also find the magnitude of the apparent velocity with which B appears to be moving sideways or normal to its centerline. Would the results be dif-ferent if the two airplanes were flying at different but constant altitudes?
Problem 2/200 2/201 In Prob. 2/200 if aircraft A is accelerating in its northward direction at the rate of 3 km/h each sec-ond while aircraft B is slowing down at the rate of 4 km/h each second in its southwesterly direction, determine the acceleration in which B appears to have to an observer in A and specify its direction ( ) measured clockwise from the north.
m/s2 45° A B vA vB N vn vr 2/202 The shuttle orbiter A is in a circular orbit of alti-tude 200 mi, while spacecraft B is in a geosynchro-nous circular orbit of altitude 22,300 mi. Determine the acceleration of B relative to a nonrotating observer in the shuttle A. Use for the surface-level gravitational acceleration and mi for the radius of the earth.
Problem 2/202 2/203 After starting from the position marked with the “x”, a football receiver B runs the slant-in pattern shown, making a cut at P and thereafter running with a constant speed in the direction shown. The quarterback releases the ball with a horizontal velocity of 100 ft/sec at the instant the receiver passes point P. Determine the angle at which the quarterback must throw the ball, and the velocity of the ball relative to the receiver when the ball is caught. Neglect any vertical mo-tion of the ball.
Problem 2/203 15 yd 15 yd 30° B Q A P x y vB vA α vB 7 yd/sec y B A 22,300 mi 200 mi x R 3959 g0 32.23 ft/sec2 96 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 96 Article 2/8 Problems 97 2/206 A batter hits the baseball A with an initial velocity of directly toward fielder B at an angle of 30° to the horizontal; the initial position of the ball is 3 ft above ground level. Fielder B re-quires sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder B chooses his running speed so that he arrives at the “catch position” simultaneously with the base-ball. The catch position is the field location at which the ball altitude is 7 ft. Determine the veloc-ity of the ball relative to the fielder at the instant the catch is made.
Problem 2/206 220′ B y x 3′ v0 A 30° 1 4 v0 100 ft/sec 2/204 The aircraft A with radar detection equipment is flying horizontally at an altitude of 12 km and is in-creasing its speed at the rate of 1.2 m/s each sec-ond. Its radar locks onto an aircraft B flying in the same direction and in the same vertical plane at an altitude of 18 km. If A has a speed of 1000 km/h at the instant when , determine the values of and at this same instant if B has a constant speed of 1500 km/h.
Problem 2/204 2/205 At a certain instant after jumping from the air-plane A, a skydiver B is in the position shown and has reached a terminal (constant) speed . The airplane has the same constant speed , and after a period of level flight is just beginning to follow the circular path shown of radius . (a) Determine the velocity and acceleration of the airplane relative to the sky-diver. (b) Determine the time rate of change of the speed of the airplane and the radius of curvature of its path, both as observed by the nonrotating skydiver.
Problem 2/205 A B r A = 2000 m ρ vA vB y x 500 m 350 m r vr A 2000 m vA 50 m/s vB 50 m/s θ 18 km A B 12 km r ¨ r ¨ 30 c02.qxd 2/8/12 7:11 PM Page 97 98 Chapter 2 Kinematics of Particles 2/9 Constrained Motion of Connected Particles Sometimes the motions of particles are interrelated because of the constraints imposed by interconnecting members. In such cases it is necessary to account for these constraints in order to determine the re-spective motions of the particles.
One Degree of Freedom Consider first the very simple system of two interconnected parti-cles A and B shown in Fig. 2/19. It should be quite evident by inspec-tion that the horizontal motion of A is twice the vertical motion of B.
Nevertheless we will use this example to illustrate the method of analy-sis which applies to more complex situations where the results cannot be easily obtained by inspection. The motion of B is clearly the same as that of the center of its pulley, so we establish position coordinates y and x measured from a convenient fixed datum. The total length of the cable is With L, r2, r1, and b all constant, the first and second time derivatives of the equation give The velocity and acceleration constraint equations indicate that, for the coordinates selected, the velocity of A must have a sign which is op-posite to that of the velocity of B, and similarly for the accelerations.
The constraint equations are valid for the motion of the system in either direction. We emphasize that vA is positive to the left and that vB is positive down.
Because the results do not depend on the lengths or pulley radii, we should be able to analyze the motion without considering them. In the lower-left portion of Fig. 2/19 is shown an enlarged view of the horizon-tal diameter ABC of the lower pulley at an instant of time. Clearly, A and A have the same motion magnitudes, as do B and B. During an infinitesimal motion of A, it is easy to see from the triangle that B moves half as far as A because point C as a point on the fixed portion of the cable momentarily has no motion. Thus, with differentiation by time in mind, we can obtain the velocity and acceleration magnitude re-lationships by inspection. The pulley, in effect, is a wheel which rolls on the fixed vertical cable. (The kinematics of a rolling wheel will be treated more extensively in Chapter 5 on rigid-body motion.) The sys-tem of Fig. 2/19 is said to have one degree of freedom since only one variable, either x or y, is needed to specify the positions of all parts of the system.
y ˙ x ˙ 0 x ¨ 2y ¨ or 0 aA 2aB 0 x ˙ 2y ˙ or 0 vA 2vB L x r2 2 2y r1 b b y x A A′ C B′ B r1 r2 Figure 2/19 c02.qxd 2/8/12 7:11 PM Page 98 Article 2/9 Constrained Motion of Connected Particles 99 Two Degrees of Freedom The system with two degrees of freedom is shown in Fig. 2/20. Here the positions of the lower cylinder and pulley C depend on the separate specifications of the two coordinates yA and yB. The lengths of the cables attached to cylinders A and B can be written, respectively, as and their time derivatives are Eliminating the terms in and gives It is clearly impossible for the signs of all three terms to be positive si-multaneously. So, for example, if both A and B have downward (posi-tive) velocities, then C will have an upward (negative) velocity.
These results can also be found by inspection of the motions of the two pulleys at C and D. For an increment dyA (with yB held fixed), the center of D moves up an amount dyA/2, which causes an upward move-ment dyA/4 of the center of C. For an increment dyB (with yA held fixed), the center of C moves up a distance dyB/2. A combination of the two movements gives an upward movement so that vC vA/4 vB/2 as before. Visualization of the actual geometry of the motion is an important ability.
A second type of constraint where the direction of the connecting member changes with the motion is illustrated in the second of the two sample problems which follow.
dyC dyA 4 dyB 2 y ¨A 2y ¨B 4y ¨C 0 or aA 2aB 4aC 0 y ˙A 2y ˙B 4y ˙C 0 or vA 2vB 4vC 0 y ¨D y ˙D 0 y ¨A 2y ¨D and 0 y ¨B 2y ¨C y ¨D 0 y ˙A 2y ˙D and 0 y ˙B 2y ˙C y ˙D LB yB yC (yC yD) constant LA yA 2yD constant Figure 2/20 A B C D yA yB yC yD c02.qxd 2/8/12 7:11 PM Page 99 SAMPLE PROBLEM 2/15 In the pulley configuration shown, cylinder A has a downward velocity of 0.3 m/s. Determine the velocity of B. Solve in two ways.
Solution (I).
The centers of the pulleys at A and B are located by the coordi-nates yA and yB measured from fixed positions. The total constant length of cable in the pulley system is where the constants account for the fixed lengths of cable in contact with the cir-cumferences of the pulleys and the constant vertical separation between the two upper left-hand pulleys. Differentiation with time gives Substitution of vA 0.3 m/s and vB gives Ans.
Solution (II).
An enlarged diagram of the pulleys at A, B, and C is shown.
During a differential movement dsA of the center of pulley A, the left end of its horizontal diameter has no motion since it is attached to the fixed part of the cable. Therefore, the right-hand end has a movement of 2dsA as shown. This movement is transmitted to the left-hand end of the horizontal diameter of the pulley at B. Further, from pulley C with its fixed center, we see that the displace-ments on each side are equal and opposite. Thus, for pulley B, the right-hand end of the diameter has a downward displacement equal to the upward displace-ment dsB of its center. By inspection of the geometry, we conclude that Dividing by dt gives Ans.
SAMPLE PROBLEM 2/16 The tractor A is used to hoist the bale B with the pulley arrangement shown. If A has a forward velocity vA, determine an expression for the upward velocity vB of the bale in terms of x.
Solution.
We designate the position of the tractor by the coordinate x and the position of the bale by the coordinate y, both measured from a fixed reference.
The total constant length of the cable is Differentiation with time yields Substituting vA and vB gives Ans.
vB 1 2 xvA h2 x2 y ˙ x ˙ 0 2y ˙ xx ˙ h2 x2 L 2(h y) l 2(h y) h2 x2 vB 2 3 vA 2 3 (0.3) 0.2 m/s (upward) 2dsA 3dsB or dsB 2 3dsA 0 3(vB) 2(0.3) or vB 0.2 m/s y ˙B y ˙A 0 3y ˙B 2y ˙A L 3yB 2yA constants 100 Chapter 2 Kinematics of Particles C B yA yB A dsA 2dsA 2dsA dsB (c) (a) (b) dsB dsB dsB h y B l A x Helpful Hints We neglect the small angularity of the cables between B and C.
The negative sign indicates that the velocity of B is upward.
Helpful Hint Differentiation of the relation for a right triangle occurs frequently in mechanics.
c02.qxd 2/8/12 7:11 PM Page 100 Article 2/9 Problems 101 2/209 Cylinder B has a downward velocity in feet per sec-ond given by , where t is in seconds.
Calculate the acceleration of A when sec.
Problem 2/209 2/210 Determine the constraint equation which relates the accelerations of bodies A and B. Assume that the upper surface of A remains horizontal.
Problem 2/210 A B B A t 2 vB t2/2 t3/6 PROBLEMS Introductory Problems 2/207 If block B has a leftward velocity of 1.2 m/s, deter-mine the velocity of cylinder A.
Problem 2/207 2/208 At a certain instant, the velocity of cylinder B is 1.2 m/s down and its acceleration is up.
Determine the corresponding velocity and accelera-tion of block A.
Problem 2/208 B A 2 m/s2 B A c02.qxd 2/8/12 7:11 PM Page 101 2/211 Determine the vertical rise h of the load W during 5 seconds if the hoisting drum wraps cable around it at the constant rate of 320 mm/s.
Problem 2/211 2/212 A truck equipped with a power winch on its front end pulls itself up a steep incline with the cable and pulley arrangement shown. If the cable is wound up on the drum at the constant rate of 40 mm/s, how long does it take for the truck to move 4 m up the incline?
Problem 2/212 W 2/213 For the pulley system shown, each of the cables at A and B is given a velocity of 2 m/s in the direction of the arrow. Determine the upward velocity v of the load m.
Problem 2/213 Representative Problems 2/214 Determine the relationship which governs the velocities of the two cylinders A and B. Express all velocities as positive down. How many degrees of freedom are present?
Problem 2/214 B A B A m 102 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 102 Article 2/9 Problems 103 2/217 Determine an expression for the velocity of the cart A down the incline in terms of the upward velocity of cylinder B.
Problem 2/217 2/218 Under the action of force P, the constant accelera-tion of block B is up the incline. For the in-stant when the velocity of B is 3 ft/sec up the incline, determine the velocity of B relative to A, the acceleration of B relative to A, and the absolute velocity of point C of the cable.
Problem 2/218 A 20° B C P 6 ft/sec2 B A C h x vB vA 2/215 The pulley system of the previous problem is modi-fied as shown with the addition of a fourth pulley and a third cylinder C. Determine the relationship which governs the velocities of the three cylinders, and state the number of degrees of freedom.
Express all velocities as positive down.
Problem 2/215 2/216 Neglect the diameters of the small pulleys and establish the relationship between the velocity of A and the velocity of B for a given value of y.
Problem 2/216 A B b b y B A C c02.qxd 2/8/12 7:11 PM Page 103 2/219 The small sliders A and B are connected by the rigid slender rod. If the velocity of slider B is 2 m/s to the right and is constant over a certain interval of time, determine the speed of slider A when the system is in the position shown.
Problem 2/219 2/220 The power winches on the industrial scaffold en-able it to be raised or lowered. For rotation in the senses indicated, the scaffold is being raised. If each drum has a diameter of 200 mm and turns at the rate of 40 rev/min. determine the upward velocity v of the scaffold.
Problem 2/220 vB R A B 2R 60° 2/221 Collars A and B slide along the fixed right-angle rods and are connected by a cord of length L.
Determine the acceleration of collar B as a function of y if collar A is given a constant upward velocity .
Problem 2/221 2/222 Collars A and B slide along the fixed rods and are connected by a cord of length L. If collar A has a velocity to the right, express the velocity of B in terms of , , and s.
Problem 2/222 s B A L 45° x vA x vB s ˙ vA x ˙ y A B L x y vA ax 104 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 104 Article 2/9 Problems 105 2/225 With all conditions of Prob. remaining the same, determine the acceleration of slider B at the instant when .
2/226 Neglect the diameter of the small pulley attached to body A and determine the magnitude of the total velocity of B in terms of the velocity which body A has to the right. Assume that the cable between B and the pulley remains vertical and solve for a given value of x.
Problem 2/226 h x A B vA sA 425 mm 2/224 2/223 The particle A is mounted on a light rod pivoted at O and therefore is constrained to move in a circular arc of radius r. Determine the velocity of A in terms of the downward velocity of the counterweight for any angle .
Problem 2/223 2/224 The rod of the fixed hydraulic cylinder is moving to the left with a constant speed .
Determine the corresponding velocity of slider B when . The length of the cord is 1050 mm, and the effects of the radius of the small pulley A may be neglected.
Problem 2/224 vA C B A sA 250 mm sA 425 mm vA 25 mm/s O r A r θ x y B vB c02.qxd 2/8/12 8:26 PM Page 105 106 Chapter 2 Kinematics of Particles 2/10 Chapter Review In Chapter 2 we have developed and illustrated the basic methods for describing particle motion. The concepts developed in this chapter form the basis for much of dynamics, and it is important to review and master this material before proceeding to the following chapters.
By far the most important concept in Chapter 2 is the time deriva-tive of a vector. The time derivative of a vector depends on direction change as well as magnitude change. As we proceed in our study of dy-namics, we will need to examine the time derivatives of vectors other than position and velocity vectors, and the principles and procedures de-veloped in Chapter 2 will be useful for this purpose.
Categories of Motion The following categories of motion have been examined in this chapter: 1. Rectilinear motion (one coordinate) 2. Plane curvilinear motion (two coordinates) 3. Space curvilinear motion (three coordinates) In general, the geometry of a given problem enables us to identify the category readily. One exception to this categorization is encountered when only the magnitudes of the motion quantities measured along the path are of interest. In this event, we can use the single distance coordi-nate measured along the curved path, together with its scalar time de-rivatives giving the speed and the tangential acceleration Plane motion is easier to generate and control, particularly in ma-chinery, than space motion, and thus a large fraction of our motion problems come under the plane curvilinear or rectilinear categories.
Use of Fixed Axes We commonly describe motion or make motion measurements with respect to fixed reference axes (absolute motion) and moving axes (rela-tive motion). The acceptable choice of the fixed axes depends on the problem. Axes attached to the surface of the earth are sufficiently “fixed” for most engineering problems, although important exceptions include earth–satellite and interplanetary motion, accurate projectile trajecto-ries, navigation, and other problems. The equations of relative motion discussed in Chapter 2 are restricted to translating reference axes.
Choice of Coordinates The choice of coordinates is of prime importance. We have devel-oped the description of motion using the following coordinates: 1. Rectangular (Cartesian) coordinates (x-y) and (x-y-z) 2. Normal and tangential coordinates (n-t) 3. Polar coordinates (r-) s ¨.
s ˙ c02.qxd 2/8/12 7:11 PM Page 106 Article 2/10 Chapter Review 107 4. Cylindrical coordinates (r--z) 5. Spherical coordinates (R--) When the coordinates are not specified, the appropriate choice usually depends on how the motion is generated or measured. Thus, for a parti-cle which slides radially along a rotating rod, polar coordinates are the natural ones to use. Radar tracking calls for polar or spherical coordi-nates. When measurements are made along a curved path, normal and tangential coordinates are indicated. An x-y plotter clearly involves rec-tangular coordinates.
Figure 2/21 is a composite representation of the x-y, n-t, and r- co-ordinate descriptions of the velocity v and acceleration a for curvilinear motion in a plane. It is frequently essential to transpose motion descrip-tion from one set of coordinates to another, and Fig. 2/21 contains the information necessary for that transition.
Approximations Making appropriate approximations is one of the most important abilities you can acquire. The assumption of constant acceleration is valid when the forces which cause the acceleration do not vary apprecia-bly. When motion data are acquired experimentally, we must utilize the nonexact data to acquire the best possible description, often with the aid of graphical or numerical approximations.
Choice of Mathematical Method We frequently have a choice of solution using scalar algebra, vector algebra, trigonometric geometry, or graphical geometry. All of these methods have been illustrated, and all are important to learn. The choice of method will depend on the geometry of the problem, how the motion data are given, and the accuracy desired. Mechanics by its very nature is geometric, so you are encouraged to develop facility in sketching vector relationships, both as an aid to the disclosure of appropriate geometric and trigonometric relations and as a means of solving vector equations graphically. Geometric portrayal is the most direct representation of the vast majority of mechanics problems.
Path t r r x ax = x ·· an = v2/ ar = r ·· – r 2 x y y n ay a at ar ax an θ θ ρ aθ t r n vy vr vx v θ vθ Path r x x y y θ θ · ay = y ·· at = v · a = r + 2r · θ ·· θ · θ vx = x · vn = 0 vr = r · vy = y · vt = v v = rθ · θ (a) Velocity components (b) Acceleration components Figure 2/21 c02.qxd 2/8/12 7:11 PM Page 107 REVIEW PROBLEMS 2/227 The position s of a particle along a straight line is given by , where s is in meters and t is the time in seconds. Determine the velocity v when the acceleration is .
2/228 While scrambling directly toward the sideline, the football quarterback Q throws a pass toward the stationary receiver R. At what speed should the quarterback run if the direction of the velocity of the ball relative to the quarterback is to be di-rectly down the field as indicated? The speed of the ball relative to the quarterback is 60 ft/sec. What is the absolute speed of the ball? Treat the problem as two-dimensional.
Problem 2/228 2/229 A golfer is out of bounds and in a gulley. For the initial conditions shown, determine the coordinates of the point of first impact of the golf ball. The cam-era platform B is in the plane of the trajectory.
Problem 2/229 60′ 40′ 10′ 20′ 20′ 210′ A B 105 ft/sec 40° x y 20 yd 5 yd Q R vQ v vQ 3 m/s2 s 8e0.4t 6t t2 2/230 At time a small ball is projected from point A with a velocity of 200 ft/sec at the angle. Ne-glect atmospheric resistance and determine the two times and when the velocity of the ball makes an angle of with the horizontal x-axis.
Problem 2/230 2/231 The third stage of a rocket is injected by its booster with a velocity u of 15 000 km/h at A into an un-powered coasting flight to B. At B its rocket motor is ignited when the trajectory makes an angle of with the horizontal. Operation is effectively above the atmosphere, and the gravitational acceleration during this interval may be taken as , con-stant in magnitude and direction. Determine the time t to go from A to B. (This quantity is needed in the design of the ignition control system.) Also de-termine the corresponding increase h in altitude.
Problem 2/231 Horiz.
Horiz.
20° x y A B u 45° 9 m/s2 20 60° u = 200 ft/sec A x 45 t2 t1 60 t 0 108 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 108 Article 2/10 Review Problems 109 2/234 In case (a), the baseball player stands relatively stationary and throws the ball with the initial con-ditions shown. In case (b), he runs with speed as he launches the ball with the same conditions relative to himself. What is the addi-tional range of the ball in case (b)? Compare the two flight times.
Problem 2/234 2/235 A small projectile is fired from point O with an ini-tial velocity at the angle of from the horizontal as shown. Neglect atmospheric resis-tance and any change in g and compute the radius of curvature of the path of the projectile 30 sec-onds after the firing.
Problem 2/235 O u = 500 m/s = 60° θ 60 u 500 m/s v v0 = 100 ft/sec (a) = 20° 6′ θ (b) v 15 ft/sec 2/232 The small cylinder is made to move along the ro-tating rod with a motion between and given by , where t is the time counted from the instant the cylinder passes the position and is the period of the oscilla-tion (time for one complete oscillation). Simultane-ously, the rod rotates about the vertical at the constant angular rate . Determine the value of r for which the radial (r-direction) acceleration is zero.
Problem 2/232 2/233 Rotation of the arm PO is controlled by the hori-zontal motion of the vertical slotted link. If and when in., determine and for this instant.
Problem 2/233 x 4″ P O A θ ¨ ˙ x 2 x ¨ 30 ft/sec2 ft/sec x ˙ 4 b r0 r b · ˙ r r0 r r0 b sin 2t r r0 b r r0 b c02.qxd 2/20/12 3:14 PM Page 109 2/236 The motion of pin P is controlled by the two mov-ing slots A and B in which the pin slides. If B has a velocity to the right while A has an upward velocity , determine the mag-nitude of the velocity of the pin.
Problem 2/236 2/237 The angular displacement of the centrifuge is given by rad, where t is in sec-onds and is the startup time. If the person loses consciousness at an acceleration level of 10g, determine the time t at which this would occur.
Verify that the tangential acceleration is negligible as the normal acceleration approaches 10g.
Problem 2/237 30′ O θ t 0 4[t 30e0.03t 30] 60° vA B A P vB vP vA 2 m/s vB 3 m/s 2/238 For the instant represented the particle P has a velocity in the direction shown and has acceleration components and . Determine , , , , and the ra-dius of curvature of the path for this position.
(Hint: Draw the related acceleration components of the total acceleration of the particle and take ad-vantage of the simplified geometry for your calcula-tions.) Problem 2/238 2/239 As part of a training exercise, the pilot of aircraft A adjusts her airspeed (speed relative to the wind) to 220 km/h while in the level portion of the approach path and thereafter holds her absolute speed con-stant as she negotiates the glide path. The ab-solute speed of the aircraft carrier is 30 km/h and that of the wind is 48 km/h. What will be the angle of the glide path with respect to the horizontal as seen by an observer on the ship?
Problem 2/239 10 θ θ = 30° y x P r t 30° v = 3′ r an at ay ar a 15 ft/sec2 ax 15 ft/sec2 v 6 ft/sec 110 Chapter 2 Kinematics of Particles 10° 30 km/h 48 km/h A C c02.qxd 2/8/12 7:11 PM Page 110 Article 2/10 Review Problems 111 2/242 Particle P moves along the curved path shown. At the instant represented, , , and the velocity v makes an angle with the horizon-tal x-axis and has a magnitude of 3.2 m/s. If the y- and r-components of the acceleration of P are and , respectively, at this posi-tion, determine the corresponding radius of curva-ture of the path and the x-component of the acceleration of the particle. Solve graphically or an-alytically.
Problem 2/242 2/243 At the instant depicted, assume that the particle P, which moves on a curved path, is 80 m from the pole O and has the velocity v and acceleration a as indicated. Determine the instantaneous values of , , , , the n- and t-components of acceleration, and the radius of curvature .
Problem 2/243 r P O r = 80 m 30° 30° 60° a = 8 m/s2 v = 30 m/s θ θ ¨ ˙ r ¨ r ˙ β θ v P r y x 1.83 m/s2 5 m/s2 60 30 r 2 m 2/240 A small aircraft is moving in a horizontal circle with a constant speed of 130 ft/sec. At the instant represented, a small package A is ejected from the right side of the aircraft with a horizontal velocity of 20 ft/sec relative to the aircraft. Neglect aerody-namic effects and calculate the coordinates of the point of impact on the ground.
Problem 2/240 2/241 Car A negotiates a curve of 60-m radius at a con-stant speed of 50 km/h. When A passes the position shown, car B is 30 m from the intersection and is accelerating south toward the intersection at the rate of . Determine the acceleration which A appears to have when observed by an occupant of B at this instant.
Problem 2/241 30 m 30° N B A 60 m 1.5 m/s2 1000′ 1500′ z y x A O c02.qxd 2/8/12 7:11 PM Page 111 2/244 The radar tracking antenna oscillates about its vertical axis according to , where is the constant circular frequency and is the dou-ble amplitude of oscillation. Simultaneously, the angle of elevation is increasing at the constant rate . Determine the expression for the mag-nitude a of the acceleration of the signal horn (a) as it passes position A and (b) as it passes the top posi-tion B, assuming that at this instant.
Problem 2/244 2/245 The rod of the fixed hydraulic cylinder is moving to the left with a constant speed .
Determine the corresponding velocity of slider B when . The length of the cord is 1600 mm, and the effects of the radius of the small pulley at A may be neglected.
Problem 2/245 vA C B A sA 250 mm sA 425 mm vA 25 mm/s θ θ B A z b 2 0 φ 0 ˙ K 20 0 cos t Computer-Oriented Problems 2/246 With all conditions of Prob. 2/245 remaining the same, determine the acceleration of slider B at the instant when .
2/247 Two particles A and B start from rest at and move along parallel paths according to and , where and are in meters and t is in seconds counted from the start.
Determine the time t (where ) when both par-ticles have the same displacement and calculate this displacement x.
2/248 A baseball is dropped from an altitude and is found to be traveling at 85 ft/sec when it strikes the ground. In addition to gravitational acceleration, which may be assumed constant, air resistance causes a deceleration component of mag-nitude , where v is the speed and k is a constant.
Determine the value of the coefficient k. Plot the speed of the baseball as a function of altitude y. If the baseball were dropped from a high altitude, but one at which g may still be assumed constant, what would be the terminal velocity ? (The terminal ve-locity is that speed at which the acceleration of gravity and that due to air resistance are equal and opposite, so that the baseball drops at a constant speed.) If the baseball were dropped from at what speed would it strike the ground if air resistance were neglected?
v h 200 ft, vt kv2 h 200 ft t 0 xB xA xB 0.08t 0.16 sin t 2 xA x 0 sA 425 mm 112 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 7:11 PM Page 112 Article 2/10 Review Problems 113 2/251 A low-flying cropduster A is moving with a con-stant speed of 40 m/s in the horizontal circle of ra-dius 300 m. As it passes the twelve-o’clock position shown at time , car B starts from rest from the position shown and accelerates along the straight road at the constant rate of until it reaches a speed of 30 m/s, after which it maintains that constant speed. Determine the velocity and acceler-ation of A with respect to B and plot the magni-tudes of both these quantities over the time period s as functions of both time and displace-ment of the car. Determine the maximum and minimum values of both quantities and state the values of the time t and the displacement at which they occur.
Problem 2/251 2/252 A projectile is launched from point A with speed . Determine the value of the launch angle which maximizes the range R indicated in the figure. Determine the corresponding value R.
Problem 2/252 A B α v0 = 30 m/s 50 m R 10 m v0 30 m/s 1000 m 350 m 300 m x y sB B A sB sB 0 t 50 3 m/s2 t 0 2/249 The slotted arm is fixed and the four-lobe cam rotates counterclockwise at the constant speed of 2 revolutions per second. The distance 12 , where r is millimeters and is in radians.
Plot the radial velocity and the radial accelera-tion of pin P versus from to .
State the acceleration of pin P for (a) , (b) , and (c) .
Problem 2/249 2/250 At time , the 1.8-lb particle P is given an ini-tial velocity at the position and subsequently slides along the circular path of ra-dius . Because of the viscous fluid and the effect of gravitational acceleration, the tangential acceleration is , where the con-stant is a drag parameter. Deter-mine and plot both and as functions of the time t over the range . Determine the maxi-mum values of and and the corresponding val-ues of t. Also determine the first time at which .
Problem 2/250 r P O 90 ˙ 0 t 5 sec ˙ k 0.2 lb-sec/ft at g cos k m v r 1.5 ft 0 v0 1 ft/sec t 0 r P O θ /4 /8 0 /2 0 ar vr cos 4 r 80 c02.qxd 2/8/12 7:11 PM Page 113 2/253 By means of the control unit M, the pendulum OA is given an oscillatory motion about the vertical given by , where is the maximum angular displacement in radians, g is the accelera-tion of gravity, l is the pendulum length, and t is the time in seconds measured from an instant when OA is vertical. Determine and plot the magni-tude a of the acceleration of A as a function of time and as a function of over the first quarter cycle of motion. Determine the minimum and maximum values of a and the corresponding values of t and .
Use the values radians, , and . (Note: The prescribed motion is not precisely that of a freely swinging pendulum for large amplitudes.) Problem 2/253 θ A O l M g 9.81 m/s2 l 0.8 m 0 /3 0 0 sin g l t 2/254 The guide with the vertical slot is given a horizon-tal oscillatory motion according to , where x is in inches and t is in seconds. The oscilla-tion causes the pin P to move in the fixed parabolic slot whose shape is given by , with y also in inches. Plot the magnitude v of the velocity of the pin as a function of time during the interval required for pin P to go from the center to the extremity in. Find and locate the maximum value of v and verify your results analytically.
Problem 2/254 y x x P x 4 y x2/4 x 4 sin 2t 114 Chapter 2 Kinematics of Particles c02.qxd 2/8/12 8:26 PM Page 114 c02.qxd 2/8/12 7:11 PM Page 115 The designers of amusement-park rides such as this roller coaster must not rely upon the principles of equilib-rium alone as they develop specifications for the cars and the supporting structure. The particle kinetics of each car must be considered in estimating the involved forces so that a safe system can be designed.
Jupiterimages/GettyImages c03.qxd 2/9/12 7:38 PM Page 116 117 3/1 Introduction According to Newton’s second law, a particle will accelerate when it is subjected to unbalanced forces. Kinetics is the study of the relations between unbalanced forces and the resulting changes in motion. In Chapter 3 we will study the kinetics of particles. This topic requires that we combine our knowledge of the properties of forces, which we developed in statics, and the kinematics of particle motion just covered in Chapter 2.
With the aid of Newton’s second law, we can combine these two topics and solve engineering problems involving force, mass, and motion.
The three general approaches to the solution of kinetics problems are: (A) direct application of Newton’s second law (called the force-mass-acceleration method), (B) use of work and energy principles, and 3/1 Introduction Section A Force, Mass, and Acceleration 3/2 Newton’s Second Law 3/3 Equation of Motion and Solution of Problems 3/4 Rectilinear Motion 3/5 Curvilinear Motion Section B Work and Energy 3/6 Work and Kinetic Energy 3/7 Potential Energy Section C Impulse and Momentum 3/8 Introduction 3/9 Linear Impulse and Linear Momentum 3/10 Angular Impulse and Angular Momentum Section D Special Applications 3/11 Introduction 3/12 Impact 3/13 Central-Force Motion 3/14 Relative Motion 3/15 Chapter Review CHAPTER OUTLINE 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 117 (C) solution by impulse and momentum methods. Each approach has its special characteristics and advantages, and Chapter 3 is subdivided into Sections A, B, and C, according to these three methods of solution. In addition, a fourth section, Section D, treats special applications and combinations of the three basic approaches. Before proceeding, you should review carefully the definitions and concepts of Chapter 1, be-cause they are fundamental to the developments which follow.
SECTION A FORCE, MASS, AND ACCELERATION 118 Chapter 3 Kinetics of Particles 3/2 Newton’s Second Law The basic relation between force and acceleration is found in New-ton’s second law, Eq. 1/1, the verification of which is entirely experi-mental. We now describe the fundamental meaning of this law by considering an ideal experiment in which force and acceleration are as-sumed to be measured without error. We subject a mass particle to the action of a single force F1, and we measure the acceleration a1 of the particle in the primary inertial system. The ratio F1/a1 of the magni-tudes of the force and the acceleration will be some number C1 whose value depends on the units used for measurement of force and accelera-tion. We then repeat the experiment by subjecting the same particle to a different force F2 and measuring the corresponding acceleration a2.
The ratio F2/a2 of the magnitudes will again produce a number C2. The experiment is repeated as many times as desired.
We draw two important conclusions from the results of these exper-iments. First, the ratios of applied force to corresponding acceleration all equal the same number, provided the units used for measurement are not changed in the experiments. Thus, We conclude that the constant C is a measure of some invariable property of the particle. This property is the inertia of the particle, which is its resistance to rate of change of velocity. For a particle of high inertia (large C), the acceleration will be small for a given force F. On the other hand, if the inertia is small, the acceleration will be large. The mass m is used as a quantitative measure of inertia, and therefore, we may write the expression C km, where k is a constant introduced to account for the units used. Thus, we may express the relation obtained from the experiments as (3/1) F kma F1 a1 F2 a2 F a C, a constant The primary inertial system or astronomical frame of reference is an imaginary set of ref-erence axes which are assumed to have no translation or rotation in space. See Art. 1/2, Chapter 1.
c03.qxd 2/9/12 7:38 PM Page 118 where F is the magnitude of the resultant force acting on the particle of mass m, and a is the magnitude of the resulting acceleration of the particle.
The second conclusion we draw from this ideal experiment is that the acceleration is always in the direction of the applied force. Thus, Eq. 3/1 becomes a vector relation and may be written (3/2) Although an actual experiment cannot be performed in the ideal manner described, the same conclusions have been drawn from countless accurately performed experiments. One of the most accurate checks is given by the precise prediction of the motions of planets based on Eq. 3/2.
Inertial System Although the results of the ideal experiment are obtained for mea-surements made relative to the “fixed” primary inertial system, they are equally valid for measurements made with respect to any nonrotating reference system which translates with a constant velocity with respect to the primary system. From our study of relative motion in Art. 2/8, we know that the acceleration measured in a system translating with no ac-celeration is the same as that measured in the primary system. Thus, Newton’s second law holds equally well in a nonaccelerating system, so that we may define an inertial system as any system in which Eq. 3/2 is valid.
If the ideal experiment described were performed on the surface of the earth and all measurements were made relative to a reference sys-tem attached to the earth, the measured results would show a slight dis-crepancy from those predicted by Eq. 3/2, because the measured acceleration would not be the correct absolute acceleration. The discrep-ancy would disappear when we introduced the correction due to the ac-celeration components of the earth. These corrections are negligible for most engineering problems which involve the motions of structures and machines on the surface of the earth. In such cases, the accelerations measured with respect to reference axes attached to the surface of the earth may be treated as “absolute,” and Eq. 3/2 may be applied with negligible error to experiments made on the surface of the earth. An increasing number of problems occur, particularly in the fields of rocket and spacecraft design, where the acceleration components of the earth are of primary concern. For this work it is essential that the F kma Article 3/2 Newton’s Second Law 119 As an example of the magnitude of the error introduced by neglect of the motion of the earth, consider a particle which is allowed to fall from rest (relative to earth) at a height h above the ground. We can show that the rotation of the earth gives rise to an eastward ac-celeration (Coriolis acceleration) relative to the earth and, neglecting air resistance, that the particle falls to the ground a distance east of the point on the ground directly under that from which it was dropped. The angular velocity of the earth is 0.729(104) rad/s, and the latitude, north or south, is . At a lat-itude of 45 and from a height of 200 m, this eastward deflection would be x 43.9 mm.
x 2 3 2h3 g cos c03.qxd 2/9/12 7:38 PM Page 119 fundamental basis of Newton’s second law be thoroughly understood and that the appropriate absolute acceleration components be employed.
Before 1905 the laws of Newtonian mechanics had been verified by innumerable physical experiments and were considered the final de-scription of the motion of bodies. The concept of time, considered an ab-solute quantity in the Newtonian theory, received a basically different interpretation in the theory of relativity announced by Einstein in 1905.
The new concept called for a complete reformulation of the accepted laws of mechanics. The theory of relativity was subjected to early ridicule, but has been verified by experiment and is now universally ac-cepted by scientists. Although the difference between the mechanics of Newton and that of Einstein is basic, there is a practical difference in the results given by the two theories only when velocities of the order of the speed of light (300 106 m/s) are encountered. Important prob-lems dealing with atomic and nuclear particles, for example, require cal-culations based on the theory of relativity.
Systems of Units It is customary to take k equal to unity in Eq. 3/2, thus putting the relation in the usual form of Newton’s second law [1/1] A system of units for which k is unity is known as a kinetic system.
Thus, for a kinetic system the units of force, mass, and acceleration are not independent. In SI units, as explained in Art. 1/4, the units of force (newtons, N) are derived by Newton’s second law from the base units of mass (kilograms, kg) times acceleration (meters per second squared, m/s2). Thus, N This system is known as an absolute system since the unit for force is dependent on the absolute value of mass.
In U.S. customary units, on the other hand, the units of mass (slugs) are derived from the units of force (pounds force, lb) divided by acceleration (feet per second squared, ft/sec2). Thus, the mass units are slugs lb-sec2/ft. This system is known as a gravitational system since mass is derived from force as determined from gravitational attraction.
For measurements made relative to the rotating earth, the relative value of g should be used. The internationally accepted value of g rela-tive to the earth at sea level and at a latitude of 45 is 9.806 65 m/s2. Ex-cept where greater precision is required, the value of 9.81 m/s2 will be used for g. For measurements relative to a nonrotating earth, the ab-solute value of g should be used. At a latitude of 45 and at sea level, the absolute value is 9.8236 m/s2. The sea-level variation in both the absolute and relative values of g with latitude is shown in Fig. 1/1 of Art. 1/5.
kgm/s2.
F ma 120 Chapter 3 Kinetics of Particles The theory of relativity demonstrates that there is no such thing as a preferred primary inertial system and that measurements of time made in two coordinate systems which have a velocity relative to one another are different. On this basis, for example, the principles of relativity show that a clock carried by the pilot of a spacecraft traveling around the earth in a circular polar orbit of 644 km altitude at a velocity of 27 080 km/h would be slow com-pared with a clock at the pole by 0.000 001 85 s for each orbit.
c03.qxd 2/9/12 7:38 PM Page 120 In the U.S. customary system, the standard value of g relative to the rotating earth at sea level and at a latitude of 45 is 32.1740 ft/sec2.
The corresponding value relative to a nonrotating earth is 32.2230 ft/sec2.
Force and Mass Units We need to use both SI units and U.S. customary units, so we must have a clear understanding of the correct force and mass units in each system. These units were explained in Art. 1/4, but it will be helpful to illustrate them here using simple numbers before applying Newton’s second law. Consider, first, the free-fall experiment as depicted in Fig.
3/1a where we release an object from rest near the surface of the earth.
We allow it to fall freely under the influence of the force of gravitational attraction W on the body. We call this force the weight of the body. In SI units for a mass m 1 kg, the weight is W 9.81 N, and the corre-sponding downward acceleration a is g 9.81 m/s2. In U.S. customary units for a mass m 1 lbm (1/32.2 slug), the weight is W 1 lbf and the resulting gravitational acceleration is g 32.2 ft/sec2. For a mass m 1 slug (32.2 lbm), the weight is W 32.2 lbf and the acceleration, of course, is also g 32.2 ft/sec2.
In Fig. 3/1b we illustrate the proper units with the simplest example where we accelerate an object of mass m along the horizontal with a force F. In SI units (an absolute system), a force F 1 N causes a mass m 1 kg to accelerate at the rate a 1 m/s2. Thus, 1 N 1 In the U.S. customary system (a gravitational system), a force F 1 lbf kgm/s2.
Article 3/2 Newton’s Second Law 121 1 —— 32.2 W = 9.81 N m = 1 kg m = 1 lbm slug ( ) m = 1 slug (32.2 lbm) a = g = 9.81 m/s2 a = 1 m/s2 F = 1 N (a) Gravitational Free-Fall (b) Newton’s Second Law a = g = 32.2 ft/sec2 W = 1 lbf W = 32.2 lbf m = 1 kg a = 32.2 ft/sec2 SI ___ U.S. Customary __ SI ___ U.S. Customary __ F = 1 lbf m=1lbm a = 1 ft/sec2 F = 1 lbf m = 1 slug 1 —— 32.2 slug ( ) (32.2 lbm) Figure 3/1 c03.qxd 2/9/12 7:38 PM Page 121 causes a mass m 1 lbm (1/32.2 slug) to accelerate at the rate a 32.2 ft/sec2, whereas a force F 1 lbf causes a mass m 1 slug (32.2 lbm) to accelerate at the rate a 1 ft/sec2.
We note that in SI units where the mass is expressed in kilograms (kg), the weight W of the body in newtons (N) is given by W mg, where g 9.81 m/s2. In U.S. customary units, the weight W of a body is expressed in pounds force (lbf), and the mass in slugs (lbf-sec2/ft) is given by m W/g, where g 32.2 ft/sec2.
In U.S. customary units, we frequently speak of the weight of a body when we really mean mass. It is entirely proper to specify the mass of a body in pounds (lbm) which must be converted to mass in slugs be-fore substituting into Newton’s second law. Unless otherwise stated, the pound (lb) is normally used as the unit of force (lbf).
3/3 Equation of Motion and Solution of Problems When a particle of mass m is subjected to the action of concurrent forces F1, F2, F3, . . . whose vector sum is ΣF, Eq. 1/1 becomes (3/3) When applying Eq. 3/3 to solve problems, we usually express it in scalar component form with the use of one of the coordinate systems developed in Chapter 2. The choice of an appropriate coordinate system depends on the type of motion involved and is a vital step in the formulation of any problem. Equation 3/3, or any one of the component forms of the force-mass-acceleration equation, is usually called the equation of motion. The equation of motion gives the instantaneous value of the acceleration cor-responding to the instantaneous values of the forces which are acting.
Two Types of Dynamics Problems We encounter two types of problems when applying Eq. 3/3. In the first type, the acceleration of the particle is either specified or can be de-termined directly from known kinematic conditions. We then determine the corresponding forces which act on the particle by direct substitution into Eq. 3/3. This problem is generally quite straightforward.
In the second type of problem, the forces acting on the particle are specified and we must determine the resulting motion. If the forces are constant, the acceleration is also constant and is easily found from Eq.
3/3. When the forces are functions of time, position, or velocity, Eq. 3/3 becomes a differential equation which must be integrated to determine the velocity and displacement.
Problems of this second type are often more formidable, as the inte-gration may be difficult to carry out, particularly when the force is a mixed function of two or more motion variables. In practice, it is fre-quently necessary to resort to approximate integration techniques, ei-ther numerical or graphical, particularly when experimental data are involved. The procedures for a mathematical integration of the accelera-tion when it is a function of the motion variables were developed in Art.
ΣF ma 122 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 122 2/2, and these same procedures apply when the force is a specified func-tion of these same parameters, since force and acceleration differ only by the constant factor of the mass.
Constrained and Unconstrained Motion There are two physically distinct types of motion, both described by Eq. 3/3. The first type is unconstrained motion where the particle is free of mechanical guides and follows a path determined by its initial motion and by the forces which are applied to it from external sources. An air-plane or rocket in flight and an electron moving in a charged field are examples of unconstrained motion.
The second type is constrained motion where the path of the parti-cle is partially or totally determined by restraining guides. An ice-hockey puck is partially constrained to move in the horizontal plane by the surface of the ice. A train moving along its track and a collar sliding along a fixed shaft are examples of more fully constrained motion. Some of the forces acting on a particle during constrained motion may be ap-plied from outside sources, and others may be the reactions on the parti-cle from the constraining guides. All forces, both applied and reactive, which act on the particle must be accounted for in applying Eq. 3/3.
The choice of an appropriate coordinate system is frequently indi-cated by the number and geometry of the constraints. Thus, if a particle is free to move in space, as is the center of mass of the airplane or rocket in free flight, the particle is said to have three degrees of freedom since three independent coordinates are required to specify the position of the particle at any instant. All three of the scalar components of the equa-tion of motion would have to be integrated to obtain the space coordi-nates as a function of time.
If a particle is constrained to move along a surface, as is the hockey puck or a marble sliding on the curved surface of a bowl, only two coor-dinates are needed to specify its position, and in this case it is said to have two degrees of freedom. If a particle is constrained to move along a fixed linear path, as is the collar sliding along a fixed shaft, its position may be specified by the coordinate measured along the shaft. In this case, the particle would have only one degree of freedom.
Article 3/3 Equation of Motion and Solution of Problems 123 Free-Body Diagram When applying any of the force-mass-acceleration equations of mo-tion, you must account correctly for all forces acting on the particle. The only forces which we may neglect are those whose magnitudes are negli-gible compared with other forces acting, such as the forces of mutual at-traction between two particles compared with their attraction to a celestial body such as the earth. The vector sum ΣF of Eq. 3/3 means the vector sum of all forces acting on the particle in question. Likewise, the corresponding scalar force summation in any one of the component di-rections means the sum of the components of all forces acting on the particle in that particular direction.
KEY CONCEPTS c03.qxd 2/9/12 7:38 PM Page 123 The only reliable way to account accurately and consistently for every force is to isolate the particle under consideration from all con-tacting and influencing bodies and replace the bodies removed by the forces they exert on the particle isolated. The resulting free-body dia-gram is the means by which every force, known and unknown, which acts on the particle is represented and thus accounted for. Only after this vital step has been completed should you write the appropriate equation or equations of motion.
The free-body diagram serves the same key purpose in dynamics as it does in statics. This purpose is simply to establish a thoroughly reli-able method for the correct evaluation of the resultant of all actual forces acting on the particle or body in question. In statics this resultant equals zero, whereas in dynamics it is equated to the product of mass and acceleration. When you use the vector form of the equation of mo-tion, remember that it represents several scalar equations and that every equation must be satisfied.
Careful and consistent use of the free-body method is the most im-portant single lesson to be learned in the study of engineering mechan-ics. When drawing a free-body diagram, clearly indicate the coordinate axes and their positive directions. When you write the equations of mo-tion, make sure all force summations are consistent with the choice of these positive directions. As an aid to the identification of external forces which act on the body in question, these forces are shown as heavy red vectors in the illustrations in this book. Sample Problems 3/1 through 3/5 in the next article contain five examples of free-body dia-grams. You should study these to see how the diagrams are constructed.
In solving problems, you may wonder how to get started and what sequence of steps to follow in arriving at the solution. This difficulty may be minimized by forming the habit of first recognizing some rela-tionship between the desired unknown quantity in the problem and other quantities, known and unknown. Then determine additional rela-tionships between these unknowns and other quantities, known and un-known. Finally, establish the dependence on the original data and develop the procedure for the analysis and computation. A few minutes spent organizing the plan of attack through recognition of the depen-dence of one quantity on another will be time well spent and will usually prevent groping for the answer with irrelevant calculations.
3/4 Rectilinear Motion We now apply the concepts discussed in Arts. 3/2 and 3/3 to prob-lems in particle motion, starting with rectilinear motion in this article and treating curvilinear motion in Art. 3/5. In both articles, we will ana-lyze the motions of bodies which can be treated as particles. This simpli-fication is possible as long as we are interested only in the motion of the mass center of the body. In this case we may treat the forces as concur-rent through the mass center. We will account for the action of noncon-current forces on the motions of bodies when we discuss the kinetics of rigid bodies in Chapter 6.
124 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 124 If we choose the x-direction, for example, as the direction of the rec-tilinear motion of a particle of mass m, the acceleration in the y- and z-directions will be zero and the scalar components of Eq. 3/3 become (3/4) For cases where we are not free to choose a coordinate direction along the motion, we would have in the general case all three compo-nent equations (3/5) where the acceleration and resultant force are given by ΣF (ΣFx)2 (ΣFy)2 (ΣFz)2 ΣF ΣFxi ΣFy j ΣFzk a ax 2 ay 2 az 2 a axi ay j azk ΣFz maz ΣFy may ΣFx max ΣFz 0 ΣFy 0 ΣFx max Article 3/4 Rectilinear Motion 125 This view of a car-collision test suggests that very large accelerations and accompanying large forces occur throughout the system of the two cars.
The crash dummies are also subjected to large forces, primarily by the shoulder-harness/seat-belt restraints.
© CTK/Alamy c03.qxd 2/9/12 7:38 PM Page 125 126 Chapter 3 Kinetics of Particles SAMPLE PROBLEM 3/1 A 75-kg man stands on a spring scale in an elevator. During the first 3 sec-onds of motion from rest, the tension T in the hoisting cable is 8300 N. Find the reading R of the scale in newtons during this interval and the upward velocity v of the elevator at the end of the 3 seconds. The total mass of the elevator, man, and scale is 750 kg.
Solution.
The force registered by the scale and the velocity both depend on the acceleration of the elevator, which is constant during the interval for which the forces are constant. From the free-body diagram of the elevator, scale, and man taken together, the acceleration is found to be The scale reads the downward force exerted on it by the man’s feet. The equal and opposite reaction R to this action is shown on the free-body diagram of the man alone together with his weight, and the equation of motion for him gives Ans.
The velocity reached at the end of the 3 seconds is Ans.
SAMPLE PROBLEM 3/2 A small inspection car with a mass of 200 kg runs along the fixed overhead cable and is controlled by the attached cable at A. Determine the acceleration of the car when the control cable is horizontal and under a tension T 2.4 kN.
Also find the total force P exerted by the supporting cable on the wheels.
Solution.
The free-body diagram of the car and wheels taken together and treated as a particle discloses the 2.4-kN tension T, the weight W mg 200(9.81) 1962 N, and the force P exerted on the wheel assembly by the cable.
The car is in equilibrium in the y-direction since there is no acceleration in this direction. Thus, Ans.
In the x-direction the equation of motion gives Ans.
[ΣFx max] 2400(12 13 ) 1962( 5 13 ) 200a a 7.30 m/s2 [ΣFy 0] P 2.4( 5 13 ) 1.962(12 13 ) 0 P 2.73 kN [v a dt] v 0 3 0 1.257 dt v 3.77 m/s [ΣFy may] R 736 75(1.257) R 830 N [ΣFy may] 8300 7360 750ay ay 1.257 m/s2 T = 8300 N y 750(9.81) = 7360 N 75(9.81) = 736 N y R ay ay 5 12 T A T = 2.4 kN W = mg = 1962 N x a y P 5 5 12 12 G Helpful Hint If the scale were calibrated in kilo-grams, it would read 830/9.81 84.6 kg which, of course, is not his true mass since the measurement was made in a noninertial (accelerat-ing) system. Suggestion: Rework this problem in U.S. customary units.
Helpful Hint By choosing our coordinate axes along and normal to the direction of the acceleration, we are able to solve the two equa-tions independently. Would this be so if x and y were chosen as horizontal and vertical?
126 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 126 SAMPLE PROBLEM 3/3 The 250-lb concrete block A is released from rest in the position shown and pulls the 400-lb log up the 30 ramp. If the coefficient of kinetic friction between the log and the ramp is 0.5, determine the velocity of the block as it hits the ground at B.
Solution.
The motions of the log and the block A are clearly dependent. Al-though by now it should be evident that the acceleration of the log up the incline is half the downward acceleration of A, we may prove it formally. The constant total length of the cable is L 2sC sA constant, where the constant accounts for the cable portions wrapped around the pulleys. Differentiating twice with re-spect to time gives 0 or We assume here that the masses of the pulleys are negligible and that they turn with negligible friction. With these assumptions the free-body diagram of the pulley C discloses force and moment equilibrium. Thus, the tension in the cable attached to the log is twice that applied to the block. Note that the acceler-ations of the log and the center of pulley C are identical.
The free-body diagram of the log shows the friction force kN for motion up the plane. Equilibrium of the log in the y-direction gives and its equation of motion in the x-direction gives For the block in the positive downward direction, we have Solving the three equations in aC, aA, and T gives us For the 20-ft drop with constant acceleration, the block acquires a velocity Ans.
vA 2(5.83)(20) 15.27 ft/sec [v2 2ax] aA 5.83 ft/sec2 aC 2.92 ft/sec2 T 205 lb 250 T 250 32.2 aA [ b ΣF ma] 0.5(346) 2T 400 sin 30 400 32.2 aC [ΣFx max] N 400 cos 30 0 N 346 lb [ΣFy 0] 0 2aC aA s ¨A, 2s ¨C 20′ 250 lb B A C 400 lb 30° k = 0.5 μ C sC 250 lb T T 2T 2T T C + 400 lb 0.5N N y x A sA Helpful Hints The coordinates used in expressing the final kinematic constraint rela-tionship must be consistent with those used for the kinetic equations of motion.
We can verify that the log will in-deed move up the ramp by calculat-ing the force in the cable necessary to initiate motion from the equilib-rium condition. This force is 2T 0.5N 400 sin 30 373 lb or T 186.5 lb, which is less than the 250-lb weight of block A. Hence, the log will move up.
Note the serious error in assuming that T 250 lb, in which case, block A would not accelerate.
Because the forces on this system re-main constant, the resulting acceler-ations also remain constant.
Article 3/4 Rectilinear Motion 127 c03.qxd 2/9/12 7:38 PM Page 127 SAMPLE PROBLEM 3/4 The design model for a new ship has a mass of 10 kg and is tested in an exper-imental towing tank to determine its resistance to motion through the water at various speeds. The test results are plotted on the accompanying graph, and the resistance R may be closely approximated by the dashed parabolic curve shown. If the model is released when it has a speed of 2 m/s, determine the time t required for it to reduce its speed to 1 m/s and the corresponding travel distance x.
Solution.
We approximate the resistance-velocity relation by R kv2 and find k by substituting R 8 N and v 2 m/s into the equation, which gives k 8/22 2 Thus, R 2v2.
The only horizontal force on the model is R, so that We separate the variables and integrate to obtain Thus, when v v0/2 1 m/s, the time is t 2.5 s.
Ans.
The distance traveled during the 2.5 seconds is obtained by integrating v dx/dt. Thus, v 10/(5 2t) so that Ans.
SAMPLE PROBLEM 3/5 The collar of mass m slides up the vertical shaft under the action of a force F of constant magnitude but variable direction. If kt where k is a constant and if the collar starts from rest with 0, determine the magnitude F of the force which will result in the collar coming to rest as reaches /2. The coeffi-cient of kinetic friction between the collar and shaft is k.
Solution.
After drawing the free-body diagram, we apply the equation of mo-tion in the y-direction to get where equilibrium in the horizontal direction requires N F sin . Substituting kt and integrating first between general limits give which becomes For /2 the time becomes t /2k, and v 0 so that Ans.
F k [1 k(0 1)] mg 2k 0 and F mg 2(1 k) F k [sin kt k(cos kt 1)] mgt mv t 0 (F cos kt k F sin kt mg) dt m v 0 dv F cos k N mg m dv dt [ΣFy may] x 0 dx 2.5 0 10 5 2t dt x 10 2 ln (5 2t) 2.5 0 3.47 m 5(1 1 1 2 ) t 0 dt 5 v 2 dv v2 t 5 1 v 1 2 s R max or 2v2 10 dv dt [ΣFx max] N s2/m2.
1 2 0 2 4 6 8 0 v, m/s R, N R v0 = 2 m/s v x B = W W m μk F θ N θ F mg μk N Helpful Hints Be careful to observe the minus sign for R.
Suggestion: Express the distance x after release in terms of the velocity v and see if you agree with the re-sulting relation x 5 ln (v0/v).
Helpful Hints If were expressed as a function of the vertical displacement y instead of the time t, the acceleration would become a function of the displace-ment and we would use v dv a dy.
We see that the results do not de-pend on k, the rate at which the force changes direction.
128 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 128 PROBLEMS Introductory Problems 3/1 The 50-kg crate is projected along the floor with an initial speed of 7 m/s at . The coefficient of ki-netic friction is 0.40. Calculate the time required for the crate to come to rest and the corresponding dis-tance x traveled.
Problem 3/1 3/2 The 50-kg crate of Prob. 3/1 is now projected down an incline as shown with an initial speed of 7 m/s. Inves-tigate the time t required for the crate to come to rest and the corresponding distance x traveled if (a) and (b) .
Problem 3/2 3/3 The 100-lb crate is carefully placed with zero velocity on the incline. Describe what happens if (a) and (b) Problem 3/3 100 lb μs = 0.30 μk = 0.25 μ μ θ 20 15 50 kg v0 = 7 m/s x k = 0.40 μ θ 30 15 50 kg v0 = 7 m/s x k = 0.40 μ x 0 3/4 A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine the force R which the elevator floor exerts on her feet and the lifting force L which she exerts on the package during the acceleration in-terval. If the elevator support cables suddenly and completely fail, what values would R and L acquire?
Problem 3/4 3/5 During a brake test, the rear-engine car is stopped from an initial speed of 100 km/h in a distance of 50 m. If it is known that all four wheels contribute equally to the braking force, determine the braking force F at each wheel. Assume a constant decelera-tion for the 1500-kg car.
Problem 3/5 50 m v1 = 100 km/h v2 = 0 9 kg 60 kg g –– 4 Article 3/4 Problems 129 c03.qxd 2/9/12 7:38 PM Page 129 3/9 A man pulls himself up the incline by the method shown. If the combined mass of the man and cart is 100 kg, determine the acceleration of the cart if the man exerts a pull of 250 N on the rope. Neglect all friction and the mass of the rope, pulleys, and wheels.
Problem 3/9 3/10 A car is climbing the hill of slope at a constant speed v. If the slope decreases abruptly to at point A, determine the acceleration a of the car just after passing point A if the driver does not change the throttle setting or shift into a different gear.
Problem 3/10 3/11 Calculate the vertical acceleration a of the 100-lb cylinder for each of the two cases illustrated. Ne-glect friction and the mass of the pulleys.
Problem 3/11 100 lb 150 lb (a) 100 lb 150 lb (b) A a v = const.
1 θ 2 θ 2 1 15° 15 3/6 What fraction n of the weight of the jet airplane is the net thrust (nozzle thrust T minus air resistance R) required for the airplane to climb at an angle with the horizontal with an acceleration a in the di-rection of flight?
Problem 3/6 3/7 The 300-Mg jet airliner has three engines, each of which produces a nearly constant thrust of 240 kN dur-ing the takeoff roll. Determine the length s of runway required if the takeoff speed is 220 km/h. Compute s first for an uphill takeoff direction from A to B and sec-ond for a downhill takeoff from B to A on the slightly inclined runway. Neglect air and rolling resistance.
Problem 3/7 3/8 The 180-lb man in the bosun’s chair exerts a pull of 50 lb on the rope for a short interval. Find his acceler-ation. Neglect the mass of the chair, rope, and pulleys.
Problem 3/8 Horizontal 0.5° A B T R θ 130 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 130 3/12 A driver finds that her car will descend the slope at a certain constant speed with no brakes or throttle required. The slope decreases fairly abruptly to at point A. If the driver takes no ac-tion but continues to coast, determine the accelera-tion a of the car just after it passes point A for the conditions (a) and (b) .
Problem 3/12 3/13 By itself, the 2500-kg pickup truck executes a 0–100 km/h acceleration run in 10 s along a level road.
What would be the corresponding time when pulling the 500-kg trailer? Assume constant acceleration and neglect all retarding forces.
Problem 3/13 3/14 Reconsider the pickup-truck/trailer combination of the previous problem. If the unit uniformly acceler-ates from rest to a speed of 25 m/s in a distance of 150 m, determine the tension T in the towing tongue OA. Neglect all effects of the tongue angle, i.e., assume that OA is horizontal.
Representative Problems 3/15 A train consists of a 400,000-lb locomotive and one hundred 200,000-lb hopper cars. If the locomotive exerts a friction force of 40,000 lb on the rails in starting the train from rest, compute the forces in couplers 1 and 100. Assume no slack in the couplers and neglect friction associated with the hopper cars.
Problem 3/15 100 99 98 3 2 1 5 2500 kg 500 kg 5° A O v = constant A θ1 θ2 2 0 2 1.5 2 1 3 3/16 The collar A is free to slide along the smooth shaft B mounted in the frame. The plane of the frame is vertical. Determine the horizontal acceleration a of the frame necessary to maintain the collar in a fixed position on the shaft.
Problem 3/16 3/17 The 5-oz pinewood-derby car is released from rest at the starting line A and crosses the finish line C 2.75 sec later. The transition at B is small and smooth.
Assume that the net retarding force is constant throughout the run and find this force.
Problem 3/17 3/18 The beam and attached hoisting mechanism together weigh 2400 lb with center of gravity at G. If the ini-tial acceleration a of point P on the hoisting cable is 20 ft/sec2, calculate the corresponding reaction at the support A.
Problem 3/18 1000 lb a P G A B 8′ 10′ 8′ 12″ 20° A B C 15′ 10′ A B 30° a Article 3/4 Problems 131 c03.qxd 2/9/12 7:38 PM Page 131 3/23 Small objects are delivered to the 72-in. inclined chute by a conveyor belt A which moves at a speed ft/sec. If the conveyor belt B has a speed ft/sec and the objects are delivered to this belt with no slipping, calculate the coefficient of fric-tion between the objects and the chute.
Problem 3/23 3/24 If the coefficients of static and kinetic friction be-tween the 20-kg block A and the 100-kg cart B are both essentially the same value of 0.50, determine the acceleration of each part for (a) N and (b) N.
Problem 3/24 3/25 A simple pendulum is pivoted at O and is free to swing in the vertical plane of the plate. If the plate is given a constant acceleration a up the incline , write an expression for the steady angle assumed by the pendulum after all initial start-up oscilla-tions have ceased. Neglect the mass of the slender supporting rod.
Problem 3/25 a O β θ P A 20 kg B 100 kg P 40 P 60 72″ v1 v2 A 30° B k v2 3.0 v1 1.2 3/19 The 10-kg steel sphere is suspended from the 15-kg frame which slides down the incline. If the coef-ficient of kinetic friction between the frame and in-cline is 0.15, compute the tension in each of the supporting wires A and B.
Problem 3/19 3/20 The block shown is observed to have a velocity 20 ft/sec as it passes point A and a velocity ft/sec as it passes point B on the incline.
Calculate the coefficient of kinetic friction be-tween the block and the incline if ft and .
Problem 3/20 3/21 Determine the initial acceleration of the 15-kg block if (a) N and (b) N. The system is ini-tially at rest with no slack in the cable, and the mass and friction of the pulleys are negligible.
Problem 3/21 3/22 The system of the previous problem starts from rest with no slack in the cable. What value of the tension T will result in an initial block acceleration of 0.8 m/s2 to the right?
μ μ 30° T 15 kg s = 0.50 k = 0.40 T 26 T 23 v1 v2 x A B θ 15 x 30 k v2 10 v1 45° 45° A B 20° k = 0.15 μ 20 132 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 132 3/26 The tractor-trailer unit is moving down the incline with a speed of 5 mi/hr when the driver brakes the tractor to a stop in a distance of 4 ft. Estimate the percent increase n in the hitch-force component which is parallel to the incline, compared with the force present at steady speed. The cart and its load combined weigh 500 lb. State any assumptions.
Problem 3/26 3/27 The device shown is used as an accelerometer and consists of a 4-oz plunger A which deflects the spring as the housing of the unit is given an upward acceleration a. Specify the necessary spring stiffness k which will permit the plunger to deflect 1/4 in. be-yond the equilibrium position and touch the electri-cal contact when the steadily but slowly increasing upward acceleration reaches 5g. Friction may be neglected.
Problem 3/27 ″ 1 — 4 A a 12 4 A 3/28 The acceleration of the 50-kg carriage A in its smooth vertical guides is controlled by the tension T exerted on the control cable which passes around the two circular pegs fixed to the carriage. Determine the value of T required to limit the downward accel-eration of the carriage to 1.2 m/s2 if the coefficient of friction between the cable and the pegs is 0.20. (Re-call the relation between the tensions in a flexible cable which is slipping on a fixed peg: ) Problem 3/28 3/29 The system is released from rest with the cable taut.
For the friction coefficients and , calculate the acceleration of each body and the ten-sion T in the cable. Neglect the small mass and fric-tion of the pulleys.
Problem 3/29 B 20 kg 30° A 60 kg μs μk , k 0.20 s 0.25 A T T2 T1 β T2 T1e.
Article 3/4 Problems 133 c03.qxd 2/9/12 7:38 PM Page 133 3/32 The sliders A and B are connected by a light rigid bar of length m and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position where m, the veloc-ity of A is m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.
Problem 3/32 3/33 The sliders A and B are connected by a light rigid bar and move with negligible friction in the slots, both of which lie in a horizontal plane. For the posi-tion shown, the hydraulic cylinder imparts a veloc-ity and acceleration to slider A of 0.4 m/s and 2 m/s2, respectively, both to the right. Determine the accel-eration of slider B and the force in the bar at this instant.
Problem 3/33 A B 0.5 m 60° 30° 3 kg 2 kg P = 40 N 2 kg 0.5 m 3 kg B A xA vA 0.9 xA 0.4 l 0.5 3/30 A jet airplane with a mass of 5 Mg has a touchdown speed of 300 km/h, at which instant the braking parachute is deployed and the power shut off. If the total drag on the aircraft varies with velocity as shown in the accompanying graph, calculate the dis-tance x along the runway required to reduce the speed to 150 km/h. Approximate the variation of the drag by an equation of the form , where k is a constant.
Problem 3/30 3/31 A heavy chain with a mass per unit length is pulled by the constant force P along a horizontal surface consisting of a smooth section and a rough section. The chain is initially at rest on the rough surface with . If the coefficient of kinetic fric-tion between the chain and the rough surface is , determine the velocity v of the chain when .
The force P is greater than in order to initi-ate motion.
Problem 3/31 L Smooth Rough k P x μ kgL x L k x 0 300 200 100 0 20 40 60 80 100 120 0 Velocity v, km/h v Drag D, kN D kv2 134 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 134 3/34 The 4-lb collar is released from rest against the light elastic spring, which has a stiffness of 10 lb/in. and has been compressed a distance of 6 in. Determine the acceleration a of the collar as a function of the vertical displacement x of the collar measured in feet from the point of release. Find the velocity v of the collar when ft. Friction is negligible.
Problem 3/34 3/35 The nonlinear spring has a tensile force-deflection relationship given by , where x is in meters and Fs is in newtons. Determine the accel-eration of the 6-kg block if it is released from rest at (a) mm and (b) mm.
Problem 3/35 Undeformed spring position 6 kg x s = 0.30 μ k = 0.25 μ x 100 x 50 Fs 150x 400x2 x x 0.5 3/36 Two configurations for raising an elevator are shown.
Elevator A with attached hoisting motor and drum has a total mass of 900 kg. Elevator B without motor and drum also has a mass of 900 kg. If the motor supplies a constant torque of 600 to its 250-mm-diameter drum for 2 s in each case, select the configuration which results in the greater up-ward acceleration and determine the corresponding velocity v of the elevator 1.2 s after it starts from rest. The mass of the motorized drum is small, thus permitting it to be analyzed as though it were in equilibrium. Neglect the mass of cables and pulleys and all friction.
Problem 3/36 3/37 Compute the acceleration of block A for the instant depicted. Neglect the masses of the pulleys.
Problem 3/37 μ 40 kg 30° T = 100 N A s = 0.50 μk = 0.40 ⎫ ⎬ ⎭ 250 mm (a) (b) A B 250 mm Nm Article 3/4 Problems 135 c03.qxd 2/9/12 7:38 PM Page 135 3/40 A shock absorber is a mechanical device which pro-vides resistance to compression or extension given by , where c is a constant and v is the time rate of change of the length of the absorber. An ab-sorber of constant N s/m is shown being tested with a 100-kg cylinder suspended from it.
The system is released with the cable taut at and allowed to extend. Determine (a) the steady-state velocity vs of the lower end of the absorber and (b) the time t and displacement y of the lower end when the cylinder has reached 90 percent of its steady-state speed. Neglect the mass of the piston and attached rod.
Problem 3/40 3/41 The design of a lunar mission calls for a 1200-kg spacecraft to lift off from the surface of the moon and travel in a straight line from point A and pass point B. If the spacecraft motor has a constant thrust of 2500 N, determine the speed of the space-craft as it passes point B. Use Table D/2 and the gravitational law from Chapter 1 as needed.
Problem 3/41 R R B A O m = 100 kg y c = 3000 N•s m y 0 c 3000 R cv 3/38 The inclined block A is given a constant rightward acceleration a. Determine the range of values of for which block B will not slip relative to block A, regardless of how large the acceleration a is. The co-efficient of static friction between the blocks is .
Problem 3/38 3/39 A spring-loaded device imparts an initial vertical ve-locity of 50 m/s to a 0.15-kg ball. The drag force on the ball is , where FD is in newtons when the speed v is in meters per second. Determine the maximum altitude h attained by the ball (a) with drag considered and (b) with drag neglected.
Problem 3/39 0.15 kg v0 = 50 m/s FD 0.002v2 A a B θ s μ s 136 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:38 PM Page 136 3/42 For what value(s) of the angle will the acceleration of the 80-lb block be 26 ft/sec2 to the right?
Problem 3/42 3/43 With the blocks initially at rest, the force P is in-creased slowly from zero to 60 lb. Plot the accelera-tions of both masses as functions of P.
Problem 3/43 3/44 An object projected vertically up from the surface of the earth with a sufficiently high velocity v0 can es-cape from the earth’s gravitational field. Calculate this velocity on the basis of the absence of an atmos-phere to offer resistance due to air friction. To elim-inate the effect of the earth’s rotation on the velocity measurement, consider the launch to be from the north or south pole. Use the mean radius of the earth and the absolute value of g as cited in Art. 1/5 and compare your answer with the value cited in Table D/2.
P A B s = 0.20 μ k = 0.15 μ s = 0.15 μ k = 0.10 μ 80 lb 100 lb θ μ 80 lb P = 100 lb s = 0.6, μk = 0.5 3/45 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a of the 30-kg block. The small pulley attached to the block has negligible mass and fric-tion. (Suggestion: First establish the kinematic rela-tionship between the accelerations of the two bodies.) Problem 3/45 3/46 The rod of the fixed hydraulic cylinder is moving to the left with a speed of 100 mm/s and this speed is momentarily increasing at a rate of 400 mm/s each sec-ond at the instant when Determine the tension in the cord at that instant. The mass of slider B is 0.5 kg, the length of the cord is 1050 mm, and the effects of the radius and friction of the small pulley at A are negligible. Find results for cases (a) negligible friction at slider B and (b) at slider B. The action is in a vertical plane.
Problem 3/46 250 mm C B A 0.5 kg sA k 0.40 sA 425 mm.
15 kg 30 kg s = k = = 0.25 μ μ μ 4 3 Article 3/4 Problems 137 c03.qxd 2/9/12 7:38 PM Page 137 3/5 Curvilinear Motion We turn our attention now to the kinetics of particles which move along plane curvilinear paths. In applying Newton’s second law, Eq. 3/3, we will make use of the three coordinate descriptions of acceleration in curvilinear motion which we developed in Arts. 2/4, 2/5, and 2/6.
The choice of an appropriate coordinate system depends on the con-ditions of the problem and is one of the basic decisions to be made in solving curvilinear-motion problems. We now rewrite Eq. 3/3 in three ways, the choice of which depends on which coordinate system is most appropriate.
Rectangular coordinates (Art. 2/4, Fig. 2/7) (3/6) where Normal and tangential coordinates (Art. 2/5, Fig. 2/10) (3/7) where Polar coordinates (Art. 2/6, Fig. 2/15) (3/8) where In applying these motion equations to a body treated as a particle, you should follow the general procedure established in the previous ar-ticle on rectilinear motion. After you identify the motion and choose the coordinate system, draw the free-body diagram of the body. Then obtain the appropriate force summations from this diagram in the usual way. The free-body diagram should be complete to avoid incor-rect force summations.
Once you assign reference axes, you must use the expressions for both the forces and the acceleration which are consistent with that as-signment. In the first of Eqs. 3/7, for example, the positive sense of the n-axis is toward the center of curvature, and so the positive sense of our force summation ΣFn must also be toward the center of curvature to agree with the positive sense of the acceleration an v2/.
ar r ¨ r ˙2 and a r ¨ 2r ˙ ˙ ΣF ma ΣFr mar an ˙2 v2/ v ˙, at v ˙, and v ˙ ΣFt mat ΣFn man ax x ¨ and ay y ¨ ΣFy may ΣFx max 138 Chapter 3 Kinetics of Particles Because of the banking in the turn of this track, the normal reaction force provides most of the normal acceler-ation of the bobsled.
Arno Balzarini/EPA/NewsCom At the highest point of the swing, this child experiences tangential acceleration. An instant later, when she has acquired velocity, she will experience normal acceleration as well.
© David Wall/Alamy c03.qxd 2/9/12 7:39 PM Page 138 SAMPLE PROBLEM 3/6 Determine the maximum speed v which the sliding block may have as it passes point A without losing contact with the surface.
Solution.
The condition for loss of contact is that the normal force N which the surface exerts on the block goes to zero. Summing forces in the normal direc-tion gives Ans.
If the speed at A were less than , then an upward normal force exerted by the surface on the block would exist. In order for the block to have a speed at A which is greater than , some type of constraint, such as a second curved sur-face above the block, would have to be introduced to provide additional down-ward force.
SAMPLE PROBLEM 3/7 Small objects are released from rest at A and slide down the smooth circular surface of radius R to a conveyor B. Determine the expression for the normal contact force N between the guide and each object in terms of and specify the correct angular velocity of the conveyor pulley of radius r to prevent any slid-ing on the belt as the objects transfer to the conveyor.
Solution.
The free-body diagram of the object is shown together with the co-ordinate directions n and t. The normal force N depends on the n-component of the acceleration which, in turn, depends on the velocity. The velocity will be cu-mulative according to the tangential acceleration at. Hence, we will find at first for any general position.
Now we can find the velocity by integrating We obtain the normal force by summing forces in the positive n-direction, which is the direction of the n-component of acceleration.
Ans.
The conveyor pulley must turn at the rate v r for /2, so that Ans.
2gR/r N mg sin m v 2 R N 3mg sin [ΣFn man] v 0 v dv 0 g cos d(R) v2 2gR sin [v dv at ds] mg cos mat at g cos [ΣFt mat] g g mg m v2 v g [ΣFn man] Helpful Hint It is essential here that we recognize the need to express the tangential ac-celeration as a function of position so that v may be found by integrating the kinematical relation v dv at ds, in which all quantities are measured along the path.
Article 3/5 Curvilinear Motion 139 c03.qxd 2/9/12 7:39 PM Page 139 SAMPLE PROBLEM 3/8 A 1500-kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h at A to a speed of 50 km/h as it passes C. The radius of curvature of the road at A is 400 m and at C is 80 m. Determine the total horizontal force exerted by the road on the tires at positions A, B, and C. Point B is the inflection point where the curvature changes direction.
Solution.
The car will be treated as a particle so that the effect of all forces ex-erted by the road on the tires will be treated as a single force. Since the motion is described along the direction of the road, normal and tangential coordinates will be used to specify the acceleration of the car. We will then determine the forces from the accelerations.
The constant tangential acceleration is in the negative t-direction, and its magnitude is given by The normal components of acceleration at A, B, and C are Application of Newton’s second law in both the n- and t-directions to the free-body diagrams of the car gives Thus, the total horizontal force acting on the tires becomes Ans.
Ans.
Ans.
At C, F Fn 2 Ft 2 (3620)2 (2170)2 4220 N At B, F Ft 2170 N At A, F Fn 2 Ft 2 (2890)2 (2170)2 3620 N At C, Fn 1500(2.41) 3620 N At B, Fn 0 At A, Fn 1500(1.929) 2890 N [ΣFn man] Ft 1500(1.447) 2170 N [ΣFt mat] At C, an (50/3.6)2 80 2.41 m/s2 At B, an 0 At A, an (100/3.6)2 400 1.929 m/s2 [an v2/] [vC 2 vA 2 2at s] at (50/3.6)2 (100/3.6)2 2(200) 1.447 m/s2 The angle made by a and F with the direction of the path can be com-puted if desired.
Note that the direction of Fn must agree with that of an.
Helpful Hints Recognize the numerical value of the conversion factor from km/h to m/s as 1000/3600 or 1/3.6.
Note that an is always directed to-ward the center of curvature.
140 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 140 SAMPLE PROBLEM 3/9 Compute the magnitude v of the velocity required for the spacecraft S to maintain a circular orbit of altitude 200 mi above the surface of the earth.
Solution.
The only external force acting on the spacecraft is the force of gravi-tational attraction to the earth (i.e., its weight), as shown in the free-body dia-gram. Summing forces in the normal direction yields where the substitution gR2 Gme has been made. Substitution of numbers gives Ans.
SAMPLE PROBLEM 3/10 Tube A rotates about the vertical O-axis with a constant angular rate and contains a small cylindrical plug B of mass m whose radial position is con-trolled by the cord which passes freely through the tube and shaft and is wound around the drum of radius b. Determine the tension T in the cord and the hori-zontal component F of force exerted by the tube on the plug if the constant an-gular rate of rotation of the drum is 0 first in the direction for case (a) and second in the direction for case (b). Neglect friction.
Solution.
With r a variable, we use the polar-coordinate form of the equations of motion, Eqs. 3/8. The free-body diagram of B is shown in the horizontal plane and discloses only T and F. The equations of motion are Case (a). With and , the forces become Ans.
Case (b). With and , the forces become Ans.
T mr2 F 2mb0 ¨ 0 r ˙ b0, r ¨ 0, T mr2 F 2mb0 ¨ 0 r ˙ b0, r ¨ 0, F m(r ¨ 2r ˙ ˙) [ΣF ma] T m(r ¨ r ˙2) [ΣFr mar] ˙ Helpful Hint Note that, for observations made within an inertial frame of reference, there is no such quantity as “centrifugal force” act-ing in the minus n-direction. Note also that neither the spacecraft nor its occupants are “weightless,” because the weight in each case is given by Newton’s law of gravitation. For this altitude, the weights are only about 10 percent less than the earth-surface values. Finally, the term “zero-g” is also misleading. It is only when we make our observations with respect to a coordinate system which has an acceleration equal to the gravitational acceleration (such as in an orbiting spacecraft) that we appear to be in a “zero-g” environment. The quantity which does go to zero aboard orbiting spacecraft is the fa-miliar normal force associated with, for example, an object in contact with a horizontal surface within the spacecraft.
Helpful Hint The minus sign shows that F is in the direction opposite to that shown on the free-body diagram.
Article 3/5 Curvilinear Motion 141 [ΣFn man] G mme (R h)2 m v2 (R h) , v Gme (R h) R g (R h) v (3959)(5280) 32.234 (3959 200)(5280) 25,326 ft/sec c03.qxd 2/9/12 7:39 PM Page 141 3/49 The 0.1-kg particle has a speed m/s as it passes the position shown. The coefficient of ki-netic friction between the particle and the vertical-plane track is . Determine the magnitude of the total force exerted by the track on the parti-cle. What is the deceleration of the particle?
Problem 3/49 3/50 The 4-oz slider has a speed ft/sec as it passes point A of the smooth guide, which lies in a horizontal plane. Determine the magnitude R of the force which the guide exerts on the slider (a) just before it passes point A of the guide and (b) as it passes point B.
Problem 3/50 A B 8″ v v 3 = 5 m v 30° ρ k 0.20 30 v 10 PROBLEMS Introductory Problems 3/47 The small 0.6-kg block slides with a small amount of friction on the circular path of radius 3 m in the ver-tical plane. If the speed of the block is 5 m/s as it passes point A and 4 m/s as it passes point B, deter-mine the normal force exerted on the block by the surface at each of these two locations.
Problem 3/47 3/48 A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to have a speed of 10 ft/sec as it passes position B, determine (a) the magnitude N of the force exerted by the fixed rod on the slider and (b) the rate at which the speed of the slider is de-creasing. Assume that friction is negligible.
Problem 3/48 30° 2′ A B 3 m v 30° B A 142 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 142 3/51 Determine the proper bank angle for the airplane flying at 400 mi/hr and making a turn of 2-mile ra-dius. Note that the force exerted by the air is nor-mal to the supporting wing surface.
Problem 3/51 3/52 The slotted arm rotates about its center in a hori-zontal plane at the constant angular rate rad/sec and carries a 3.22-lb spring-mounted slider which oscillates freely in the slot. If the slider has a speed of 24 in./sec relative to the slot as it crosses the center, calculate the horizontal side thrust P ex-erted by the slotted arm on the slider at this in-stant. Determine which side, A or B, of the slot is in contact with the slider.
Problem 3/52 24 in./sec A = 10 rad/sec θ .
B ˙ 10 3/53 The hollow tube is pivoted about a horizontal axis through point O and is made to rotate in the verti-cal plane with a constant counterclockwise angular velocity rad/sec. If a 0.2-lb particle is sliding in the tube toward O with a velocity of 4 ft/sec rela-tive to the tube when the position is passed, calculate the magnitude N of the normal force exerted by the wall of the tube on the particle at this instant.
Problem 3/53 3/54 The member OA rotates about a horizontal axis through O with a constant counterclockwise angu-lar velocity rad/sec. As it passes the position , a small block of mass m is placed on it at a radial distance in. If the block is observed to slip at , determine the coefficient of static friction between the block and the member.
Problem 3/54 3/55 In the design of a space station to operate outside the earth’s gravitational field, it is desired to give the structure a rotational speed N which will simu-late the effect of the earth’s gravity for members of the crew. If the centers of the crew’s quarters are to be located 12 m from the axis of rotation, calculate the necessary rotational speed N of the space sta-tion in revolutions per minute.
Problem 3/55 12 m N s 50 r 18 0 3 30 ˙ 3 Article 3/5 Problems 143 c03.qxd 2/9/12 7:39 PM Page 143 3/58 In order to simulate a condition of apparent “weight-lessness” experienced by astronauts in an orbiting spacecraft, a jet transport can change its direction at the top of its flight path by dropping its flight-path direction at a prescribed rate for a short in-terval of time. Specify if the aircraft has a speed km/h.
Problem 3/58 3/59 The standard test to determine the maximum lat-eral acceleration of a car is to drive it around a 200-ft-diameter circle painted on a level asphalt sur-face. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If this maximum speed is 35 mi/hr for a 3000-lb car, determine its lateral ac-celeration capability an in g’s and compute the mag-nitude F of the total friction force exerted by the pavement on the car tires.
Problem 3/59 v 100 ft θ .
600 km/h v 600 ˙ ˙ 3/56 A “swing ride” is shown in the figure. Calculate the necessary angular velocity for the swings to as-sume an angle with the vertical. Neglect the mass of the cables and treat the chair and person as one particle.
Problem 3/56 3/57 A Formula-1 car encounters a hump which has a circular shape with smooth transitions at either end. (a) What speed vB will cause the car to lose con-tact with the road at the topmost point B? (b) For a speed km/h, what is the normal force ex-erted by the road on the 640-kg car as it passes point A?
Problem 3/57 ρ = 300 m 10° A B vA 190 3.2 m θ ω 5 m 35 144 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 144 3/60 The car of Prob. 3/59 is traveling at 25 mi/hr when the driver applies the brakes, and the car continues to move along the circular path. What is the maxi-mum deceleration possible if the tires are limited to a total horizontal friction force of 2400 lb?
Representative Problems 3/61 The concept of variable banking for racetrack turns is shown in the figure. If the two radii of curvature are ft and ft for cars A and B, re-spectively, determine the maximum speed for each car. The coefficient of static friction is for both cars.
Problem 3/61 3/62 The small ball of mass m and its supporting wire be-come a simple pendulum when the horizontal cord is severed. Determine the ratio k of the tension T in the supporting wire immediately after the cord is cut to that in the wire before the cord is cut.
Problem 3/62 Wire m Cord θ A B 27° 22° s 0.90 B 320 A 300 3/63 A small object is given an initial horizontal velocity v0 at the bottom of a smooth slope. The angle made by the slope with the horizontal varies accord-ing to sin , where k is a constant and s is the distance measured along the slope from the bottom.
Determine the maximum distance s which the ob-ject slides up the slope.
3/64 A 3220-lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform rate in a distance of 300 ft measured along the curve from A to B. The radius of curvature of the path of the car at B is 600 ft. Cal-culate the total friction force exerted by the road on the tires at B. The road at B lies in a horizontal plane.
Problem 3/64 3/65 A pilot flies an airplane at a constant speed of 600 km/h in the vertical circle of radius 1000 m. Calcu-late the force exerted by the seat on the 90-kg pilot at point A and at point B.
Problem 3/65 300′ 600′ A B ks Article 3/5 Problems 145 c03.qxd 2/9/12 7:39 PM Page 145 3/68 A flatbed truck going 100 km/h rounds a horizontal curve of 300-m radius inwardly banked at . The coefficient of static friction between the truck bed and the 200-kg crate it carries is 0.70. Calculate the friction force F acting on the crate.
Problem 3/68 3/69 Explain how to utilize the graduated pendulum to measure the speed of a vehicle traveling in a hori-zontal circular arc of known radius r.
Problem 3/69 O m l θ 10° ρ 10 3/66 The 30-Mg aircraft is climbing at the angle under a jet thrust T of 180 kN. At the instant repre-sented, its speed is 300 km/h and is increasing at the rate of 1.96 m/s2. Also is decreasing as the aircraft begins to level off. If the radius of curvature of the path at this instant is 20 km, compute the lift L and drag D. (Lift L and drag D are the aerodynamic forces normal to and opposite to the flight direction, respectively.) Problem 3/66 3/67 The hollow tube assembly rotates about a vertical axis with angular velocity rad/s and rad/s2. A small 0.2-kg slider P moves in-side the horizontal tube portion under the control of the string which passes out the bottom of the assem-bly. If m, m/s, and m/s2, deter-mine the tension T in the string and the horizontal force exerted on the slider by the tube.
Problem 3/67 ω P r T F r ¨ 4 r ˙ 2 r 0.8 ˙ ¨ 2 ˙ 4 T θ 15 146 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 146 3/70 The bowl-shaped device rotates about a vertical axis with a constant angular velocity . If the particle is observed to approach a steady-state position in the presence of a very small amount of friction, determine . The value of r is 0.2 m.
Problem 3/70 3/71 The 2-kg slider fits loosely in the smooth slot of the disk, which rotates about a vertical axis through point O. The slider is free to move slightly along the slot before one of the wires becomes taut. If the disk starts from rest at time and has a constant clockwise angular acceleration of 0.5 rad/s2, plot the tensions in wires 1 and 2 and the magnitude N of the force normal to the slot as functions of time t for the interval .
Problem 3/71 45° 1 2 100 mm O ·· θ 0 t 5 s t 0 ω r m θ r 40 3/72 A 2-kg sphere S is being moved in a vertical plane by a robotic arm. When the angle is , the angu-lar velocity of the arm about a horizontal axis through O is 50 deg/s clockwise and its angular ac-celeration is 200 deg/s2 counterclockwise. In addi-tion, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping sur-faces is 0.50. Compare P with the minimum gripping force Ps required to hold the sphere in static equilib-rium in the position.
Problem 3/72 30 30 Article 3/5 Problems 147 c03.qxd 2/9/12 7:39 PM Page 147 3/75 A stretch of highway includes a succession of evenly spaced dips and humps, the contour of which may be represented by the relation .
What is the maximum speed at which the car A can go over a hump and still maintain contact with the road? If the car maintains this critical speed, what is the total reaction N under its wheels at the bottom of a dip? The mass of the car is m.
Problem 3/75 3/76 Determine the speed v at which the race car will have no tendency to slip sideways on the banked track, that is, the speed at which there is no reliance on friction. In addition, determine the minimum and maximum speeds, using the coefficient of static friction . State any assumptions.
Problem 3/76 s 0.90 y x b A L y b sin (2x/L) 3/73 The rocket moves in a vertical plane and is being propelled by a thrust T of 32 kN. It is also subjected to an atmospheric resistance R of 9.6 kN. If the rocket has a velocity of 3 km/s and if the gravita-tional acceleration is 6 m/s2 at the altitude of the rocket, calculate the radius of curvature of its path for the position described and the time-rate-of-change of the magnitude v of the velocity of the rocket. The mass of the rocket at the instant consid-ered is 2000 kg.
Problem 3/73 3/74 The robot arm is elevating and extending simulta-neously. At a given instant, , deg/s, deg/s2, m, m/s, and m/s2. Compute the radial and transverse forces Fr and that the arm must exert on the gripped part P, which has a mass of 1.2 kg. Compare with the case of static equilibrium in the same position.
Problem 3/74 F l ¨ 0.3 l ˙ 0.4 l 0.5 ¨ 120 ˙ 40 30 T R Vertical 30° 148 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 148 3/77 Small steel balls, each with a mass of 65 g, enter the semicircular trough in the vertical plane with a hori-zontal velocity of 4.1 m/s at A. Find the force R ex-erted by the trough on each ball in terms of and the velocity vB of the balls at B. Friction is negligible.
Problem 3/77 3/78 The flat circular disk rotates about a vertical axis through O with a slowly increasing angular velocity . Prior to rotation, each of the 0.5-kg sliding blocks has the position mm with no force in its at-tached spring. Each spring has a stiffness of 400 N/m. Determine the value of x for each spring for a steady speed of 240 rev/min. Also calculate the nor-mal force N exerted by the side of the slot on the block. Neglect any friction between the blocks and the slots, and neglect the mass of the springs. (Hint: Sum forces along and normal to the slot.) Problem 3/78 80 mm 80 mm x O A A ω x x 25 vA = 4.1 m/s vB 320 mm A B θ 3/79 The spring-mounted 0.8-kg collar A oscillates along the horizontal rod, which is rotating at the constant angular rate rad/s. At a certain instant, r is increasing at the rate of 800 mm/s. If the coefficient of kinetic friction between the collar and the rod is 0.40, calculate the friction force F exerted by the rod on the collar at this instant.
Problem 3/79 3/80 The slotted arm revolves in the horizontal plane about the fixed vertical axis through point O. The 3-lb slider C is drawn toward O at the constant rate of 2 in./sec by pulling the cord S. At the instant for which in., the arm has a counterclockwise an-gular velocity rad/sec and is slowing down at the rate of 2 rad/sec2. For this instant, determine the tension T in the cord and the magnitude N of the force exerted on the slider by the sides of the smooth radial slot. Indicate which side, A or B, of the slot contacts the slider.
Problem 3/80 6 r 9 Vertical A θ r .
˙ 6 Article 3/5 Problems 149 c03.qxd 2/9/12 7:39 PM Page 149 3/84 At the instant when , the horizontal guide is given a constant upward velocity m/s. For this instant calculate the force N exerted by the fixed circular slot and the force P exerted by the horizon-tal slot on the 0.5-kg pin A. The width of the slots is slightly greater than the diameter of the pin, and friction is negligible.
Problem 3/84 3/85 The particle P is released at time from the po-sition inside the smooth tube with no velocity relative to the tube, which is driven at the constant angular velocity about a vertical axis. Determine the radial velocity vr, the radial position r, and the transverse velocity as functions of time t. Explain why the radial velocity increases with time in the absence of radial forces. Plot the absolute path of the particle during the time it is inside the tube for m, m, and rad/s.
Problem 3/85 0 1 l 1 r0 0.1 v 0 r r0 t 0 A B v0 θ 250 mm v0 2 30 3/81 A small coin is placed on the horizontal surface of the rotating disk. If the disk starts from rest and is given a constant angular acceleration , deter-mine an expression for the number of revolutions N through which the disk turns before the coin slips.
The coefficient of static friction between the coin and the disk is .
Problem 3/81 3/82 The rotating drum of a clothes dryer is shown in the figure. Determine the angular velocity of the drum which results in loss of contact between the clothes and the drum at . Assume that the small vanes prevent slipping until loss of contact.
Problem 3/82 3/83 A body at rest relative to the surface of the earth ro-tates with the earth and therefore moves in a circu-lar path about the polar axis of the earth considered fixed. Derive an expression for the ratio k of the ap-parent weight of such a body as measured by a spring scale at the equator (calibrated to read the actual force applied) to the true weight of the body, which is the absolute gravitational attraction to the earth. The absolute acceleration due to gravity at the equator is m/s2. The radius of the earth at the equator is km, and the angu-lar velocity of the earth is rad/s. If the true weight is 100 N, what is the apparent mea-sured weight ?
W 0.729(104) R 6378 g 9.815 50 Vertical θ .
.
r s ¨ 150 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 150 3/86 The small 5-oz slider A moves without appreciable friction in the hollow tube, which rotates in a hori-zontal plane with a constant angular speed rad/sec. The slider is launched with an initial speed ft/sec relative to the tube at the inertial co-ordinates in. and . Determine the magni-tude P of the horizontal force exerted on the slider by the tube just before the slider exits the tube.
Problem 3/86 3/87 The two 0.2-kg sliders A and B move without fric-tion in the horizontal-plane circular slot. Determine the acceleration of each slider and the normal reac-tion force exerted on each when the system starts from rest in the position shown and is acted upon by the 4-N force P. Also find the tension in the inexten-sible connecting cord AB.
Problem 3/87 0.8 m A B O P = 4 N v y 0 x 6 r ˙0 60 7 3/88 Repeat the questions of the previous problem for the revised system configuration shown in the figure.
Problem 3/88 3/89 The 3000-lb car is traveling at 60 mi/hr on the straight portion of the road, and then its speed is re-duced uniformly from A to C, at which point it comes to rest. Compute the magnitude F of the total friction force exerted by the road on the car (a) just before it passes point B, (b) just after it passes point B, and (c) just before it stops at point C.
Problem 3/89 0.8 m A B O P = 4 N 45° Article 3/5 Problems 151 c03.qxd 2/9/12 7:39 PM Page 151 3/92 The small pendulum of mass m is suspended from a trolley which runs on a horizontal rail. The trolley and pendulum are initially at rest with . If the trolley is given a constant acceleration , deter-mine the maximum angle through which the pendulum swings. Also find the tension T in the cord in terms of .
Problem 3/92 3/93 A small object is released from rest at A and slides with friction down the circular path. If the coeffi-cient of friction is 0.20, determine the velocity of the object as it passes B. (Hint: Write the equations of motion in the n- and t-directions, eliminate N, and substitute . The resulting equation is a linear nonhomogeneous differential equation of the form , the solution of which is well known.) Problem 3/93 3 m m θ k = 0.20 μ A B dy/dx ƒ(x)y g(x) v dv atr d θ m l a max a g 0 3/90 The spacecraft P is in the elliptical orbit shown. At the instant represented, its speed is ft/sec.
Determine the corresponding values of , , , and . Use ft/sec2 as the acceleration of grav-ity on the surface of the earth and mi as the radius of the earth.
Problem 3/90 3/91 The slotted arm OA rotates about a horizontal axis through point O. The 0.2-kg slider P moves with negligible friction in the slot and is controlled by the inextensible cable BP. For the instant under consideration, , rad/s, , and m. Determine the corresponding values of the tension in cable BP and the force reaction R perpendicular to the slot. Which side of the slot contacts the slider?
Problem 3/91 θ ω 0.3 m A P B O r = 0.6 m r 0.6 ¨ 0 ˙ 4 30 v R 3959 g 32.23 ¨ r ¨ ˙ r ˙ v 13,244 152 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 152 3/94 The slotted arm OB rotates in a horizontal plane about point O of the fixed circular cam with constant angular velocity rad/s. The spring has a stiff-ness of 5 kN/m and is uncompressed when .
The smooth roller A has a mass of 0.5 kg. Determine the normal force N which the cam exerts on A and also the force R exerted on A by the sides of the slot when . All surfaces are smooth. Neglect the small diameter of the roller.
Problem 3/94 3/95 A small collar of mass m is given an initial velocity of magnitude v0 on the horizontal circular track fab-ricated from a slender rod. If the coefficient of ki-netic friction is , determine the distance traveled before the collar comes to rest. (Hint: Recognize that the friction force depends on the net normal force.) Problem 3/95 v0 k 0.1 m 0.1 m B A O θ 45 0 ˙ 15 3/96 The small cart is nudged with negligible velocity from its horizontal position at A onto the parabolic path, which lies in a vertical plane. Neglect friction and show that the cart maintains contact with the path for all values of k.
Problem 3/96 x y y = kx2 A Article 3/5 Problems 153 c03.qxd 2/9/12 7:39 PM Page 153 SECTION B WORK AND ENERGY 3/6 Work and Kinetic Energy In the previous two articles, we applied Newton’s second law F ma to various problems of particle motion to establish the instantaneous re-lationship between the net force acting on a particle and the resulting ac-celeration of the particle. When we needed to determine the change in velocity or the corresponding displacement of the particle, we integrated the computed acceleration by using the appropriate kinematic equations.
There are two general classes of problems in which the cumulative effects of unbalanced forces acting on a particle are of interest to us.
These cases involve (1) integration of the forces with respect to the dis-placement of the particle and (2) integration of the forces with respect to the time they are applied. We may incorporate the results of these inte-grations directly into the governing equations of motion so that it be-comes unnecessary to solve directly for the acceleration. Integration with respect to displacement leads to the equations of work and energy, which are the subject of this article. Integration with respect to time leads to the equations of impulse and momentum, discussed in Section C.
Definition of Work We now develop the quantitative meaning of the term “work.” Fig-ure 3/2a shows a force F acting on a particle at A which moves along the path shown. The position vector r measured from some convenient ori-gin O locates the particle as it passes point A, and dr is the differential displacement associated with an infinitesimal movement from A to A.
The work done by the force F during the displacement dr is defined as The magnitude of this dot product is dU F ds cos , where is the angle between F and dr and where ds is the magnitude of dr. This ex-pression may be interpreted as the displacement multiplied by the force component Ft F cos in the direction of the displacement, as repre-sented by the dashed lines in Fig. 3/2b. Alternatively, the work dU may be interpreted as the force multiplied by the displacement component ds cos in the direction of the force, as represented by the full lines in Fig. 3/2b.
With this definition of work, it should be noted that the component Fn F sin normal to the displacement does no work. Thus, the work dU may be written as Work is positive if the working component Ft is in the direction of the displacement and negative if it is in the opposite direction. Forces which dU Ft ds dU Fdr 154 Chapter 3 Kinetics of Particles The concept of work was also developed in the study of virtual work in Chapter 7 of Vol. 1 Statics.
O A A′ F r r + dr dr α α (a) (b) α Ft = F cos ds = |dr| α ds cos F Fn Figure 3/2 c03.qxd 2/9/12 7:39 PM Page 154 do work are termed active forces. Constraint forces which do no work are termed reactive forces.
Units of Work The SI units of work are those of force (N) times displacement (m) or This unit is given the special name joule (J), which is defined as the work done by a force of 1 N acting through a distance of 1 m in the direction of the force. Consistent use of the joule for work (and energy) rather than the units will avoid possible ambiguity with the units of moment of a force or torque, which are also written In the U.S. customary system, work has the units of ft-lb. Dimen-sionally, work and moment are the same. In order to distinguish be-tween the two quantities, it is recommended that work be expressed as foot pounds (ft-lb) and moment as pound feet (lb-ft). It should be noted that work is a scalar as given by the dot product and involves the prod-uct of a force and a distance, both measured along the same line. Mo-ment, on the other hand, is a vector as given by the cross product and involves the product of force and distance measured at right angles to the force.
Calculation of Work During a finite movement of the point of application of a force, the force does an amount of work equal to or In order to carry out this integration, it is necessary to know the rela-tions between the force components and their respective coordinates or the relation between Ft and s. If the functional relationship is not known as a mathematical expression which can be integrated but is specified in the form of approximate or experimental data, then we can compute the work by carrying out a numerical or graphical integration as represented by the area under the curve of Ft versus s, as shown in Fig. 3/3.
Examples of Work When work must be calculated, we may always begin with the defin-ition of work, U insert appropriate vector expressions for the force F and the differential displacement vector dr, and carry out the re-quired integration. With some experience, simple work calculations, such as those associated with constant forces, may be performed by in-spection. We now formally compute the work associated with three fre-quently occurring forces: constant forces, spring forces, and weights.
Fdr, U s2 s1 Ft ds U 2 1 Fdr 2 1 (Fx dx Fy dy Fz dz) Nm.
N m Nm.
Article 3/6 Work and Kinetic Energy 155 s Ft s1 s2 dU = Ft ds Figure 3/3 c03.qxd 2/9/12 7:39 PM Page 155 (1) Work Associated with a Constant External Force. Consider the constant force P applied to the body as it moves from position 1 to position 2, Fig. 3/4. With the force P and the differential displacement dr written as vectors, the work done on the body by the force is (3/9) As previously discussed, this work expression may be interpreted as the force component P cos times the distance L traveled. Should be be-tween 90 and 270, the work would be negative. The force component P sin normal to the displacement does no work.
(2) Work Associated with a Spring Force. We consider here the common linear spring of stiffness k where the force required to stretch or compress the spring is proportional to the deformation x, as shown in Fig. 3/5a. We wish to determine the work done on the body by the spring force as the body undergoes an arbitrary displacement from an initial position x1 to a final position x2. The force exerted by the spring on the body is F kxi, as shown in Fig. 3/5b. From the definition of work, we have (3/10) If the initial position is the position of zero spring deformation so that x1 0, then the work is negative for any final position x2 0. This is verified by recognizing that if the body begins at the undeformed spring position and then moves to the right, the spring force is to the left; if the body begins at x1 0 and moves to the left, the spring force is to the right. On the other hand, if we move from an arbitrary initial po-sition x1 0 to the undeformed final position x2 0, we see that the work is positive. In any movement toward the undeformed spring posi-tion, the spring force and the displacement are in the same direction.
In the general case, of course, neither x1 nor x2 is zero. The magni-tude of the work is equal to the shaded trapezoidal area of Fig. 3/5a. In calculating the work done on a body by a spring force, care must be U1-2 2 1 Fdr 2 1 (kxi) dx i x2 x1 kx dx 1 2 k(x1 2 x2 2) x2 x1 P cos dx P cos (x2 x1) PL cos U1-2 2 1 Fdr 2 1 [(P cos )i (P sin )j] dx i 156 Chapter 3 Kinetics of Particles P dr x y α L 1 2 Figure 3/4 c03.qxd 2/9/12 7:39 PM Page 156 Article 3/6 Work and Kinetic Energy 157 taken to ensure that the units of k and x are consistent. If x is in meters (or feet), k must be in N/m (or lb/ft). In addition, be sure to recognize that the variable x represents a deformation from the unstretched spring length and not the total length of the spring.
The expression F kx is actually a static relationship which is true only when elements of the spring have no acceleration. The dynamic be-havior of a spring when its mass is accounted for is a fairly complex problem which will not be treated here. We shall assume that the mass of the spring is small compared with the masses of other accelerating parts of the system, in which case the linear static relationship will not involve appreciable error.
(3) Work Associated with Weight. Case (a) g constant. If the al-titude variation is sufficiently small so that the acceleration of gravity g may be considered constant, the work done by the weight mg of the body shown in Fig. 3/6a as the body is displaced from an arbitrary alti-tude y1 to a final altitude y2 is (3/11) mg y2 y1 dy mg( y2 y1) U1-2 2 1 Fdr 2 1 (mgj) (dxi dyj) x x1 x2 x F = kx kx Undeformed position dr Force F required to stretch or compress spring (a) (b) Figure 3/5 c03.qxd 2/9/12 7:39 PM Page 157 We see that horizontal movement does not contribute to this work. We also note that if the body rises (perhaps due to other forces not shown), then ( y2 y1) 0 and this work is negative. If the body falls, ( y2 y1) 0 and the work is positive.
Case (b) g constant. If large changes in altitude occur, then the weight (gravitational force) is no longer constant. We must therefore use the gravitational law (Eq. 1/2) and express the weight as a variable force of magnitude F as indicated in Fig. 3/6b. Using the radial coordinate shown in the figure allows the work to be expressed as (3/12) where the equivalence Gme gR2 was established in Art. 1/5, with g representing the acceleration of gravity at the earth’s surface and R rep-resenting the radius of the earth. The student should verify that if a body rises to a higher altitude (r2 r1), this work is negative, as it was in case (a). If the body falls to a lower altitude (r2 r1), the work is posi-tive. Be sure to realize that r represents a radial distance from the cen-ter of the earth and not an altitude h r R above the surface of the earth. As in case (a), had we considered a transverse displacement in ad-dition to the radial displacement shown in Fig. 3/6b, we would have con-cluded that the transverse displacement, because it is perpendicular to the weight, does not contribute to the work. Gmem 1 r2 1 r1 mgR2 1 r2 1 r1 U1-2 2 1 Fdr 2 1 Gmem r2 erdrer Gmem r2 r1 dr r2 Gmem r 2 , 158 Chapter 3 Kinetics of Particles Earth me 1 2 2 1 r1 r2 y2 y1 r dr dr er m mg m (a) (b) Gmem —–— r2 R x y Figure 3/6 c03.qxd 2/9/12 7:39 PM Page 158 Article 3/6 Work and Kinetic Energy 159 Work and Curvilinear Motion We now consider the work done on a particle of mass m, Fig. 3/7, moving along a curved path under the action of the force F, which stands for the resultant ΣF of all forces acting on the particle. The posi-tion of m is specified by the position vector r, and its displacement along its path during the time dt is represented by the change dr in its posi-tion vector. The work done by F during a finite movement of the parti-cle from point 1 to point 2 is where the limits specify the initial and final end points of the motion.
When we substitute Newton’s second law F ma, the expression for the work of all forces becomes But at ds, where at is the tangential component of the accelera-tion of m. In terms of the velocity v of the particle, Eq. 2/3 gives at ds v dv. Thus, the expression for the work of F becomes (3/13) where the integration is carried out between points 1 and 2 along the curve, at which points the velocities have the magnitudes v1 and v2, respectively.
Principle of Work and Kinetic Energy The kinetic energy T of the particle is defined as (3/14) and is the total work which must be done on the particle to bring it from a state of rest to a velocity v. Kinetic energy T is a scalar quantity with the units of or joules (J) in SI units and ft-lb in U.S. customary units. Kinetic energy is always positive, regardless of the direction of the velocity.
Equation 3/13 may be restated as (3/15) which is the work-energy equation for a particle. The equation states that the total work done by all forces acting on a particle as it moves from point 1 to point 2 equals the corresponding change in kinetic en-ergy of the particle. Although T is always positive, the change T may U1-2 T2 T1 T Nm T 1 2 mv2 U1-2 2 1 Fdr v2 v1 mv dv 1 2 m(v2 2 v1 2) a dr U1-2 2 1 Fdr 2 1 ma dr U1-2 2 1 F dr s2 s1 Ft ds Ft Fn F = ΣF Path 1 2 x s1 s2 z t n m dr r α y O Figure 3/7 c03.qxd 2/9/12 7:39 PM Page 159 be positive, negative, or zero. When written in this concise form, Eq.
3/15 tells us that the work always results in a change of kinetic energy.
Alternatively, the work-energy relation may be expressed as the ini-tial kinetic energy T1 plus the work done U1-2 equals the final kinetic en-ergy T2, or (3/15a) When written in this form, the terms correspond to the natural se-quence of events. Clearly, the two forms 3/15 and 3/15a are equivalent.
Advantages of the Work-Energy Method We now see from Eq. 3/15 that a major advantage of the method of work and energy is that it avoids the necessity of computing the acceler-ation and leads directly to the velocity changes as functions of the forces which do work. Further, the work-energy equation involves only those forces which do work and thus give rise to changes in the magnitude of the velocities.
We consider now a system of two particles joined together by a con-nection which is frictionless and incapable of any deformation. The forces in the connection are equal and opposite, and their points of ap-plication necessarily have identical displacement components in the di-rection of the forces. Therefore, the net work done by these internal forces is zero during any movement of the system. Thus, Eq. 3/15 is ap-plicable to the entire system, where U1-2 is the total or net work done on the system by forces external to it and T is the change, T2 T1, in the total kinetic energy of the system. The total kinetic energy is the sum of the kinetic energies of both elements of the system. We thus see that another advantage of the work-energy method is that it enables us to analyze a system of particles joined in the manner described without dismembering the system.
Application of the work-energy method requires isolation of the par-ticle or system under consideration. For a single particle you should draw a free-body diagram showing all externally applied forces. For a system of particles rigidly connected without springs, draw an active-force diagram showing only those external forces which do work (active forces) on the entire system. Power The capacity of a machine is measured by the time rate at which it can do work or deliver energy. The total work or energy output is not a measure of this capacity since a motor, no matter how small, can deliver a large amount of energy if given sufficient time. On the other hand, a large and powerful machine is required to deliver a large amount of en-ergy in a short period of time. Thus, the capacity of a machine is rated by its power, which is defined as the time rate of doing work.
T1 U1-2 T2 160 Chapter 3 Kinetics of Particles The active-force diagram was introduced in the method of virtual work in statics. See Chapter 7 of Vol. 1 Statics.
c03.qxd 2/9/12 7:39 PM Page 160 Article 3/6 Work and Kinetic Energy 161 Accordingly, the power P developed by a force F which does an amount of work U is P dU/dt Because dr/dt is the velocity v of the point of application of the force, we have (3/16) Power is clearly a scalar quantity, and in SI it has the units of J/s.
The special unit for power is the watt (W), which equals one joule per sec-ond (J/s). In U.S. customary units, the unit for mechanical power is the horsepower (hp). These units and their numerical equivalences are Efficiency The ratio of the work done by a machine to the work done on the machine during the same time interval is called the mechanical effi-ciency em of the machine. This definition assumes that the machine op-erates uniformly so that there is no accumulation or depletion of energy within it. Efficiency is always less than unity since every device operates with some loss of energy and since energy cannot be created within the machine. In mechanical devices which involve moving parts, there will always be some loss of energy due to the negative work of kinetic fric-tion forces. This work is converted to heat energy which, in turn, is dis-sipated to the surroundings. The mechanical efficiency at any instant of time may be expressed in terms of mechanical power P by (3/17) In addition to energy loss by mechanical friction, there may also be electrical and thermal energy loss, in which case, the electrical efficiency ee and thermal efficiency et are also involved. The overall efficiency e in such instances is e emeeet em Poutput Pinput 1 hp 746 W 0.746 kW 1 hp 550 ft-lb/sec 33,000 ft-lb/min 1 W 1 J/s Nm/s P Fv Fdr/dt.
The power which must be produced by a bike rider depends on the bicy-cle speed and the propulsive force which is exerted by the supporting surface on the rear wheel. The dri-ving force depends on the slope being negotiated.
Media Bakery c03.qxd 2/9/12 7:39 PM Page 161 162 Chapter 3 Kinetics of Particles SAMPLE PROBLEM 3/11 Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coef-ficient of kinetic friction is 0.30.
Solution.
The free-body diagram of the crate is drawn and includes the nor-mal force R and the kinetic friction force F calculated in the usual manner. The work done by the weight is positive, whereas that done by the friction force is negative. The total work done on the crate during the motion is The work-energy equation gives Ans.
Since the net work done is negative, we obtain a decrease in the kinetic energy.
SAMPLE PROBLEM 3/12 The flatbed truck, which carries an 80-kg crate, starts from rest and attains a speed of 72 km/h in a distance of 75 m on a level road with constant accelera-tion. Calculate the work done by the friction force acting on the crate during this interval if the static and kinetic coefficients of friction between the crate and the truck bed are (a) 0.30 and 0.28, respectively, or (b) 0.25 and 0.20, respectively.
Solution.
If the crate does not slip on the bed, its acceleration will be that of the truck, which is Case (a). This acceleration requires a friction force on the block of which is less than the maximum possible value of sN 0.30(80)(9.81) = 235 N.
Therefore, the crate does not slip and the work done by the actual static friction force of 213 N is Ans.
Case (b). For s 0.25, the maximum possible friction force is 0.25(80)(9.81) 196.2 N, which is slightly less than the value of 213 N required for no slipping.
Therefore, we conclude that the crate slips, and the friction force is governed by the kinetic coefficient and is F 0.20(80)(9.81) = 157.0 N. The acceleration becomes The distances traveled by the crate and the truck are in proportion to their ac-celerations. Thus, the crate has a displacement of (1.962/2.67)75 = 55.2 m, and the work done by kinetic friction is Ans.
U1-2 157.0(55.2) 8660 J or 8.66 kJ [U Fs] a F/m 157.0/80 1.962 m/s2 [F ma] U1-2 213(75) 16 000 J or 16 kJ [U Fs] F 80(2.67) 213 N [F ma] a v2 2s (72/3.6)2 2(75) 2.67 m/s2 [v2 2as] v2 3.15 m/s 1 2 (50)(4)2 151.9 1 2 (50)v2 2 1 2 mv1 2 U1-2 1 2 mv2 2 [T1 U1-2 T2] U1-2 50(9.81)(10 sin 15) 142.1(10) 151.9 J [U Fs] 15° 50 kg B A 10 m 50(9.81) N R = 474 N kR = 142.1 N μ 80(9.81) N a F 80(9.81) N Helpful Hint The work due to the weight depends only on the vertical distance traveled.
Helpful Hints We note that static friction forces do no work when the contacting sur-faces are both at rest. When they are in motion, however, as in this prob-lem, the static friction force acting on the crate does positive work and that acting on the truck bed does negative work.
This problem shows that a kinetic friction force can do positive work when the surface which supports the object and generates the friction force is in motion. If the supporting surface is at rest, then the kinetic friction force acting on the moving part always does negative work.
c03.qxd 2/9/12 7:39 PM Page 162 Article 3/6 Work and Kinetic Energy 163 SAMPLE PROBLEM 3/13 The 50-kg block at A is mounted on rollers so that it moves along the fixed horizontal rail with negligible friction under the action of the constant 300-N force in the cable. The block is released from rest at A, with the spring to which it is attached extended an initial amount x1 0.233 m. The spring has a stiffness k 80 N/m. Calculate the velocity v of the block as it reaches position B.
Solution.
It will be assumed initially that the stiffness of the spring is small enough to allow the block to reach position B. The active-force diagram for the system composed of both block and cable is shown for a general position. The spring force 80x and the 300-N tension are the only forces external to this sys-tem which do work on the system. The force exerted on the block by the rail, the weight of the block, and the reaction of the small pulley on the cable do no work on the system and are not included on the active-force diagram.
As the block moves from x1 0.233 m to x2 0.233 1.2 1.433 m, the work done by the spring force acting on the block is The work done on the system by the constant 300-N force in the cable is the force times the net horizontal movement of the cable over pulley C, which is 0.9 0.6 m. Thus, the work done is 300(0.6) 180 J. We now apply the work-energy equation to the system and get Ans.
We take special note of the advantage to our choice of system. If the block alone had constituted the system, the horizontal component of the 300-N cable tension on the block would have to be integrated over the 1.2-m displacement.
This step would require considerably more effort than was needed in the solu-tion as presented. If there had been appreciable friction between the block and its guiding rail, we would have found it necessary to isolate the block alone in order to compute the variable normal force and, hence, the variable friction force. Integration of the friction force over the displacement would then be re-quired to evaluate the negative work which it would do.
0 80.0 180 1 2 (50)v2 v 2.00 m/s [T1 U1-2 T2] (1.2)2 (0.9)2 80.0 J U1-2 1 2 80[0.2332 (0.233 1.2)2] [U1-2 1 2 k(x1 2 x2 2)] 0.9 m 300 N B C A 1.2 m x1 300 N 80x System x Helpful Hint Recall that this general formula is valid for any initial and final spring deflections x1 and x2, positive (spring in tension) or negative (spring in compression). In deriving the spring-work formula, we assumed the spring to be linear, which is the case here.
c03.qxd 2/9/12 7:39 PM Page 163 SAMPLE PROBLEM 3/14 The power winch A hoists the 800-lb log up the 30 incline at a constant speed of 4 ft/sec. If the power output of the winch is 6 hp, compute the coefficient of ki-netic friction k between the log and the incline. If the power is suddenly increased to 8 hp, what is the corresponding instantaneous acceleration a of the log?
Solution.
From the free-body diagram of the log, we get N 800 cos 30 693 lb, and the kinetic friction force becomes 693k. For constant speed, the forces are in equilibrium so that The power output of the winch gives the tension in the cable Substituting T gives Ans.
When the power is increased, the tension momentarily becomes and the corresponding acceleration is given by Ans.
SAMPLE PROBLEM 3/15 A satellite of mass m is put into an elliptical orbit around the earth. At point A, its distance from the earth is h1 500 km and it has a velocity v1 30 000 km/h. Determine the velocity v2 of the satellite as it reaches point B, a distance h2 1200 km from the earth.
Solution.
The satellite is moving outside of the earth’s atmosphere so that the only force acting on it is the gravitational attraction of the earth. For the large change in altitude of this problem, we cannot assume that the acceleration due to gravity is constant. Rather, we must use the work expression, derived in this article, which accounts for variation in the gravitational acceleration with alti-tude. Put another way, the work expression accounts for the variation of the weight F with altitude. This work expression is The work-energy equation T1 U1-2 T2 gives Substituting the numerical values gives Ans.
v2 7663 m/s or v2 7663(3.6) 27 590 km/h 69.44(106) 10.72(106) 58.73(106) (m/s)2 v2 2 30 000 3.6 2 2(9.81)[(6371)(103)]2 103 6371 1200 103 6371 500 1 2 mv1 2 mgR2 1 r2 1 r1 1 2 mv2 2 v2 2 v1 2 2gR2 1 r2 1 r1 U1-2 mgR2 1 r2 1 r1 Gmme r2 a 11.07 ft/sec2 1100 693(0.613) 800 sin 30 800 32.2 a [ΣFx max] T P/v 8(550)/4 1100 lb [P Tv] 825 693k 400 k 0.613 T P/v 6(550)/4 825 lb [P Tv] T 693k 800 sin 30 0 T 693k 400 [ΣFx 0] A 4 ft/sec 30° T 800 lb 30° kN μ N x A B h1 v1 v2 h2 O R A B r O F r2 r1 Helpful Hints Note the conversion from horse-power to ft-lb/sec.
As the speed increases, the accelera-tion will drop until the speed stabi-lizes at a value higher than 4 ft/sec.
Helpful Hints Note that the result is independent of the mass m of the satellite.
Consult Table D/2, Appendix D, to find the radius R of the earth.
164 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 164 PROBLEMS Introductory Problems 3/97 The spring is unstretched when . If the body moves from the initial position to the final position , (a) determine the work done by the spring on the body and (b) determine the work done on the body by its weight.
Problem 3/97 3/98 The small body has a speed at point A.
Neglecting friction, determine its speed at point B after it has risen 0.8 m. Is knowledge of the shape of the track necessary?
Problem 3/98 3/99 The 64.4-lb crate slides down the curved path in the vertical plane. If the crate has a velocity of 3 ft/sec down the incline at A and a velocity of 25 ft/sec at B, compute the work Uƒ done on the crate by friction during the motion from A to B.
Problem 3/99 20′ 3 ft/sec A B 25 ft/sec 30′ B A 5 m/s x 7 kg 4 kN/m 20° x2 200 mm x1 100 mm x 0 3/100 The 1.5-lb collar slides with negligible friction on the fixed rod in the vertical plane. If the collar starts from rest at A under the action of the con-stant 2-lb horizontal force, calculate its velocity v as it hits the stop at B.
Problem 3/100 3/101 In the design of a spring bumper for a 3500-lb car, it is desired to bring the car to a stop from a speed of 5 mi/hr in a distance equal to 6 in. of spring de-formation. Specify the required stiffness k for each of the two springs behind the bumper. The springs are undeformed at the start of impact.
Problem 3/101 3/102 A two-engine jet transport has a loaded weight of 90,000 lb and a forward thrust of 9800 lb per engine during takeoff. If the transport requires 4800 ft of level runway starting from rest to become airborne at a speed of 140 knots , de-termine the average resistance R to motion over the runway length due to drag (air resistance) and me-chanical retardation by the landing gear.
(1 knot 1.151 mi/hr) 5 mi/hr 2 lb 30″ 15″ A B Article 3/6 Problems 165 5 m/s 0.8 m B A c03.qxd 2/9/12 7:39 PM Page 165 3/105 The two small 0.2-kg sliders are connected by a light rigid bar and are constrained to move with-out friction in the circular slot. The force is constant in magnitude and direction and is ap-plied to the moving slider A. The system starts from rest in the position shown. Determine the speed of slider A as it passes the initial position of slider B if (a) the circular track lies in a horizontal plane and if (b) the circular track lies in a vertical plane. The value of R is 0.8 m.
Problem 3/105 3/106 The man and his bicycle together weigh 200 lb.
What power P is the man developing in riding up a 5-percent grade at a constant speed of 15 mi/hr?
Problem 3/106 5 100 15 mi/hr 30° R A B O P P 12 N 3/103 The small collar of mass m is released from rest at A and slides down the curved rod in the vertical plane with negligible friction. Express the velocity v of the collar as it strikes the base at B in terms of the given conditions.
Problem 3/103 3/104 For the sliding collar of Prob. 3/103, if , , and , and if the velocity of the collar as it strikes the base B is 4.70 m/s after re-lease of the collar from rest at A, calculate the work Q of friction. What happens to the energy which is lost?
h 1.5 m b 0.8 m m 0.5 kg A b h B 166 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 166 3/107 The system is released from rest with no slack in the cable and with the spring unstretched. Deter-mine the distanced s traveled by the 10-kg cart be-fore it comes to rest (a) if m approaches zero and (b) if . Assume no mechanical interference.
Problem 3/107 3/108 The system is released from rest with no slack in the cable and with the spring stretched 200 mm.
Determine the distance s traveled by the 10-kg cart before it comes to rest (a) if m approaches zero and (b) if . Assume no mechanical interference.
Problem 3/108 m k = 125 N/m 25° 10 kg m 2 kg m k = 125 N/m 25° 10 kg m 2 kg 3/109 The 2-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane.
The coefficient of kinetic friction is 0.40. Calculate (a) the velocity v of the collar as it strikes the spring and (b) the maximum deflection x of the spring.
Problem 3/109 3/110 Each of the two systems is released from rest. Cal-culate the velocity v of each 50-lb cylinder after the 40-lb cylinder has dropped 6 ft. The 20-lb cylinder of case (a) is replaced by a 20-lb force in case (b).
Problem 3/110 50 lb 40 lb 20 lb 50 lb 40 lb 20 lb (b) (a) 0.5 m 2 kg A k = 1.6 kN/m = 0.40 k μ 60° Article 3/6 Problems 167 c03.qxd 2/9/12 7:39 PM Page 167 Representative Problems 3/113 An escalator handles a steady load of 30 people per minute in elevating them from the first to the sec-ond floor through a vertical rise of 24 ft. The aver-age person weighs 140 lb. If the motor which drives the unit delivers 4 hp, calculate the mechan-ical efficiency e of the system.
Problem 3/113 3/114 A 1200-kg car enters an 8-percent downhill grade at a speed of 100 km/h. The driver applies her brakes to bring the car to a speed of 25 km/h in a distance of 0.5 km measured along the road. Calcu-late the energy loss Q dissipated from the brakes in the form of heat. Neglect any friction losses from other causes such as air resistance.
3/115 The 15-lb cylindrical collar is released from rest in the position shown and drops onto the spring. Cal-culate the velocity v of the cylinder when the spring has been compressed 2 in.
Problem 3/115 18″ 15 lb A B k = 80 lb/in.
3/111 The 120-lb woman jogs up the flight of stairs in 5 seconds. Determine her average power output.
Convert all given information to SI units and re-peat your calculation.
Problem 3/111 3/112 The 4-kg ball and the attached light rod rotate in the vertical plane about the fixed axis at O. If the assembly is released from rest at and moves under the action of the 60-N force, which is main-tained normal to the rod, determine the velocity v of the ball as approaches . Treat the ball as a particle.
Problem 3/112 O A 4 kg 60 N θ 200 mm 300 mm 90 0 9′ 168 Chapter 3 Kinetics of Particles 24′ c03.qxd 2/9/12 7:39 PM Page 168 3/116 Determine the constant force P required to cause the 0.5-kg slider to have a speed at po-sition 2. The slider starts from rest at position 1 and the unstretched length of the spring of modu-lus is 200 mm. Neglect friction.
Problem 3/116 3/117 In a design test of penetration resistance, a 12-g bullet is fired through a 400-mm stack of fibrous plates with an entering velocity of 600 m/s. If the bullet emerges with a velocity of 300 m/s, calculate the average resistance R to penetration. What is the loss of energy and where does it go?
Problem 3/117 300 m/s 600 m/s 400 mm Q P k m 250 mm 250 mm 200 mm 1 2 200 mm 15° k 250 N/m v2 0.8 m/s 3/118 The motor unit A is used to elevate the 300-kg cylinder at a constant rate of 2 m/s. If the power meter B registers an electrical input of 2.20 kW, calculate the combined electrical and mechanical efficiency e of the system.
Problem 3/118 3/119 A 1700-kg car starts from rest at position A and ac-celerates uniformly up the incline, reaching a speed of 100 km/h at position B. Determine the power required just before the car reaches position B and also the power required when the car is halfway between positions A and B. Calculate the net tractive force F required.
Problem 3/119 A B 100 6 110 m vB = 100 km/h vA = 0 B A 100 kg 300 kg 2 m/s Article 3/6 Problems 169 c03.qxd 2/9/12 7:39 PM Page 169 3/122 A projectile is launched from the north pole with an initial vertical velocity . What value of will result in a maximum altitude of R/2? Neglect aero-dynamic drag and use as the surface-level acceleration due to gravity.
Problem 3/122 3/123 The spring is compressed an amount and the system is released from rest. Determine the power supplied by the spring to the 4-kg cart (a) just after release, (b) as the cart passes the po-sition for which the spring is compressed an amount /2, and (c) as the cart passes the equilib-rium position.
Problem 3/123 k = 3.5 kN/m Unstretched position δ 4 kg 80 mm R v0 g 9.825 m/s2 v0 v0 3/120 Two 425,000-lb locomotives pull 50 200,000-lb coal hoppers. The train starts from rest and accelerates uniformly to a speed of 40 mi/hr over a distance of 8000 ft on a level track. The constant rolling resis-tance of each car is 0.005 times its weight. Neglect all other retarding forces and assume that each lo-comotive contributes equally to the tractive force.
Determine (a) the tractive force exerted by each lo-comotive at 20 mi/hr, (b) the power required from each locomotive at 20 mi/hr, (c) the power required from each locomotive as the train speed ap-proaches 40 mi/hr, and (d) the power required from each locomotive if the train cruises at a steady 40 mi/hr.
Problem 3/120 3/121 The 0.6-lb slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 1.3-lb tension in the cord. If the slider is released from rest at A, calcu-late its velocity v as it reaches B.
Problem 3/121 6″ 1.3 lb A 10″ 24″ B 50 coal hoppers 170 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 170 3/124 In a test to determine the crushing characteristics of a packing material, a steel cone of mass m is re-leased, falls a distance h, and then penetrates the material. The radius of the cone is proportional to the square of the distance from its tip. The resis-tance R of the material to penetration depends on the cross-sectional area of the penetrating object and thus is proportional to the fourth power of the cone penetration distance x, or . If the cone comes to rest at a distance , determine the constant k in terms of the test conditions and re-sults. Utilize a single application of the work-energy equation.
Problem 3/124 3/125 The small slider of mass m is released from rest while in position A and then slides along the verti-cal-plane track. The track is smooth from A to D and rough (coefficient of kinetic friction ) from point D on. Determine (a) the normal force NB ex-erted by the track on the slider just after it passes point B, (b) the normal force NC exerted by the track on the slider as it passes the bottom point C, and (c) the distance s traveled along the incline past point D before the slider stops.
Problem 3/125 2R 30° A B s C D m R k μ k h x x d R kx4 3/126 The 0.5-kg collar slides with negligible friction along the fixed spiral rod, which lies in the vertical plane. The rod has the shape of the spiral , where r is in meters and is in radians. The collar is released from rest at A and slides to B under the action of a constant radial force . Calcu-late the velocity v of the slider as it reaches B.
Problem 3/126 3/127 The 300-lb carriage has an initial velocity of 9 ft/sec down the incline at A, when a constant force of 110 lb is applied to the hoisting cable as shown. Calcu-late the velocity of the carriage when it reaches B.
Show that in the absence of friction this velocity is independent of whether the initial velocity of the carriage at A was up or down the incline.
Problem 3/127 10′ 300 lb vA = 9 ft/sec A 12 110 lb B 5 y x T T T A B r θ T 10 N r 0.3 Article 3/6 Problems 171 c03.qxd 2/9/12 7:39 PM Page 171 3/130 The two 0.2-kg sliders A and B are connected by a light rigid bar of length . If the system is released from rest while in the position shown with the spring undeformed, determine the maxi-mum compression of the spring. Note the pres-ence of a constant 0.14-MPa air pressure acting on one 500-mm2 side of slider A. Neglect friction. The motion occurs in a vertical plane.
Problem 3/130 3/131 Once under way at a steady speed, the 1000-kg ele-vator A rises at the rate of 1 story (3 m) per sec-ond. Determine the power input Pin into the motor unit M if the combined mechanical and electrical efficiency of the system is .
Problem 3/131 A M e 0.8 B A 60° 30° k = 1.2 kN/m L L 0.5 m 3/128 Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its respective guide, with y being in the vertical direction. A 20-N horizontal force is applied to the midpoint of the connecting link of negligible mass, and the assem-bly is released from rest with . Calculate the velocity with which A strikes the horizontal guide when .
Problem 3/128 3/129 The ball is released from position A with a velocity of 3 m/s and swings in a vertical plane. At the bot-tom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc.
Calculate the velocity of the ball as it passes po-sition C.
Problem 3/129 60° B C A 1.2 m 3 m/s 0.8 m vC 0.2 m 0.2 m 20 N A B x y θ 90 vA 0 172 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 172 3/132 The 7-kg collar A slides with negligible friction on the fixed vertical shaft. When the collar is released from rest at the bottom position shown, it moves up the shaft under the action of the constant force applied to the cable. Calculate the stiff-ness k which the spring must have if its maximum compression is to be limited to 75 mm. The posi-tion of the small pulley at B is fixed.
Problem 3/132 3/133 Calculate the horizontal velocity v with which the 48-lb carriage must strike the spring in order to compress it a maximum of 4 in. The spring is known as a “hardening” spring, since its stiffness increases with deflection as shown in the accompa-nying graph.
Problem 3/133 x 48 lb F, lb 60x 3x2 x, in.
x 4 0 0 225 mm 75 mm 450 mm A B F k F 200 N 3/134 The spring attached to the 10-kg mass is nonlin-ear, having the force–deflection relationship shown in the figure, and is unstretched when . If the mass is moved to the position and re-leased from rest, determine its velocity v when . Determine the corresponding velocity if the spring were linear according to , where x is in meters and the force F is in kilonewtons.
Problem 3/134 3/135 The 6-kg cylinder is released from rest in the posi-tion shown and falls on the spring, which has been initially precompressed 50 mm by the light strap and restraining wires. If the stiffness of the spring is 4 kN/m, compute the additional deflection of the spring produced by the falling cylinder before it rebounds.
Problem 3/135 100 mm 6 kg δ 10 kg s = 0.25 μ k = 0.20 μ x Force F, kN Deflection x, m Linear, F = 4x Nonlinear, F = 4x – 120x3 F 4x v x 0 x 100 mm x 0 Article 3/6 Problems 173 c03.qxd 2/9/12 7:39 PM Page 173 3/138 The 50-lb slider in the position shown has an ini-tial velocity on the inclined rail and slides under the influence of gravity and friction.
The coefficient of kinetic friction between the slider and the rail is 0.50. Calculate the velocity of the slider as it passes the position for which the spring is compressed a distance . The spring offers a compressive resistance C and is known as a “hardening” spring, since its stiffness increases with deflection as shown in the accompa-nying graph.
Problem 3/138 3′ v0 = 2 ft/sec x 50 lb k = 0.50 60° C, lb 9x2 100x x, in.
x μ x 4 in v0 2 ft/sec 3/136 Extensive testing of an experimental 2000-lb auto-mobile reveals the aerodynamic drag force FD and the total nonaerodynamic rolling-resistance force FR to be as shown in the plot. Determine (a) the power required for steady speeds of 30 and 60 mi/hr on a level road, (b) the power required for a steady speed of 60 mi/hr both up and down a 6-percent in-cline, and (c) the steady speed at which no power is required going down the 6-percent incline.
Problem 3/136 3/137 The three springs of equal moduli are unstretched when the cart is released from rest in the position . If and , determine (a) the speed v of the cart when , (b) the maximum displacement xmax of the cart, and (c) the steady-state displacement xss that would exist after all oscillations cease.
Problem 3/137 x k m 20° k k x 50 mm m 10 kg k 120 N/m x 0 0 20 40 60 80 0 20 40 60 80 Speed v, mi/hr Force, lb FR (constant) FD (parabolic) 174 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 174 3/7 Potential Energy In the previous article on work and kinetic energy, we isolated a particle or a combination of joined particles and determined the work done by gravity forces, spring forces, and other externally applied forces acting on the particle or system. We did this to evaluate U in the work-energy equation. In the present article we will introduce the concept of potential energy to treat the work done by gravity forces and by spring forces. This concept will simplify the analysis of many problems.
Gravitational Potential Energy We consider first the motion of a particle of mass m in close proxim-ity to the surface of the earth, where the gravitational attraction (weight) mg is essentially constant, Fig. 3/8a. The gravitational poten-tial energy Vg of the particle is defined as the work mgh done against the gravitational field to elevate the particle a distance h above some arbi-trary reference plane (called a datum), where Vg is taken to be zero.
Thus, we write the potential energy as (3/18) This work is called potential energy because it may be converted into energy if the particle is allowed to do work on a supporting body while it returns to its lower original datum plane. In going from one level at h h1 to a higher level at h h2, the change in potential energy becomes The corresponding work done by the gravitational force on the particle is mgh. Thus, the work done by the gravitational force is the nega-tive of the change in potential energy.
When large changes in altitude in the field of the earth are encoun-tered, Fig. 3/8b, the gravitational force Gmme/r2 mgR2/r2 is no longer constant. The work done against this force to change the radial position of the particle from r1 to r2 is the change (Vg)2 (Vg)1 in gravitational potential energy, which is It is customary to take (Vg)2 0 when r2 , so that with this datum we have (3/19) In going from r1 to r2, the corresponding change in potential energy is Vg mgR2 1 r1 1 r2 Vg mgR2 r r2 r1 mgR2 dr r2 mgR2 1 r1 1 r2 (Vg)2 (Vg)1 Vg mg(h2 h1) mgh Vg mgh Earth me r m R (a) (b) Vg = 0 Vg = mgh mg h Vg = – mgR2 —–— r mgR2 —–— r2 Figure 3/8 Article 3/7 Potential Energy 175 c03.qxd 2/9/12 7:39 PM Page 175 which, again, is the negative of the work done by the gravitational force.
We note that the potential energy of a given particle depends only on its position, h or r, and not on the particular path it followed in reaching that position.
Elastic Potential Energy The second example of potential energy occurs in the deformation of an elastic body, such as a spring. The work which is done on the spring to deform it is stored in the spring and is called its elastic potential energy Ve.
This energy is recoverable in the form of work done by the spring on the body attached to its movable end during the release of the deformation of the spring. For the one-dimensional linear spring of stiffness k, which we discussed in Art. 3/6 and illustrated in Fig. 3/5, the force supported by the spring at any deformation x, tensile or compressive, from its undeformed position is F kx. Thus, we define the elastic potential energy of the spring as the work done on it to deform it an amount x, and we have (3/20) If the deformation, either tensile or compressive, of a spring in-creases from x1 to x2 during the motion, then the change in potential en-ergy of the spring is its final value minus its initial value or which is positive. Conversely, if the deformation of a spring decreases during the motion interval, then the change in potential energy of the spring becomes negative. The magnitude of these changes is repre-sented by the shaded trapezoidal area in the F-x diagram of Fig. 3/5a.
Because the force exerted on the spring by the moving body is equal and opposite to the force F exerted by the spring on the body, it follows that the work done on the spring is the negative of the work done on the body. Therefore, we may replace the work U done by the spring on the body by Ve, the negative of the potential energy change for the spring, provided the spring is now included within the system.
Work-Energy Equation With the elastic member included in the system, we now modify the work-energy equation to account for the potential-energy terms. If stands for the work of all external forces other than gravitational forces and spring forces, we may write Eq. 3/15 as (Vg) (Ve) T or (3/21) where V is the change in total potential energy, gravitational plus elastic.
This alternative form of the work-energy equation is often far more convenient to use than Eq. 3/15, since the work of both gravity and spring forces is accounted for by focusing attention on the end-point positions of U 1-2 T V U 1-2 U 1-2 Ve 1 2 k(x2 2 x1 2) Ve x 0 kx dx 1 2 kx2 176 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 176 the particle and on the end-point lengths of the elastic spring. The path followed between these end-point positions is of no consequence in the evaluation of Vg and Ve.
Note that Eq. 3/21 may be rewritten in the equivalent form (3/21a) To help clarify the difference between the use of Eqs. 3/15 and 3/21, Fig. 3/9 shows schematically a particle of mass m constrained to move along a fixed path under the action of forces F1 and F2, the gravitational force W mg, the spring force F, and the normal reaction N. In Fig.
3/9b, the particle is isolated with its free-body diagram. The work done by each of the forces F1, F2, W, and the spring force F kx is evaluated, say, from A to B, and equated to the change T in kinetic energy using Eq. 3/15. The constraint reaction N, if normal to the path, will do no work. The alternative approach is shown in Fig. 3/9c, where the spring is included as a part of the isolated system. The work done during the interval by F1 and F2 is the -term of Eq. 3/21 with the changes in elastic and gravitational potential energies included on the energy side of the equation.
We note with the first approach that the work done by F kx could require a somewhat awkward integration to account for the changes in magnitude and direction of F as the particle moves from A U 1-2 T1 V1 U 1-2 T2 V2 Article 3/7 Potential Energy 177 System (a) (c) h N F = kx W = mg U1-2 = ΔT U′1-2 = ΔT + ΔV A B F1 F2 F1 F2 (b) F1 F2 Vg = 0 Vg = mgh Figure 3/9 c03.qxd 2/9/12 7:39 PM Page 177 to B. With the second approach, however, only the initial and final lengths of the spring are required to evaluate Ve. This greatly simpli-fies the calculation.
For problems where the only forces are gravitational, elastic, and nonworking constraint forces, the U-term of Eq. 3/21a is zero, and the energy equation becomes (3/22) where E T V is the total mechanical energy of the particle and its attached spring. When E is constant, we see that transfers of energy be-tween kinetic and potential may take place as long as the total mechani-cal energy T V does not change. Equation 3/22 expresses the law of conservation of dynamical energy.
Conservative Force Fields We have observed that the work done against a gravitational or an elastic force depends only on the net change of position and not on the particular path followed in reaching the new position. Forces with this characteristic are associated with conservative force fields, which possess an important mathematical property.
Consider a force field where the force F is a function of the coordi-nates, Fig. 3/10. The work done by F during a displacement dr of its point of application is dU . The total work done along its path from 1 to 2 is The integral is a line integral which depends, in general, on the particular path followed between any two points 1 and 2 in space. If, however, is an exact differential† dV of some scalar function V of the coordinates, then (3/23) which depends only on the end points of the motion and which is thus independent of the path followed. The minus sign before dV is arbitrary but is chosen to agree with the customary designation of the sign of po-tential energy change in the gravity field of the earth.
If V exists, the differential change in V becomes dV V x dx V y dy V z dz U1-2 V2 V1 dV (V2 V1) Fdr Fdr U Fdr (Fx dx Fy dy Fz dz) Fdr T1 V1 T2 V2 or E1 E2 178 Chapter 3 Kinetics of Particles Optional.
†Recall that a function d P dx Q dy R dz is an exact differential in the coordinates x-y-z if P y Q x P z R x Q z R y 1 2 y x dr r F z Figure 3/10 c03.qxd 2/9/12 7:39 PM Page 178 Comparison with dV = Fx dx Fy dy Fz dz gives us The force may also be written as the vector (3/24) where the symbol stands for the vector operator “del”, which is The quantity V is known as the potential function, and the expression V is known as the gradient of the potential function.
When force components are derivable from a potential as described, the force is said to be conservative, and the work done by F between any two points is independent of the path followed.
i x j y k z F V Fx V x Fy V y Fz V z F dr Article 3/7 Potential Energy 179 c03.qxd 2/9/12 7:39 PM Page 179 SAMPLE PROBLEM 3/16 The 6-lb slider is released from rest at position 1 and slides with negligible friction in a vertical plane along the circular rod. The attached spring has a stiff-ness of 2 lb/in. and has an unstretched length of 24 in. Determine the velocity of the slider as it passes position 2.
Solution.
The work done by the weight and the spring force on the slider will be treated using potential-energy methods. The reaction of the rod on the slider is normal to the motion and does no work. Hence, 0. We define the datum to be at the level of position 1, so that the gravitational potential energies are The initial and final elastic (spring) potential energies are Substitution into the alternative work-energy equation yields Ans.
SAMPLE PROBLEM 3/17 The 10-kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250-N force is constant and the pulley offers negligible resistance to the motion of the cord. Calculate the velocity vC of the slider as it passes point C.
Solution.
The slider and inextensible cord together with the attached spring will be analyzed as a system, which permits the use of Eq. 3/21a. The only non-potential force doing work on this system is the 250-N tension applied to the cord. While the slider moves from A to C, the point of application of the 250-N force moves a distance of or 1.5 0.9 0.6 m.
We define a datum at position A so that the initial and final gravitational poten-tial energies are The initial and final elastic potential energies are Substitution into the alternative work-energy equation 3/21a gives Ans.
vC 0.974 m/s [TA VA U A-C TC VC] 0 0 10.8 150 1 2 (10)vC 2 58.9 97.2 VC 1 2 kxB 2 1 2 60(0.6 1.2)2 97.2 J VA 1 2 kxA 2 1 2 (60)(0.6)2 10.8 J VA 0 VC mgh 10(9.81)(1.2 sin 30) 58.9 J U A-C 250(0.6) 150 J AB BC v2 23.6 ft/sec [T1 V1 U 1-2 T2 V2] 0 48 0 1 2 6 32.2 v2 2 12 8.24 V2 1 2 kx2 2 1 2 (2)(12) 242 12 24 12 2 8.24 ft-lb V1 1 2 kx1 2 1 2 (2)(12) 24 12 2 48 ft-lb V2 mgh 6 24 12 12 ft-lb V1 0 U 1-2 24″ k = 2 lb/in.
6 lb 1 2 v2 24″ 250 N vC 0.9 m B 30° A C 1.2 m Helpful Hints Do not hesitate to use subscripts tai-lored to the problem at hand. Here we use A and C rather than 1 and 2.
The reactions of the guides on the slider are normal to the direction of motion and do no work.
Helpful Hint Note that if we evaluated the work done by the spring force acting on the slider by means of the integral , it would necessitate a lengthy computation to account for the change in the magnitude of the force, along with the change in the angle between the force and the tangent to the path. Note further that v2 de-pends only on the end conditions of the motion and does not require knowledge of the shape of the path.
Fdr 180 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 180 SAMPLE PROBLEM 3/18 The system shown is released from rest with the lightweight slender bar OA in the vertical position shown. The torsional spring at O is undeflected in the initial position and exerts a restoring moment of magnitude k on the bar, where is the counterclockwise angular deflection of the bar. The string S is attached to point C of the bar and slips without friction through a vertical hole in the support surface.
For the values mA 2 kg, mB 4 kg, L 0.5 m, and k 13 : (a) Determine the speed vA of particle A when reaches 90.
(b) Plot vA as a function of over the range 0 90. Identify the maximum value of vA and the value of at which this maximum occurs.
Solution (a). We begin by establishing a general relationship for the potential energy associated with the deflection of a torsional spring. Recalling that the change in potential energy is the work done on the spring to deform it, we write We also need to establish the relationship between vA and vB when 90. Not-ing that the speed of point C is always vA/2, and further noting that the speed of cylinder B is one-half the speed of point C at 90, we conclude that at 90, Establishing datums at the initial altitudes of bodies A and B, and with state 1 at 0 and state 2 at 90, we write With numbers: Solving, Ans.
(b). We leave our definition of the initial state 1 as is, but now redefine state 2 to be associated with an arbitrary value of . From the accompanying diagram constructed for an arbitrary value of , we see that the speed of cylinder B can be written as Finally, because vA L , Upon substitution of the given quantities, we vary to produce the plot of vA versus . The maximum value of vA is seen to be Ans.
(vA)max 1.400 m/s at 56.4 mB g 1 2 L2 2 2 L 2 sin 90 2 1 2 k2 0 0 0 1 2 mAvA 2 1 2 mB vA 4 cos 90 2 2 mA gL(1 cos ) [T1 V1 U 1-2 T2 V2] vB vA 4 cos 90 2 ˙ 1 2L ˙ 2 cos 90 2 L ˙ 4 cos 90 2 vB 1 2 d dt (CC ) 1 2 d dt 2 L 2 sin 90 2 vA 0.794 m/s 0 1 2 (2)vA 2 1 2 (4) vA 4 2 2(9.81)(0.5) 4(9.81) 0.52 4 1 2 (13) 2 2 0 0 0 1 2 mAvA 2 1 2 mBvB 2 mA gL mB g L2 4 1 2 k 2 2 [T1 V1 U 1-2 T2 V2] vB 1 4 vA Ve 0 k d 1 2 k2 Nm/rad B O S k C mB A mA θ θ L –– 2 L –– 2 O C″ (top of hole) C′ C θ θ 90° – —–—– 2 θ 90° – —–—– 2 L –– 2 L –– 2 L –– 2 0 0.5 1 1.5 0 10 20 30 40 50 60 70 80 90 , deg vA, m/s θ (vA)max = 1.400 m/s at = 56.4° θ Helpful Hints Note that mass B will move down-ward by one-half of the length of string initially above the supporting surface. This downward distance is .
The absolute-value signs reflect the fact that vB is known to be positive.
1 2 L 22 L2 4 Article 3/7 Potential Energy 181 c03.qxd 2/9/12 8:23 PM Page 181 3/141 The 1.2-kg slider is released from rest in position A and slides without friction along the vertical-plane guide shown. Determine (a) the speed of the slider as it passes position B and (b) the maximum deflection of the spring.
Problem 3/141 3/142 The 1.2-kg slider of the system of Prob. 3/141 is released from rest in position A and slides without friction along the vertical-plane guide. Determine the normal force exerted by the guide on the slider (a) just before it passes point C, (b) just after it passes point C, and (c) just before it passes point E.
3/143 Point P on the 2-kg cylinder has an initial velocity as it passes position A. Neglect the mass of the pulleys and cable and determine the distance y of point P below A when the 3-kg cylin-der has acquired an upward velocity of 0.6 m/s.
Problem 3/143 3 kg 2 kg A P v0 v0 0.8 m/s 3 m 1.2 kg k = 24 kN/m E D C B A 1.5 m 30° 30° vB PROBLEMS Introductory Problems 3/139 The 2-lb collar is released from rest at A and slides freely up the inclined rod, striking the stop at B with a velocity v. The spring of stiffness lb/ft has an unstretched length of 15 in. Calculate v.
Problem 3/139 3/140 The 4-kg slider is released from rest at A and slides with negligible friction down the circular rod in the vertical plane. Determine (a) the velocity v of the slider as it reaches the bottom at B and (b) the maximum deformation x of the spring.
Problem 3/140 0.6 m 4 kg k = 20 kN/m A B O A B k = 1.60 lb/ft 20″ 10″ 18″ k 1.60 182 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 182 3/144 The spring of constant k is unstretched when the slider of mass m passes position B. If the slider is released from rest in position A, determine its speeds as it passes points B and C. What is the nor-mal force exerted by the guide on the slider at posi-tion C? Neglect friction between the mass and the circular guide, which lies in a vertical plane.
Problem 3/144 3/145 It is desired that the 100-lb container, when released from rest in the position shown, have no velocity after dropping 7 ft to the platform below. Specify the proper weight W of the counterbalancing cylinder.
Problem 3/145 24′ W 5′ 7′ 100 lb A B m k C R R 3/146 The system is released from rest with the spring initially stretched 3 in. Calculate the velocity v of the cylinder after it has dropped 0.5 in. The spring has a stiffness of 6 lb/in. Neglect the mass of the small pulley.
Problem 3/146 3/147 The projectile of Prob. 3/122 is repeated here. By the method of this article, determine the vertical launch velocity which will result in a maximum altitude of R/2. The launch is from the north pole and aerodynamic drag can be neglected. Use as the surface-level acceleration due to gravity.
Problem 3/147 R v0 g 9.825 m/s2 v0 k = 6 lb/in.
100 lb Article 3/7 Problems 183 c03.qxd 2/9/12 7:39 PM Page 183 Representative Problems 3/150 The 0.8-kg particle is attached to the system of two light rigid bars, all of which move in a vertical plane. The spring is compressed an amount b/2 when , and the length . The system is released from rest in a position slightly above that for . (a) If the maximum value of is ob-served to be , determine the spring constant k.
(b) For , determine the speed of the particle when . Also find the corresponding value of .
Problem 3/150 3/151 The two springs, each of stiffness , are of equal length and undeformed when . If the mechanism is released from rest in the position , determine its angular velocity when . The mass m of each sphere is 3 kg. Treat the spheres as particles and neglect the masses of the light rods and springs.
Problem 3/151 O k k m m m θ 0.25 m 0 ˙ 20 0 k 1.2 kN/m C O B A k b b b m θ ˙ 25 v k 400 N/m 50 0 b 0.30 m 0 3/148 The 1.5-kg slider C moves along the fixed rod under the action of the spring whose unstretched length is 0.3 m. If the velocity of the slider is 2 m/s at point A and 3 m/s at point B, calculate the work Uƒ done by friction between these two points. Also, determine the average friction force acting on the slider between A and B if the length of the path is 0.70 m. The x-y plane is horizontal.
Problem 3/148 3/149 The light rod is pivoted at O and carries the 5- and 10-lb particles. If the rod is released from rest at and swings in the vertical plane, calculate (a) the velocity v of the 5-lb particle just before it hits the spring in the dashed position and (b) the maximum compression x of the spring. Assume that x is small so that the position of the rod when the spring is compressed is essentially horizontal.
Problem 3/149 O 10 lb 5 lb 12″ k = 200 lb/in.
θ 18″ 60 0.4 m 0.4 m 0.3 m B x y z C A v k = 800 N/m 184 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 184 3/152 If the system is released from rest, determine the speeds of both masses after B has moved 1 m. Ne-glect friction and the masses of the pulleys.
Problem 3/152 3/153 The 3-lb ball is given an initial velocity in the vertical plane at position A, where the two horizontal attached springs are unstretched. The ball follows the dashed path shown and crosses point B, which is 5 in. directly below A. Calculate the velocity of the ball at B. Each spring has a stiffness of 10 lb/in.
Problem 3/153 12″ 12″ vA vB A B 5″ vB vA 8 ft/sec 20° 40 kg 8 kg A B 3/154 The 0.75-kg particle is attached to the light slender rod OA which pivots freely about a horizontal axis through point O. The system is released from rest while in the position where the spring is un-stretched. If the particle is observed to stop mo-mentarily in the position , determine the spring constant k. For your computed value of k, what is the particle speed v at the position ?
Problem 3/154 3/155 The spring has an unstretched length of 25 in. If the system is released from rest in the position shown, determine the speed v of the ball (a) when it has dropped a vertical distance of 10 in. and (b) when the rod has rotated .
Problem 3/155 9 lb 1.2 lb/in.
O 10″ 24″ 26″ 35 O 0.75 kg A B k 0.6 m 0.6 m θ 25 50 0 Article 3/7 Problems 185 c03.qxd 2/9/12 7:39 PM Page 185 3/158 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50-N force. Calculate the velocity v of the collar as it passes position B.
Problem 3/158 3/159 The shank of the 5-lb vertical plunger occupies the dashed position when resting in equilibrium against the spring of stiffness . The upper end of the spring is welded to the plunger, and the lower end is welded to the base plate. If the plunger is lifted in. above its equilibrium po-sition and released from rest, calculate its velocity v as it strikes the button A. Friction is negligible.
Problem 3/159 k = 10 lb/in.
W = 5 lb 1 – 2 1 ″ 1 – 4 ″ A 11 2 k 10 lb/in A B 2 m k = 30 N/m 50 N 30° 1.5 m 3/156 The two 1.5-kg spheres are released from rest and gently nudged outward from the position and then rotate in a vertical plane about the fixed centers of their attached gears, thus maintaining the same angle for both rods. Determine the ve-locity v of each sphere as the rods pass the position . The spring is unstretched when , and the masses of the two identical rods and the two gear wheels may be neglected.
Problem 3/156 3/157 A rocket launches an unpowered space capsule at point A with an absolute velocity at an altitude of 25 mi. After the capsule has trav-eled a distance of 250 mi measured along its ab-solute space trajectory, its velocity at B is 7600 mi/hr and its altitude is 50 mi. Determine the aver-age resistance P to motion in the rarified atmos-phere. The earth weight of the capsule is 48 lb, and the mean radius of the earth is 3959 mi. Consider the center of the earth fixed in space.
Problem 3/157 B A vB vA 50 mi 25 mi vA 8000 mi/hr 80 mm 1.5 kg θ 1.5 kg k = 60 N/m 240 mm 240 mm θ 0 30 0 186 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 186 3/160 Upon its return voyage from a space mission, the spacecraft has a velocity of 24 000 km/h at point A, which is 7000 km from the center of the earth. De-termine the velocity of the spacecraft when it reaches point B, which is 6500 km from the center of the earth. The trajectory between these two points is outside the effect of the earth’s atmosphere.
Problem 3/160 3/161 The 5-kg cylinder is released from rest in the posi-tion shown and compresses the spring of stiffness . Determine the maximum compres-sion xmax of the spring and the maximum velocity of the cylinder along with the corresponding deflection x of the spring.
Problem 3/161 5 kg 100 mm x k = 1.8 kN/m vmax k 1.8 kN/m A B O 3/162 A 175-lb pole vaulter carrying a uniform 16-ft, 10-lb pole approaches the jump with a velocity v and manages to barely clear the bar set at a height of 18 ft. As he clears the bar, his velocity and that of the pole are essentially zero. Calculate the mini-mum possible value of required for him to make the jump. Both the horizontal pole and the center of gravity of the vaulter are 42 in. above the ground during the approach.
Problem 3/162 3/163 The cylinder of mass m is attached to the collar bracket at A by a spring of stiffness k. The collar fits loosely on the vertical shaft, which is lowering both the collar and the suspended cylinder with a constant velocity v. When the collar strikes the base B, it stops abruptly with essentially no re-bound. Determine the maximum additional deflec-tion of the spring after the impact.
Problem 3/163 m A B k v v Article 3/7 Problems 187 42″ 18′ 16′ v c03.qxd 2/9/12 7:39 PM Page 187 3/166 Calculate the maximum velocity of slider B if the system is released from rest with . Motion is in the vertical plane. Assume that friction is negli-gible. The sliders have equal masses, and the mo-tion is restricted to .
Problem 3/166 3/167 The mechanism is released from rest with where the uncompressed spring of stiffness is just touching the underside of the 4-kg collar. Determine the angle corresponding to the maximum compression of the spring. Mo-tion is in the vertical plane, and the mass of the links may be neglected.
Problem 3/167 θ 4 kg Vertical k 3 kg 200 mm 200 mm 300 mm 3 kg k 900 N/m 180, 0.9 m A B y x y 0 x y 3/164 The cars of an amusement-park ride have a speed at the lowest part of the track. Deter-mine their speed at the highest part of the track.
Neglect energy loss due to friction. (Caution: Give careful thought to the change in potential energy of the system of cars.) Problem 3/164 3/165 The two right-angle rods with attached spheres are released from rest in the position . If the sys-tem is observed to momentarily come to rest when , determine the spring constant k. The spring is unstretched when . Treat the spheres as particles and neglect friction.
Problem 3/165 60 mm 180 mm 2 kg 2 kg k θ θ 0 45 0 v2 v1 90° 15 m 90° 15 m v2 v1 90 km/h 188 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 188 3/168 A particle of mass m is attached to one end of a light slender rod which pivots about a horizontal axis through point O. The spring constant and the distance . If the system is released from rest in the horizontal position shown where the spring is unstretched, the bar is observed to deflect a maximum of clockwise. Determine (a) the particle mass m and (b) the particle speed v after a displacement of from the position shown. Neglect friction.
Problem 3/168 3/169 The 3-kg sphere is carried by the parallelogram linkage where the spring is unstretched when . If the mechanism is released from rest at , calculate the velocity v of the sphere when the position is passed. The links are in the vertical plane, and their mass is small and may be neglected.
Problem 3/169 θ θ 3 kg 500 mm 500 mm 500 mm k = 100 N/m 135 90 90 b –– 2 b O A m C B k b 15 30 b 200 mm k 200 N/m 3/170 The system is at rest with the spring unstretched when . The 3-kg particle is then given a slight nudge to the right. (a) If the system comes to mo-mentary rest at , determine the spring con-stant k. (b) For the value , find the speed of the particle when . Use the value m throughout and neglect friction.
Problem 3/170 k C B m A b O 1.25b θ b 0.40 25 k 100 N/m 40 0 Article 3/7 Problems 189 c03.qxd 2/9/12 7:39 PM Page 189 3/172 The flexible bicycle-type chain of length and mass per unit length is released from rest with in the smooth circular channel and falls through the hole in the supporting surface. Deter-mine the velocity v of the chain as the last link leaves the slot.
Problem 3/172 r θ 0 r/2 3/171 The system is released from rest with the angle . Determine when reaches . Use the values , , and m.
Neglect friction and the mass of bar OB, and treat the body B as a particle.
Problem 3/171 O C m2 m1 B A 2b b 2b θ b 0.40 m2 1.25 kg m1 1 kg 60 ˙ 90 190 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 190 SECTION C IMPULSE AND MOMENTUM 3/8 Introduction In the previous two articles, we focused attention on the equations of work and energy, which are obtained by integrating the equation of mo-tion F ma with respect to the displacement of the particle. We found that the velocity changes could be expressed directly in terms of the work done or in terms of the overall changes in energy. In the next two arti-cles, we will integrate the equation of motion with respect to time rather than displacement. This approach leads to the equations of impulse and momentum. These equations greatly facilitate the solution of many prob-lems in which the applied forces act during extremely short periods of time (as in impact problems) or over specified intervals of time.
3/9 Linear Impulse and Linear Momentum Consider again the general curvilinear motion in space of a particle of mass m, Fig. 3/11, where the particle is located by its position vector r measured from a fixed origin O. The velocity of the particle is v and is tangent to its path (shown as a dashed line). The resultant ΣF of all forces on m is in the direction of its acceleration . We may now write the basic equation of motion for the particle, Eq. 3/3, as (3/25) where the product of the mass and velocity is defined as the linear mo-mentum G mv of the particle. Equation 3/25 states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum. In SI the units of linear momentum mv are seen to be , which also equals . In U.S. customary units, the units of linear momentum mv are [lb/(ft/sec2)][ft/sec] lb-sec.
Because Eq. 3/25 is a vector equation, we recognize that, in addition to the equality of the magnitudes of ΣF and , the direction of the resultant force coincides with the direction of the rate of change in linear momen-tum, which is the direction of the rate of change in velocity. Equation 3/25 is one of the most useful and important relationships in dynamics, and it is valid as long as the mass m of the particle is not changing with time.
The case where m changes with time is discussed in Art. 4/7 of Chapter 4.
We now write the three scalar components of Eq. 3/25 as (3/26) These equations may be applied independently of one another.
The Linear Impulse-Momentum Principle All that we have done so far in this article is to rewrite Newton’s second law in an alternative form in terms of momentum. But we are now able to describe the effect of the resultant force ΣF on the linear ΣFx G ˙x ΣFy G ˙y ΣFz G ˙z G ˙ N s kg m/s ΣF mv ˙ d dt (mv) or ΣF G ˙ v ˙ r ˙ Article 3/9 Linear Impulse and Linear Momentum 191 Path ΣF r2 r1 v1 t1 v2 r y m O x v = r · G = mv G · v · z t2 Figure 3/11 c03.qxd 2/9/12 7:39 PM Page 191 momentum of the particle over a finite period of time simply by inte-grating Eq. 3/25 with respect to the time t. Multiplying the equation by dt gives ΣF dt dG, which we integrate from time t1 to time t2 to obtain (3/27) Here the linear momentum at time t2 is G2 mv2 and the linear momen-tum at time t1 is G1 mv1. The product of force and time is defined as the linear impulse of the force, and Eq. 3/27 states that the total linear impulse on m equals the corresponding change in linear momentum of m.
Alternatively, we may write Eq. 3/27 as (3/27a) which says that the initial linear momentum of the body plus the linear impulse applied to it equals its final linear momentum.
The impulse integral is a vector which, in general, may involve changes in both magnitude and direction during the time interval.
Under these conditions, it will be necessary to express ΣF and G in com-ponent form and then combine the integrated components. The compo-nents of Eq. 3/27a are the scalar equations (3/27b) These three scalar impulse-momentum equations are completely independent.
Whereas Eq. 3/27 clearly stresses that the external linear impulse causes a change in the linear momentum, the order of the terms in Eqs.
3/27a and 3/27b corresponds to the natural sequence of events. While the form of Eq. 3/27 may be best for the experienced dynamicist, the form of Eqs. 3/27a and 3/27b is very effective for the beginner.
We now introduce the concept of the impulse-momentum diagram.
Once the body to be analyzed has been clearly identified and isolated, we construct three drawings of the body as shown in Fig. 3/12. In the first drawing, we show the initial momentum mv1, or components thereof. In m(v1)z t2 t1 ΣFz dt m(v2)z m(v1)y t2 t1 ΣFy dt m(v2)y m(v1)x t2 t1 ΣFx dt m(v2)x G1 t2 t1 ΣF dt G2 t2 t1 ΣF dt G2 G1 G 192 Chapter 3 Kinetics of Particles ΣF dt t1 t2 G1 = mv1 G2 = mv2 + = Figure 3/12 c03.qxd 2/9/12 7:39 PM Page 192 the second or middle drawing, we show all the external linear impulses (or components thereof). In the final drawing, we show the final linear momentum mv2 (or its components). The writing of the impulse-momen-tum equations 3/27b then follows directly from these drawings, with a clear one-to-one correspondence between diagrams and equation terms.
We note that the center diagram is very much like a free-body dia-gram, except that the impulses of the forces appear rather than the forces themselves. As with the free-body diagram, it is necessary to in-clude the effects of all forces acting on the body, except those forces whose magnitudes are negligible.
In some cases, certain forces are very large and of short duration. Such forces are called impulsive forces. An example is a force of sharp impact.
We frequently assume that impulsive forces are constant over their time of duration, so that they can be brought outside the linear-impulse integral.
In addition, we frequently assume that nonimpulsive forces can be ne-glected in comparison with impulsive forces. An example of a nonimpulsive force is the weight of a baseball during its collision with a bat—the weight of the ball (about 5 oz) is small compared with the force (which could be several hundred pounds in magnitude) exerted on the ball by the bat.
There are cases where a force acting on a particle varies with the time in a manner determined by experimental measurements or by other approximate means. In this case a graphical or numerical integra-tion must be performed. If, for example, a force F acting on a particle in a given direction varies with the time t as indicated in Fig. 3/13, then the impulse, F dt, of this force from t1 to t2 is the shaded area under the curve.
Conservation of Linear Momentum If the resultant force on a particle is zero during an interval of time, we see that Eq. 3/25 requires that its linear momentum G remain con-stant. In this case, the linear momentum of the particle is said to be con-served. Linear momentum may be conserved in one coordinate direction, such as x, but not necessarily in the y- or z-direction. A careful examina-tion of the impulse-momentum diagram of the particle will disclose whether the total linear impulse on the particle in a particular direction is zero. If it is, the corresponding linear momentum is unchanged (con-served) in that direction.
Consider now the motion of two particles a and b which interact during an interval of time. If the interactive forces F and F between them are the only unbalanced forces acting on the particles during the interval, it follows that the linear impulse on particle a is the negative of the linear impulse on particle b. Therefore, from Eq. 3/27, the change in linear momentum Ga of particle a is the negative of the change Gb in linear momentum of particle b. So we have Ga Gb or (Ga Gb) 0. Thus, the total linear momentum G Ga Gb for the system of the two particles remains constant during the interval, and we write (3/28) Equation 3/28 expresses the principle of conservation of linear momentum.
G 0 or G1 G2 t2 t1 Article 3/9 Linear Impulse and Linear Momentum 193 Time, t F2 t1 t2 F1 Force, F Figure 3/13 The impact force exerted by the rac-quet on this tennis ball will usually be much larger than the weight of the tennis ball.
GJON MILI/GettyImages/Time & LifeCreative/Getty Images, Inc.
c03.qxd 2/9/12 7:39 PM Page 193 SAMPLE PROBLEM 3/19 A tennis player strikes the tennis ball with her racket when the ball is at the uppermost point of its trajectory as shown. The horizontal velocity of the ball just before impact with the racket is v1 50 ft/sec, and just after impact its ve-locity is v2 70 ft/sec directed at the 15 angle as shown. If the 2-oz ball is in contact with the racket for 0.02 sec, determine the magnitude of the average force R exerted by the racket on the ball. Also determine the angle made by R with the horizontal.
Solution.
We construct the impulse-momentum diagrams for the ball as follows: We can now solve for the impact forces as We note that the impact force Ry 3.64 lb is considerably larger than the 0.125-lb weight of the ball. Thus, the weight mg, a nonimpulsive force, could have been neglected as small in comparison with Ry. Had we neglected the weight, the computed value of Ry would have been 3.52 lb.
We now determine the magnitude and direction of R as Ans.
Ans.
tan1 Ry Rx tan1 3.64 22.8 9.06 R Rx 2 Ry 2 22.82 3.642 23.1 lb Ry 3.64 lb Rx 22.8 lb 2/16 32.2 (0) Ry(0.02) (2/16)(0.02) 2/16 32.2 (70 sin 15) [m(vy )1 t2 t1 ΣFy dt m(vy )2] [m(vx )1 t2 t1 ΣFx dt m(vx )2] 2/16 32.2 (50) Rx(0.02) 2/16 32.2 (70 cos 15 ) x y 15° Rx dt mv2 mv1 + = mg dt Ry dt t1 t2 t1 t2 t1 t2 15° v1 v2 For the linear impulse Rx dt, the average impact force Rx is a constant, so that it can be brought outside the integral sign, resulting in Rx dt Rx(t2 t1) Rxt. The linear impulse in the y-direction has been similarly treated.
t2 t1 t2 t1 Helpful Hints Recall that for the impulse-momentum diagrams, initial linear momentum goes in the first diagram, all exter-nal linear impulses go in the second diagram, and final linear momen-tum goes in the third diagram.
194 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 194 –2k lb F Up z y F1 F1 F2 F2 10 m/s y x 0 2 4 F, N 3 2 1 t, s 0 t = 1 s t = 2 s t = 3 s 8 8j m/s – 6i m/s 6 4 2 00 2 x, m y, m 4 6 v2 = 10 m/s x = 126.9° θ Helpful Hint Don’t forget that ΣF includes all ex-ternal forces acting on the particle, including the weight.
SAMPLE PROBLEM 3/20 A 2-lb particle moves in the vertical y-z plane (z up, y horizontal) under the action of its weight and a force F which varies with time. The linear momentum of the particle in pound-seconds is given by the expression G (t2 3)j (t3 4)k, where t is the time in seconds. Determine F and its magnitude for the instant when t 2 sec.
Solution.
The weight expressed as a vector is 2k lb. Thus, the force-momen-tum equation becomes For t 2 sec, Ans.
Thus, Ans.
SAMPLE PROBLEM 3/21 A particle with a mass of 0.5 kg has a velocity of 10 m/s in the x-direction at time t 0. Forces F1 and F2 act on the particle, and their magnitudes change with time according to the graphical schedule shown. Determine the velocity v2 of the particle at the end of the 3-s interval. The motion occurs in the horizontal x-y plane.
Solution.
First, we construct the impulse-momentum diagrams as shown.
Then the impulse-momentum equations follow as Thus, Ans.
Although not called for, the path of the particle for the first 3 seconds is plot-ted in the figure. The velocity at t 3 s is shown together with its components.
x tan1 8 6 126.9 v2 6i 8j m/s and v2 62 82 10 m/s (v2)y 8 m/s 0.5(0) [1(2) 2(3 2)] 0.5(v2)y [m(v1)y t2 t1 ΣFy dt m(v2)y] (v2)x 6 m/s 0.5(10) [4(1) 2(3 1)] 0.5(v2)x [m(v1)x t2 t1 ΣFx dt m(v2)x] m(v1)y = 0 m(v2)x m(v2)y m(v1)x = 0.5 (10) kg·m/s + = F1 dt F2 dt t1 t2 t1 t2 F 62 62 62 lb F 2k 3(2)j 2(22)k 6j 6k lb 3tj 2t2k F 2k d dt [3 2 (t2 3)j 2 3 (t3 4)k] [ΣF G ˙] 2 3 3 2 Article 3/9 Linear Impulse and Linear Momentum 195 Helpful Hint The impulse in each direction is the corresponding area under the force-time graph. Note that F1 is in the negative x-direction, so its impulse is negative.
c03.qxd 2/9/12 7:39 PM Page 195 SAMPLE PROBLEM 3/22 The loaded 150-kg skip is rolling down the incline at 4 m/s when a force P is applied to the cable as shown at time t 0. The force P is increased uniformly with the time until it reaches 600 N at t 4 s, after which time it remains con-stant at this value. Calculate (a) the time at which the skip reverses its direc-tion and (b) the velocity v of the skip at t 8 s. Treat the skip as a particle.
Solution.
The stated variation of P with the time is plotted, and the impulse-momentum diagrams of the skip are drawn.
Part (a). The skip reverses direction when its velocity becomes zero. We will assume that this condition occurs at t 4 t s. The impulse-momentum equa-tion applied consistently in the positive x-direction gives Ans.
Part (b). Applying the momentum equation to the entire 8-s interval gives Ans.
The same result is obtained by analyzing the interval from to 8 s.
SAMPLE PROBLEM 3/23 The 50-g bullet traveling at 600 m/s strikes the 4-kg block centrally and is embedded within it. If the block slides on a smooth horizontal plane with a veloc-ity of 12 m/s in the direction shown prior to impact, determine the velocity v2 of the block and embedded bullet immediately after impact.
Solution.
Since the force of impact is internal to the system composed of the block and bullet and since there are no other external forces acting on the sys-tem in the plane of motion, it follows that the linear momentum of the system is conserved. Thus, Ans.
The final velocity and its direction are given by Ans.
Ans.
tan 13.33 10.26 1.299 52.4 [tan vy /vx] v2 (10.26)2 (13.33)2 16.83 m/s [v vx 2 vy 2] v2 10.26i 13.33j m/s 0.050(600j) 4(12)(cos 30i sin 30j) (4 0.050)v2 [G1 G2] t (v2)x 4.76 m/s 150(4) 1 2 (4)(2)(600) 4(2)(600) 150(9.81) sin 30(8) 150(v2)x m(v1 )x ΣFx dt m(v2 )x t 2.46 s t 4 2.46 6.46 s 150(4) 1 2 (4)(2)(600) 2(600)t 150(9.81) sin 30(4 t) 150(0) m(v1 )x ΣFx dt m(v2 )x 150(4) kg·m/s 30° 30° 30° 150(9.81) dt x 150v2 + = 2P dt N1 dt N2 dt t v1 = 4 m/s P 30° P, N t, s 600 00 4 8 t′ Δt 30° x y 12 m/s 4 kg 600 m/s 0.050 kg x 16.83 m/s θ = 52.4° Helpful Hint Working with the vector form of the principle of conservation of linear momentum is clearly equivalent to working with the component form.
Helpful Hint The impulse-momentum diagram keeps us from making the error of using the impulse of P rather than 2P or of forgetting the impulse of the component of the weight. The first term in the linear impulse is the tri-angular area of the P-t relation for the first 4 s, doubled for the force of 2P.
196 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 196 PROBLEMS Introductory Problems 3/173 The rubber mallet is used to drive a cylindrical plug into the wood member. If the impact force varies with time as shown in the plot, determine the magnitude of the linear impulse delivered by the mallet to the plug.
Problem 3/173 3/174 The 1500-kg car has a velocity of 30 km/h up the 10-percent grade when the driver applies more power for 8 s to bring the car up to a speed of 60 km/h. Calculate the time average F of the total force tangent to the road exerted on the tires dur-ing the 8 s. Treat the car as a particle and neglect air resistance.
Problem 3/174 3/175 A 0.2-kg particle is moving with a velocity m/s at time If the single force F (5 3t)i (2 t2)j 3k N acts on the particle, determine its velocity at time .
t2 4 s v2 t1 1 s.
v1 i j 2k v 10 1 0 200 0 0.002 0.010 0.009 t, s F, N F 3/176 A 75-g projectile traveling at 600 m/s strikes and becomes embedded in the 50-kg block, which is ini-tially stationary. Compute the energy lost during the impact. Express your answer as an absolute value and as a percentage n of the original system energy E.
Problem 3/176 3/177 A jet-propelled airplane with a mass of 10 Mg is flying horizontally at a constant speed of 1000 km/h under the action of the engine thrust T and the equal and opposite air resistance R. The pilot ig-nites two rocket-assist units, each of which devel-ops a forward thrust T0 of 8 kN for 9 s. If the velocity of the airplane in its horizontal flight is 1050 km/h at the end of the 9 s, calculate the time-average increase in air resistance. The mass of the rocket fuel used is negligible compared with that of the airplane.
Problem 3/177 3/178 A 60-g bullet is fired horizontally with a velocity into the 3-kg block of soft wood initially at rest on the horizontal surface. The bullet emerges from the block with the velocity , and the block is observed to slide a distance of 2.70 m before coming to rest. Deter-mine the coefficient of kinetic friction between the block and the supporting surface.
Problem 3/178 2.70 m 400 m/s 600 m/s 3 kg 60 g k v2 400 m/s v1 600 m/s R T 2T0 R 75 g 50 kg 600 m/s E Article 3/9 Problems 197 c03.qxd 2/9/12 7:39 PM Page 197 3/182 The 90-kg man dives from the 40-kg canoe. The velocity indicated in the figure is that of the man rel-ative to the canoe just after loss of contact. If the man, woman, and canoe are initially at rest, deter-mine the horizontal component of the absolute velocity of the canoe just after separation. Neglect drag on the canoe, and assume that the 60-kg woman remains motionless relative to the canoe.
Problem 3/182 3/183 An experimental rocket sled weighs 5200 lb and is propelled by six rocket motors each with an im-pulse rating of 8600 lb-sec. The rockets are fired at 1-sec intervals, and the duration of each rocket fir-ing is 2 sec. If the velocity of the sled 10 sec from the start is 200 mi/hr, determine the time average R of the total aerodynamic and mechanical resis-tance to motion. Neglect the loss of mass due to ex-hausted fuel compared with the mass of the sled.
Problem 3/183 3/184 The 200-kg lunar lander is descending onto the moon’s surface with a velocity of 6 m/s when its retro-engine is fired. If the engine produces a thrust T for 4 s which varies with time as shown and then cuts off, calculate the velocity of the lan-der when assuming that it has not yet landed. Gravitational acceleration at the moon’s surface is Problem 3/184 T 6 m/s 800 2 4 0 0 T, N t, s 1.62 m/s2.
t 5 s, Rocket thrust R 3 m/s 40 kg 60 kg 90 kg 30° 3/179 At time , the velocity of cylinder A is 0.3 m/s down. By the methods of this article, determine the velocity of cylinder B at time Assume no mechanical interference and neglect all friction.
Problem 3/179 3/180 The resistance to motion of a certain racing tobog-gan is 2 percent of the normal force on its runners.
Calculate the time t required for the toboggan to reach a speed of 100 km/h down the slope if it starts from rest.
Problem 3/180 3/181 Freight car A with a gross weight of 150,000 lb is moving along the horizontal track in a switching yard at 2 mi/hr. Freight car B with a gross weight of 120,000 lb and moving at 3 mi/hr overtakes car A and is coupled to it. Determine (a) the common velocity v of the two cars as they move together after being cou-pled and (b) the loss of energy due to the impact.
Problem 3/181 3 mi/hr B A 2 mi/hr E 5 12 v 5 kg 4 kg A B t 2 s.
t 0 198 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 198 3/185 The slider of mass m1 is released from rest in the position shown and then slides down the right side of the contoured body of mass m2. For the condi-tions m1 0.50 kg, m2 3 kg, and r 0.25 m, de-termine the absolute velocities of both masses at the instant of separation. Neglect friction.
Problem 3/185 3/186 A supertanker with a total displacement (weight) of 140(103) long tons (one long ton equals 2240 lb) is moving forward at a speed of 2 knots when the en-gines are reversed to produce a rearward propeller thrust of 90,000 lb. How long would it take the tanker to acquire a speed of 2 knots in the reverse direction? Can you justify neglecting the impulse of water resistance of the hull? (Recall 1 knot 1.151 mi/hr.) 3/187 The 20-lb block is moving to the right with a veloc-ity of 2 ft/sec on a horizontal surface when a force P is applied to it at time t 0. Calculate the veloc-ity v of the block when t 0.4 sec. The coefficient of kinetic friction is 0.30.
Problem 3/187 P 20 lb v0 = 2 ft/sec k = 0.30 P, lb t, sec 0.4 0.2 0 16 8 0 μ k r r m1 m2 Representative Problems 3/188 The initially stationary 12-kg block is subjected to the time-varying force whose magnitude P is shown in the plot. The angle remains constant.
Determine the block speed at (a) t 1 s and (b) t 4 s.
Problem 3/188 3/189 The tow truck with attached 1200-kg car acceler-ates uniformly from 30 km/h to 70 km/h over a 15-s interval. The average rolling resistance for the car over this speed interval is 500 N. Assume that the 60 angle shown represents the time average con-figuration and determine the average tension in the tow cable.
Problem 3/189 3/190 The 140-g projectile is fired with a velocity of 600 m/s and picks up three washers, each with a mass of 100 g. Find the common velocity v of the projec-tile and washers. Determine also the loss of energy during the interaction.
Problem 3/190 600 m/s E 60° 12 kg μs = 0.50 μk = 0.40 P 30° 5 0 100 0 t, s P, N 30 Article 3/9 Problems 199 c03.qxd 2/9/12 7:39 PM Page 199 3/194 The initially stationary 100-lb block is subjected to the time-varying force whose magnitude P is shown in the plot. Determine the speed v of the block at times t 1, 3, 5, and 7 sec. Note that the force P is zero after t 6 sec.
Problem 3/194 3/195 The 900-kg motorized unit A is designed to raise and lower the 600-kg bucket B of concrete. Deter-mine the average force R which supports unit A during the 6 seconds required to slow the descent of the bucket from 3 m/s to 0.5 m/s. Analyze the entire system as a unit without finding the tension in the cable.
Problem 3/195 B v A 100 lb μs = 0.60, μk = 0.40 P 4 6 0 80 20 0 t, sec P, lb 3/191 The spring of modulus k 200 N/m is compressed a distance of 300 mm and suddenly released with the system at rest. Determine the absolute veloci-ties of both masses when the spring is unstretched.
Neglect friction.
Problem 3/191 3/192 The 4-kg cart, at rest at time t 0, is acted on by a horizontal force which varies with time t as shown.
Neglect friction and determine the velocity of the cart at t 1 s and at t 3 s.
Problem 3/192 3/193 The space shuttle launches an 800-kg satellite by ejecting it from the cargo bay as shown. The ejec-tion mechanism is activated and is in contact with the satellite for 4 s to give it a velocity of 0.3 m/s in the z-direction relative to the shuttle. The mass of the shuttle is 90 Mg. Determine the component of velocity vƒ of the shuttle in the minus z-direction resulting from the ejection. Also find the time aver-age Fav of the ejection force.
Problem 3/193 y x z v F 4 kg Force F, N 0 20 0 2 4 Parabolic Linear Time t, s 30 k = 200 N/m 7 kg 3 kg 200 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 200 3/196 The cart of mass m is subjected to the exponentially decreasing force F, which represents a shock or blast loading. If the cart is stationary at time t 0, deter-mine its velocity v and displacement s as functions of time. What is the value of v for large values of t?
Problem 3/196 3/197 Determine the time required by a diesel-electric locomotive, which produces a constant drawbar pull of 60,000 lb, to increase the speed of an 1800-ton freight train from 20 mi/hr to 30 mi/hr up a 1-percent grade. Train resistance is 10 lb per ton.
3/198 The 450-kg ram of a pile driver falls 1.4 m from rest and strikes the top of a 240-kg pile embedded 0.9 m in the ground. Upon impact the ram is seen to move with the pile with no noticeable rebound.
Determine the velocity v of the pile and ram imme-diately after impact. Can you justify using the principle of conservation of momentum even though the weights act during the impact?
Problem 3/198 1.4 m 0.9 m F m F0 0 Time t Force F 0 F0e–bt 3/199 The cart is moving down the incline with a velocity v0 20 m/s at t 0, at which time the force P begins to act as shown. After 5 seconds the force continues at the 50-N level. Determine the velocity of the cart at time t 8 s and calculate the time t at which the cart velocity is zero.
Problem 3/199 3/200 Car B is initially stationary and is struck by car A moving with initial speed v1 20 mi/hr. The cars become entangled and move together with speed after the collision. If the time duration of the colli-sion is 0.1 sec, determine (a) the common final speed , (b) the average acceleration of each car during the collision, and (c) the magnitude R of the average force exerted by each car on the other car during the impact. All brakes are released during the collision.
Problem 3/200 3/201 The 12-Mg truck drives onto the 350-Mg barge from the dock at 20 km/h and brakes to a stop on the deck. The barge is free to move in the water, which offers negligible resistance to motion at low speeds. Calculate the speed of the barge after the truck has come to rest on it.
Problem 3/201 v 20 km/h 12 Mg 350 Mg 20 mi/hr 4000 lb 2000 lb A B v v P, N 0 50 0 5 Parabolic P t, s v0 = 20 m/s 6 kg 15° Article 3/9 Problems 201 c03.qxd 2/9/12 7:39 PM Page 201 3/204 A 16.1-lb body is traveling in a horizontal straight line with a velocity of 12 ft/sec when a horizontal force P is applied to it at right angles to the initial direction of motion. If P varies according to the ac-companying graph, remains constant in direction, and is the only force acting on the body in its plane of motion, find the magnitude of the velocity of the body when t 2 sec and the angle which the velocity makes with the direction of P.
Problem 3/204 3/205 The force P, which is applied to the 10-kg block ini-tially at rest, varies linearly with time as indicated.
If the coefficients of static and kinetic friction between the block and the horizontal surface are 0.60 and 0.40, respectively, determine the velocity of the block when t 4 s.
Problem 3/205 P 10 kg P, N t, s 100 4 0 0 s = 0.60, k = 0.40 μ μ 1 1.5 2 0 0 2 4 P, lb t, sec 3/202 An 8-Mg truck is resting on the deck of a barge which displaces 240 Mg and is at rest in still water.
If the truck starts and drives toward the bow at a speed relative to the barge vrel 6 km/h, calculate the speed v of the barge. The resistance to the mo-tion of the barge through the water is negligible at low speeds.
Problem 3/202 3/203 Car B weighing 3200 lb and traveling west at 30 mi/hr collides with car A weighing 3400 lb and traveling north at 20 mi/hr as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude v of their common velocity immediately after the impact and the angle made by the velocity vector with the north direction.
Problem 3/203 30 mi/hr 20 mi/hr A B N W 8 Mg vrel = 6 km/h 240 Mg v 202 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 202 3/206 The 10-kg block is at rest on the rough incline at time t 0 and then it is subjected to the force of constant direction and time-varying magnitude P given in the plot. Determine the velocity of the block at times t 1, 3, 5, and 7 s. Note that the force P is zero after t 6 s.
Problem 3/206 3/207 The 1.62-oz golf ball is struck by the five-iron and acquires the velocity shown in a time period of 0.001 sec. Determine the magnitude R of the aver-age force exerted by the club on the ball. What ac-celeration magnitude a does this force cause, and what is the distance d over which the launch veloc-ity is achieved, assuming constant acceleration?
Problem 3/207 3/208 The 580-ton tug is towing the 1200-ton coal barge at a steady speed of 6 knots. For a short period of time, the stern winch takes in the towing cable at the rate of 2 ft/sec. Calculate the reduced speed v of the tug during this interval. Assume the tow cable to be horizontal. (Recall 1 knot 1.688 ft/sec) Problem 3/208 v = 150 ft/sec 25° 10 kg µs = 0.50 µk = 0.40 P 20° 6 4 0 100 0 t, s P, N 15° 3/209 The cylindrical plug A of mass mA is released from rest at B and slides down the smooth circular guide. The plug strikes the block C and becomes embedded in it. Write the expression for the dis-tance s which the block and plug slide before com-ing to rest. The coefficient of kinetic friction between the block and the horizontal surface is .
Problem 3/209 3/210 The baseball is traveling with a horizontal velocity of 85 mi/hr just before impact with the bat. Just after the impact, the velocity of the -oz ball is 130 mi/hr directed at to the horizontal as shown. Determine the x- and y-components of the average force R exerted by the bat on the baseball during the 0.005-sec impact. Comment on the treat-ment of the weight of the baseball (a) during the im-pact and (b) over the first few seconds after impact.
Problem 3/210 35° 85 mi/hr 130 mi/hr 35 51 8 r mA mC A B C s k μ k Article 3/9 Problems 203 c03.qxd 2/20/12 3:20 PM Page 203 3/213 The simple pendulum A of mass mA and length l is suspended from the trolley B of mass mB. If the system is released from rest at 0, determine the velocity vB of the trolley when Friction is negligible.
Problem 3/213 3/214 Two barges, each with a displacement (mass) of 500 Mg, are loosely moored in calm water. A stunt driver starts his 1500-kg car from rest at A, drives along the deck, and leaves the end of the ramp at a speed of 50 km/h relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at B. Cal-culate the velocity v2 imparted to barge 2 just after the car has come to rest on the barge. Neglect the resistance of the water to motion at the low veloci-ties involved.
Problem 3/214 A B 2 1 15° 15 θ A B l 90.
3/211 A tennis player strikes the tennis ball with her racket while the ball is still rising. The ball speed before im-pact with the racket is v1 15 m/s and after impact its speed is v2 22 m/s, with directions as shown in the figure. If the 60-g ball is in contact with the racket for 0.05 s, determine the magnitude of the av-erage force R exerted by the racket on the ball. Find the angle β made by R with the horizontal. Comment on the treatment of the ball weight during impact.
Problem 3/211 3/212 The 400-kg ram of a pile driver is designed to fall 1.5 m from rest and strike the top of a 300-kg pile partially embedded in the ground. The deeper the penetration, the greater is the tendency for the ram to rebound as a result of the impact. Calculate the velocity v of the pile immediately after impact for the following three conditions: (a) initial resistance to penetration is small at the outset, and the ram is observed to move with the pile immediately after impact; (b) resistance to penetration has increased, and the ram is seen to have zero velocity immedi-ately after impact; (c) resistance to penetration is high, and the ram is seen to rebound to a height of 100 mm above the point of impact. Why is it per-missible to neglect the impulse of the weight of the ram during impact?
Problem 3/212 1.5 m 20° v1 v2 10° 204 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 204 3/10 Angular Impulse and Angular Momentum In addition to the equations of linear impulse and linear momentum, there exists a parallel set of equations for angular impulse and angular momentum. First, we define the term angular momentum. Figure 3/14a shows a particle P of mass m moving along a curve in space. The particle is located by its position vector r with respect to a convenient origin O of fixed coordinates x-y-z. The velocity of the particle is v , and its linear momentum is G mv. The moment of the linear momentum vector mv about the origin O is defined as the angular momentum HO of P about O and is given by the cross-product relation for the moment of a vector (3/29) The angular momentum then is a vector perpendicular to the plane A defined by r and v. The sense of HO is clearly defined by the right-hand rule for cross products.
The scalar components of angular momentum may be obtained from the expansion (3/30) so that Each of these expressions for angular momentum may be checked easily from Fig. 3/15, which shows the three linear-momentum components, by taking the moments of these components about the respective axes.
To help visualize angular momentum, we show in Fig. 3/14b a two-dimensional representation in plane A of the vectors shown in part a of the figure. The motion is viewed in plane A defined by r and v. The mag-nitude of the moment of mv about O is simply the linear momentum mv times the moment arm r sin or mvr sin , which is the magnitude of the cross product HO r mv.
Angular momentum is the moment of linear momentum and must not be confused with linear momentum. In SI units, angular momentum has the units . In the U.S. customary system, angular momentum has the units [lb/(ft/sec2)][ft/sec][ft] lb-ft-sec.
Rate of Change of Angular Momentum We are now ready to relate the moment of the forces acting on the particle P to its angular momentum. If ΣF represents the resultant of all forces acting on the particle P of Fig. 3/14, the moment MO about the origin O is the vector cross product ΣMO r ΣF r mv ˙ kg (m/s) m kg m2/s Nms Hx m(vz y vy z) Hy m(vx z vz x) Hz m(vy x vx y) HO r mv m(vz y vy z)i m(vx z vz x)j m(vy x vx y)k HO r mv r ˙ mvy mvx mvz m z z y O y x x r Figure 3/15 A H O = r mv P P r r O z y x θ θ θ mv View in plane A (a) (b) HO = mvr sinθ r sinθ mv Figure 3/14 HO m i x vx j y vy k z vz Article 3/10 Angular Impulse and Angular Momentum 205 c03.qxd 2/9/12 7:39 PM Page 205 where Newton’s second law ΣF m has been substituted. We now dif-ferentiate Eq. 3/29 with time, using the rule for the differentiation of a cross product (see item 9, Art. C/7, Appendix C) and obtain The term v mv is zero since the cross product of parallel vectors is identically zero. Substitution into the expression for ΣMO gives (3/31) Equation 3/31 states that the moment about the fixed point O of all forces acting on m equals the time rate of change of angular momentum of m about O. This relation, particularly when extended to a system of particles, rigid or nonrigid, provides one of the most powerful tools of analysis in dynamics.
Equation 3/31 is a vector equation with scalar components (3/32) The Angular Impulse-Momentum Principle Equation 3/31 gives the instantaneous relation between the mo-ment and the time rate of change of angular momentum. To obtain the effect of the moment ΣMO on the angular momentum of the particle over a finite period of time, we integrate Eq. 3/31 from time t1 to time t2.
Multiplying the equation by dt gives ΣMO dt dHO, which we integrate to obtain (3/33) where (HO)2 r2 mv2 and (HO)1 r1 mv1. The product of moment and time is defined as angular impulse, and Eq. 3/33 states that the total angular impulse on m about the fixed point O equals the corre-sponding change in angular momentum of m about O.
Alternatively, we may write Eq. 3/33 as (3/33a) which states that the initial angular momentum of the particle plus the angular impulse applied to it equals its final angular momentum.
The units of angular impulse are clearly those of angular momentum, which are or in SI units and lb-ft-sec in U.S. custom-ary units.
As in the case of linear impulse and linear momentum, the equation of angular impulse and angular momentum is a vector equation where changes in direction as well as magnitude may occur during the interval of integration. Under these conditions, it is necessary to express ΣMO kgm2/s Nms (HO)1 t2 t1 ΣMO dt (HO)2 t2 t1 ΣMO dt (HO)2 (HO)1 HO ΣMOx H ˙Ox ΣMOy H ˙Oy ΣMOz H ˙Oz ΣMO H ˙ O H ˙ O r ˙ mv r mv ˙ v mv r mv ˙ v ˙ 206 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 206 and HO in component form and then combine the integrated compo-nents. The x-component of Eq. 3/33a is or (3/33b) where the subscripts 1 and 2 refer to the values of the respective quantities at times t1 and t2. Similar expressions exist for the y- and z-components of the angular impulse-momentum equation.
Plane-Motion Applications The foregoing angular-impulse and angular-momentum relations have been developed in their general three-dimensional forms. Most of the ap-plications of interest to us, however, can be analyzed as plane-motion prob-lems where moments are taken about a single axis normal to the plane of motion. In this case, the angular momentum may change magnitude and sense, but the direction of the vector remains unaltered.
Thus, for a particle of mass m moving along a curved path in the x-y plane, Fig. 3/16, the angular momenta about O at points 1 and 2 have the magnitudes mv1d1 and r2 mv2 mv2d2, respectively. In the illustration both and are repre-sented in the counterclockwise sense in accord with the direction of the moment of the linear momentum. The scalar form of Eq. 3/33a applied to the motion between points 1 and 2 during the time interval t1 to t2 becomes (HO)2 (HO)1 (HO)2 (HO)1 r1 mv1 m(vz y vy z)1 t2 t1 ΣMOx dt m(vz y vy z)2 (HOx)1 t2 t1 ΣMOx dt (HOx)2 Article 3/10 Angular Impulse and Angular Momentum 207 ΣMO = ΣFr sin θ θ ΣF r1 r2 d2 d1 O r (HO)1 = mv1d1 (HO)2 = mv2d2 mv2 x 2 mv1 1 y Figure 3/16 (HO)1 t2 t1 ΣMO dt (HO)2 or mv1d1 t2 t1 ΣFr sin dt mv2d2 c03.qxd 2/9/12 7:39 PM Page 207 This example should help clarify the relation between the scalar and vector forms of the angular impulse-momentum relations.
Whereas Eq. 3/33 clearly stresses that the external angular impulse causes a change in the angular momentum, the order of the terms in Eqs. 3/33a and 3/33b corresponds to the natural sequence of events.
Equation 3/33a is analogous to Eq. 3/27a, just as Eq. 3/31 is analogous to Eq. 3/25.
As was the case for linear-momentum problems, we encounter im-pulsive (large magnitude, short duration) and nonimpulsive forces in angular-momentum problems. The treatment of these forces was dis-cussed in Art. 3/9.
Equations 3/25 and 3/31 add no new basic information since they are merely alternative forms of Newton’s second law. We will discover in subsequent chapters, however, that the motion equations expressed in terms of the time rate of change of momentum are applicable to the motion of rigid and nonrigid bodies and provide a very general and pow-erful approach to many problems. The full generality of Eq. 3/31 is usu-ally not required to describe the motion of a single particle or the plane motion of rigid bodies, but it does have important use in the analysis of the space motion of rigid bodies introduced in Chapter 7.
Conservation of Angular Momentum If the resultant moment about a fixed point O of all forces acting on a particle is zero during an interval of time, Eq. 3/31 requires that its angular momentum HO about that point remain constant. In this case, the angular momentum of the particle is said to be conserved. Angular momentum may be conserved about one axis but not about another axis. A careful examination of the free-body diagram of the particle will disclose whether the moment of the resultant force on the particle about a fixed point is zero, in which case, the angular momentum about that point is unchanged (conserved).
Consider now the motion of two particles a and b which interact during an interval of time. If the interactive forces F and F between them are the only unbalanced forces acting on the particles during the interval, it follows that the moments of the equal and opposite forces about any fixed point O not on their line of action are equal and oppo-site. If we apply Eq. 3/33 to particle a and then to particle b and add the two equations, we obtain Ha Hb 0 (where all angular momenta are referred to point O). Thus, the total angular momentum for the sys-tem of the two particles remains constant during the interval, and we write (3/34) which expresses the principle of conservation of angular momentum.
HO 0 or (HO)1 (HO)2 208 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 208 SAMPLE PROBLEM 3/24 A small sphere has the position and velocity indicated in the figure and is acted upon by the force F. Determine the angular mo-mentum HO about point O and the time derivative .
Solution.
We begin with the definition of angular momentum and write Ans.
From Eq. 3/31, Ans.
As with moments of forces, the position vector must run from the reference point (O in this case) to the line of action of the linear momentum mv. Here r runs directly to the particle.
SAMPLE PROBLEM 3/25 A comet is in the highly eccentric orbit shown in the figure. Its speed at the most distant point A, which is at the outer edge of the solar system, is vA 740 m/s. Determine its speed at the point B of closest approach to the sun.
Solution.
Because the only significant force acting on the comet, the gravitational force exerted on it by the sun, is central (points to the sun center O), angular momentum about O is conserved.
Ans. vB 59 200 m/s vB rAvA rB 6000(106 )740 75(106 ) mrAvA mrBvB (HO)A (HO)B 60i 30j Nm (3i 6j 4k) 10k r F H ˙ O M O 40i 30k Nm/s (3i 6j 4k) 2(5j) HO r mv H ˙ O Article 3/10 Angular Impulse and Angular Momentum 209 z x y O 3 m 4 m 2 kg 6 m F = 10 N v = 5 m/s B O A 6000(106) km 75(106) km (Not to scale) c03.qxd 2/9/12 7:39 PM Page 209 SAMPLE PROBLEM 3/26 The assembly of the light rod and two end masses is at rest when it is struck by the falling wad of putty traveling with speed v1 as shown. The putty adheres to and travels with the right-hand end mass. Determine the angular velocity of the assembly just after impact. The pivot at O is frictionless, and all three masses may be assumed to be particles.
Solution.
If we ignore the angular impulses associated with the weights dur-ing the collision process, then system angular momentum about O is conserved during the impact.
Ans.
Note that each angular-momentum term is written in the form mvd, and the final transverse velocities are expressed as radial distances times the common final angular velocity SAMPLE PROBLEM 3/27 A small mass particle is given an initial velocity v0 tangent to the horizontal rim of a smooth hemispherical bowl at a radius r0 from the vertical centerline, as shown at point A. As the particle slides past point B, a distance h below A and a distance r from the vertical centerline, its velocity v makes an angle with the horizontal tangent to the bowl through B. Determine .
Solution.
The forces on the particle are its weight and the normal reaction ex-erted by the smooth surface of the bowl. Neither force exerts a moment about the axis O-O, so that angular momentum is conserved about that axis. Thus, Also, energy is conserved so that E1 E2. Thus Eliminating v and substituting r2 h2 give Ans. v0 r0 v0 2 2ghr0 2 h2 cos r0 2 v v0 2 2gh 1 2 mv0 2 mgh 1 2 mv2 0 [T1 V1 T2 V2] [(HO)1 (HO)2] mv0 r0 mvr cos ˙2.
˙2 v1 19l CW mv1l (m 2m)(l ˙2 )l 4m(2l ˙2 )2l (HO )1 (HO )2 ˙2 210 Chapter 3 Kinetics of Particles 2m 2l O m l 4m v1 A O O B h r r0 v0 v θ Helpful Hint The angle is measured in the plane tangent to the hemispherical surface at B. cos1 1 1 2gh v0 2 1 h2 r0 2 c03.qxd 2/9/12 7:39 PM Page 210 PROBLEMS Introductory Problems 3/215 Determine the magnitude HO of the angular mo-mentum of the 2-kg sphere about point O (a) by using the vector definition of angular momentum and (b) by using an equivalent scalar approach.
The center of the sphere lies in the x-y plane.
Problem 3/215 3/216 The 3-kg sphere moves in the x-y plane and has the indicated velocity at a particular instant. Deter-mine its (a) linear momentum, (b) angular momen-tum about point O, and (c) kinetic energy.
Problem 3/216 3/217 A particle with a mass of 4 kg has a position vector in meters given by r 3t2i 2tj 3tk, where t is the time in seconds. For t 3 s determine the mag-nitude of the angular momentum of the particle and the magnitude of the moment of all forces on the particle, both about the origin of coordinates.
3/218 A 0.4-kg particle is located at the position r1 2i 3j k m and has the velocity v1 i j 2k m/s at time t 0. If the particle is acted upon by a sin-gle force which has the moment MO (4 2t)i j 5k about the origin O of the coordinate system in use, determine the angular momentum about O of the particle when t 4 s.
N m 3 1 2t2 4 m /s 3 kg 60° 45° 2 m O y x 45° 12 m 5 m 2 kg 7 m/s x O y 3/219 At a certain instant, the particle of mass m has the position and velocity shown in the figure, and it is acted upon by the force F. Determine its angular momentum about point O and the time rate of change of this angular momentum.
Problem 3/219 3/220 The small spheres, which have the masses and ini-tial velocities shown in the figure, strike and be-come attached to the spiked ends of the rod, which is freely pivoted at O and is initially at rest. Deter-mine the angular velocity of the assembly after impact. Neglect the mass of the rod.
Problem 3/220 O m 2m L L v 3v z x y c b O a F m v Article 3/10 Problems 211 c03.qxd 2/9/12 7:39 PM Page 211 3/223 The assembly starts from rest and reaches an an-gular speed of 150 rev/min under the action of a 20-N force T applied to the string for t seconds. De-termine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.
Problem 3/223 Representative Problems 3/224 Just after launch from the earth, the space-shuttle orbiter is in the 37 137–mi orbit shown. At the apogee point A, its speed is 17,290 mi/hr. If nothing were done to modify the orbit, what would its speed be at the perigee P? Neglect aerodynamic drag.
(Note that the normal practice is to add speed at A, which raises the perigee altitude to a value that is well above the bulk of the atmosphere.) Problem 3/224 17,290 mi/hr A O P 37 mi 137 mi T 100 mm 400 mm 3 kg 3/221 The particle of mass m is gently nudged from the equilibrium position A and subsequently slides along the smooth circular path which lies in a ver-tical plane. Determine the magnitude of its angu-lar momentum about point O as it passes (a) point B and (b) point C. In each case, determine the time rate of change of HO.
Problem 3/221 3/222 A wad of clay of mass m1 with an initial horizontal velocity v1 hits and adheres to the massless rigid bar which supports the body of mass m2, which can be assumed to be a particle. The pendulum assem-bly is freely pivoted at O and is initially stationary.
Determine the angular velocity of the combined body just after impact. Why is linear momentum of the system not conserved?
Problem 3/222 L/2 L/2 m1 m2 O v1 ˙ B O m r A C 212 Chapter 3 Kinetics of Particles c03.qxd 2/20/12 3:20 PM Page 212 3/225 A small 4-oz particle is projected with a horizontal velocity of 6 ft/sec into the top A of the smooth cir-cular guide fixed in the vertical plane. Calculate the time rate of change B of angular momentum about point B when the particle passes the bottom of the guide at C.
Problem 3/225 3/226 The small particle of mass m and its restraining cord are spinning with an angular velocity on the horizontal surface of a smooth disk, shown in section. As the force F is slightly relaxed, r in-creases and changes. Determine the rate of change of with respect to r and show that the work done by F during a movement dr equals the change in kinetic energy of the particle. Problem 3/226 r m F ω 10″ B A C x y 6 ft /sec H ˙ 3/227 The 6-kg sphere and 4-kg block (shown in section) are secured to the arm of negligible mass which ro-tates in the vertical plane about a horizontal axis at O. The 2-kg plug is released from rest at A and falls into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity rad/s. Determine the angular velocity of the arm immediately after the plug has wedged itself in the block.
Problem 3/227 3/228 The two spheres of equal mass m are able to slide along the horizontal rotating rod. If they are ini-tially latched in position a distance r from the ro-tating axis with the assembly rotating freely with an angular velocity , determine the new angular velocity after the spheres are released and finally assume positions at the ends of the rod at a radial distance of 2r. Also find the fraction n of the initial kinetic energy of the system which is lost. Neglect the small mass of the rod and shaft.
Problem 3/228 2r r m m r 2r 0 ω 0 600 mm 2 kg 300 mm 500 mm 4 kg 6 kg O A ω 0 0 2 Article 3/10 Problems 213 c03.qxd 2/9/12 7:39 PM Page 213 3/231 Determine the magnitude HO of the angular mo-mentum about the launch point O of the projectile of mass m, which is launched with speed v0 at the angle as shown (a) at the instant of launch and (b) at the instant of impact. Qualitatively account for the two results. Neglect atmospheric resistance.
Problem 3/231 3/232 The particle of mass m is launched from point O with a horizontal velocity u at time t 0. Deter-mine its angular momentum HO relative to point O as a function of time.
Problem 3/232 3/233 A particle of mass m is released from rest in posi-tion A and then slides down the smooth vertical-plane track. Determine its angular momentum about both points A and D (a) as it passes position B and (b) as it passes position C.
Problem 3/233 A B C ρ ρ D m 30° x m u O y O A θ v0 3/229 The speed of Mercury at its point A of maximum distance from the sun is 38 860 m/s. Determine its speeds at points B and P.
Problem 3/229 3/230 A small 0.1-kg particle is given a velocity of 2 m/s on the horizontal x-y plane and is guided by the fixed curved rail. Friction is negligible. As the par-ticle crosses the y-axis at A, its velocity is in the x-direction, and as it crosses the x-axis at B, its velocity makes a 60 angle with the x-axis. The ra-dius of curvature of the path at B is 500 mm. De-termine the time rate of change of the angular momentum of the particle about the z-axis through O at both A and B.
Problem 3/230 HO vA = 38 860 m/s O B P A 69.82(106) km 46(106) km 56.70(106) km Mercury Sun 214 Chapter 3 Kinetics of Particles x y A O B 60° z 300 mm 200 mm c03.qxd 2/9/12 7:39 PM Page 214 3/234 At the point A of closest approach to the sun, a comet has a velocity vA 188,500 ft/sec. Deter-mine the radial and transverse components of its velocity vB at point B, where the radial distance from the sun is 75(106) mi.
Problem 3/234 3/235 A pendulum consists of two 3.2-kg concentrated masses positioned as shown on a light but rigid bar.
The pendulum is swinging through the vertical po-sition with a clockwise angular velocity 6 rad/s when a 50-g bullet traveling with velocity 300 m/s in the direction shown strikes the lower mass and becomes embedded in it. Calculate the angular velocity which the pendulum has immediately after impact and find the maximum angular deflec-tion of the pendulum.
Problem 3/235 400 mm 200 mm 20° O v θ ω v A v vr B 50 (106) mi 75 (106) mi v θ S S 3/236 The 1.5-lb sphere moves in a horizontal plane and is controlled by a cord which is reeled in and out below the table in such a way that the center of the sphere is confined to the path given by where x and y are in feet. If the speed of the sphere is vA 8 ft/sec as it passes point A, determine the tension TB in the cord as the sphere passes point B. Friction is negligible.
Problem 3/236 z y B A x O T vA (y2/16) 1 (x2/25) Article 3/10 Problems 215 c03.qxd 2/9/12 7:39 PM Page 215 3/238 The assembly of two 5-kg spheres is rotating freely about the vertical axis at 40 rev/min with .
If the force F which maintains the given position is increased to raise the base collar and reduce to , determine the new angular velocity . Also de-termine the work U done by F in changing the con-figuration of the system. Assume that the mass of the arms and collars is negligible.
Problem 3/238 100 mm 300 mm 5 kg 5 kg F ω θ 300 mm 300 mm 60 90 3/237 A particle is launched with a horizontal velocity v0 0.55 m/s from the position shown and then slides without friction along the funnel-like surface. Deter-mine the angle which its velocity vector makes with the horizontal as the particle passes level O-O.
The value of r is 0.9 m.
Problem 3/237 r m m O O 30° v0 0.15r 30 216 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 216 SECTION D SPECIAL APPLICATIONS 3/11 Introduction The basic principles and methods of particle kinetics were devel-oped and illustrated in the first three sections of this chapter. This treatment included the direct use of Newton’s second law, the equations of work and energy, and the equations of impulse and momentum. We paid special attention to the kind of problem for which each of the ap-proaches was most appropriate.
Several topics of specialized interest in particle kinetics will be briefly treated in Section D: 1. Impact 2. Central-force motion 3. Relative motion These topics involve further extension and application of the fundamen-tal principles of dynamics, and their study will help to broaden your background in mechanics.
3/12 Impact The principles of impulse and momentum have important use in de-scribing the behavior of colliding bodies. Impact refers to the collision between two bodies and is characterized by the generation of relatively large contact forces which act over a very short interval of time. It is im-portant to realize that an impact is a very complex event involving ma-terial deformation and recovery and the generation of heat and sound.
Small changes in the impact conditions may cause large changes in the impact process and thus in the conditions immediately following the im-pact. Therefore, we must be careful not to rely heavily on the results of impact calculations.
Direct Central Impact As an introduction to impact, we consider the collinear motion of two spheres of masses m1 and m2, Fig. 3/17a, traveling with velocities v1 and v2. If v1 is greater than v2, collision occurs with the contact forces di-rected along the line of centers. This condition is called direct central impact.
Following initial contact, a short period of increasing deformation takes place until the contact area between the spheres ceases to in-crease. At this instant, both spheres, Fig. 3/17b, are moving with the same velocity v0. During the remainder of contact, a period of restora-tion occurs during which the contact area decreases to zero. In the final condition shown in part c of the figure, the spheres now have new veloc-ities and where must be less than All velocities are arbi-trarily assumed positive to the right, so that with this scalar notation a velocity to the left would carry a negative sign. If the impact is not v2 .
v1 v2 , v1 Article 3/12 Impact 217 m2 m2 m1 m2 m1 m1 v1 v0 v2 Before impact (a) Maximum deformation during impact (b) After impact (c) > < v1′ v2′ Figure 3/17 c03.qxd 2/9/12 7:39 PM Page 217 overly severe and if the spheres are highly elastic, they will regain their original shape following the restoration. With a more severe impact and with less elastic bodies, a permanent deformation may result.
Because the contact forces are equal and opposite during impact, the linear momentum of the system remains unchanged, as discussed in Art. 3/9. Thus, we apply the law of conservation of linear momentum and write (3/35) We assume that any forces acting on the spheres during impact, other than the large internal forces of contact, are relatively small and pro-duce negligible impulses compared with the impulse associated with each internal impact force. In addition, we assume that no appreciable change in the positions of the mass centers occurs during the short du-ration of the impact.
Coefficient of Restitution For given masses and initial conditions, the momentum equation contains two unknowns, and Clearly, we need an additional rela-tionship to find the final velocities. This relationship must reflect the ca-pacity of the contacting bodies to recover from the impact and can be expressed by the ratio e of the magnitude of the restoration impulse to the magnitude of the deformation impulse. This ratio is called the coeffi-cient of restitution.
Let Fr and Fd represent the magnitudes of the contact forces dur-ing the restoration and deformation periods, respectively, as shown in Fig. 3/18. For particle 1 the definition of e together with the impulse-momentum equation give us Similarly, for particle 2 we have We are careful in these equations to express the change of momentum (and therefore v) in the same direction as the impulse (and thus the force). The time for the deformation is taken as t0 and the total time of contact is t. Eliminating v0 between the two expressions for e gives us (3/36) e v2 v1 v1 v2 relative velocity of separation relative velocity of approach e t t0 Fr dt t0 0 Fd dt m2(v2 v0 ) m2(v0 v2 ) v2 v0 v0 v2 e t t0 Fr dt t0 0 Fd dt m1[v1 (v0 )] m1[v0 (v1 )] v0 v1 v1 v0 v2 .
v1 m1v1 m2v2 m1v1 m2v2 218 Chapter 3 Kinetics of Particles Fd Deformation period v1 v2 m1 m2 Fr Restoration period v0 v0 v1′ v2′ m1 m2 ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ Figure 3/18 c03.qxd 2/9/12 7:39 PM Page 218 If the two initial velocities v1 and v2 and the coefficient of restitution e are known, then Eqs. 3/35 and 3/36 give us two equations in the two unknown final velocities and Energy Loss During Impact Impact phenomena are almost always accompanied by energy loss, which may be calculated by subtracting the kinetic energy of the system just after impact from that just before impact. Energy is lost through the generation of heat during the localized inelastic deformation of the material, through the generation and dissipation of elastic stress waves within the bodies, and through the generation of sound energy.
According to this classical theory of impact, the value e 1 means that the capacity of the two particles to recover equals their tendency to deform. This condition is one of elastic impact with no energy loss. The value e 0, on the other hand, describes inelastic or plastic impact where the particles cling together after collision and the loss of energy is a maxi-mum. All impact conditions lie somewhere between these two extremes.
Also, it should be noted that a coefficient of restitution must be as-sociated with a pair of contacting bodies. The coefficient of restitution is frequently considered a constant for given geometries and a given com-bination of contacting materials. Actually, it depends on the impact ve-locity and approaches unity as the impact velocity approaches zero as shown schematically in Fig. 3/19. A handbook value for e is generally unreliable.
Oblique Central Impact We now extend the relationships developed for direct central impact to the case where the initial and final velocities are not parallel, Fig.
3/20. Here spherical particles of mass m1 and m2 have initial velocities v1 and v2 in the same plane and approach each other on a collision course, as shown in part a of the figure. The directions of the velocity vectors are measured from the direction tangent to the contacting surfaces, Fig. 3/20b. Thus, the initial velocity components along the t- and n-axes are (v1)n v1 sin 1, (v1)t v1 cos 1, (v2)n v2 sin 2, v2 .
v1 Article 3/12 Impact 219 (b) (c) (d) (e) (a) n n m1 v1 v2 v2′ F 00 t0 Time, t t |F| –F v1′ m1 m2 2 θ 1 θ 2′ θ 1′ θ m2 t Figure 3/20 Perfectly elastic Glass on glass Steel on steel Lead on lead Perfectly plastic Relative impact velocity 1 00 Coefficient of restitution, e Figure 3/19 c03.qxd 2/9/12 7:39 PM Page 219 and (v2)t v2 cos 2. Note that (v1)n is a negative quantity for the partic-ular coordinate system and initial velocities shown.
The final rebound conditions are shown in part c of the figure. The impact forces are F and F, as seen in part d of the figure. They vary from zero to their peak value during the deformation portion of the im-pact and back again to zero during the restoration period, as indicated in part e of the figure where t is the duration of the impact interval.
For given initial conditions of m1, m2, (v1)n, (v1)t, (v2)n, and (v2)t, there will be four unknowns, namely, and The four needed equations are obtained as follows: (1) Momentum of the system is conserved in the n-direction. This gives (2) and (3) The momentum for each particle is conserved in the t-direction since there is no impulse on either particle in the t-direction.
Thus, (4) The coefficient of restitution, as in the case of direct central im-pact, is the positive ratio of the recovery impulse to the deformation im-pulse. Equation 3/36 applies, then, to the velocity components in the n-direction. For the notation adopted with Fig. 3/20, we have Once the four final velocity components are found, the angles and of Fig. 3/20 may be easily determined.
2 1 e (v2 )n (v1 )n (v1 )n (v2 )n m2(v2 )t m2(v2 )t m1(v1 )t m1(v1 )t m1(v1 )n m2(v2 )n m1(v1 )n m2(v2 )n (v2 )t .
(v2 )n, (v1 )t , (v1 )n, Pool balls about to undergo impact.
220 Chapter 3 Kinetics of Particles © Silvestre Machado/SuperStock c03.qxd 2/9/12 7:39 PM Page 220 SAMPLE PROBLEM 3/28 The ram of a pile driver has a mass of 800 kg and is released from rest 2 m above the top of the 2400-kg pile. If the ram rebounds to a height of 0.1 m after impact with the pile, calculate (a) the velocity of the pile immediately after impact, (b) the coefficient of restitution e, and (c) the percentage loss of energy due to the impact.
Solution.
Conservation of energy during free fall gives the initial and final ve-locities of the ram from v Thus, (a) Conservation of momentum (G1 G2) for the system of the ram and pile gives Ans.
(b) The coefficient of restitution yields Ans.
(c) The kinetic energy of the system just before impact is the same as the poten-tial energy of the ram above the pile and is The kinetic energy T just after impact is The percentage loss of energy is, therefore, Ans.
SAMPLE PROBLEM 3/29 A ball is projected onto the heavy plate with a velocity of 50 ft/sec at the 30 angle shown. If the effective coefficient of restitution is 0.5, compute the rebound velocity v and its angle .
Solution.
Let the ball be denoted body 1 and the plate body 2. The mass of the heavy plate may be considered infinite and its corresponding velocity zero after impact. The coefficient of restitution is applied to the velocity components nor-mal to the plate in the direction of the impact force and gives Momentum of the ball in the t-direction is unchanged since, with assumed smooth surfaces, there is no force acting on the ball in that direction. Thus, The rebound velocity v and its angle are then Ans.
Ans.
tan1 (v1 )n (v1 )t tan1 12.5 43.3 16.10 v (v1 )n 2 (v1 )t 2 12.52 43.32 45.1 ft/sec m(v1 )t m(v1 )t (v1 )t (v1)t 50 cos 30 43.3 ft/sec e (v2 )n (v1 )n (v1 )n (v2 )n 0.5 0 (v1 )n 50 sin 30 0 (v1 )n 12.5 ft/sec 15 700 8620 15 700 (100) 45.1% T 1 2 (800)(1.401)2 1 2 (2400)(2.55)2 8620 J T Vg mgh 800(9.81)(2) 15 700 J e rel. vel. separation rel. vel. approach e 2.55 1.401 6.26 0 0.631 800(6.26) 0 800(1.401) 2400vp vp 2.55 m/s vr 2(9.81)(2) 6.26 m/s vr 2(9.81)(0.1) 1.401 m/s 2gh.
vp 2 m drop 0.1 m rebound Before impact ram pile y Immediately after impact vr vp = 0 vr′ vp′ Fimpact W << Fimpact Helpful Hint The impulses of the weights of the ram and pile are very small com-pared with the impulses of the im-pact forces and thus are neglected during the impact.
Helpful Hint We observe here that for infinite mass there is no way of applying the principle of conservation of momen-tum for the system in the n-direc-tion. From the free-body diagram of the ball during impact, we note that the impulse of the weight W is ne-glected since W is very small com-pared with the impact force.
Article 3/12 Impact 221 2 30° v′ θ 1 50 ft/sec t n ′ c03.qxd 2/9/12 7:39 PM Page 221 SAMPLE PROBLEM 3/30 Spherical particle 1 has a velocity v1 6 m/s in the direction shown and col-lides with spherical particle 2 of equal mass and diameter and initially at rest. If the coefficient of restitution for these conditions is e 0.6, determine the result-ing motion of each particle following impact. Also calculate the percentage loss of energy due to the impact.
Solution.
The geometry at impact indicates that the normal n to the contact-ing surfaces makes an angle 30 with the direction of v1, as indicated in the figure. Thus, the initial velocity components are (v1)n v1 cos 30 6 cos 30 5.20 m/s, (v1)t v1 sin 30 6 sin 30 3 m/s, and (v2)n (v2)t 0.
Momentum conservation for the two-particle system in the n-direction gives or, with m1 m2, (a) The coefficient-of-restitution relationship is (b) Simultaneous solution of Eqs. a and b yields Conservation of momentum for each particle holds in the t-direction because, with assumed smooth surfaces, there is no force in the t-direction. Thus for par-ticles 1 and 2, we have The final speeds of the particles are Ans.
Ans.
The angle which makes with the t-direction is Ans.
The kinetic energies just before and just after impact, with m m1 m2, are The percentage energy loss is then Ans.
E E (100) T T T (100) 18m 13.68m 18m (100) 24.0% T 1 2 m1v1 2 1 2 m2v2 2 1 2 m(3.17)2 1 2 m(4.16)2 13.68m T 1 2 m1v1 2 1 2 m2v2 2 1 2 m(6)2 0 18m tan1 (v1 )n (v1 )t tan1 1.039 3 19.11 v1 v2 (v2 )n 2 (v2 )t 2 (4.16)2 02 4.16 m/s v1 (v1 )n 2 (v1 )t 2 (1.039)2 32 3.17 m/s m2(v2 )t m2(v2 )t (v2 )t (v2 )t 0 m1(v1 )t m1(v1 )t (v1 )t (v1 )t 3 m/s (v1 )n 1.039 m/s (v2 )n 4.16 m/s e (v2 )n (v1 )n (v1 )n (v2 )n 0.6 (v2 )n (v1 )n 5.20 0 5.20 0 (v1 )n (v2 )n m1(v1 )n m2(v2 )n m1(v1 )n m2(v2 )n 1 2 v1 n r v1 2r θ n 30° 1 2 2′ 1′ t v1 v2′ v1′ θ1′ n 1 2 t F F Helpful Hints Be sure to set up n- and t-coordinates which are, respectively, normal to and tangent to the contacting sur-faces. Calculation of the 30 angle is critical to all that follows.
Note that, even though there are four equations in four unknowns for the standard problem of oblique cen-tral impact, only one pair of the equations is coupled.
We note that particle 2 has no initial or final velocity component in the t-direction. Hence, its final velocity is restricted to the n-direction.
v2 222 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 222 PROBLEMS Introductory Problems 3/239 Tennis balls are usually rejected if they fail to re-bound to waist level when dropped from shoulder level. If a ball just passes the test as indicated in the figure, determine the coefficient of restitution e and the percentage n of the original energy lost during the impact.
Problem 3/239 3/240 If the tennis ball of Prob. 3/239 has a coefficient of restitution e 0.8 during impact with the court surface, determine the velocity v0 with which the ball must be thrown downward from the 1600-mm shoulder level if it is return to the same level after bouncing once on the court surface.
3/241 Compute the final velocities and after collision of the two cylinders which slide on the smooth hori-zontal shaft. The coefficient of restitution is e 0.6.
Problem 3/241 v1 = 7 m/s m1 = 2 kg v2 = 5 m/s m2 = 3 kg v2 v1 1100 mm 1600 mm 3/242 The two bodies have the masses and initial veloci-ties shown in the figure. The coefficient of restitu-tion for the collision is e 0.3, and friction is negligible. If the time duration of the collision is 0.025 s, determine the average impact force which is exerted on the 3-kg body.
Problem 3/242 3/243 The sphere of mass m1 travels with an initial ve-locity v1 directed as shown and strikes the sphere of mass m2. For a given coefficient of restitution e, determine the mass ratio m1/m2 which results in m1 being motionless after the impact.
Problem 3/243 3/244 Three identical steel cylinders are free to slide on the fixed horizontal shaft. Cylinders 2 and 3 are at rest and are approached by cylinder 1 at a speed u.
Express the final speed v of cylinder 3 in terms of u and the coefficient of restitution e.
Problem 3/244 u 1 2 3 v1 m2 m1 3 kg 4 kg 0.5 m/s 0.7 m/s Article 3/12 Problems 223 c03.qxd 2/9/12 7:39 PM Page 223 3/248 If the center of the ping-pong ball is to clear the net as shown, at what height h should the ball be horizontally served? Also determine h2. The coeffi-cient of restitution for the impacts between ball and table is e 0.9, and the radius of the ball is r 0.75 in.
Problem 3/248 3/249 In the selection of the ram of a pile driver, it is de-sired that the ram lose all of its kinetic energy at each blow. Hence, the velocity of the ram is zero immediately after impact. The mass of each pile to be driven is 300 kg, and experience has shown that a coefficient of restitution of 0.3 can be expected.
What should be the mass m of the ram? Compute the velocity v of the pile immediately after impact if the ram is dropped from a height of 4 m onto the pile. Also compute the energy loss due to im-pact at each blow.
Problem 3/249 4 m m 300 kg 3 m E v0 h 9″ h2 3/245 Cylinder A is moving to the right with speed v when it impacts the initially stationary cylinder B.
Both cylinders have mass m, and the coefficient of restitution for the collision is e. Determine the maximum deflection of the spring of modulus k.
Neglect friction.
Problem 3/245 3/246 Car B is initially stationary and is struck by car A, which is moving with speed v. The mass of car B is pm, where m is the mass of car A and p is a positive constant. If the coefficient or restitution is e 0.1, express the speeds and of the two cars at the end of the impact in terms of p and v. Evaluate your expressions for p 0.5.
Problem 3/246 3/247 Determine the coefficient of restitution e for a steel ball dropped from rest at a height h above a heavy horizontal steel plate if the height of the second re-bound is h2.
Problem 3/247 h h2 A B m pm v vB vA v k B A m m 224 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 224 3/250 Freight car A of mass mA is rolling to the right when it collides with freight car B of mass mB ini-tially at rest. If the two cars are coupled together at impact, show that the fractional loss of energy equals mB/(mA mB).
Problem 3/250 Representative Problems 3/251 Pool ball B is to be shot into the side pocket D by banking it off the cushion at C. Specify the location x of the cushion impact for coefficients of restitu-tion (a) e 1 and (b) e 0.8.
Problem 3/251 x v d B D C d/2 d/2 A A B 3/252 Determine the value of the coefficient of restitu-tion e which results in the final velocity being perpendicular to the initial velocity v.
Problem 3/252 3/253 Determine the value of the coefficient of restitu-tion e for which the outgoing angle is one-half of the incoming angle as shown. Evaluate your gen-eral expression for Problem 3/253 v v′ θ θ –– 2 40.
v v′ 60° v Article 3/12 Problems 225 c03.qxd 2/9/12 7:39 PM Page 225 3/257 A basketball traveling with the velocity shown in the figure strikes the backboard at A. If the coeffi-cient of restitution for this impact is e 0.84, de-termine the required distance h above the hoop if the ball is to arrive at the center B of the hoop.
Carry out two solutions: (a) an approximate solu-tion obtained by neglecting the effects of gravity from A to B and (b) a solution which accounts for gravity from A to B. Neglect the diameter of the ball compared with h.
Problem 3/257 24 ft/sec 40° 15″ B A h 3/254 The figure shows n spheres of equal mass m sus-pended in a line by wires of equal length so that the spheres are almost touching each other. If sphere 1 is released from the dashed position and strikes sphere 2 with a velocity v1, write an expres-sion for the velocity vn of the nth sphere immedi-ately after being struck by the one adjacent to it.
The common coefficient of restitution is e.
Problem 3/254 3/255 The ball is released from position A and drops 0.75 m to the incline. If the coefficient of restitu-tion in the impact is e 0.85, determine the slant range R.
Problem 3/255 3/256 A projectile is launched from point A and has a horizontal range L1 as shown. If the coefficient of restitution at B is e, determine the distance L2.
Problem 3/256 L1 A B L2 0.75 m R C 20° A vn v1 1 2 3 n 226 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 226 3/258 The two cars collide at right angles in the intersec-tion of two icy roads. Car A has a mass of 1200 kg and car B has a mass of 1600 kg. The cars become entangled and move off together with a common velocity in the direction indicated. If car A was traveling 50 km/h at the instant of impact, com-pute the corresponding velocity of car B just before impact.
Problem 3/258 3/259 The two identical steel balls moving with initial ve-locities vA and vB collide as shown. If the coefficient of restitution is determine the velocity of each ball just after impact and the percentage loss n of system kinetic energy.
Problem 3/259 y x A B vB = 8 ft/sec vA = 6 ft/sec 30° e 0.7, y x 30° v′ vB vA = 50 km/h B A v 3/260 A 0.1-kg meteor and a 1000-kg spacecraft have the indicated absolute velocities just before colliding.
The meteor punches a hole entirely through the spacecraft. Instruments indicate that the velocity of the meteor relative to the spacecraft just after the collision is vm/s 1880i 6898j m/s. Deter-mine the direction of the absolute velocity of the spacecraft after the collision.
Problem 3/260 3/261 Two identical hockey pucks moving with initial ve-locities vA and vB collide as shown. If the coefficient of restitution is , determine the velocity (magnitude and direction with respect to the posi-tive x-axis) of each puck just after impact. Also calculate the percentage loss n of system kinetic energy.
Problem 3/261 vA = 6 m/s vB = 10 m/s 30° A y x B e 0.75 Article 3/12 Problems 227 z x y vm = 7000 m/s vs = 2000 m/s vs′ θ c03.qxd 2/9/12 7:39 PM Page 227 3/264 During a pregame warmup period, two basketballs collide above the hoop when in the positions shown. Just before impact, ball 1 has a velocity v1 which makes a 30 angle with the horizontal. If the velocity v2 of ball 2 just before impact has the same magnitude as v1, determine the two possible values of the angle , measured from the horizontal, which will cause ball 1 to go directly through the center of the basket. The coefficient of restitution is e 0.8.
Problem 3/264 30° 2 1 v2 v1 v1′ θ 3/262 Sphere A collides with sphere B as shown in the figure. If the coefficient of restitution is e 0.5, de-termine the x- and y-components of the velocity of each sphere immediately after impact. Motion is confined to the x-y plane.
Problem 3/262 3/263 Determine the coefficient of restitution e which will allow the ball to bounce down the steps as shown. The tread and riser dimensions, d and h, respectively, are the same for every step, and the ball bounces the same distance above each step.
What horizontal velocity vx is required so that the ball lands in the center of each tread?
Problem 3/263 d h h′ h y x vA = 3 m/s vB = 12 m/s B A 30° 20° 45° 10 kg 2 kg 228 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 228 3/265 The 0.5-kg cylinder A is released from rest from the position shown and drops the distance h1 0.6 m.
It then collides with the 0.4-kg block B; the coeffi-cient of restitution is e 0.8. Determine the maxi-mum downward displacement h2 of block B.
Neglect all friction and assume that block B is ini-tially held in place by a hidden mechanism until the collision begins. The two springs of modulus k 500 N/m are initially unstretched, and the dis-tance d 0.8 m.
Problem 3/265 k k B mB A mA d h2 d h1 3/266 A child throws a ball from point A with a speed of 50 ft/sec. It strikes the wall at point B and then re-turns exactly to point A. Determine the necessary angle if the coefficient of restitution in the wall impact is e 0.5.
Problem 3/266 A B α 10′ Article 3/12 Problems 229 c03.qxd 2/9/12 7:39 PM Page 229 3/13 Central-Force Motion When a particle moves under the influence of a force directed to-ward a fixed center of attraction, the motion is called central-force mo-tion. The most common example of central-force motion is the orbital movement of planets and satellites. The laws which govern this motion were deduced from observation of the motions of the planets by J. Kepler (1571–1630). An understanding of central-force motion is required to design high-altitude rockets, earth satellites, and space vehicles.
Motion of a Single Body Consider a particle of mass m, Fig. 3/21, moving under the action of the central gravitational attraction where m0 is the mass of the attracting body, which is assumed to be fixed, G is the universal gravitational constant, and r is the distance be-tween the centers of the masses. The particle of mass m could represent the earth moving about the sun, the moon moving about the earth, or a satellite in its orbital motion about the earth above the atmosphere.
The most convenient coordinate system to use is polar coordinates in the plane of motion since F will always be in the negative r-direction and there is no force in the -direction.
Equations 3/8 may be applied directly for the r- and -directions to give (3/37) The second of the two equations when multiplied by r/m is seen to be the same as d(r2 )/dt 0, which is integrated to give (3/38) The physical significance of Eq. 3/38 is made clear when we note that the angular momentum r mv of m about m0 has the magnitude mr2 . Thus, Eq. 3/38 merely states that the angular momentum of m about m0 remains constant (is conserved). This statement is easily de-duced from Eq. 3/31, which shows that the angular momentum HO re-mains constant (is conserved) if there is no moment acting on the particle about a fixed point O.
We observe that during time dt, the radius vector sweeps out an area, shaded in Fig. 3/21, equal to dA (r d). Therefore, the rate at which area is swept by the radius vector is , which is constant ac-cording to Eq. 3/38. This conclusion is expressed in Kepler’s second law of planetary motion, which states that the areas swept through in equal times are equal.
A ˙ 1 2 r2 ˙ (1 2 r) ˙ r2 ˙ h, a constant ˙ 0 m(r ¨ 2r ˙ ˙) G mm0 r2 m(r ¨ r ˙2) F G mm0 r2 230 Chapter 3 Kinetics of Particles Path Focus Directrix m m0 r x d v F θ dθ Figure 3/21 c03.qxd 2/9/12 7:39 PM Page 230 The shape of the path followed by m may be obtained by solving the first of Eqs. 3/37, with the time t eliminated through combination with Eq. 3/38. To this end the mathematical substitution r 1/u is useful.
Thus, (1/u2) , which from Eq. 3/38 becomes h( ) or h(du/d). The second time derivative is h(d2u/d2) , which by combining with Eq. 3/38, becomes h2u2(d2u/d2). Substitution into the first of Eqs. 3/37 now gives or (3/39) which is a nonhomogeneous linear differential equation.
The solution of this familiar second-order equation may be verified by direct substitution and is where C and are the two integration constants. The phase angle may be eliminated by choosing the x-axis so that r is a minimum when 0.
Thus, (3/40) Conic Sections The interpretation of Eq. 3/40 requires a knowledge of the equa-tions for conic sections. We recall that a conic section is formed by the locus of a point which moves so that the ratio e of its distance from a point (focus) to a line (directrix) is constant. Thus, from Fig. 3/21, e r/(d r cos ), which may be rewritten as (3/41) which is the same form as Eq. 3/40. Thus, we see that the motion of m is along a conic section with d 1/C and ed h2/(Gm0), or (3/42) The three cases to be investigated correspond to e 1 (ellipse), e 1 (parabola), and e 1 (hyperbola). The trajectory for each of these cases is shown in Fig. 3/22.
e h2 C Gm0 1 r 1 d cos 1 ed 1 r C cos Gm0 h2 u 1 r C cos ( ) Gm0 h2 d2 u d2 u Gm0 h2 Gm0u2 h2 u2 d2 u d2 1 u h2u4 r ¨ ˙ r ¨ r ˙ u ˙/ ˙ r ˙ u ˙ r ˙ Article 3/13 Central-Force Motion 231 Circle Ellipse e < 1 Parabola e = 1 Hyperbola e > 1 Apogee 2b 2a Perigee e = 0 r m v m0 a(1 – e) a(1 + e) θ Figure 3/22 c03.qxd 2/9/12 7:39 PM Page 231 Case 1: ellipse (e 1). From Eq. 3/41 we deduce that r is a mini-mum when 0 and is a maximum when . Thus, With the distance d expressed in terms of a, Eq. 3/41 and the maximum and minimum values of r may be written as (3/43) In addition, the relation b a , which comes from the geometry of the ellipse, gives the expression for the semiminor axis. We see that the ellipse becomes a circle with r a when e 0. Equation 3/43 is an expression of Kepler’s first law, which says that the planets move in el-liptical orbits around the sun as a focus.
The period for the elliptical orbit is the total area A of the ellipse divided by the constant rate at which the area is swept through.
Thus, from Eq. 3/38, We can eliminate reference to or h in the expression for by substi-tuting Eq. 3/42, the identity d 1/C, the geometric relationships a ed/(1 e2) and b a for the ellipse, and the equivalence Gm0 gR2. The result after simplification is (3/44) In this equation note that R is the mean radius of the central attracting body and g is the absolute value of the acceleration due to gravity at the surface of the attracting body.
Equation 3/44 expresses Kepler’s third law of planetary motion which states that the square of the period of motion is proportional to the cube of the semimajor axis of the orbit.
Case 2: parabola (e 1). Equations 3/41 and 3/42 become The radius vector becomes infinite as approaches , so the dimension a is infinite.
Case 3: hyperbola (e 1). From Eq. 3/41 we see that the radial distance r becomes infinite for the two values of the polar angle 1 and 1 r 1 d (1 cos ) and h2C Gm0 2 a3/2 Rg 1 e2 ˙ A A ˙ ab 1 2 r2 ˙ or 2ab h A ˙ 1 e2 rmin a(1 e) rmax a(1 e) 1 r 1 e cos a(1 e2) 2a rmin rmax ed 1 e ed 1 e or a ed 1 e2 232 Chapter 3 Kinetics of Particles Artist conception of the Mars Recon-naissance Orbiter, which arrived at Mars in March 2006.
Courtesy NASA/JPL-Caltech c03.qxd 2/9/12 7:39 PM Page 232 1 defined by cos 1 1/e. Only branch I corresponding to 1 1, Fig. 3/23, represents a physically possible motion. Branch II corresponds to angles in the remaining sector (with r negative). For this branch, pos-itive r’s may be used if is replaced by and r by r. Thus, Eq. 3/41 becomes But this expression contradicts the form of Eq. 3/40 where Gm0/h2 is neces-sarily positive. Thus branch II does not exist (except for repulsive forces).
Energy Analysis Now consider the energies of particle m. The system is conservative, and the constant energy E of m is the sum of its kinetic energy T and potential energy V. The kinetic energy is T mv2 and the potential energy from Eq. 3/19 is V mgR2/r.
Recall that g is the absolute acceleration due to gravity measured at the surface of the attracting body, R is the radius of the attracting body, and Gm0 gR2. Thus, This constant value of E can be determined from its value at 0, where 0, 1/r C gR2/h2 from Eq. 3/40, and h/r from Eq.
3/38. Substituting this into the expression for E and simplifying yield Now C is eliminated by substitution of Eq. 3/42, which may be written as h2C egR2, to obtain (3/45) The plus value of the radical is mandatory since by definition e is posi-tive. We now see that for the These conclusions, of course, depend on the arbitrary selection of the datum condition for zero potential energy (V 0 when r ).
The expression for the velocity v of m may be found from the energy equation, which is 1 2 mv2 mgR2 r E hyperbolic orbit e 1, E is positive parabolic orbit e 1, E is zero elliptical orbit e 1, E is negative 2E m h2C2 g2R4 h2 r ˙ r ˙ E 1 2 m(r ˙2 r2 ˙2) mgR2 r 1 2 m(r ˙2 r2 ˙2) 1 2 1 r 1 d cos ( ) 1 ed or 1 r 1 ed cos d Article 3/13 Central-Force Motion 233 I 1 θ 1 –θ II Figure 3/23 e 1 2Eh2 mg2R4 c03.qxd 2/9/12 7:39 PM Page 233 The total energy E is obtained from Eq. 3/45 by combining Eq. 3/42 and 1/C d a(1 e2)/e to give for the elliptical orbit (3/46) Substitution into the energy equation yields (3/47) from which the magnitude of the velocity may be computed for a partic-ular orbit in terms of the radial distance r.
Next, combining the expressions for rmin and rmax corresponding to perigee and apogee, Eq. 3/43, with Eq. 3/47 results in a pair of expres-sions for the respective velocities at these two positions for the elliptical orbit: (3/48) Selected numerical data pertaining to the solar system are included in Appendix D and are useful in applying the foregoing relationships to problems in planetary motion.
Summary of Assumptions The foregoing analysis is based on three assumptions: 1. The two bodies possess spherical mass symmetry so that they may be treated as if their masses were concentrated at their centers, that is, as if they were particles.
2. There are no forces present except the gravitational force which each mass exerts on the other.
3. Mass m0 is fixed in space.
Assumption (1) is excellent for bodies which are distant from the cen-tral attracting body, which is the case for most heavenly bodies. A sig-nificant class of problems for which assumption (1) is poor is that of artificial satellites in the very near vicinity of oblate planets. As a comment on assumption (2), we note that aerodynamic drag on a low-altitude earth satellite is a force which usually cannot be ignored in the orbital analysis. For an artificial satellite in earth orbit, the error of assumption (3) is negligible because the ratio of the mass of the satellite to that of the earth is very small. On the other hand, for the earth–moon system, a small but significant error is introduced if as-sumption (3) is invoked—note that the lunar mass is about 1/81 times that of the earth.
v2 2gR2 1 r 1 2a E gR2m 2a 234 Chapter 3 Kinetics of Particles vA R g a 1 e 1 e R g a rmin rmax vP R g a 1 e 1 e R g a rmax rmin c03.qxd 2/9/12 7:39 PM Page 234 Perturbed Two-Body Problem We now account for the motion of both masses and allow the pres-ence of other forces in addition to those of mutual attraction by consid-ering the perturbed two-body problem. Figure 3/24 depicts the major mass m0, the minor mass m, their respective position vectors r1 and r2 measured relative to an inertial frame, the gravitation forces F and F, and a non-two-body force P which is exerted on mass m. The force P may be due to aerodynamic drag, solar pressure, the presence of a third body, on-board thrusting activities, a nonspherical gravitational field, or a combination of these and other sources.
Application of Newton’s second law to each mass results in Dividing the first equation by m0, the second equation by m, and sub-tracting the first equation from the second give or (3/49) Equation 3/49 is a second-order differential equation which, when solved, yields the relative position vector r as a function of time. Numer-ical techniques are usually required for the integration of the scalar dif-ferential equations which are equivalent to the vector equation 3/49, especially if P is nonzero.
Restricted Two-Body Problem If m0 m and P 0, we have the restricted two-body problem, the equation of motion of which is (3/49a) With r and expressed in polar coordinates, Eq. 3/49a becomes When we equate coefficients of like unit vectors, we recover Eqs. 3/37.
Comparison of Eq. 3/49 (with P 0) and Eq. 3/49a enables us to relax the assumption that mass m0 is fixed in space. If we replace m0 by (m0 m) in the expressions derived with the assumption of m0 fixed, then we obtain expressions which account for the motion of m0. For ex-ample, the corrected expression for the period of elliptical motion of m about m0 is, from Eq. 3/44, (3/49b) where the equality R2g Gm0 has been used.
2 a3/2 G(m0 m) (r ¨ r ˙2)er (r ¨ 2r ˙ ˙)e G m0 r3 (rer) 0 r ¨ r ¨ G m0 r3 r 0 r ¨ G (m0 m) r3 r P m G (m0 m) r3 r P m r ¨2 r ¨1 r ¨ G mm0 r3 r m0r ¨1 and G mm0 r3 r P mr ¨2 Article 3/13 Central-Force Motion 235 P r r1 r2 P F = G r – F mm0 ——– r3 m0 m Figure 3/24 c03.qxd 2/9/12 7:39 PM Page 235 SAMPLE PROBLEM 3/31 An artificial satellite is launched from point B on the equator by its carrier rocket and inserted into an elliptical orbit with a perigee altitude of 2000 km. If the apogee altitude is to be 4000 km, compute (a) the necessary perigee velocity vP and the corresponding apogee velocity vA, (b) the velocity at point C where the altitude of the satellite is 2500 km, and (c) the period for a complete orbit.
Solution. (a) The perigee and apogee velocities for specified altitudes are given by Eqs. 3/48, where Thus, Ans.
Ans.
(b) For an altitude of 2500 km the radial distance from the center of the earth is r 6371 2500 8871 km. From Eq. 3/47 the velocity at point C becomes Ans.
(c) The period of the orbit is given by Eq. 3/44, which becomes Ans.
or 2.507 h 2 a3/2 Rg 2 [(9371)(103)]3/2 (6371)(103)9.825 9026 s vC 6881 m/s or 24 773 km/h 47.353(106)(m/s)2 vC 2 2gR2 1 r 1 2a 2(9.825)[(6371)(103)]2 1 8871 1 18 742 1 103 5861 m/s or 21 099 km/h 7261 m/s or 26 140 km/h a (rmin rmax )/2 9371 km rmin 6371 2000 8371 km rmax 6371 4000 10 371 km A P vP vA C B 12 742 km 4000 km 2000 km 2a 2500 km θ O R Helpful Hints The mean radius of 12 742/2 6371 km from Table D/2 in Appendix D is used. Also the absolute acceleration due to gravity g 9.825 m/s2 from Art. 1/5 will be used.
236 Chapter 3 Kinetics of Particles We must be careful with units. It is often safer to work in base units, meters in this case, and convert later.
We should observe here that the time interval between successive overhead transits of the satellite as recorded by an observer on the equator is longer than the period calculated here since the observer will have moved in space due to the counterclockwise rotation of the earth, as seen looking down on the north pole.
vP R g a rmax rmin 6371(103) 9.825 9371(103) 10 371 8371 vA R g a rmin rmax 6371(103) 9.825 9371(103) 8371 10 371 c03.qxd 2/9/12 7:39 PM Page 236 PROBLEMS (Unless otherwise indicated, the velocities mentioned in the problems which follow are measured from a nonrotat-ing reference frame moving with the center of the attract-ing body. Also, aerodynamic drag is to be neglected unless stated otherwise. Use g 9.825 m/s2 (32.23 ft/sec2) for the absolute gravitational acceleration at the surface of the earth and treat the earth as a sphere of radius R 6371 km (3959 mi).) Introductory Problems 3/267 Determine the speed v of the earth in its orbit about the sun. Assume a circular orbit of radius 93(106) miles.
3/268 What velocity v must the space shuttle have in order to release the Hubble space telescope in a cir-cular earth orbit 590 km above the surface of the earth?
Problem 3/268 3/269 Show that the path of the moon is concave toward the sun at the position shown. Assume that the sun, earth, and moon are in the same line.
Problem 3/269 Earth Moon Sunlight 590 km 3/270 A spacecraft is orbiting the earth in a circular orbit of altitude H. If its rocket engine is activated to produce a sudden burst of speed, determine the in-crease necessary to allow the spacecraft to es-cape from the earth’s gravity field. Calculate if H 200 mi.
3/271 Determine the apparent velocity vrel of a satellite moving in a circular equatorial orbit 200 mi above the earth as measured by an observer on the equa-tor (a) for a west-to-east orbit and (b) for an east-to-west orbit. Why is the west-to-east orbit more easily achieved?
3/272 A spacecraft is in an initial circular orbit with an altitude of 350 km. As it passes point P, onboard thrusters give it a velocity boost of 25 m/s. Deter-mine the resulting altitude gain at point A.
Problem 3/272 3/273 If the perigee altitude of an earth satellite is 240 km and the apogee altitude is 400 km, compute the eccentricity e of the orbit and the period of one complete orbit in space.
3/274 In one of the orbits of the Apollo spacecraft about the moon, its distance from the lunar surface var-ied from 60 mi to 180 mi. Compute the maximum velocity of the spacecraft in this orbit.
Δh A P 350 km h v v Article 3/13 Problems 237 c03.qxd 2/9/12 7:39 PM Page 237 3/277 Determine the speed v required of an earth satellite at point A for (a) a circular orbit, (b) an elliptical orbit of eccentricity e 0.1, (c) an elliptical orbit of eccentricity e 0.9, and (d) a parabolic orbit. In cases (b), (c), and (d), A is the orbit perigee.
Problem 3/277 Representative Problems 3/278 Initially in the 240-km circular orbit, the space-craft S receives a velocity boost at P which will take it to with no speed at that point. Deter-mine the required velocity increment v at point P and also determine the speed when r 2rP. At what value of does r become 2rP?
Problem 3/278 P r O S 240 km θ r l R A v 0.1R 3/275 A satellite is in a circular earth orbit of radius 2R, where R is the radius of the earth. What is the minimum velocity boost necessary to reach point B, which is a distance 3R from the center of the earth? At what point in the original circular orbit should the velocity increment be added?
Problem 3/275 3/276 The Mars orbiter for the Viking mission was de-signed to make one complete trip around the planet in exactly the same time that it takes Mars to revolve once about its own axis. This time is 24 h, 37 min, 23 s. In this way, it is possible for the or-biter to pass over the landing site of the lander capsule at the same time in each Martian day at the orbiter’s minimum (periapsis) altitude. For the Viking I mission, the periapsis altitude of the or-biter was 1508 km. Make use of the data in Table D/2 in Appendix D and compute the maximum (apoapsis) altitude ha for the orbiter in its elliptical path.
Problem 3/276 ha hp = 1508 km R B 2R 3R v 238 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 238 3/279 Satellite A moving in the circular orbit and satel-lite B moving in the elliptical orbit collide and be-come entangled at point C. If the masses of the satellites are equal, determine the maximum alti-tude hmax of the resulting orbit.
Problem 3/279 3/280 If the earth were suddenly deprived of its orbital velocity around the sun, find the time t which it would take for the earth to “fall” to the location of the center of the sun. (Hint: The time would be one-half the period of a degenerate elliptical orbit around the sun with the semiminor axis approach-ing zero.) Refer to Table D/2 for the exact period of the earth around the sun.
3/281 Just after launch from the earth, the space-shuttle orbiter is in the 37 137-mi orbit shown. The first time that the orbiter passes the apogee A, its two orbital-maneuvering-system (OMS) engines are fired to circularize the orbit. If the weight of the orbiter is 175,000 lb and the OMS engines have a thrust of 6000 lb each, determine the required time duration t of the burn.
Problem 3/281 A O P 37 mi 137 mi B A C 800 mi 800 mi 200 mi 3/282 After launch from the earth, the 85 000-kg space-shuttle orbiter is in the elliptical orbit shown. If the orbit is to be circularized at the apogee altitude of 320 km, determine the necessary time duration during which its two orbital-maneuvering-system (OMS) engines, each of which has a thrust of 30 kN, must be fired when the apogee position C is reached.
Problem 3/282 3/283 Just before separation of the lunar module, the Apollo 17 command module was in the lunar orbit shown in the figure. Determine the spacecraft speeds at points P and A, which are called perilune and apolune, respectively. Later in the mission, with the lunar module on the surface of the moon, the orbit of the command module was to be circu-larized. Determine the speed increment re-quired if circularization is to be performed at A.
Problem 3/283 P A 28 km 109 km v 320 km 240 km B C t Article 3/13 Problems 239 c03.qxd 2/9/12 7:39 PM Page 239 3/286 Determine the angle made by the velocity vector v with respect to the -direction for an earth satel-lite traveling in an elliptical orbit of eccentricity e.
Express in terms of the angle measured from perigee.
Problem 3/286 3/287 Two satellites B and C are in the same circular orbit of altitude 500 miles. Satellite B is 1000 mi ahead of satellite C as indicated. Show that C can catch up to B by “putting on the brakes.” Specifi-cally, by what amount should the circular-orbit velocity of C be reduced so that it will rendezvous with B after one period in its new elliptical orbit?
Check to see that C does not strike the earth in the elliptical orbit.
Problem 3/287 3/288 Determine the necessary amount by which the circular-orbit velocity of satellite C should be re-duced if the catch-up maneuver of Prob. 3/287 is to be accomplished with not one but two periods in a new elliptical orbit.
v 1 0 0 0 m i B R C 500 mi v v r m Perigee β θ θ 3/284 Determine the required velocity vB in the direction indicated so that the spacecraft path will be tan-gent to the circular orbit at point C. What must be the distance b so that this path is possible?
Problem 3/284 3/285 An earth satellite A is in a circular west-to-east equatorial orbit a distance 300 km above the sur-face of the earth as indicated. An observer B on the equator who sees the satellite directly overhead will see it directly overhead in the next orbit at po-sition because of the rotation of the earth. The radial line to the satellite will have rotated through the angle and the observer will measure the apparent period as a value slightly greater than the true period . Calculate and Problem 3/285 Equ ato r θ ω A N O B A′ B′ .
2 , B C B b 16,000 mi 3959 mi 4000 mi vB 240 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:39 PM Page 240 3/289 The spacecraft S is to be injected into a circular orbit of altitude 400 km. Because of equipment malfunction, the injection speed v is correct for the circular orbit, but the initial velocity v makes an angle with the intended direction. What is the maximum permissible error in order that the spacecraft not strike the earth? Neglect atmos-pheric resistance.
Problem 3/289 3/290 The 175,000-lb space-shuttle orbiter is in a circu-lar orbit of altitude 200 miles. The two orbital-maneuvering-system (OMS) engines, each of which has a thrust of 6000 lb, are fired in retrothrust for 150 seconds. Determine the angle which locates the intersection of the shuttle trajectory with the earth’s surface. Assume that the shuttle position B corresponds to the completion of the OMS burn and that no loss of altitude occurs during the burn.
Problem 3/290 3/291 Compare the orbital period of the moon calculated with the assumption of a fixed earth with the pe-riod calculated without this assumption.
B C 200 mi β 400 km S v α 3/292 A satellite is placed in a circular polar orbit a dis-tance H above the earth. As the satellite goes over the north pole at A, its retro-rocket is activated to produce a burst of negative thrust which reduces its velocity to a value which will ensure an equato-rial landing. Derive the expression for the required reduction of velocity at A. Note that A is the apogee of the elliptical path.
Problem 3/292 3/293 A spacecraft moving in a west-to-east equatorial orbit is observed by a tracking station located on the equator. If the spacecraft has a perigee altitude H 150 km and velocity v when directly over the station and an apogee altitude of 1500 km, determine an expression for the angular rate p (relative to the earth) at which the antenna dish must be ro-tated when the spacecraft is directly overhead.
Compute p. The angular velocity of the earth is 0.7292( ) rad/s.
Problem 3/293 v H East West p ω N R 104 B A N R H S vA Article 3/13 Problems 241 c03.qxd 2/9/12 7:39 PM Page 241 3/296 In 1995 a spacecraft called the Solar and Helio-spheric Observatory (SOHO) was placed into a cir-cular orbit about the sun and inside that of the earth as shown. Determine the distance h so that the period of the spacecraft orbit will match that of the earth, with the result that the spacecraft will remain between the earth and the sun in a “halo” orbit.
Problem 3/296 Sun Earth h S 3/294 Sometime after launch from the earth, a spacecraft S is in the orbital path of the earth at some dis-tance from the earth at position P. What velocity boost at P is required so that the spacecraft ar-rives at the orbit of Mars at A as shown?
Problem 3/294 3/295 A spacecraft with a mass of 800 kg is traveling in a circular orbit 6000 km above the earth. It is de-sired to change the orbit to an elliptical one with a perigee altitude of 3000 km as shown. The transi-tion is made by firing the retro-engine at A with a reverse thrust of 2000 N. Calculate the required time t for the engine to be activated.
Problem 3/295 A 6000 km 3000 km A P Sun S Mars Earth v 242 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 242 3/297 A space vehicle moving in a circular orbit of radius r1 transfers to a larger circular orbit of radius r2 by means of an elliptical path between A and B. (This transfer path is known as the Hohmann transfer ellipse.) The transfer is accomplished by a burst of speed at A and a second burst of speed at B. Write expressions for and in terms of the radii shown and the value of g of the accelera-tion due to gravity at the earth’s surface. If each is positive, how can the velocity for path 2 be less than the velocity for path 1? Compute each if r1 (6371 500) km and r2 (6371 35 800) km. Note that r2 has been chosen as the radius of a geosynchronous orbit.
Problem 3/297 A B r2 r1 1 2 v v vB vA vB vA 3/298 At the instant represented in the figure, a small experimental satellite A is ejected from the shuttle orbiter with a velocity vr 100 m/s relative to the shuttle, directed toward the center of the earth.
The shuttle is in a circular orbit of altitude h 200 km. For the resulting elliptical orbit of the satellite, determine the semimajor axis a and its orientation, the period , eccentricity e, apogee speed , perigee speed vp, rmax, and rmin. Sketch the satellite orbit.
Problem 3/298 vr h A x y va Article 3/13 Problems 243 c03.qxd 2/9/12 7:40 PM Page 243 3/14 Relative Motion Up to this point in our development of the kinetics of particle mo-tion, we have applied Newton’s second law and the equations of work-energy and impulse-momentum to problems where all measurements of motion were made with respect to a reference system which was consid-ered fixed. The nearest we can come to a “fixed” reference system is the primary inertial system or astronomical frame of reference, which is an imaginary set of axes attached to the fixed stars. All other reference sys-tems then are considered to have motion in space, including any refer-ence system attached to the moving earth.
The acceleration of points attached to the earth as measured in the primary system are quite small, however, and we normally neglect them for most earth-surface measurements. For example, the acceleration of the center of the earth in its near-circular orbit around the sun consid-ered fixed is 0.00593 m/s2 (or 0.01946 ft/sec2), and the acceleration of a point on the equator at sea level with respect to the center of the earth considered fixed is 0.0339 m/s2 (or 0.1113 ft/sec2). Clearly, these acceler-ations are small compared with g and with most other significant accel-erations in engineering work. Thus, we make only a small error when we assume that our earth-attached reference axes are equivalent to a fixed reference system.
Relative-Motion Equation We now consider a particle A of mass m, Fig. 3/25, whose motion is observed from a set of axes x-y-z which translate with respect to a fixed reference frame X-Y-Z. Thus, the x-y-z directions always remain parallel to the X-Y-Z directions. We postpone discussion of motion relative to a rotating reference system until Arts. 5/7 and 7/7. The acceleration of the origin B of x-y-z is aB. The acceleration of A as observed from or relative to x-y-z is arel aA/B A/B, and by the relative-motion principle of Art.
2/8, the absolute acceleration of A is Thus, Newton’s second law ΣF maA becomes (3/50) We can identify the force sum ΣF, as always, by a complete free-body diagram. This diagram will appear the same to an observer in x-y-z or to one in X-Y-Z as long as only the real forces acting on the particle are rep-resented. We can conclude immediately that Newton’s second law does not hold with respect to an accelerating system since ΣF marel.
D’Alembert’s Principle The particle acceleration we measure from a fixed set of axes X-Y-Z, Fig. 3/26a, is its absolute acceleration a. In this case the familiar rela-tion ΣF ma applies. When we observe the particle from a moving ΣF m(aB arel) aA aB arel r ¨ 244 Chapter 3 Kinetics of Particles ΣF Z Y X z y x m A rA rA/B = rrel rB B O Figure 3/25 y x Y X ΣF a m (a) (b) m – ma Y X ΣF a Figure 3/26 c03.qxd 2/9/12 7:40 PM Page 244 system x-y-z attached to the particle, Fig. 3/26b, the particle necessar-ily appears to be at rest or in equilibrium in x-y-z. Thus, the observer who is accelerating with x-y-z concludes that a force ma acts on the particle to balance ΣF. This point of view, which allows the treatment of a dynamics problem by the methods of statics, was an outgrowth of the work of D’Alembert contained in his Traité de Dynamique pub-lished in 1743.
This approach merely amounts to rewriting the equation of motion as ΣF ma 0, which assumes the form of a zero force summation if ma is treated as a force. This fictitious force is known as the inertia force, and the artificial state of equilibrium created is known as dynamic equilibrium. The apparent transformation of a problem in dynamics to one in statics has become known as D’Alembert’s principle.
Opinion differs concerning the original interpretation of D’Alem-bert’s principle, but the principle in the form in which it is generally known is regarded in this book as being mainly of historical interest. It evolved when understanding and experience with dynamics were ex-tremely limited and was a means of explaining dynamics in terms of the principles of statics, which were more fully understood. This excuse for using an artificial situation to describe a real one is no longer justified, as today a wealth of knowledge and experience with dynamics strongly supports the direct approach of thinking in terms of dynamics rather than statics. It is somewhat difficult to understand the long persistence in the acceptance of statics as a way of understanding dynamics, partic-ularly in view of the continued search for the understanding and de-scription of physical phenomena in their most direct form.
We cite only one simple example of the method known as D’Alem-bert’s principle. The conical pendulum of mass m, Fig. 3/27a, is swinging in a horizontal circle, with its radial line r having an angular velocity . In the straightforward application of the equation of motion ΣF man in the direction n of the acceleration, the free-body diagram in part b of the figure shows that T sin mr2. When we apply the equilibrium require-ment in the y-direction, T cos mg 0, we can find the unknowns T and . But if the reference axes are attached to the particle, the particle will appear to be in equilibrium relative to these axes. Accordingly, the in-ertia force ma must be added, which amounts to visualizing the applica-tion of mr2 in the direction opposite to the acceleration, as shown in part c of the figure. With this pseudo free-body diagram, a zero force summa-tion in the n-direction gives T sin mr2 0 which, of course, gives us the same result as before.
We may conclude that no advantage results from this alternative formulation. The authors recommend against using it since it intro-duces no simplification and adds a nonexistent force to the diagram. In the case of a particle moving in a circular path, this hypothetical inertia force is known as the centrifugal force since it is directed away from the center and is opposite to the direction of the acceleration. You are urged to recognize that there is no actual centrifugal force acting on the particle. The only actual force which may properly be called cen-trifugal is the horizontal component of the tension T exerted by the particle on the cord.
Article 3/14 Relative Motion 245 y mg (b) n T l (a) h r θ ω θ y mg (c) n T θ mr 2 ω Figure 3/27 c03.qxd 2/9/12 7:40 PM Page 245 Constant-Velocity, Nonrotating Systems In discussing particle motion relative to moving reference systems, we should note the special case where the reference system has a con-stant velocity and no rotation. If the x-y-z axes of Fig. 3/25 have a con-stant velocity, then aB 0 and the acceleration of the particle is aA arel. Therefore, we may write Eq. 3/50 as (3/51) which tells us that Newton’s second law holds for measurements made in a system moving with a constant velocity. Such a system is known as an inertial system or as a Newtonian frame of reference. Observers in the moving system and in the fixed system will also agree on the desig-nation of the resultant force acting on the particle from their identical free-body diagrams, provided they avoid the use of any so-called “inertia forces.” We will now examine the parallel question concerning the validity of the work-energy equation and the impulse-momentum equation relative to a constant-velocity, nonrotating system. Again, we take the x-y-z axes of Fig. 3/25 to be moving with a constant velocity vB relative to the fixed axes X-Y-Z. The path of the particle A relative to x-y-z is governed by rrel and is represented schematically in Fig. 3/28. The work done by ΣF rela-tive to x-y-z is dUrel . But ΣF maA marel since aB 0. Also for the same reason that at ds v dv in Art. 2/5 on curvilinear motion. Thus, we have We define the kinetic energy relative to x-y-z as Trel so that we now have (3/52) which shows that the work-energy equation holds for measurements made relative to a constant-velocity, nonrotating system.
Relative to x-y-z, the impulse on the particle during time dt is ΣF dt maA dt marel dt. But marel dt m dvrel d(mvrel) so We define the linear momentum of the particle relative to x-y-z as Grel mvrel, which gives us ΣF dt dGrel. Dividing by dt and integrating give (3/53) Thus, the impulse-momentum equations for a fixed reference system also hold for measurements made relative to a constant-velocity, nonro-tating system.
Finally, we define the relative angular momentum of the parti-cle about a point in x-y-z, such as the origin B, as the moment of the ΣF G ˙ rel and ΣF dt Grel ΣF dt d(mvrel) dUrel dTrel or Urel Trel 1 2 mvrel 2 dUrel mareldrrel mvrel dvrel d(1 2 mvrel 2) areldrrel vreldvrel ΣFdrrel r ˙B ΣF marel 246 Chapter 3 Kinetics of Particles vrel rrel aA = arel drrel z y m Path relative to x-y-z x Z Y O B X ΣF Figure 3/28 c03.qxd 2/9/12 7:40 PM Page 246 relative linear momentum. Thus, . The time derivative gives . The first term is nothing more than vrel mvrel 0, and the second term becomes rrel ΣF ΣMB, the sum of the moments about B of all forces on m. Thus, we have (3/54) which shows that the moment-angular momentum relation holds with respect to a constant-velocity, nonrotating system.
Although the work-energy and impulse-momentum equations hold relative to a system translating with a constant velocity, the individual expressions for work, kinetic energy, and momentum differ between the fixed and the moving systems. Thus, Equations 3/51 through 3/54 are formal proof of the validity of the Newtonian equations of kinetics in any constant-velocity, nonrotating system. We might have surmised these conclusions from the fact that ΣF ma depends on acceleration and not velocity. We are also ready to conclude that there is no experiment which can be conducted in and rel-ative to a constant-velocity, nonrotating system (Newtonian frame of reference) which discloses its absolute velocity. Any mechanical experi-ment will achieve the same results in any Newtonian system. (G mvA) (Grel mvrel) (T 1 2 mvA 2) (Trel 1 2 mvrel 2) (dU ΣFdrA) (dUrel ΣFdrrel) ΣMB (H ˙ B)rel (H ˙ B)rel r ˙rel Grel rrel G ˙ rel (HB)rel rrel Grel Article 3/14 Relative Motion 247 Relative motion is a critical issue during aircraft-carrier landings.
Giovanni Colla/StocktrekImages, Inc.
c03.qxd 2/9/12 7:40 PM Page 247 SAMPLE PROBLEM 3/32 A simple pendulum of mass m and length r is mounted on the flatcar, which has a constant horizontal acceleration a0 as shown. If the pendulum is released from rest relative to the flatcar at the position 0, determine the expression for the tension T in the supporting light rod for any value of . Also find T for /2 and .
Solution.
We attach our moving x-y coordinate system to the translating car with origin at O for convenience. Relative to this system, n- and t-coordinates are the natural ones to use since the motion is circular within x-y. The acceleration of m is given by the relative-acceleration equation where arel is the acceleration which would be measured by an observer riding with the car. He would measure an n-component equal to and a t-component equal to . The three components of the absolute acceleration of m are shown in the separate view.
First, we apply Newton’s second law to the t-direction and get Integrating to obtain as a function of yields We now apply Newton’s second law to the n-direction, noting that the n-component of the absolute acceleration is a0 cos .
Ans.
For /2 and , we have Ans.
Ans. T m[3g(0) a0(2 3)] 5ma0 T/2 m[3g(1) a0(2 0)] m(3g 2a0) T m[3g sin a0(2 3 cos )] m[2g sin 2a0(1 cos ) a0 cos ] [ΣFn man] T mg sin m(r ˙2 a0 cos ) r ˙2 ˙2 2 1 r [g sin a0(1 cos )] [ ˙ d ˙ ¨ d] ˙ 0 ˙ d ˙ 0 1 r ( g cos a0 sin ) d ˙ r ¨ g cos a0 sin mg cos m(r ¨ a0 sin ) [ΣFt mat] r ¨ r ˙2 a a0 arel O r m a0 θ y O x a0 t T mg Free-body diagram Acceleration components θ θ n r 2 · θ r ·· θ Helpful Hints We choose the t-direction first since the n-direction equation, which con-tains the unknown T, will involve , which, in turn, is obtained from an integration of .
¨ ˙2 Be sure to recognize that may be obtained from v dv at ds by dividing by r2.
¨ d ˙ d ˙ 248 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 248 SAMPLE PROBLEM 3/33 The flatcar moves with a constant speed v0 and carries a winch which pro-duces a constant tension P in the cable attached to the small carriage. The car-riage has a mass m and rolls freely on the horizontal surface starting from rest relative to the flatcar at x 0, at which instant X x0 b. Apply the work-energy equation to the carriage, first, as an observer moving with the frame of reference of the car and, second, as an observer on the ground. Show the compat-ibility of the two expressions.
Solution.
To the observer on the flatcar, the work done by P is The change in kinetic energy relative to the car is The work-energy equation for the moving observer becomes To the observer on the ground, the work done by P is The change in kinetic energy for the ground measurement is The work-energy equation for the fixed observer gives To reconcile this equation with that for the moving observer, we can make the following substitutions: Thus, and The work-energy equation for the fixed observer now becomes which is merely Px , as concluded by the moving observer. We see, there-fore, that the difference between the two work-energy expressions is U Urel T Trel mv0x ˙ 1 2 mx ˙2 Px mv0x ˙ 1 2 mx ˙2 mv0x ˙ X ˙ 2 v0 2 (v0 2 x ˙2 2v0x ˙ v0 2) x ˙2 2v0x ˙ Px mx ¨v0t Px mv0x ˙ P(X b) Px P(x0 b) Px mx ¨(x0 b) X x0 x, X ˙ v0 x ˙, X ¨ x ¨ P(X b) 1 2 m(X ˙2 v0 2) [U T] T 1 2 m(X ˙2 v0 2) U X b P dX P(X b) Px 1 2 mx ˙2 [Urel Trel] Trel 1 2 m(x ˙2 0) Urel x 0 P dx Px for constant P v0 m P x0 x X x0 x x = 0 X b Helpful Hints The only coordinate which the mov-ing observer can measure is x.
Article 3/14 Relative Motion 249 To the ground observer, the initial velocity of the carriage is v0, so its initial kinetic energy is .
1 2 mv0 2 The symbol t stands for the time of motion from x 0 to x x. The displacement x0 b of the carriage is its velocity v0 times the time t or x0 b v0t. Also, since the constant acceleration times the time equals the velocity change, .
x ¨t x ˙ c03.qxd 2/9/12 7:40 PM Page 249 3/301 The cart with attached x-y axes moves with an ab-solute speed v 2 m/s to the right. Simultaneously, the light arm of length l 0.5 m rotates about point B of the cart with angular velocity 2 rad/s. The mass of the sphere is m 3 kg. Deter-mine the following quantities for the sphere when 0: G, Grel, T, Trel, HO, where the sub-script “rel” indicates measurement relative to the x-y axes. Point O is an inertially fixed point coinci-dent with point B at the instant under consideration.
Problem 3/301 3/302 The aircraft carrier is moving at a constant speed and launches a jet plane with a mass of 3 Mg in a distance of 75 m along the deck by means of a steam-driven catapult. If the plane leaves the deck with a velocity of 240 km/h relative to the carrier and if the jet thrust is constant at 22 kN during takeoff, compute the constant force P exerted by the catapult on the airplane during the 75-m travel of the launch carriage.
Problem 3/302 75 m v θ O, B l m x y (HB)rel ˙ PROBLEMS Introductory Problems 3/299 If the spring of constant k is compressed a distance as indicated, calculate the acceleration arel of the block of mass m1 relative to the frame of mass m2 upon release of the spring. The system is initially stationary.
Problem 3/299 3/300 The flatbed truck is traveling at the constant speed of 60 km/h up the 15-percent grade when the 100-kg crate which it carries is given a shove which im-parts to it an initial relative velocity 3 m/s toward the rear of the truck. If the crate slides a distance x 2 m measured on the truck bed before coming to rest on the bed, compute the coefficient of kinetic friction k between the crate and the truck bed.
Problem 3/300 60 km/h 15 100 x x ˙ δ k m2 m1 250 Chapter 3 Kinetics of Particles c03.qxd 2/10/12 12:16 AM Page 250 3/303 The 4000-lb van is driven from position A to posi-tion B on the barge, which is towed at a constant speed v0 10 mi/hr. The van starts from rest rela-tive to the barge at A, accelerates to v 15 mi/hr relative to the barge over a distance of 80 ft, and then stops with a deceleration of the same magni-tude. Determine the magnitude of the net force F between the tires of the van and the barge during this maneuver.
Problem 3/303 Representative Problems 3/304 The launch catapult of the aircraft carrier gives the 7-Mg jet airplane a constant acceleration and launches the airplane in a distance of 100 m mea-sured along the angled takeoff ramp. The carrier is moving at a steady speed vC 16 m/s. If an ab-solute aircraft speed of 90 m/s is desired for takeoff, determine the net force F supplied by the catapult and the aircraft engines.
Problem 3/304 3/305 The coefficients of friction between the flatbed of the truck and crate are s 0.80 and k 0.70.
The coefficient of kinetic friction between the truck tires and the road surface is 0.90. If the truck stops from an initial speed of 15 m/s with maximum braking (wheels skidding), determine where on the bed the crate finally comes to rest or the velocity vrel relative to the truck with which the crate strikes the wall at the forward edge of the bed.
v0 = 10 mi/hr v = 15 mi/hr 80′ A B 80′ vC 15° Problem 3/305 3/306 A boy of mass m is standing initially at rest relative to the moving walkway, which has a constant hori-zontal speed u. He decides to accelerate his progress and starts to walk from point A with a steadily in-creasing speed and reaches point B with a speed v relative to the walkway. During his accelera-tion he generates an average horizontal force F between his shoes and the walkway. Write the work-energy equations for his absolute and relative motions and explain the meaning of the term muv.
Problem 3/306 3/307 The block of mass m is attached to the frame by the spring of stiffness k and moves horizontally with negligible friction within the frame. The frame and block are initially at rest with x x0, the uncom-pressed length of the spring. If the frame is given a constant acceleration a0, determine the maximum velocity (vrel)max of the block relative to the frame.
Problem 3/307 x k m a0 x ˙max x ˙ 3.2 m s x xA A B u Article 3/14 Problems 251 c03.qxd 2/10/12 12:16 AM Page 251 3/310 Consider the system of Prob. 3/309 where the mass of the ball is m 10 kg and the length of the light rod is l 0.8 m. The ball–rod assembly is free to rotate about a vertical axis through O. The carriage, rod, and ball are initially at rest with 0 when the car-riage is given a constant acceleration aO 3 m/s2.
Write an expression for the tension T in the rod as a function of and calculate T for the position /2.
3/311 A simple pendulum is placed on an elevator, which accelerates upward as shown. If the pendulum is displaced an amount 0 and released from rest rela-tive to the elevator, find the tension T0 in the sup-porting light rod when 0. Evaluate your result for 0 /2.
Problem 3/311 3/312 A boy of mass m is standing initially at rest relative to the moving walkway inclined at the angle and moving with a constant speed u. He decides to accel-erate his progress and starts to walk from point A with a steadily increasing speed and reaches point B with a speed vr relative to the walkway. During his acceleration he generates a constant average force F tangent to the walkway between his shoes and the walkway surface. Write the work-energy equations for the motion between A and B for his absolute mo-tion and his relative motion and explain the mean-ing of the term muvr. If the boy weighs 150 lb and if u 2 ft/sec, s 30 ft, and 10, calculate the power Prel developed by the boy as he reaches the speed of 2.5 ft/sec relative to the walkway.
Problem 3/312 s x B A x0 u u θ O θ l m a0 3/308 The slider A has a mass of 2 kg and moves with negligible friction in the 30 slot in the vertical slid-ing plate. What horizontal acceleration a0 should be given to the plate so that the absolute accelera-tion of the slider will be vertically down? What is the value of the corresponding force R exerted on the slider by the slot?
Problem 3/308 3/309 The ball A of mass 10 kg is attached to the light rod of length l 0.8 m. The mass of the carriage alone is 250 kg, and it moves with an acceleration aO as shown. If 3 rad/s when 90, find the kinetic energy T of the system if the carriage has a velocity of 0.8 m/s (a) in the direction of aO and (b) in the di-rection opposite to aO. Treat the ball as a particle.
Problem 3/309 A O aO θ l ˙ a0 A 30° 252 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 252 3/313 A ball is released from rest relative to the elevator at a distance h1 above the floor. The speed of the el-evator at the time of ball release is v0. Determine the bounce height h2 of the ball (a) if v0 is constant and (b) if an upward elevator acceleration a g/4 begins at the instant the ball is released. The coeffi-cient of restitution for the impact is e.
Problem 3/313 3/314 The small slider A moves with negligible friction down the tapered block, which moves to the right with constant speed v v0. Use the principle of work-energy to determine the magnitude vA of the absolute velocity of the slider as it passes point C if it is released at point B with no velocity relative to the block. Apply the equation, both as an observer fixed to the block and as an observer fixed to the ground, and reconcile the two relations.
h2 h1 v0 a = g – 4 Problem 3/314 3/315 When a particle is dropped from rest relative to the surface of the earth at a latitude , the initial ap-parent acceleration is the relative acceleration due to gravity grel. The absolute acceleration due to gravity g is directed toward the center of the earth.
Derive an expression for grel in terms of g, R, , and , where R is the radius of the earth treated as a sphere and is the constant angular velocity of the earth about the polar axis considered fixed. (Al-though axes x-y-z are attached to the earth and hence rotate, we may use Eq. 3/50 as long as the particle has no velocity relative to x-y-z). (Hint: Use the first two terms of the binomial expansion for the approximation.) Problem 3/315 N B θ y x ω O R aB g grel γ l B A C θ v Article 3/14 Problems 253 c03.qxd 2/9/12 7:40 PM Page 253 Problem 3/316 θ O O S S r r r x x θ t, θ, t r y (a) (b) y S S t t Elliptical Orbit Circular Orbit P F F θ m m x x 3/316 The figure represents the space shuttle S, which is (a) in a circular orbit about the earth and (b) in an elliptical orbit where P is its perigee position. The exploded views on the right represent the cabin space with its x-axis oriented in the direction of the orbit. The astronauts conduct an experiment by ap-plying a known force F in the x-direction to a small mass m. Explain why F does or does not hold in each case, where x is measured within the space-craft. Assume that the shuttle is between perigee and apogee in the elliptical orbit so that the orbital speed is changing with time. Note that the t- and x-axes are tangent to the path, and the -axis is normal to the radial r-direction.
mx ¨ 254 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 254 3/15 CHAPTER REVIEW In Chapter 3 we have developed the three basic methods of solution to problems in particle kinetics. This experience is central to the study of dy-namics and lays the foundation for the subsequent study of rigid-body and nonrigid-body dynamics. These three methods are summarized as follows: 1. Direct Application of Newton’s Second Law First, we applied Newton’s second law ΣF ma to determine the instantaneous relation between forces and the acceleration they pro-duce. With the background of Chapter 2 for identifying the kind of mo-tion and with the aid of our familiar free-body diagram to be certain that all forces are accounted for, we were able to solve a large variety of problems using x-y, n-t, and r- coordinates for plane-motion problems and x-y-z, r--z, and R-- coordinates for space problems.
2. Work-Energy Equations Next, we integrated the basic equation of motion ΣF ma with re-spect to displacement and derived the scalar equations for work and en-ergy. These equations enable us to relate the initial and final velocities to the work done during an interval by forces external to our defined system.
We expanded this approach to include potential energy, both elastic and gravitational. With these tools we discovered that the energy approach is especially valuable for conservative systems, that is, systems wherein the loss of energy due to friction or other forms of dissipation is negligible.
3. Impulse-Momentum Equations Finally, we rewrote Newton’s second law in the form of force equals time rate of change of linear momentum and moment equals time rate of change of angular momentum. Then we integrated these relations with respect to time and derived the impulse and momentum equations.
These equations were then applied to motion intervals where the forces were functions of time. We also investigated the interactions between particles under conditions where the linear momentum is conserved and where the angular momentum is conserved.
In the final section of Chapter 3, we employed these three basic methods in specific application areas as follows: 1. We noted that the impulse-momentum method is convenient in de-veloping the relations governing particle impacts.
2. We observed that the direct application of Newton’s second law en-ables us to determine the trajectory properties of a particle under central-force attraction.
3. Finally, we saw that all three basic methods may be applied to parti-cle motion relative to a translating frame of reference.
Successful solution of problems in particle kinetics depends on knowledge of the prerequisite particle kinematics. Furthermore, the principles of particle kinetics are required to analyze particle systems and rigid bodies, which are covered in the remainder of Dynamics.
Article 3/15 Chapter Review 255 c03.qxd 2/9/12 7:40 PM Page 255 3/320 Collar A is free to slide with negligible friction on the circular guide mounted in the vertical frame.
Determine the angle assumed by the collar if the frame is given a constant horizontal acceleration a to the right.
Problem 3/320 3/321 The position of the small 0.5-kg blocks in the smooth radial slots in the disk which rotates about a vertical axis at O is used to activate a speed-control mechanism. If each block moves from r 150 mm to r 175 mm while the speed of the disk changes slowly from 300 rev/min to 400 rev/min, design the spring by calculating the spring con-stant k of each spring. The springs are attached to the inner ends of the slots and to the blocks.
Problem 3/321 k k r r = θ ω O A a r θ REVIEW PROBLEMS 3/317 The 4-kg slider is released from rest in position A and slides down the vertical-plane guide. If the maximum compression of the spring is observed to be 40 mm, determine the work Uƒ done by friction.
Problem 3/317 3/318 The crate is at rest at point A when it is nudged down the incline. If the coefficient of kinetic fric-tion between the crate and the incline is 0.30 from A to B and 0.22 from B to C, determine its speeds at points B and C.
Problem 3/318 3/319 An 88-kg sprinter starts from rest and reaches his maximum speed of 11 m/s in 2.5 s with uniform ac-celeration. What is his power output when his speed is 5 m/s? Comment on the conditions stated in this problem.
A C B 7 m 7 m s = 0.28 μ k = 0.22 μ s = 0.40 μ k = 0.30 μ 20° 10° 0.6 m 4 kg k = 20 kN/m A 256 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 256 3/322 The simple 2-kg pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at B and continues in the smaller arc in the vertical plane. Calculate the magnitude of the force R supported by the pin at B when the pendu-lum passes the position 30 .
Problem 3/322 3/323 For the elliptical orbit of a spacecraft around the earth, determine the speed vA at point A which re-sults in a perigee altitude at B of 200 km. What is the eccentricity e of the orbit?
Problem 3/323 A B vA 600 km 200 km θ 800 mm 400 mm 2 kg A B 90° 3/324 The spring of stiffness k is compressed and sud-denly released, sending the particle of mass m slid-ing along the track. Determine the minimum spring compression for which the particle will not lose contact with the loop-the-loop track. The sliding surface is smooth except for the rough por-tion of length s equal to R, where the coefficient of kinetic friction is µk.
Problem 3/324 3/325 The last two appearances of Comet Halley were in 1910 and 1986. The distance of its closest approach to the sun averages about one-half of the distance between the earth and the sun. Determine its max-imum distance from the sun. Neglect the gravita-tional effects of the planets.
3/326 A small sphere of mass m is connected by a string to a swivel at O and moves in a circle of radius r on the smooth plane inclined at an angle with the horizontal. If the sphere has a velocity u at the top position A, determine the tension in the string as the sphere passes the 90 position B and the bot-tom position C.
Problem 3/326 θ r m u A C O B m k δ R s = R B A Rough area k μ Article 3/15 Review Problems 257 c03.qxd 2/9/12 7:40 PM Page 257 3/329 A 3600-lb car is traveling with a speed of 60 mi/hr as it approaches point A. Beginning at A, it decel-erates uniformly to a speed of 25 mi/hr as it passes point C of the horizontal and unbanked ramp.
Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Problem 3/329 3/330 After release from rest at B, the 2-lb cylindrical plug A slides down the smooth path and embeds it-self in the 4-lb block C. Determine the velocity v of the block and embedded plug immediately after engagement and find the maximum deflection x of the spring. Neglect any friction under block C.
What fraction n of the original energy of the sys-tem is lost?
Problem 3/330 B A C k = 80 lb/ft 6′ 200′ A B C 300′ 3/327 The quarter-circular hollow tube of circular cross section starts from rest at time t 0 and rotates about point O in a horizontal plane with a con-stant counterclockwise angular acceleration 2 rad/s2. At what time t will the 0.5-kg particle P slip relative to the tube? The coefficient of static fric-tion between the particle and the tube is µs 0.80.
Problem 3/327 3/328 A person rolls a small ball with speed u along the floor from point A. If x 3R, determine the re-quired speed u so that the ball returns to A after rolling on the circular surface in the vertical plane from B to C and becoming a projectile at C. What is the minimum value of x for which the game could be played if contact must be maintained to point C? Neglect friction.
Problem 3/328 R C u x A B 0.75 m O P 60° ·· θ ¨ 258 Chapter 3 Kinetics of Particles c03.qxd 2/20/12 3:20 PM Page 258 3/331 The pickup truck is used to hoist the 40-kg bale of hay as shown. If the truck has reached a constant velocity v 5 m/s when x 12 m, compute the corresponding tension T in the rope.
Problem 3/331 3/332 A slider C has a speed of 3 m/s as it passes point A of the guide, which lies in a horizontal plane. The coefficient of kinetic friction between the slider and the guide is Compute the tangential deceleration at of the slider just after it passes point A if (a) the slider hole and guide cross section are both circular and (b) the slider hole and guide cross section are both square. In case (b), the sides of the square are vertical and horizontal. Assume a slight clearance between the slider and the guide.
Problem 3/332 0.6 m A C (a) (b) k 0.60.
16 m x v 3/333 The frame of mass 6m is initially at rest. A particle of mass m is attached to the end of the light rod, which pivots freely at A. If the rod is released from rest in the horizontal position shown, determine the velocity vrel of the particle with respect to the frame when the rod is vertical.
Problem 3/333 3/334 The object of the pinball-type game is to project the particle so that it enters the hole at E. When the spring is compressed and suddenly released, the particle is projected along the track, which is smooth except for the rough portion between points B and C, where the coefficient of kinetic friction is The particle becomes a projectile at point D. Determine the correct spring compression so that the particle enters the hole at E. State any necessary conditions relating the lengths d and .
k.
6m m l A Article 3/15 Review Problems 259 ρ d E Hole Rough area D C B m A k ρ ρ δ k μ ρ ρ Problem 3/334 c03.qxd 2/9/12 7:40 PM Page 259 3/337 The 3-kg block A is released from rest in the position shown and subsequently strikes the 1-kg cart B. If the coefficient of restitution for the colli-sion is e 0.7, determine the maximum displace-ment s of cart B beyond point C. Neglect friction.
Problem 3/337 3/338 One of the functions of the space shuttle is to re-lease communications satellites at low altitude. A booster rocket is fired at B, placing the satellite in an elliptical transfer orbit, the apogee of which is at the altitude necessary for a geosynchronous orbit.
(A geosynchronous orbit is an equatorial-plane cir-cular orbit whose period is equal to the absolute ro-tational period of the earth. A satellite in such an orbit appears to remain stationary to an earth-fixed observer.) A second booster rocket is then fired at C, and the final circular orbit is achieved. On one of the early space-shuttle missions, a 1500-lb satellite was released from the shuttle at B, where h1 170 miles. The booster rocket was to fire for t 90 sec-onds, forming a transfer orbit with h2 22,300 miles. The rocket failed during its burn. Radar ob-servations determined the apogee altitude of the transfer orbit to be only 700 miles. Determine the actual time which the rocket motor operated be-fore failure. Assume negligible mass change during the booster rocket firing.
Problem 3/338 t 0.6 m 1.8 m 3 kg 1 kg 60° 30° A B C s 60 3/335 The 2-lb piece of putty is dropped 6 ft onto the 18-lb block initially at rest on the two springs, each with a stiffness k 3 lb/in. Calculate the addi-tional deflection of the springs due to the impact of the putty, which adheres to the block upon contact.
Problem 3/335 3/336 A baseball pitcher delivers a fastball with a near-horizontal velocity of 90 mi/hr. The batter hits a home run over the center-field fence. The 5-oz ball travels a horizontal distance of 350 ft, with an initial velocity in the 45 direction shown. Deter-mine the magnitude Fav of the average force ex-erted by the bat on the ball during the 0.005 seconds of contact between the bat and the ball.
Neglect air resistance during the flight of the ball.
Problem 3/336 45° 90 mi/hr v x Bat y 6′ 18 lb k = 3 lb/in.
2 lb k k δ 260 Chapter 3 Kinetics of Particles C B h1 = 170 mi h2 = 22,300 mi c03.qxd 2/9/12 7:40 PM Page 260 3/339 The system is released from rest while in the posi-tion shown. If m1 0.5 kg, m2 4 kg, d 0.5 m, and , determine the speeds of both bodies just after the block leaves the incline (before strik-ing the horizontal surface). Neglect all friction.
Problem 3/339 3/340 The retarding forces which act on the race car are the drag force FD and a nonaerodynamic force FR.
The drag force is where CD is the drag coefficient, is the air density, v is the car speed, and S 30 ft2 is the projected frontal area of the car. The nonaerodynamic force FR is constant at 200 lb. With its sheet metal in good condition, the race car has a drag coefficient CD 0.3 and it has a corresponding top speed v 200 mi/hr. After a minor collision, the damaged front-end sheet metal causes the drag coefficient to be CD 0.4. What is the corresponding top speed of the race car?
Problem 3/340 v FD CD(1 2 v2)S, θ m1 d m2 20 3/341 Extensive wind-tunnel and coast-down studies of a 2000-lb automobile reveal the aerodynamic drag force FD and the total nonaerodynamic rolling re-sistance force FR to vary with speed as shown in the plot. Determine (a) the power P required for steady speeds of 30 mi/hr and 60 mi/hr and (b) the time t and the distance s required for the car to coast down to a speed of 5 mi/hr from an initial speed of 60 mi/hr. Assume a straight, level road and no wind.
Problem 3/341 3/342 The hollow tube rotates with a constant angular velocity about a horizontal axis through end O.
At time t 0 the tube passes the vertical position 0, at which instant the small ball of mass m is released with r essentially zero. Determine r as a function of .
Problem 3/342 θ m r O 0 ω 0 60 40 20 00 20 40 60 80 Force, lb Speed v, mi/hr FR (linear) FD (parabolic) Article 3/15 Review Problems 261 c03.qxd 2/9/12 7:40 PM Page 261 3/345 The system of Prob. 3/130 is repeated here. The two 0.2-kg sliders are connected by a light rigid bar of length L 0.5 m. If the system is released from rest in the position shown with the spring un-stretched, plot the speeds of A and B as functions of the displacement of B (with zero being the ini-tial position). The 0.14-MPa air pressure acting on one 500-mm2 side of slider A is constant. The mo-tion occurs in a vertical plane. Neglect friction.
State the maximum values of vA and vB and the position of B at which each occurs.
Problem 3/345 3/346 The square plate is at rest in position A at time t 0 and subsequently translates in a vertical circle according to kt2, where k 1 rad/s2, the dis-placement is in radians, and time t is in seconds.
A small 0.4-kg instrument P is temporarily fixed to the plate with adhesive. Plot the required shear force F vs. time t for . If the adhesive fails when the shear force F reaches 30 N, deter-mine the time t and angular position when fail-ure occurs.
Problem 3/346 P A r = 1.5 m O θ 0 t 5 s B A 60° 30° k = 1.2 kN/m L Computer-Oriented Problems 3/343 The bowl-shaped device from Prob. 3/70 rotates about a vertical axis with a constant angular veloc-ity 6 rad/s. The value of r is 0.2 m. Determine the range of the position angle for which a sta-tionary value is possible if the coefficient of static friction between the particle and the surface is 0.20.
Problem 3/343 3/344 If the vertical frame starts from rest with a con-stant acceleration a and the smooth sliding collar A is initially at rest in the bottom position 0, plot as a function of θ and find the maximum po-sition angle max reached by the collar. Use the val-ues a g/2 and r 0.3 m.
Problem 3/344 A a r θ ˙ ω r m θ r s = 0.20 μ s 262 Chapter 3 Kinetics of Particles c03.qxd 2/9/12 7:40 PM Page 262 3/347 The system of Prob. 3/171 is repeated here. The system is released from rest with . Deter-mine and plot as a function of . Determine the maximum magnitude of in the ensuing motion and the value of at which it occurs. Also find the minimum value of . Use the values m1 1 kg, m2 1.25 kg, and b 0.4 m. Neglect friction and the mass of bar OB, and treat the body B as a particle.
Problem 3/347 3/348 The 26-in. drum rotates about a horizontal axis with a constant angular velocity 7.5 rad/sec.
The small block A has no motion relative to the drum surface as it passes the bottom position 0.
Determine the coefficient of static friction which would result in block slippage at an angular position ; plot your expression for Determine the minimum required coefficient value min which would allow the block to remain fixed relative to the drum throughout a full revolution.
For a friction coefficient slightly less than min, at what angular position would slippage occur?
Problem 3/348 θ O A r = 13″ Ω = 7.5 rad/sec 0 180.
s O C m2 m1 B A 2b b 2b θ ˙ ˙ 90 3/349 A 20-lb sphere A is held at the 60 angle shown and released. It strikes the 10-lb sphere B. The coeffi-cient of restitution for this collision is e 0.75.
Sphere B is attached to the end of a light rod that pivots freely about point O. If the spring of con-stant k 100 lb/ft is initially unstretched, deter-mine the maximum rotation angle of the light rod after impact.
Problem 3/349 3/350 A particle of mass m is introduced with zero veloc-ity at r 0 when 0. It slides outward through the smooth hollow tube, which is driven at the con-stant angular velocity 0 about a horizontal axis through point O. If the length l of the tube is 1 m and 0 0.5 rad/s, determine the time t after re-lease and the angular displacement for which the particle exits the tube.
Problem 3/350 θ m r O 0 ω 18″ 24″ 60° θ A B O 24″ Article 3/15 Review Problems 263 c03.qxd 2/9/12 7:40 PM Page 263 3/353 The simple pendulum of length l 0.5 m has an angular velocity 0.2 rad/s at time t 0 when 0. Derive an integral expression for the time t required to reach an arbitrary angle . Plot t vs. for and state the value of t for Problem 3/353 3/354 A 1.8-lb particle P is given an initial velocity v0 1 ft/sec at the position 0 and subsequently slides along the circular path of radius r 1.5 ft. A drag force of magnitude kv acts in the direction opposite to the velocity. If the drag parameter k 0.2 lb-sec/ft, determine and plot the particle speed v and the normal force N exerted on the particle by the surface as functions of over the range . Determine the maximum values of v and N and the values of at which these maxima occur. Neglect friction between the particle and the circular surface.
Problem 3/354 θ r P O 0 90 θ l θ0 = 0.2 rad/s · θ 2.
0 2 ˙ 0 3/351 The tennis player practices by hitting the ball against the wall at A. The ball bounces off the court surface at B and then up to its maximum height at C. For the conditions shown in the figure, plot the location of point C for values of the coeffi-cient of restitution in the range 0.5 e 0.9. (The value of e is common to both A and B.) For what value of e is x 0 at point C, and what is the corre-sponding value of y?
Problem 3/351 3/352 The system of Prob. 3/154 is repeated here. If the 0.75-kg particle is released from rest when in the position 0, where the spring is unstretched, de-termine and plot its speed v as a function of over the range where max is the value of at which the system momentarily comes to rest.
The value of the spring modulus k is 100 N/m, and friction can be neglected. State the maximum speed and the angle at which it occurs.
Problem 3/352 O 0.75 kg A B k 0.6 m 0.6 m θ 0 max, 30′ B C y x 5° 3′ 80 ft/sec A 264 Chapter 3 Kinetics of Particles c03.qxd 2/20/12 3:20 PM Page 264 c03.qxd 2/9/12 7:40 PM Page 265 The forces of interaction between the rotating blades of this Harrier jumpjet engine and the air which passes over them is a subject which is introduced in this chapter.
© Frits van Gansewinkel/Alamy c04.qxd 2/8/12 7:33 PM Page 266 267 4/1 Introduction In the previous two chapters, we have applied the principles of dy-namics to the motion of a particle. Although we focused primarily on the kinetics of a single particle in Chapter 3, we mentioned the motion of two particles, considered together as a system, when we discussed work-energy and impulse-momentum.
Our next major step in the development of dynamics is to extend these principles, which we applied to a single particle, to describe the motion of a general system of particles. This extension will unify the re-maining topics of dynamics and enable us to treat the motion of both rigid bodies and nonrigid systems.
Recall that a rigid body is a solid system of particles wherein the distances between particles remain essentially unchanged. The overall motions found with machines, land and air vehicles, rockets and space-craft, and many moving structures provide examples of rigid-body prob-lems. On the other hand, we may need to study the time-dependent changes in the shape of a nonrigid, but solid, body due to elastic or in-elastic deformations. Another example of a nonrigid body is a defined mass of liquid or gaseous particles flowing at a specified rate. Examples are the air and fuel flowing through the turbine of an aircraft engine, the burned gases issuing from the nozzle of a rocket motor, or the water passing through a rotary pump.
4/1 Introduction 4/2 Generalized Newton’s Second Law 4/3 Work-Energy 4/4 Impulse-Momentum 4/5 Conservation of Energy and Momentum 4/6 Steady Mass Flow 4/7 Variable Mass 4/8 Chapter Review CHAPTER OUTLINE 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:33 PM Page 267 268 Chapter 4 Kinetics of Systems of Particles Although we can extend the equations for single-particle motion to a general system of particles without much difficulty, it is difficult to un-derstand the generality and significance of these extended principles without considerable problem experience. For this reason, you should frequently review the general results obtained in the following articles during the remainder of your study of dynamics. In this way, you will understand how these broader principles unify dynamics.
mi F1 f1 F2 f2 F3 f3 mi ρi ri O G System boundary r – Figure 4/1 4/2 Generalized Newton’s Second Law We now extend Newton’s second law of motion to cover a general mass system which we model by considering n mass particles bounded by a closed surface in space, Fig. 4/1. This bounding envelope, for exam-ple, may be the exterior surface of a given rigid body, the bounding sur-face of an arbitrary portion of the body, the exterior surface of a rocket containing both rigid and flowing particles, or a particular volume of fluid particles. In each case, the system to be considered is the mass within the envelope, and that mass must be clearly defined and isolated.
Figure 4/1 shows a representative particle of mass mi of the system isolated with forces F1, F2, F3, . . . acting on mi from sources external to the envelope, and forces f1, f2, f3, . . . acting on mi from sources internal to the system boundary. The external forces are due to contact with ex-ternal bodies or to external gravitational, electric, or magnetic effects.
The internal forces are forces of reaction with other mass particles within the boundary. The particle of mass mi is located by its position vector ri measured from the nonaccelerating origin O of a Newtonian set of reference axes. The center of mass G of the isolated system of particles is located by the position vector which, from the definition of the mass center as covered in statics, is given by where the total system mass is m Σmi. The summation sign Σ repre-sents the summation over all n particles.
Newton’s second law, Eq. 3/3, when applied to mi gives where is the acceleration of mi. A similar equation may be written for each of the particles of the system. If these equations written for all par-ticles of the system are added together, the result is The term ΣF then becomes the vector sum of all forces acting on all particles of the isolated system from sources external to the system, and ΣF Σf Σmir ¨i r ¨i F1 F2 F3 f1 f2 f3 mir ¨i Σn i1 mr Σmiri r KEY CONCEPTS It was shown in Art. 3/14 that any nonrotating and nonaccelerating set of axes constitutes a Newtonian reference system in which the principles of Newtonian mechanics are valid.
c04.qxd 2/8/12 7:33 PM Page 268 Article 4/3 Work-Energy 269 If m is a function of time, a more complex situation develops; this situation is discussed in Art. 4/7 on variable mass.
Σf becomes the vector sum of all forces on all particles produced by the internal actions and reactions between particles. This last sum is identi-cally zero since all internal forces occur in pairs of equal and opposite actions and reactions. By differentiating the equation defining twice with time, we have where m has a zero time derivative as long as mass is not entering or leaving the system. Substitution into the summation of the equations of motion gives or (4/1) where is the acceleration of the center of mass of the system.
Equation 4/1 is the generalized Newton’s second law of motion for a mass system and is called the equation of motion of m. The equation states that the resultant of the external forces on any system of masses equals the total mass of the system times the acceleration of the center of mass. This law expresses the so-called principle of motion of the mass center.
Observe that is the acceleration of the mathematical point which represents instantaneously the position of the mass center for the given n particles. For a nonrigid body, this acceleration need not represent the acceleration of any particular particle. Note also that Eq. 4/1 holds for each instant of time and is therefore an instantaneous relationship.
Equation 4/1 for the mass system had to be proved, as it cannot be in-ferred directly from Eq. 3/3 for a single particle.
Equation 4/1 may be expressed in component form using x-y-z coor-dinates or whatever coordinate system is most convenient for the prob-lem at hand. Thus, (4/1a) Although Eq. 4/1, as a vector equation, requires that the accelera-tion vector have the same direction as the resultant external force ΣF, it does not follow that ΣF necessarily passes through G. In general, in fact, ΣF does not pass through G, as will be shown later.
4/3 Work-Energy In Art. 3/6 we developed the work-energy relation for a single parti-cle, and we noted that it applies to a system of two joined particles.
Now consider the general system of Fig. 4/1, where the work-energy re-lation for the representative particle of mass mi is (U1-2)i Ti. Here (U1-2)i is the work done on mi during an interval of motion by all forces F1 F2 F3 applied from sources external to the system and by all forces f1 f2 f3 applied from sources internal to the system.
The kinetic energy of mi is Ti , where vi is the magnitude of the particle velocity vi .
r ˙i 1 2 mivi 2 a ΣFx max ΣFy may ΣFz maz a r ¨ a F ma ΣF mr ¨ r ¨i Σmi mr ¨ r c04.qxd 2/8/12 7:33 PM Page 269 Work-Energy Relation For the entire system, the sum of the work-energy equations writ-ten for all particles is Σ(U1-2)i ΣTi, which may be represented by the same expressions as Eqs. 3/15 and 3/15a of Art. 3/6, namely, (4/2) where U1-2 Σ(U1-2)i, the work done by all forces, external and internal, on all particles, and T is the change in the total kinetic energy T ΣTi of the system.
For a rigid body or a system of rigid bodies joined by ideal friction-less connections, no net work is done by the internal interacting forces or moments in the connections. We see that the work done by all pairs of internal forces, labeled here as fi and fi, at a typical connection, Fig.
4/2, in the system is zero since their points of application have identical displacement components while the forces are equal but opposite. For this situation U1-2 becomes the work done on the system by the external forces only.
For a nonrigid mechanical system which includes elastic members capable of storing energy, a part of the work done by the external forces goes into changing the internal elastic potential energy Ve. Also, if the work done by the gravity forces is excluded from the work term and is accounted for instead by the changes in gravitational potential energy Vg, then we may equate the work done on the system during an in-terval of motion to the change E in the total mechanical energy of the system. Thus, E or (4/3) or (4/3a) which are the same as Eqs. 3/21 and 3/21a. Here, as in Chapter 3, V Vg Ve represents the total potential energy.
Kinetic Energy Expression We now examine the expression T for the kinetic energy of the mass system in more detail. By our principle of relative motion discussed in Art. 2/8, we may write the velocity of the representative particle as where is the velocity of the mass center G and is the velocity of mi with respect to a translating reference frame moving with the mass ˙i v vi v ˙i Σ 1 2 mivi 2 T1 V1 U 1-2 T2 V2 U 1-2 T V U 1-2 U 1-2 U1-2 T or T1 U1-2 T2 270 Chapter 4 Kinetics of Systems of Particles fi –fi Figure 4/2 c04.qxd 2/8/12 7:33 PM Page 270 center G. We recall the identity and write the kinetic energy of the system as Because i is measured from the mass center, Σmii 0 and the third term is Σ(mii) 0. Also .
Therefore, the total kinetic energy becomes (4/4) This equation expresses the fact that the total kinetic energy of a mass system equals the kinetic energy of mass-center translation of the sys-tem as a whole plus the kinetic energy due to motion of all particles rel-ative to the mass center.
4/4 Impulse-Momentum We now develop the concepts of momentum and impulse as applied to a system of particles.
Linear Momentum From our definition in Art. 3/8, the linear momentum of the repre-sentative particle of the system depicted in Fig. 4/1 is Gi mivi where the velocity of mi is vi .
The linear momentum of the system is defined as the vector sum of the linear momenta of all of its particles, or G Σmivi. By substituting the relative-velocity relation vi and noting again that Σmii m 0, we obtain or (4/5) Thus, the linear momentum of any system of constant mass is the prod-uct of the mass and the velocity of its center of mass.
The time derivative of G is , which by Eq. 4/1 is the resul-tant external force acting on the system. Thus, we have (4/6) ΣF G ˙ ma mv ˙ G mv v Σmi d dt (0) G Σmi(v ˙i) Σmiv d dt Σmi i v ˙i r ˙i T 1 2 mv 2 Σ 1 2 mi ˙i2 1 2 mv 2 1 2 v 2 Σmi Σ 1 2 miv 2 v Σmi ˙i v d dt Σ 1 2 miv 2 Σ 1 2 mi ˙i2 Σmiv ˙i T Σ 1 2 mivi vi Σ 1 2 mi(v ˙i) (v ˙i) vi 2 vi vi Article 4/4 Impulse-Momentum 271 c04.qxd 2/8/12 7:33 PM Page 271 which has the same form as Eq. 3/25 for a single particle. Equation 4/6 states that the resultant of the external forces on any mass system equals the time rate of change of the linear momentum of the system. It is an alternative form of the generalized second law of motion, Eq. 4/1.
As was noted at the end of the last article, ΣF, in general, does not pass through the mass center G. In deriving Eq. 4/6, we differentiated with respect to time and assumed that the total mass is constant. Thus, the equation does not apply to systems whose mass changes with time.
Angular Momentum We now determine the angular momentum of our general mass system about the fixed point O, about the mass center G, and about an arbitrary point P, shown in Fig. 4/3, which may have an acceleration aP .
About a Fixed Point O. The angular momentum of the mass system about the point O, fixed in the Newtonian reference system, is defined as the vector sum of the moments of the linear momenta about O of all particles of the system and is The time derivative of the vector product is . The first summation vanishes since the cross product of two parallel vectors and mivi is zero. The second summation is Σ(ri miai) Σ(ri Fi), which is the vector sum of the moments about O of all forces acting on all particles of the system. This moment sum ΣMO represents only the moments of forces external to the system, since the internal forces cancel one another and their moments add up to zero. Thus, the moment sum is (4/7) which has the same form as Eq. 3/31 for a single particle.
ΣMO H ˙ O r ˙i Σ(ri miv ˙i) Σ(r ˙i mivi) H ˙ O HO Σ(ri mivi) r ¨P 272 Chapter 4 Kinetics of Systems of Particles O (fixed) P (arbitrary) rP ri f1 f2 f3 F3 F2 F1 mi mi r – – G System boundary i ρ ρ ' i ρ Figure 4/3 c04.qxd 2/20/12 3:21 PM Page 272 Equation 4/7 states that the resultant vector moment about any fixed point of all external forces on any system of mass equals the time rate of change of angular momentum of the system about the fixed point. As in the linear-momentum case, Eq. 4/7 does not apply if the total mass of the system is changing with time.
About the Mass Center G. The angular momentum of the mass system about the mass center G is the sum of the moments of the linear momenta about G of all particles and is (4/8) We may write the absolute velocity as so that HG becomes The first term on the right side of this equation may be rewritten as , which is zero because Σmii 0 by definition of the mass center. Thus, we have (4/8a) The expression of Eq. 4/8 is called the absolute angular momentum because the absolute velocity is used. The expression of Eq. 4/8a is called the relative angular momentum because the relative velocity is used. With the mass center G as a reference, the absolute and relative angular momenta are seen to be identical. We will see that this identity does not hold for an arbitrary reference point P; there is no distinction for a fixed reference point O.
Differentiating Eq. 4/8 with respect to time gives The first summation is expanded as . The first term may be rewritten as , which is zero from the definition of the mass center. The second term is zero because the cross product of parallel vectors is zero. With Fi representing the sum of all external forces acting on mi and fi the sum of all internal forces acting on mi, the second summation by Newton’s second law be-comes Σi (Fi fi) Σi Fi ΣMG, the sum of all external mo-ments about point G. Recall that the sum of all internal moments Σi fi is zero. Thus, we are left with (4/9) where we may use either the absolute or the relative angular momentum.
Equations 4/7 and 4/9 are among the most powerful of the govern-ing equations in dynamics and apply to any defined system of constant mass—rigid or nonrigid.
ΣMG H ˙ G d dt Σmii r ˙ Σmi ˙i r ˙ mi ˙i Σ ˙i mir ˙ Σ ˙i Σi mir ¨i ˙i) mi(r ˙ H ˙ G Σ ˙i ˙i r ˙i HG Σi mi ˙i Σmi i r ˙ Σi mi ˙i ˙i) Σi mir ˙ HG Σi mi(r ˙ ˙i) (r ˙ r ˙i HG Σi mir ˙i Article 4/4 Impulse-Momentum 273 c04.qxd 2/8/12 7:33 PM Page 273 About an Arbitrary Point P. The angular momentum about an ar-bitrary point P (which may have an acceleration ) will now be ex-pressed with the notation of Fig. 4/3. Thus, The first term may be written as . The second term is Σi HG. Thus, rearranging gives (4/10) Equation 4/10 states that the absolute angular momentum about any point P equals the angular momentum about G plus the moment about P of the linear momentum of the system considered concentrated at G.
We now make use of the principle of moments developed in our study of statics where we represented a force system by a resultant force through any point, such as G, and a corresponding couple. Figure 4/4 represents the resultants of the external forces acting on the system ex-pressed in terms of the resultant force ΣF through G and the corre-sponding couple ΣMG. We see that the sum of the moments about P of all forces external to the system must equal the moment of their resul-tants. Therefore, we may write which, by Eqs. 4/9 and 4/6, becomes (4/11) Equation 4/11 enables us to write the moment equation about any con-venient moment center P and is easily visualized with the aid of Fig. 4/4.
This equation forms a rigorous basis for much of our treatment of pla-nar rigid-body kinetics in Chapter 6.
We may also develop similar momentum relationships by using the momentum relative to P. Thus, from Fig. 4/3 where is the velocity of mi relative to P. With the substitution and we may write The first summation is . The second summation is and the third summation is where both are zero by defini-tion of the mass center. The fourth summation is (HG)rel. Rearranging gives us (4/12) (HP)rel (HG)rel mvrel Σmi i ˙ d dt Σmi i mvrel Σi mi ˙i Σ mi ˙i Σi mi ˙ (HP)rel Σ mi ˙ ˙i ˙ ˙ i i i ˙ i (HP)rel Σ i mi ˙ i ΣMP H ˙ G ma ΣMP ΣMG ΣF mv HP HG mv mir ˙i Σmir ˙i Σmivi mv HP Σ i mir ˙i Σ( i) mir ˙i r ¨P 274 Chapter 4 Kinetics of Systems of Particles G P ΣF = ma – – ρ ΣMG = G · H Figure 4/4 c04.qxd 2/8/12 7:33 PM Page 274 where (HG)rel is the same as HG (see Eqs. 4/8 and 4/8a). Note the simi-larity of Eqs. 4/12 and 4/10.
The moment equation about P may now be expressed in terms of the angular momentum relative to P. We differentiate the definition (HP)rel with time and make the substitution to obtain The first summation is identically zero, and the second summation is the sum ΣMP of the moments of all external forces about P. The third summa-tion becomes . Sub-stituting and rearranging terms give (4/13) The form of Eq. 4/13 is convenient when a point P whose acceleration is known is used as a moment center. The equation reduces to the simpler form 4/5 Conservation of Energy and Momentum Under certain common conditions, there is no net change in the total mechanical energy of a system during an interval of motion. Under other conditions, there is no net change in the momentum of a system.
These conditions are treated separately as follows.
Conservation of Energy A mass system is said to be conservative if it does not lose energy by virtue of internal friction forces which do negative work or by virtue of in-elastic members which dissipate energy upon cycling. If no work is done on a conservative system during an interval of motion by external forces (other than gravity or other potential forces), then none of the energy of the system is lost. For this case, and we may write Eq. 4/3 as (4/14) or (4/14a) which expresses the law of conservation of dynamical energy. The total energy E T V is a constant, so that E1 E2. This law holds only in the ideal case where internal kinetic friction is sufficiently small to be neglected.
T1 V1 T2 V2 T V 0 U 1-2 0 ΣMP (H ˙ P)rel maP Σ i miaP aP Σmi i aP m maP (H ˙ p)rel Σ ˙ i mi ˙ i Σ i mir ¨i Σ i mir ¨P ¨ i r ¨P r ¨i Σ i mi ˙ i Article 4/5 Conservation of Energy and Momentum 275 c04.qxd 2/8/12 7:33 PM Page 275 Conservation of Momentum If, for a certain interval of time, the resultant external force ΣF act-ing on a conservative or nonconservative mass system is zero, Eq. 4/6 re-quires that , so that during this interval (4/15) which expresses the principle of conservation of linear momentum.
Thus, in the absence of an external impulse, the linear momentum of a system remains unchanged.
Similarly, if the resultant moment about a fixed point O or about the mass center G of all external forces on any mass system is zero, Eq.
4/7 or 4/9 requires, respectively, that (4/16) These relations express the principle of conservation of angular momen-tum for a general mass system in the absence of an angular impulse.
Thus, if there is no angular impulse about a fixed point (or about the mass center), the angular momentum of the system about the fixed point (or about the mass center) remains unchanged. Either equation may hold without the other.
We proved in Art. 3/14 that the basic laws of Newtonian mechanics hold for measurements made relative to a set of axes which translate with a constant velocity. Thus, Eqs. 4/1 through 4/16 are valid provided all quantities are expressed relative to the translating axes.
Equations 4/1 through 4/16 are among the most important of the basic derived laws of mechanics. In this chapter we have derived these laws for the most general system of constant mass to establish the generality of these laws. Common applications of these laws are specific mass systems such as rigid and nonrigid solids and certain fluid systems, which are dis-cussed in the following articles. Study these laws carefully and compare them with their more restricted forms encountered earlier in Chapter 3.
(HO)1 (HO)2 or (HG)1 (HG)2 G1 G2 G ˙ 0 276 Chapter 4 Kinetics of Systems of Particles The principles of particle-system kinetics form the foundation for the study of the forces associated with the water-spraying equipment of these firefighting boats at the site of the Deep-water Horizon fire in the Gulf of Mexico.
Photo Researchers, Inc.
c04.qxd 2/8/12 7:33 PM Page 276 Using Eq. 4/10 with P replaced by O is more expedient than using Eq. 4/8 or 4/8a. The m in Eq. 4/10 is the total mass, which is 10m in this ex-ample. The quantity in Eq. 4/10, with P replaced by O, is We again recognize that here and that the mass of this system is 10m.
r r.
Because of the simple geometry, the cross products are performed by inspection.
Helpful Hints All summation signs are from i = 1 to 4, and all are performed in order of the mass numbers in the given figure.
Article 4/5 Conservation of Energy and Momentum 277 SAMPLE PROBLEM 4/1 The system of four particles has the indicated particle masses, positions, velocities, and external forces. Determine , , , T, G, , , , and .
Solution.
The position of the mass center of the system is Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
For we use Eq. 4/10: Ans.
For we could use Eq. 4/9 or Eq. 4/11 with P replaced by O. Using the latter, we have Ans.
Fd(0.2i 0.8j 1.4k) 10m F 10m(i j) H .
G Fd(j 2k) d(0.4i 0.2j 0.2k) [H .
G MO ma] H .
G, 10mv(0.3i 0.3j 0.3k) mvd(2i 4.2j 2.2k) HG mvd(2i 6j 4k) d(0.4i 0.2j 0.2k) [HG HO mv] HG, H .
O MO 2dFk Fdj Fd( j 2k) mvd(2i 6j 4k) HO ri mi r .
i 0 2mvdi 3mv(2d)j 4mvdk G (mi)r . 10m(v)(0.3i 0.3j 0.3k) mv(3i 3j 3k) T 1 2 mivi 2 1 2 [m(2v)2 2mv2 3mv2 4mv2] 11 2 mv2 r ¨ F mi Fi Fj 10m F 10m (i j) v(0.3i 0.3j 0.3k) m(vi vj) 2m(vj) 3m(vk) 4m(vi) 10m r . mir ˙i mi d(0.4i 0.2j 0.2k) m(2di 2dj) 2m(dk) 3m(2di) 4m(dj) m 2m 3m 4m r miri mi H ˙ G HG H ˙ O HO r ¨ r ˙ r z x y O F F 1 4 3 2 v v v d d 2d 2d 4m 3m 2m 2v 2d m c04.qxd 2/8/12 7:33 PM Page 277 278 Chapter 4 Kinetics of Systems of Particles SAMPLE PROBLEM 4/2 Each of the three balls has a mass m and is welded to the rigid equiangular frame of negligible mass. The assembly rests on a smooth horizontal surface. If a force F is suddenly applied to one bar as shown, determine (a) the acceleration of point O and (b) the angular acceleration of the frame.
Solution. (a) Point O is the mass center of the system of the three balls, so that its acceleration is given by Eq. 4/1.
Ans.
(b) We determine from the moment principle, Eq. 4/9. To find HG we note that the velocity of each ball relative to the mass center O as measured in the nonrotating axes x-y is , where is the common angular velocity of the spokes. The angular momentum of the system about O is the sum of the mo-ments of the relative linear momenta as shown by Eq. 4/8, so it is expressed by Equation 4/9 now gives Ans.
SAMPLE PROBLEM 4/3 Consider the same conditions as for Sample Problem 4/2, except that the spokes are freely hinged at O and so do not constitute a rigid system. Explain the difference between the two problems.
Solution.
The generalized Newton’s second law holds for any mass system, so that the acceleration of the mass center G is the same as with Sample Problem 4/1, namely, Ans.
Although G coincides with O at the instant represented, the motion of the hinge O is not the same as the motion of G since O will not remain the center of mass as the angles between the spokes change.
Both ΣMG and have the same values for the two problems at the instant represented. However, the angular motions of the spokes in this problem are all different and are not easily determined.
H ˙G a F 3m i a Fb d dt (3mr2 ˙) 3mr2 ¨ so ¨ Fb 3mr2 [ΣMG H ˙ G] HO HG 3(mr ˙)r 3mr2 ˙ ˙ r ˙ ¨ Fi 3ma a aO F 3m i [ΣF ma ] ¨ F y x r r r m m m 120° 120° b O Weld Grel O · θ · θ · θ F y x r r r m m m 120° 120° b O Hinge Helpful Hint This present system could be dismem-bered and the motion equations writ-ten for each of the parts, with the unknowns eliminated one by one. Or a more sophisticated method using the equations of Lagrange could be employed. (See the first author’s Dy-namics, 2nd Edition SI Version, 1975, for a discussion of this approach.) Although is initially zero, we need the expression for HO HG in order to get . We observe also that is independent of the motion of O.
¨ H ˙G ˙ Helpful Hints We note that the result depends only on the magnitude and direction of F and not on b, which locates the line of action of F.
c04.qxd 2/8/12 7:33 PM Page 278 SAMPLE PROBLEM 4/4 A shell with a mass of 20 kg is fired from point O, with a velocity u 300 m/s in the vertical x-z plane at the inclination shown. When it reaches the top of its trajectory at P, it explodes into three fragments A, B, and C. Immedi-ately after the explosion, fragment A is observed to rise vertically a distance of 500 m above P, and fragment B is seen to have a horizontal velocity vB and eventually lands at point Q. When recovered, the masses of the fragments A, B, and C are found to be 5, 9, and 6 kg, respectively. Calculate the velocity which fragment C has immediately after the explosion. Neglect atmospheric resistance.
Solution.
From our knowledge of projectile motion, the time required for the shell to reach P and its vertical rise are The velocity of A has the magnitude With no z-component of velocity initially, fragment B requires 24.5 s to return to the ground. Thus, its horizontal velocity, which remains constant, is Since the force of the explosion is internal to the system of the shell and its three fragments, the linear momentum of the system remains unchanged during the explosion. Thus, Ans. vC (427)2 (173.4)2 (82.5)2 468 m/s vC 427i 173.4j 82.5k m/s 6vC 2560i 1040j 495k 20(300)(3 5)i 5(99.0k) 9(163.5)(i cos 45 j sin 45) 6vC mv mAvA mBvB mCvC [G1 G2] vB s/t 4000/24.5 163.5 m/s vA 2ghA 2(9.81)(500) 99.0 m/s h uz 2 2g [(300)(4/5)]2 2(9.81) 2940 m t uz /g 300(4/5)/9.81 24.5 s z y x h 45° 4000 m Q O 3 4 u = 300 m/s C B A vB vC vA P Helpful Hints The velocity v of the shell at the top of its trajectory is, of course, the con-stant horizontal component of its ini-tial velocity u, which becomes u(3/5).
We note that the mass center of the three fragments while still in flight continues to follow the same trajec-tory which the shell would have fol-lowed if it had not exploded.
Article 4/5 Conservation of Energy and Momentum 279 c04.qxd 2/8/12 7:33 PM Page 279 SAMPLE PROBLEM 4/5 The 32.2-lb carriage A moves horizontally in its guide with a speed of 4 ft/sec and carries two assemblies of balls and light rods which rotate about a shaft at O in the carriage. Each of the four balls weighs 3.22 lb. The assembly on the front face rotates counterclockwise at a speed of 80 rev/min, and the assem-bly on the back side rotates clockwise at a speed of 100 rev/min. For the entire system, calculate (a) the kinetic energy T, (b) the magnitude G of the linear mo-mentum, and (c) the magnitude HO of the angular momentum about point O.
Solution. (a) Kinetic energy.
The velocities of the balls with respect to O are The kinetic energy of the system is given by Eq. 4/4. The translational part is The rotational part of the kinetic energy depends on the squares of the relative velocities and is The total kinetic energy is Ans.
(b) Linear momentum.
The linear momentum of the system by Eq. 4/5 is the total mass times vO, the velocity of the center of mass. Thus, Ans.
(c) Angular momentum about O.
The angular momentum about O is due to the moments of the linear momenta of the balls. Taking counterclockwise as positive, we have Ans. 3.77 2.09 1.676 ft-lb-sec HO 2 3.22 32.2 18 12(12.57)(1,2) 2 3.22 32.2 12 12(10.47)(3,4) HO Σri mivi G 32.2 32.2 4 3.22 32.2 (4) 5.6 lb-sec [G mv] T 1 2 mv 2 Σ 1 2 mi ˙i2 11.20 26.8 38.0 ft-lb 15.80 10.96 26.8 ft-lb Σ 1 2 mi ˙i2 2 1 2 3.22 32.2 (12.57)2(1,2) 2 1 2 3.22 32.2 (10.47)2(3,4) 1 2 mv 2 1 2 32.2 32.2 4 3.22 32.2(42) 11.20 ft-lb (vrel)3,4 12 12 100(2) 60 10.47 ft/sec (vrel)1,2 18 12 80(2) 60 12.57 ft/sec [ ˙i vrel r ˙] 4 ft/sec 18″ 18″ 1 2 4 3 12″ 12″ A O 100 rev/min 80 rev/min Contrary to the case of kinetic en-ergy where the direction of rotation was immaterial, angular momentum is a vector quantity and the direction of rotation must be accounted for.
There is a temptation to overlook the contribution of the balls since their linear momenta relative to O in each pair are in opposite direc-tions and cancel. However, each ball also has a velocity component and hence a momentum component .
miv v Note that the direction of rotation, clockwise or counterclockwise, makes no difference in the calculation of ki-netic energy, which depends on the square of the velocity.
Helpful Hints Note that the mass m is the total mass, carriage plus the four balls, and that is the velocity of the mass cen-ter O, which is the carriage velocity.
v 280 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:33 PM Page 280 Article 4/5 Problems 281 4/5 The two 2-kg balls are initially at rest on the horizon-tal surface when a vertical force F 60 N is applied to the junction of the attached wires as shown. Com-pute the vertical component ay of the initial accelera-tion of each ball by considering the system as a whole.
Problem 4/5 4/6 Three monkeys A, B, and C weighing 20, 25, and 15 lb, respectively, are climbing up and down the rope suspended from D. At the instant represented, A is descending the rope with an acceleration of 5 ft/sec2, and C is pulling himself up with an acceleration of 3 ft/sec2. Monkey B is climbing up with a constant speed of 2 ft/sec. Treat the rope and monkeys as a complete system and calculate the tension T in the rope at D.
Problem 4/6 A B C D F y 2 kg 2 kg θ θ PROBLEMS Introductory Problems 4/1 The system of three particles has the indicated parti-cle masses, velocities, and external forces. Determine , , , T, , and for this two-dimensional system.
Problem 4/1 4/2 For the particle system of Prob. 4/1, determine and .
4/3 The system of three particles has the indicated parti-cle masses, velocities, and external forces. Determine , T, HO, and for this three–dimensional system.
Problem 4/3 4/4 For the particle system of Prob. 4/3, determine HG and .
H ˙ G F v 2v 3v 1.5d 2d d 4m 2m m O z y x H ˙ O r ¨ r ˙ , r, H .
G HG y O x F 2v v 2d (stationary) 4m m 2m d H ˙ O HO r ¨ r ˙ r c04.qxd 2/13/12 2:22 PM Page 281 F F F F Hinge Hinge Weld 282 Chapter 4 Kinetics of Systems of Particles 4/7 The three small spheres are connected by the cords and spring and are supported by a smooth horizontal surface. If a force F 6.4 N is applied to one of the cords, find the acceleration of the mass center of the spheres for the instant depicted.
Problem 4/7 4/8 The two spheres, each of mass m, are connected by the spring and hinged bars of negligible mass. The spheres are free to slide in the smooth guides up the incline . Determine the acceleration aC of the center C of the spring.
Problem 4/8 4/9 Calculate the acceleration of the center of mass of the system of the four 10-kg cylinders. Neglect friction and the mass of the pulleys and cables.
Problem 4/9 10 kg 10 kg 500 N 250 N 10 kg 10 kg C m m L F b θ F 0.5 kg 0.8 kg 0.3 kg a 4/10 The four systems slide on a smooth horizontal sur-face and have the same mass m. The configurations of mass in the two pairs are identical. What can be said about the acceleration of the mass center for each system? Explain any difference in the accelera-tions of the members.
Problem 4/10 4/11 The total linear momentum of a system of five parti-cles at time t 2.2 s is given by G2.2 3.4i 2.6j 4.6k . At time t 2.4 s, the linear momentum has changed to G2.4 3.7i 2.2j 4.9k .
Calculate the magnitude F of the time average of the resultant of the external forces acting on the system during the interval.
4/12 The two small spheres, each of mass m, are rigidly connected by a rod of negligible mass and are released from rest in the position shown and slide down the smooth circular guide in the vertical plane. Determine their common velocity v as they reach the horizontal dashed position. Also find the force R between sphere 1 and the supporting surface an instant before the sphere reaches the bottom position A.
Problem 4/12 45° 2 1 2 1 y x m m r A kgm/s kg m/s c04.qxd 2/8/12 7:33 PM Page 282 Representative Problems 4/13 The two small spheres, each of mass m, and their connecting rod of negligible mass are rotating about their mass center G with an angular velocity . At the same instant the mass center has a velocity v in the x-direction. Determine the angular momentum of the assembly at the instant when G has coor-dinates x and y.
Problem 4/13 4/14 Each of the five connected particles has a mass of 0.6 kg, with G as the center of mass of the system.
At a certain instant the angular momentum of the system about G is 1.20k , and the x- and y-components of the velocity of G are 3 m/s and 4 m/s, respectively. Calculate the angular momen-tum HO of the system about O for this instant.
Problem 4/14 4/15 The three identical bars, each weighing 8 lb, are con-nected by the two freely pinned links of negligible weight and are resting on a smooth horizontal sur-face. Calculate the initial acceleration a of the center of the middle bar when the 10-lb force is applied to the connecting link as shown.
y x O G 0.4 m 0.3 m kg m2/s ω y O x m m r r G v HO Article 4/5 Problems 283 Problem 4/15 4/16 A centrifuge consists of four cylindrical containers, each of mass m, at a radial distance r from the rota-tion axis. Determine the time t required to bring the centrifuge to an angular velocity from rest under a constant torque M applied to the shaft. The diame-ter of each container is small compared with r, and the mass of the shaft and supporting arms is small compared with m.
Problem 4/16 4/17 The three small spheres are welded to the light rigid frame which is rotating in a horizontal plane about a vertical axis through O with an angular velocity 20 rad/s. If a couple MO 30 is applied to the frame for 5 seconds, compute the new angular velocity .
Problem 4/17 0.4 m 0.5 m 0.6 m 3 kg 3 kg 4 kg O MO = 30 N·m θ · ˙ Nm ˙ m M r 10 lb c04.qxd 2/8/12 7:33 PM Page 283 4/20 The 300-kg and 400-kg mine cars are rolling in oppo-site directions along the horizontal track with the respective speeds of 0.6 m/s and 0.3 m/s. Upon im-pact the cars become coupled together. Just prior to impact, a 100-kg boulder leaves the delivery chute with a velocity of 1.2 m/s in the direction shown and lands in the 300-kg car. Calculate the velocity v of the system after the boulder has come to rest rela-tive to the car. Would the final velocity be the same if the cars were coupled before the boulder dropped?
Problem 4/20 4/21 The three freight cars are rolling along the horizon-tal track with the velocities shown. After the impacts occur, the three cars become coupled together and move with a common velocity v. The weights of the loaded cars A, B, and C are 130,000, 100,000, and 150,000 lb, respectively. Determine v and calculate the percentage loss n of energy of the system due to coupling.
Problem 4/21 4/22 The man of mass m1 and the woman of mass m2 are standing on opposite ends of the platform of mass m0 which moves with negligible friction and is initially at rest with s 0. The man and woman begin to ap-proach each other. Derive an expression for the dis-placement s of the platform when the two meet in terms of the displacement x1 of the man relative to the platform.
2 mi/hr A B C 1 mi/hr 1.5 mi/hr 30° 100 kg 1.2 m/s 0.6 m/s 0.3 m/s 300 kg 400 kg 284 Chapter 4 Kinetics of Systems of Particles 4/18 The four 3-kg balls are rigidly mounted to the rotat-ing frame and shaft, which are initially rotating freely about the vertical z-axis at the angular rate of 20 rad/s clockwise when viewed from above. If a con-stant torque M 30 is applied to the shaft, cal-culate the time t to reverse the direction of rotation and reach an angular velocity 20 rad/s in the same sense as M.
Problem 4/18 4/19 Billiard ball A is moving in the y-direction with a ve-locity of 2 m/s when it strikes ball B of identical size and mass initially at rest. Following the impact, the balls are observed to move in the directions shown.
Calculate the velocities vA and vB which the balls have immediately after the impact. Treat the balls as particles and neglect any friction forces acting on the balls compared with the force of impact.
Problem 4/19 vA vB y B 2 m/s A 30° 50° M z 3 kg 3 kg 0.3 m 0.3 m 0.5 m 0.5 m 3 kg 3 kg θ · ˙ Nm c04.qxd 2/8/12 7:33 PM Page 284 Problem 4/22 4/23 The woman A, the captain B, and the sailor C weigh 120, 180, and 160 lb, respectively, and are sitting in the 300-lb skiff which is gliding through the water with a speed of 1 knot. If the three people change their positions as shown in the second figure, find the distance x from the skiff to the position where it would have been if the people had not moved. Ne-glect any resistance to motion afforded by the water.
Does the sequence or timing of the change in posi-tions affect the final result?
Problem 4/23 4/24 The two spheres are rigidly connected to the rod of negligible mass and are initially at rest on the smooth horizontal surface. A force F is suddenly ap-plied to one sphere in the y-direction and imparts an impulse of 10 during a negligibly short period of time. As the spheres pass the dashed position, calcu-late the velocity of each one.
Problem 4/24 1.5 kg 1.5 kg 600 mm F x ω vy y N s 1 knot A B C A C B 6′ 8′ 2′ 4′ 6′ x 4′ A s l x2 m0 m1 m2 x1 Article 4/5 Problems 285 4/25 The three small spheres, each of mass m, are se-cured to the light rods to form a rigid unit supported in the vertical plane by the smooth circular surface.
The force of constant magnitude P is applied perpen-dicular to one rod at its midpoint. If the unit starts from rest at 0, determine (a) the minimum force Pmin which will bring the unit to rest at 60 and (b) the common velocity v of spheres 1 and 2 when 60 if P 2Pmin.
Problem 4/25 4/26 The three small steel balls, each of mass 2.75 kg, are connected by the hinged links of negligible mass and equal length. They are released from rest in the posi-tions shown and slide down the quarter-circular guide in the vertical plane. When the upper sphere reaches the bottom position, the spheres have a hori-zontal velocity of 1.560 m/s. Calculate the energy loss Q due to friction and the total impulse Ix on the system of three spheres during this interval.
Problem 4/26 x 360 mm 60° 2 1 P 60° r/2 r/2 θ c04.qxd 2/8/12 7:33 PM Page 285 4/29 The cars of a roller-coaster ride have a speed of 30 km/h as they pass over the top of the circular track.
Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 18 m, and all six cars have the same mass.
Problem 4/29 4/30 The two small spheres, each of mass m, are connected by a cord of length 2b (measured to the centers of the spheres) and are initially at rest on a smooth horizon-tal surface. A projectile of mass m0 with a velocity v0 perpendicular to the cord hits it in the middle, caus-ing the deflection shown in part b of the figure. Deter-mine the velocity v of m0 as the two spheres near contact, with approaching 90 as indicated in part c of the figure. Also find for this condition.
Problem 4/30 m m0 m0 v0 v m b b b b θ θ b b (a) (b) (c) m m m m ˙ 30 km/h v 18 m 18 m 286 Chapter 4 Kinetics of Systems of Particles 4/27 Two steel balls, each of mass m, are welded to a light rod of length L and negligible mass and are initially at rest on a smooth horizontal surface. A horizontal force of magnitude F is suddenly applied to the rod as shown. Determine (a) the instantaneous accelera-tion of the mass center G and (b) the correspond-ing rate at which the angular velocity of the assembly about G is changing with time.
Problem 4/27 4/28 The small car, which has a mass of 20 kg, rolls freely on the horizontal track and carries the 5-kg sphere mounted on the light rotating rod with r 0.4 m. A geared motor drive maintains a constant angular speed 4 rad/s of the rod. If the car has a velocity v 0.6 m/s when 0, calculate v when 60.
Neglect the mass of the wheels and any friction.
Problem 4/28 O v θ θ r ˙ b F x m m y L –– 2 L –– 2 G ¨ a c04.qxd 2/8/12 7:33 PM Page 286 4/31 The carriage of mass 2m is free to roll along the hor-izontal rails and carries the two spheres, each of mass m, mounted on rods of length l and negligible mass. The shaft to which the rods are secured is mounted in the carriage and is free to rotate. If the system is released from rest with the rods in the ver-tical position where 0, determine the velocity vx of the carriage and the angular velocity of the rods for the instant when 180. Treat the carriage and the spheres as particles and neglect any friction.
Problem 4/31 4/32 The 50,000-lb flatcar supports a 15,000-lb vehicle on a 5 ramp built on the flatcar. If the vehicle is re-leased from rest with the flatcar also at rest, deter-mine the velocity v of the flatcar when the vehicle has rolled s 40 ft down the ramp just before hit-ting the stop at B. Neglect all friction and treat the vehicle and the flatcar as particles.
Problem 4/32 s A B 5° x l l m 2m m θ θ ˙ Article 4/5 Problems 287 4/33 A flexible nonextensible rope of mass per unit length and length equal to 1/4 of the circumference of the fixed drum of radius r is released from rest in the horizontal dashed position, with end B secured to the top of the drum. When the rope finally comes to rest with end A at C, determine the loss of energy Q of the system. What becomes of the lost energy?
Problem 4/33 4/34 A horizontal bar of mass and small diameter is suspended by two wires of length l from a carriage of mass which is free to roll along the horizontal rails. If the bar and carriage are released from rest with the wires making an angle with the vertical, determine the velocity of the bar relative to the carriage and the velocity of the carriage at the in-stant when . Neglect all friction and treat the carriage and the bar as particles in the vertical plane of motion.
Problem 4/34 θ m1 m2 l 0 vc vb/c m2 m1 A C B r c04.qxd 2/8/12 7:33 PM Page 287 4/6 Steady Mass Flow The momentum relation developed in Art. 4/4 for a general system of mass provides us with a direct means of analyzing the action of mass flow where a change of momentum occurs. The dynamics of mass flow is of great importance in the description of fluid machinery of all types in-cluding turbines, pumps, nozzles, air-breathing jet engines, and rockets.
The treatment of mass flow in this article is not intended to take the place of a study of fluid mechanics, but merely to present the basic prin-ciples and equations of momentum which find important use in fluid mechanics and in the general flow of mass whether the form be liquid, gaseous, or granular.
One of the most important cases of mass flow occurs during steady-flow conditions where the rate at which mass enters a given volume equals the rate at which mass leaves the same volume. The volume in question may be enclosed by a rigid container, fixed or moving, such as the nozzle of a jet aircraft or rocket, the space between blades in a gas turbine, the volume within the casing of a centrifugal pump, or the vol-ume within the bend of a pipe through which a fluid is flowing at a steady rate. The design of such fluid machines depends on the analysis of the forces and moments associated with the corresponding momen-tum changes of the flowing mass.
Analysis of Flow Through a Rigid Container Consider a rigid container, shown in section in Fig. 4/5a, into which mass flows in a steady stream at the rate m through the entrance sec-tion of area A1. Mass leaves the container through the exit section of area A2 at the same rate, so that there is no accumulation or depletion of the total mass within the container during the period of observation.
The velocity of the entering stream is v1 normal to A1 and that of the leaving stream is v2 normal to A2. If 1 and 2 are the respective densi-ties of the two streams, conservation of mass requires that (4/17) To describe the forces which act, we isolate either the mass of fluid within the container or the entire container and the fluid within it. We would use the first approach if the forces between the container and the fluid were to be described, and we would adopt the second approach when the forces external to the container are desired.
The latter situation is our primary interest, in which case, the sys-tem isolated consists of the fixed structure of the container and the fluid within it at a particular instant of time. This isolation is described by a free-body diagram of the mass within a closed volume defined by the ex-terior surface of the container and the entrance and exit surfaces. We must account for all forces applied externally to this system, and in Fig.
4/5a the vector sum of this external force system is denoted by ΣF. In-cluded in ΣF are 1. the forces exerted on the container at points of its attachment to other structures, including attachments at A1 and A2, if present, 1A1v1 2 A2v2 m 288 Chapter 4 Kinetics of Systems of Particles d2 d1 O A1 A2 v2 v1 ΣF ΣF (a) (b) Δm Time t ΣF Δm Time t + Δt Figure 4/5 c04.qxd 2/8/12 7:33 PM Page 288 Article 4/6 Steady Mass Flow 289 2. the forces acting on the fluid within the container at A1 and A2 due to any static pressure which may exist in the fluid at these positions, and 3. the weight of the fluid and structure if appreciable.
The resultant ΣF of all of these external forces must equal , the time rate of change of the linear momentum of the isolated system. This statement follows from Eq. 4/6, which was developed in Art. 4/4 for any systems of constant mass, rigid or nonrigid.
Incremental Analysis The expression for may be obtained by an incremental analysis.
Figure 4/5b illustrates the system at time t when the system mass is that of the container, the mass within it, and an increment m about to enter during time t. At time t t the same total mass is that of the container, the mass within it, and an equal increment m which leaves the container in time t. The linear momentum of the container and mass within it between the two sections A1 and A2 remains unchanged during t so that the change in momentum of the system in time t is Division by t and passage to the limit yield mv, where Thus, by Eq. 4/6 (4/18) Equation 4/18 establishes the relation between the resultant force on a steady-flow system and the corresponding mass flow rate and vector ve-locity increment. Alternatively, we may note that the time rate of change of linear momentum is the vector difference between the rate at which linear mo-mentum leaves the system and the rate at which linear momentum en-ters the system. Thus, we may write mv2 mv1 mv, which agrees with the foregoing result.
We can now see one of the powerful applications of our general force-momentum equation which we derived for any mass system. Our system here includes a body which is rigid (the structural container for the mass stream) and particles which are in motion (the flow of mass).
By defining the boundary of the system, the mass within which is con-stant for steady-flow conditions, we are able to utilize the generality of Eq. 4/6. However, we must be very careful to account for all external G ˙ ΣF mv m lim tl0 m t dm dt G ˙ G (m)v2 (m)v1 m(v2 v1) G ˙ G ˙ We must be careful not to interpret dm/dt as the time derivative of the mass of the isolated system. That derivative is zero since the system mass is constant for a steady-flow process.
To help avoid confusion, the symbol m rather than dm/dt is used to represent the steady mass flow rate.
The jet exhaust of this VTOL aircraft can be vectored downward for verti-cal takeoffs and landings.
Stocktrek Images, Inc.
c04.qxd 2/8/12 7:33 PM Page 289 forces acting on the system, and they become clear if our free-body dia-gram is correct.
Angular Momentum in Steady-Flow Systems A similar formulation is obtained for the case of angular momentum in steady-flow systems. The resultant moment of all external forces about some fixed point O on or off the system, Fig. 4/5a, equals the time rate of change of angular momentum of the system about O. This fact was established in Eq. 4/7 which, for the case of steady flow in a single plane, becomes (4/19) When the velocities of the incoming and outgoing flows are not in the same plane, the equation may be written in vector form as (4/19a) where d1 and d2 are the position vectors to the centers of A1 and A2 from the fixed reference O. In both relations, the mass center G may be used alternatively as a moment center by virtue of Eq. 4/9.
Equations 4/18 and 4/19a are very simple relations which find im-portant use in describing relatively complex fluid actions. Note that these equations relate external forces to the resultant changes in mo-mentum and are independent of the flow path and momentum changes internal to the system.
The foregoing analysis may also be applied to systems which move with constant velocity by noting that the basic relations ΣF and ΣMO or ΣMG apply to systems moving with constant veloc-ity as discussed in Arts. 3/12 and 4/4. The only restriction is that the mass within the system remain constant with respect to time.
Three examples of the analysis of steady mass flow are given in the following sample problems, which illustrate the application of the prin-ciples embodied in Eqs. 4/18 and 4/19a.
H ˙ G H ˙ O G ˙ ΣMO m(d2 v2 d1 v1) ΣMO m(v2 d2 v1d1) 290 Chapter 4 Kinetics of Systems of Particles The principles of steady mass flow are critical to the design of this hovercraft.
© Robin Weaver/Alamy c04.qxd 2/8/12 7:33 PM Page 290 SAMPLE PROBLEM 4/6 The smooth vane shown diverts the open stream of fluid of cross-sectional area A, mass density , and velocity v. (a) Determine the force components R and F required to hold the vane in a fixed position. (b) Find the forces when the vane is given a constant velocity u less than v and in the direction of v.
Solution. Part (a). The free-body diagram of the vane together with the fluid portion undergoing the momentum change is shown. The momentum equation may be applied to the isolated system for the change in motion in both the x- and y-directions. With the vane stationary, the magnitude of the exit velocity v equals that of the entering velocity v with fluid friction neglected. The changes in the velocity components are then and The mass rate of flow is m Av, and substitution into Eq. 4/18 gives Ans.
Ans.
Part (b). In the case of the moving vane, the final velocity v of the fluid upon exit is the vector sum of the velocity u of the vane plus the velocity of the fluid relative to the vane v u. This combination is shown in the velocity diagram to the right of the figure for the exit conditions. The x-component of v is the sum of the components of its two parts, so (v u) cos u. The change in x-velocity of the stream is The y-component of v is (v u) sin , so that the change in the y-velocity of the stream is vy (v u) sin .
The mass rate of flow m is the mass undergoing momentum change per unit of time. This rate is the mass flowing over the vane per unit time and not the rate of issuance from the nozzle. Thus, The impulse-momentum principle of Eq. 4/18 applied in the positive coordi-nate directions gives Ans.
Ans. R A(v u)2 sin [ΣFy mvy] F A(v u)2(1 cos ) F A(v u)[(v u)( 1 cos )] [ΣFx mvx] m A(v u) vx (v u) cos (u v) (v u)(1 cos ) v x R Av2 sin R Av[v sin ] [ΣFy mvy] F Av2(1 cos ) F Av[v(1 cos )] [ΣFx mvx] vy v sin 0 v sin vx v cos v v(1 cos ) v θ y x F R Fixed vane v′ v θ y F x u R Moving vane θ v – u v – u u v′ Helpful Hints Be careful with algebraic signs when using Eq. 4/18. The change in vx is the final value minus the initial value measured in the positive x-direction. Also we must be careful to write F for ΣFx.
Observe that for given values of u and v, the angle for maximum force F is 180.
Article 4/6 Steady Mass Flow 291 c04.qxd 2/8/12 7:33 PM Page 291 SAMPLE PROBLEM 4/7 For the moving vane of Sample Problem 4/6, determine the optimum speed u of the vane for the generation of maximum power by the action of the fluid on the vane.
Solution.
The force R shown with the figure for Sample Problem 4/6 is normal to the velocity of the vane so it does no work. The work done by the force F shown is negative, but the power developed by the force (reaction to F) exerted by the fluid on the moving vane is The velocity of the vane for maximum power for the one blade in the stream is specified by Ans.
The second solution u v gives a minimum condition of zero power. An angle 180 completely reverses the flow and clearly produces both maximum force and maximum power for any value of u.
SAMPLE PROBLEM 4/8 The offset nozzle has a discharge area A at B and an inlet area A0 at C. A liquid enters the nozzle at a static gage pressure p through the fixed pipe and is-sues from the nozzle with a velocity v in the direction shown. If the constant density of the liquid is , write expressions for the tension T, shear Q, and bend-ing moment M in the pipe at C.
Solution.
The free-body diagram of the nozzle and the fluid within it shows the tension T, shear Q, and bending moment M acting on the flange of the nozzle where it attaches to the fixed pipe. The force pA0 on the fluid within the nozzle due to the static pressure is an additional external force.
Continuity of flow with constant density requires that where v0 is the velocity of the fluid at the entrance to the nozzle. The momentum principle of Eq. 4/18 applied to the system in the two coordinate directions gives Ans.
Ans.
The moment principle of Eq. 4/19 applied in the clockwise sense gives Ans. M Av2(a cos b sin ) M Av(va cos vb sin 0) [ΣMO m(v2 d2 v1d1)] Q Av2 sin Q Av(v sin 0) [ΣFy mvy] T pA0 Av2 A A0 cos pA0 T Av(v cos v0) [ΣFx mvx] Av A0v0 (v 3u)(v u) 0 u v 3 A(1 cos )(v2 4uv 3u2) 0 dP du 0 P A(v u)2u(1 cos ) [P Fu] 292 Chapter 4 Kinetics of Systems of Particles a θ v B C b O x y Q T M pA0 Helpful Hints Again, be careful to observe the cor-rect algebraic signs of the terms on both sides of Eqs. 4/18 and 4/19.
The forces and moment acting on the pipe are equal and opposite to those shown acting on the nozzle.
Helpful Hint The result here applies to a single vane only. In the case of multiple vanes, such as the blades on a tur-bine disk, the rate at which fluid is-sues from the nozzles is the same rate at which fluid is undergoing momentum change. Thus, m Av rather than A(v u). With this change, the optimum value of u turns out to be u v/2.
c04.qxd 2/8/12 7:33 PM Page 292 SAMPLE PROBLEM 4/9 An air-breathing jet aircraft of total mass m flying with a constant speed v consumes air at the mass rate and exhausts burned gas at the mass rate with a velocity u relative to the aircraft. Fuel is consumed at the constant rate .
The total aerodynamic forces acting on the aircraft are the lift L, normal to the di-rection of flight, and the drag D, opposite to the direction of flight. Any force due to the static pressure across the inlet and exhaust surfaces is assumed to be included in D. Write the equation for the motion of the aircraft and identify the thrust T.
Solution.
The free-body diagram of the aircraft together with the air, fuel, and exhaust gas within it is given and shows only the weight, lift, and drag forces as defined. We attach axes x-y to the aircraft and apply our momentum equation relative to the moving system.
The fuel will be treated as a steady stream entering the aircraft with no ve-locity relative to the system and leaving with a relative velocity u in the exhaust stream. We now apply Eq. 4/18 relative to the reference axes and treat the air and fuel flows separately. For the air flow, the change in velocity in the x-direction relative to the moving system is and for the fuel flow the x-change in velocity relative to x-y is Thus, we have where the substitution has been made. Changing signs gives which is the equation of motion of the system.
If we modify the boundaries of our system to expose the interior surfaces on which the air and gas act, we will have the simulated model shown, where the air exerts a force on the interior of the turbine and the exhaust gas reacts against the interior surfaces with the force .
The commonly used model is shown in the final diagram, where the net ef-fect of air and exhaust momentum changes is replaced by a simulated thrust Ans.
applied to the aircraft from a presumed external source.
Inasmuch as is generally only 2 percent or less of , we can use the ap-proximation and express the thrust as Ans.
We have analyzed the case of constant velocity. Although our Newtonian principles do not generally hold relative to accelerating axes, it can be shown that we may use the F ma equation for the simulated model and write T mg sin D m with virtually no error.
v ˙ T m g(u v) m g m a m a m ƒ T m gu m av m gu m av m gu m av mg sin D m g m a m ƒ m gu m av mg sin D m a(u v) m ƒu [ΣFx mvx] vƒ u (0) u va u (v) (u v) m ƒ m g m a x y v D L θ mg T x y mg m′ gu m′ av L L mg D D Helpful Hints Note that the boundary of the sys-tem cuts across the air stream at the entrance to the air scoop and across the exhaust stream at the nozzle.
We are permitted to use moving axes which translate with constant velocity. See Arts. 3/14 and 4/2.
Riding with the aircraft, we ob-serve the air entering our system with a velocity v measured in the plus x-direction and leaving the system with an x-velocity of u.
The final value minus the initial one gives the expression cited, namely, u (v) (u v).
We now see that the “thrust” is, in reality, not a force external to the entire airplane shown in the first fig-ure but can be modeled as an exter-nal force.
Article 4/6 Steady Mass Flow 293 c04.qxd 2/8/12 7:33 PM Page 293 PROBLEMS Introductory Problems 4/35 The jet aircraft has a mass of 4.6 Mg and a drag (air resistance) of 32 kN at a speed of 1000 km/h at a particular altitude. The aircraft consumes air at the rate of 106 kg/s through its intake scoop and uses fuel at the rate of 4 kg/s. If the exhaust has a rear-ward velocity of 680 m/s relative to the exhaust noz-zle, determine the maximum angle of elevation at which the jet can fly with a constant speed of 1000 km/h at the particular altitude in question.
Problem 4/35 4/36 A jet of air issues from the nozzle with a velocity of 300 ft/sec at the rate of 6.50 ft3/sec and is deflected by the right-angle vane. Calculate the force F re-quired to hold the vane in a fixed position. The spe-cific weight of the air is 0.0753 lb/ft3.
Problem 4/36 y x F v α 4/37 Fresh water issues from the nozzle with a velocity of 30 m/s at the rate of 0.05 m3/s and is split into two equal streams by the fixed vane and deflected through 60 as shown. Calculate the force F required to hold the vane in place. The density of water is 1000 kg/m3.
Problem 4/37 4/38 The jet water ski has reached its maximum velocity of 70 km/h when operating in salt water. The water intake is in the horizontal tunnel in the bottom of the hull, so the water enters the intake at the veloc-ity of 70 km/h relative to the ski. The motorized pump discharges water from the horizontal exhaust nozzle of 50-mm diameter at the rate of 0.082 m3/s.
Calculate the resistance R of the water to the hull at the operating speed.
Problem 4/38 F A 60° 60° 294 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:33 PM Page 294 Article 4/6 Problems 295 4/42 The 90 vane moves to the left with a constant veloc-ity of 10 m/s against a stream of fresh water issuing with a velocity of 20 m/s from the 25-mm-diameter nozzle. Calculate the forces Fx and Fy on the vane required to support the motion.
Problem 4/42 Representative Problems 4/43 A jet of fluid with cross-sectional area A and mass density issues from the nozzle with a velocity v and impinges on the inclined trough shown in section.
Some of the fluid is diverted in each of the two direc-tions. If the trough is smooth, the velocity of both di-verted streams remains v, and the only force which can be exerted on the trough is normal to the bottom surface. Hence, the trough will be held in position by forces whose resultant is F normal to the trough. By writing impulse-momentum equations for the direc-tions along and normal to the trough, determine the force F required to support the trough. Also find the volume rates of flow and for the two streams.
Problem 4/43 θ F v 2 1 Q2 Q1 Fx Fy 20 m/s 10 m/s y x 4/39 The fire tug discharges a stream of salt water (den-sity 1030 kg/m3) with a nozzle velocity of 40 m/s at the rate of 0.080 m3/s. Calculate the propeller thrust T which must be developed by the tug to maintain a fixed position while pumping.
Problem 4/39 4/40 The figure shows the top view of an experimental rocket sled which is traveling at a speed of 1000 ft/sec when its forward scoop enters a water channel to act as a brake. The water is diverted at right angles rela-tive to the motion of the sled. If the frontal flow area of the scoop is 15 in.2, calculate the initial braking force. The specific weight of water is 62.4 lb/ft3.
Problem 4/40 4/41 A jet-engine noise suppressor consists of a movable duct which is secured directly behind the jet exhaust by cable A and deflects the blast directly upward.
During a ground test, the engine sucks in air at the rate of 43 kg/s and burns fuel at the rate of 0.8 kg/s.
The exhaust velocity is 720 m/s. Determine the ten-sion T in the cable.
Problem 4/41 A 15° v Rails Scoop Water channel 30° c04.qxd 2/8/12 7:33 PM Page 295 4/44 The 8-oz ball is supported by the vertical stream of fresh water which issues from the 1/2-in.-diameter nozzle with a velocity of 35 ft/sec. Calculate the height h of the ball above the nozzle. Assume that the stream remains intact and there is no energy lost in the jet stream.
Problem 4/44 4/45 A jet-engine thrust reverser to reduce an aircraft speed of 200 km/h after landing employs folding vanes which deflect the exhaust gases in the direc-tion indicated. If the engine is consuming 50 kg of air and 0.65 kg of fuel per second, calculate the brak-ing thrust as a fraction n of the engine thrust with-out the deflector vanes. The exhaust gases have a velocity of 650 m/s relative to the nozzle.
Problem 4/45 200 km/h 30° 30° h lb 1 — 2 4/46 Salt water is being discharged into the atmosphere from the two 30° outlets at the total rate of Each of the discharge nozzles has a flow diameter of 100 mm, and the inside diameter of the pipe at the connecting section A is 250 mm. The pressure of the water at section A-A is 550 kPa. If each of the six bolts at the flange A-A is tightened to a tension of 10 kN, calculate the average pressure p on the flange gasket, which has an area of . The pipe above the flange and the water within it have a mass of 60 kg.
Problem 4/46 4/47 The axial-flow fan C pumps air through the duct of circular cross section and exhausts it with a velocity v at B. The air densities at A and B are A and B, respectively, and the corresponding pressures are pA and pB. The fixed deflecting blades at D restore axial flow to the air after it passes through the propeller blades C. Write an expression for the resultant hori-zontal force R exerted on the fan unit by the flange and bolts at A.
Problem 4/47 A B C D Dia.
= d Dia.
= d E A A 30° 30° 24(103) mm2 30 m3/min.
296 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:33 PM Page 296 Article 4/6 Problems 297 4/50 The sump pump has a net mass of 310 kg and pumps fresh water against a 6-m head at the rate of 0.125 m3/s. Determine the vertical force R between the supporting base and the pump flange at A during operation. The mass of water in the pump may be taken as the equivalent of a 200-mm-diameter col-umn 6 m in height.
Problem 4/50 4/51 In a test of the operation of a “cherry-picker” fire truck, the equipment is free to roll with its brakes released. For the position shown, the truck is ob-served to deflect the spring of stiffness k 15 kN/m a distance of 150 mm because of the action of the horizontal stream of water issuing from the nozzle when the pump is activated. If the exit diameter of the nozzle is 30 mm, calculate the velocity v of the stream as it leaves the nozzle. Also determine the added moment M which the joint at A must resist when the pump is in operation with the nozzle in the position shown.
Problem 4/51 v 75° 90° 15 m A 4.8 m 6 m 45° 200 mm 100 mm A 250 mm 4/48 Air is pumped through the stationary duct A with a velocity of 50 ft/sec and exhausted through an exper-imental nozzle section BC. The average static pres-sure across section B is 150 lb/in.2 gage, and the specific weight of air at this pressure and at the tem-perature prevailing is 0.840 lb/ft3. The average static pressure across the exit section C is measured to be 2 lb/in.2 gage, and the corresponding specific weight of air is 0.0760 lb/ft3. Calculate the force T exerted on the nozzle flange at B by the bolts and the gasket to hold the nozzle in place.
Problem 4/48 4/49 One of the most advanced methods for cutting metal plates uses a high-velocity water jet which carries an abrasive garnet powder. The jet issues from the 0.01-in.-diameter nozzle at A and follows the path shown through the thickness t of the plate. As the plate is slowly moved to the right, the jet makes a nar-row precision slot in the plate. The water-abrasive mixture is used at the low rate of 1/2 gal/min and has a specific weight of 68 lb/ft3. Water issues from the bottom of the plate with a velocity which is 60 percent of the impinging nozzle velocity. Calculate the horizontal force F required to hold the plate against the jet. (There are 231 in.3 in 1 gal.) Problem 4/49 F A t 45° 30° Nozzle 8″ 4″ A B C 50 ft/sec c04.qxd 2/8/12 7:34 PM Page 297 298 Chapter 4 Kinetics of Systems of Particles 4/52 The experimental ground-effect machine has a total weight of 4200 lb. It hovers 1 or 2 ft off the ground by pumping air at atmospheric pressure through the cir-cular intake duct at B and discharging it horizontally under the periphery of the skirt C. For an intake velocity v of 150 ft/sec, calculate the average air pres-sure p under the 18-ft-diameter machine at ground level. The specific weight of the air is 0.076 lb/ft3.
Problem 4/52 4/53 A commercial aircraft flying horizontally at 500 mi/hr encounters a heavy downpour of rain falling vertically at the rate of 20 ft/sec with an intensity equivalent to an accumulation of 1 in./hr on the ground. The upper surface area of the aircraft projected onto the horizon-tal plane is 2960 ft2. Calculate the negligible down-ward force F of the rain on the aircraft.
Problem 4/53 4/54 The ducted fan unit of mass m is supported in the vertical position on its flange at A. The unit draws in air with a density and a velocity u through section A and discharges it through section B with a velocity v.
Both inlet and outlet pressures are atmospheric.
Write an expression for the force R applied to the flange of the fan unit by the supporting slab.
500 mi/hr 3′ C B v 9′ Problem 4/54 4/55 The 180 return pipe discharges salt water (specific weight 64.4 lb/ft3) into the atmosphere at a constant rate of 1.6 ft3/sec. The static pressure in the water at section A is 10 lb/in.2 above atmospheric pressure.
The flow area of the pipe at A is 20 in.2 and that at each of the two outlets is 3.2 in.2 If each of the six flange bolts is tightened with a torque wrench so that it is under a tension of 150 lb, determine the av-erage pressure p on the gasket between the two flanges. The flange area in contact with the gasket is 16 in.2 Also determine the bending moment M in the pipe at section A if the left-hand discharge is blocked off and the flow rate is cut in half. Neglect the weight of the pipe and the water within it.
Problem 4/55 8″ A d u v v B B A θ θ c04.qxd 2/8/12 7:34 PM Page 298 Article 4/6 Problems 299 4/58 The industrial blower sucks in air through the axial opening A with a velocity v1 and discharges it at atmospheric pressure and temperature through the 150-mm-diameter duct B with a velocity v2. The blower handles 16 m3 of air per minute with the motor and fan running at 3450 rev/min. If the motor requires 0.32 kW of power under no load (both ducts closed), calculate the power P consumed while air is being pumped.
Problem 4/58 4/59 The feasibility of a one-passenger VTOL (vertical takeoff and landing) craft is under review. The pre-liminary design calls for a small engine with a high power-to-weight ratio driving an air pump that draws in air through the 70 ducts with an inlet velocity v 40 m/s at a static gage pressure of 1.8 kPa across the inlet areas totaling 0.1320 m2. The air is exhausted vertically down with a velocity u 420 m/s.
For a 90-kg passenger, calculate the maximum net mass m of the machine for which it can take off and hover. (See Table D/1 for air density.) Problem 4/59 70° 70° u v v A B 200 mm v1 v2 4/56 The fire hydrant is tested under a high standpipe pressure. The total flow of 10 ft3/sec is divided equally between the two outlets, each of which has a cross-sectional area of 0.040 ft2. The inlet cross-sectional area at the base is 0.75 ft2. Neglect the weight of the hydrant and water within it and com-pute the tension T, the shear V, and the bending mo-ment M in the base of the standpipe at B. The specific weight of water is 62.4 lb/ft3. The static pressure of the water as it enters the base at B is 120 lb/in.2 Problem 4/56 4/57 A rotary snow plow mounted on a large truck eats its way through a snow drift on a level road at a con-stant speed of 20 km/h. The plow discharges 60 Mg of snow per minute from its 45 chute with a velocity of 12 m/s relative to the plow. Calculate the tractive force P on the tires in the direction of motion neces-sary to move the plow and find the corresponding lateral force R between the tires and the road.
Problem 4/57 45° z y x 30° 24″ 20″ 30″ B y x c04.qxd 2/8/12 7:34 PM Page 299 4/60 The military jet aircraft has a gross weight of 24,000 lb and is poised for takeoff with brakes set while the engine is revved up to maximum power. At this con-dition, air with a specific weight of 0.0753 lb/ft3 is sucked into the intake ducts at the rate of 106 lb/sec with a static pressure of 0.30 lb/in.2 (gage) across the duct entrance. The total cross-sectional area of both intake ducts (one on each side) is 1800 in.2 The air–fuel ratio is 18, and the exhaust velocity u is 3100 ft/sec with zero back pressure (gage) across the exhaust nozzle. Compute the initial acceleration a of the aircraft upon release of the brakes.
Problem 4/60 4/61 The helicopter shown has a mass m and hovers in position by imparting downward momentum to a column of air defined by the slipstream boundary shown. Find the downward velocity v given to the air by the rotor at a section in the stream below the rotor, where the pressure is atmospheric and the stream radius is r. Also find the power P required of the engine. Neglect the rotational energy of the air, any temperature rise due to air friction, and any change in air density .
Problem 4/61 r v v0 u 4/62 The VTOL (vertical takeoff and landing) military aircraft is capable of rising vertically under the ac-tion of its jet exhaust, which can be “vectored” from 0 for takeoff and hovering to 90 for forward flight. The loaded aircraft has a mass of 8600 kg. At full takeoff power, its turbo-fan engine consumes air at the rate of 90 kg/s and has an air–fuel ratio of 18.
Exhaust-gas velocity is 1020 m/s with essentially at-mospheric pressure across the exhaust nozzles. Air with a density of 1.206 kg/m3 is sucked into the in-take scoops at a pressure of 2 kPa (gage) over the total inlet area of 1.10 m2. Determine the angle for vertical takeoff and the corresponding vertical accel-eration ay of the aircraft.
Problem 4/62 4/63 A marine terminal for unloading bulk wheat from a ship is equipped with a vertical pipe with a nozzle at A which sucks wheat up the pipe and transfers it to the storage building. Calculate the x- and y-components of the force R required to change the momentum of the flowing mass in rounding the bend. Identify all forces applied externally to the bend and mass within it. Air flows through the 14-in.-diameter pipe at the rate of 18 tons per hour under a vacuum of 9 in. of mercury ( p 4.42 lb/in.2) and carries with it 150 tons of wheat per hour at a speed of 124 ft/sec.
Problem 4/63 y x 60° B C A v0 u y θ 300 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:34 PM Page 300 Article 4/6 Problems 301 4/66 An axial section of the suction nozzle A for a bulk wheat unloader is shown here. The outer pipe is se-cured to the inner pipe by several longitudinal webs which do not restrict the flow of air. A vacuum of 9 in. of mercury ( p 4.42 lb/in.2 gage) is main-tained in the inner pipe, and the pressure across the bottom of the outer pipe is atmospheric ( p 0). Air at 0.075 lb/ft3 is drawn in through the space between the pipes at a rate of 18 tons/hr at atmospheric pres-sure and draws with it 150 tons of wheat per hour up the pipe at a velocity of 124 ft/sec. If the nozzle unit below section A-A weighs 60 lb, calculate the compression C in the connection at A-A.
Problem 4/66 16.5″ 15″ 14″ Air Air A A 4/64 The sprinkler is made to rotate at the constant an-gular velocity and distributes water at the volume rate Q. Each of the four nozzles has an exit area A.
Write an expression for the torque M on the shaft of the sprinkler necessary to maintain the given mo-tion. For a given pressure and, thus, flow rate Q, at what speed 0 will the sprinkler operate with no applied torque? Let be the density of the water.
Problem 4/64 4/65 A high-speed jet of air issues from the 40-mm-diameter nozzle A with a velocity v of 240 m/s and impinges on the vane OB, shown in its edge view. The vane and its right-angle extension have negligible mass compared with the attached 6-kg cylinder and are freely pivoted about a horizontal axis through O.
Calculate the angle assumed by the vane with the horizontal. The air density under the prevailing con-ditions is 1.206 kg/m3. State any assumptions.
Problem 4/65 θ 120 mm 240 mm 6 kg A B O v v r b ω M c04.qxd 2/8/12 7:34 PM Page 301 4/67 In the figure is shown an impulse-turbine wheel for a hydroelectric power plant which is to operate with a static head of water of 300 m at each of its six noz-zles and is to rotate at the speed of .
Each wheel and generator unit is to develop an out-put power of . The efficiency of the gener-ator may be taken to be 0.90, and an efficiency of 0.85 for the conversion of the kinetic energy of the water jets to energy delivered by the turbine may be expected. The mean peripheral speed of such a wheel for greatest efficiency will be about 0.47 times the jet velocity. If each of the buckets is to have the shape shown, determine the necessary jet diameter d and wheel diameter D. Assume that the water acts on the bucket which is at the tangent point of each jet stream.
Problem 4/67 D 10° 10° u v Bucket detail 22 000 kW 270 rev/min 4/68 A test vehicle designed for impact studies has a mass m 1.4 Mg and is accelerated from rest by the impingement of a high-velocity water jet upon its curved deflector attached to the rear of the vehicle.
The jet of fresh water is produced by the air-operated piston and issues from the 140-mm-diameter nozzle with a velocity v 150 m/s. Frictional resistance of the vehicle, treated as a particle, amounts to 10 per-cent of its weight. Determine the velocity u of the vehicle 3 seconds after release from rest. (Hint: Adapt the results of Sample Problem 4/6.) Problem 4/68 60° u m v To air supply 302 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:34 PM Page 302 Article 4/7 Variable Mass 303 4/7 Variable Mass In Art. 4/4 we extended the equations for the motion of a particle to include a system of particles. This extension led to the very general ex-pressions ΣF ΣMO and ΣMG which are Eqs. 4/6, 4/7, and 4/9, respectively. In their derivation, the summations were taken over a fixed collection of particles, so that the mass of the system to be analyzed was constant.
In Art. 4/6 these momentum principles were extended in Eqs. 4/18 and 4/19a to describe the action of forces on a system defined by a geo-metric volume through which passes a steady flow of mass. Therefore, the amount of mass within this volume was constant with respect to time and thus we were able to use Eqs. 4/6, 4/7, and 4/9. When the mass within the boundary of a system under consideration is not constant, the foregoing relationships are no longer valid. Equation of Motion We will now develop the equation for the linear motion of a system whose mass varies with time. Consider first a body which gains mass by overtaking and swallowing a stream of matter, Fig. 4/6a. The mass of the body and its velocity at any instant are m and v, respectively. The stream of matter is assumed to be moving in the same direction as m with a constant velocity v0 less than v. By virtue of Eq. 4/18, the force exerted by m on the particles of the stream to accelerate them from a ve-locity v0 to a greater velocity v is R m(v v0) where the time rate of increase of m is m and where u is the magnitude of the rel-ative velocity with which the particles approach m. In addition to R, all other forces acting on m in the direction of its motion are denoted by m ˙ m ˙u, H ˙ G, H ˙ O, G ˙, In relativistic mechanics the mass is found to be a function of velocity, and its time deriva-tive has a meaning different from that in Newtonian mechanics.
Figure 4/6 R ΣF ΣF R ΣF v0 v v0 v0 m m m expels mass (v > v0) m m0 m swallows mass (v > v0) m expels mass (v > v0) (a) (b) (c) v v c04.qxd 2/8/12 7:34 PM Page 303 ΣF. The equation of motion of m from Newton’s second law is, there-fore, ΣF R or (4/20) Similarly, if the body loses mass by expelling it rearward so that its velocity v0 is less than v, Fig. 4/6b, the force R required to decelerate the particles from a velocity v to a lesser velocity v0 is R m(v0 [v]) m(v v0). But m since m is decreasing. Also, the relative veloc-ity with which the particles leave m is u v v0. Thus, the force R be-comes R If ΣF denotes the resultant of all other forces acting on m in the direction of its motion, Newton’s second law requires that ΣF R or which is the same relationship as in the case where m is gaining mass.
We may use Eq. 4/20, therefore, as the equation of motion of m, whether it is gaining or losing mass.
A frequent error in the use of the force-momentum equation is to express the partial force sum F as From this expansion we see that the direct differentiation of the linear momentum gives the correct force ΣF only when the body picks up mass initially at rest or when it expels mass which is left with zero absolute velocity. In both instances, v0 0 and u v.
Alternative Approach We may also obtain Eq. 4/20 by a direct differentiation of the mo-mentum from the basic relation ΣF provided a proper system of constant total mass is chosen. To illustrate this approach, we take the case where m is losing mass and use Fig. 4/6c, which shows the system of m and an arbitrary portion m0 of the stream of ejected mass. The mass of this system is m m0 and is constant.
The ejected stream of mass is assumed to move undisturbed once separated from m, and the only force external to the entire system is ΣF which is applied directly to m as before. The reaction R is inter-nal to the system and is not disclosed as an external force on the system.
With constant total mass, the momentum principle ΣF is applicable and we have ΣF d dt (mv m0v0) mv ˙ m ˙v m ˙ 0v0 m0v ˙0 G ˙ m ˙u G ˙, ΣF d dt (mv) mv ˙ m ˙v ΣF mv ˙ m ˙u mv ˙ m ˙u.
m ˙ ΣF mv ˙ m ˙u mv ˙ The Super Scooper is a firefighting airplane which can quickly ingest water from a lake by skimming across the surface with just a bottom-mounted scoop entering the water. The mass within the aircraft boundary varies during the scooping operation as well as during the dumping operation shown.
304 Chapter 4 Kinetics of Systems of Particles © Patrick Forget/AgeFotostock America, Inc.
c04.qxd 2/8/12 7:34 PM Page 304 Article 4/7 Variable Mass 305 Clearly, and the velocity of the ejected mass with respect to m is u v v0. Also 0 since m0 moves undisturbed with no ac-celeration once free of m. Thus, the relation becomes which is identical to the result of the previous formulation, Eq. 4/20.
Application to Rocket Propulsion The case of m losing mass is clearly descriptive of rocket propulsion.
Figure 4/7a shows a vertically ascending rocket, the system for which is the mass within the volume defined by the exterior surface of the rocket and the exit plane across the nozzle. External to this system, the free-body diagram discloses the instantaneous values of gravitational attrac-tion mg, aerodynamic resistance R, and the force pA due to the average static pressure p across the nozzle exit plane of area A. The rate of mass flow is m Thus, we may write the equation of motion of the rocket, ΣF as pA mg R or (4/21) Equation 4/21 is of the form “ΣF ma” where the first term in “ΣF” is the thrust T mu. Thus, the rocket may be simulated as a body to which an external thrust T is applied, Fig. 4/7b, and the problem may then be analyzed like any other F ma problem, except that m is a function of time.
Observe that, during the initial stages of motion when the magni-tude of the velocity v of the rocket is less than the relative exhaust veloc-ity u, the absolute velocity v0 of the exhaust gases will be directed rearward. On the other hand, when the rocket reaches a velocity v whose magnitude is greater than u, the absolute velocity v0 of the exhaust gases will be directed forward. For a given mass rate of flow, the rocket thrust T depends only on the relative exhaust velocity u and not on the magni-tude or on the direction of the absolute velocity v0 of the exhaust gases.
In the foregoing treatment of bodies whose mass changes with time, we have assumed that all elements of the mass m of the body were mov-ing with the same velocity v at any instant of time and that the particles of mass added to or expelled from the body underwent an abrupt transi-tion of velocity upon entering or leaving the body. Thus, this velocity change has been modeled as a mathematical discontinuity. In reality, this change in velocity cannot be discontinuous even though the transi-tion may be rapid. In the case of a rocket, for example, the velocity change occurs continuously in the space between the combustion zone and the exit plane of the exhaust nozzle. A more general analysis of variable-mass dynamics removes this restriction of discontinuous veloc-ity change and introduces a slight correction to Eq. 4/20.
mu pA mg R mv ˙ m ˙u, mv ˙ m ˙u, mv ˙ m ˙.
ΣF mv ˙ m ˙u v ˙0 m ˙, m ˙ 0 For a development of the equations which describe the general motion of a time-dependent system of mass, see Art. 53 of the first author’s Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Inc.
Figure 4/7 pA mg Actual system (a) Simulated system (b) R pA T = m′u mg R c04.qxd 2/8/12 7:34 PM Page 305 306 Chapter 4 Kinetics of Systems of Particles SAMPLE PROBLEM 4/10 The end of a chain of length L and mass per unit length which is piled on a platform is lifted vertically with a constant velocity v by a variable force P. Find P as a function of the height x of the end above the platform. Also find the en-ergy lost during the lifting of the chain.
Solution I (Variable-Mass Approach).
Equation 4/20 will be used and applied to the moving part of the chain of length x which is gaining mass. The force summation ΣF includes all forces acting on the moving part except the force exerted by the particles which are being attached. From the diagram we have The velocity is constant so that 0. The rate of increase of mass is v, and the relative velocity with which the attaching particles approach the moving part is u v 0 v. Thus, Eq. 4/20 becomes Ans.
We now see that the force P consists of the two parts, gx, which is the weight of the moving part of the chain, and v2, which is the added force required to change the momentum of the links on the platform from a condition at rest to a velocity v.
Solution II (Constant-Mass Approach).
The principle of impulse and momentum for a system of particles expressed by Eq. 4/6 will be applied to the entire chain considered as the system of constant mass. The free-body diagram of the system shows the unknown force P, the total weight of all links gL, and the force g(L x) exerted by the platform on those links which are at rest on it.
The momentum of the system at any position is Gx xv and the momentum equation gives Ans.
Again the force P is seen to be equal to the weight of the portion of the chain which is off the platform plus the added term which accounts for the time rate of increase of momentum of the chain.
Energy Loss.
Each link on the platform acquires its velocity abruptly through an impact with the link above it, which lifts it off the platform. The succession of impacts gives rise to an energy loss E (negative work E) so that the work-energy equation becomes P dx E T Vg, where Substituting into the work-energy equation gives Ans.
1 2 gL2 v2L E 1 2 Lv2 1 2 gL2 E 1 2 Lv2 T 1 2 Lv2 Vg gL L 2 1 2 gL2 P dx L 0 ( gx v2) dx 1 2 gL2 v2L U 1-2 ΣFx dGx dt P g(L x) gL d dt ( xv) P (gx v2) [ΣF mv ˙ m ˙u] P gx 0 v(v) P (gx v2) m ˙ v ˙ ΣFx P gx P x g(L – x) P P Solution I Solution II gL gx ρ ρ ρ Helpful Hints The model of Fig. 4/6a shows the mass being added to the leading end of the moving part. With the chain the mass is added to the trailing end, but the effect is the same.
We must be very careful not to use ΣF for a system whose mass is changing. Thus, we have taken the total chain as the system since its mass is constant.
G ˙ Note that includes work done by internal nonelastic forces, such as the link-to-link impact forces, where this work is converted into heat and acoustical energy loss E.
U 1-2 c04.qxd 2/8/12 7:34 PM Page 306 Article 4/7 Variable Mass 307 SAMPLE PROBLEM 4/11 Replace the open-link chain of Sample Problem 4/10 by a flexible but inex-tensible rope or bicycle-type chain of length L and mass per unit length. Deter-mine the force P required to elevate the end of the rope with a constant velocity v and determine the corresponding reaction R between the coil and the platform.
Solution.
The free-body diagram of the coil and moving portion of the rope is shown in the left-hand figure. Because of some resistance to bending and some lateral motion, the transition from rest to vertical velocity v will occur over an appreciable segment of the rope. Nevertheless, assume first that all moving ele-ments have the same velocity so that Eq. 4/6 for the system gives We assume further that all elements of the coil of rope are at rest on the plat-form and transmit no force to the platform other than their weight, so that R g(L x). Substitution into the foregoing relation gives which is the same result as that for the chain in Sample Problem 4/10.
The total work done on the rope by P becomes Substitution into the work-energy equation gives which is twice the kinetic energy of vertical motion. Thus, an equal amount of kinetic energy is unaccounted for. This conclusion largely negates our assumption of one-dimensional x-motion.
In order to produce a one-dimensional model which retains the inextensibility property assigned to the rope, it is necessary to impose a physical constraint at the base to guide the rope into vertical motion and at the same time preserve a smooth transition from rest to upward velocity v without energy loss. Such a guide is in-cluded in the free-body diagram of the entire rope in the middle figure and is rep-resented schematically in the middle free-body diagram of the right-hand figure.
For a conservative system, the work-energy equation gives Substitution into the impulse-momentum equation ΣFx gives Although this force, which exceeds the weight by is unrealistic experimen-tally, it would be present in the idealized model.
Equilibrium of the vertical section requires Because it requires a force of v2 to change the momentum of the rope elements, the restraining guide must supply the balance F which, in turn, is trans-mitted to the platform.
1 2 v2 T0 P gx 1 2 v2 gx gx 1 2 v2 1 2 v2, 1 2 v2 gx R gL v2 R 1 2 v2 g(L x) G ˙x P 1 2 v2 gx [dU dT dVg] P dx d(1 2 xv2) dgx x 2 1 2 xv2 [U 1-2 T Vg] v2x 1 2 gx2 T gx x 2 T xv2 U 1-2 P dx x 0 ( v2 gx) dx v2x 1 2 gx2 P g(L x) v2 gL or P v2 gx ΣFx dGx dt P R gL d dt ( xv) P R v2 gL P R gL ρ ρ ρ ρ P R gL P gx v x T 0 T 0 F R g(L – x) r = v/r ω Helpful Hints Perfect flexibility would not permit any resistance to bending.
Remember that v is constant and equals Also note that this same relation applies to the chain of Sam-ple Problem 4/10.
This added term of unaccounted-for kinetic energy exactly equals the en-ergy lost by the chain during the im-pact of its links.
This restraining guide may be visu-alized as a canister of negligible mass rotating within the coil with an angular velocity v/r and con-nected to the platform through its shaft. As it turns, it feeds the rope from a rest position to an upward velocity v, as indicated in the accom-panying figure.
Note that the mass center of the sec-tion of length x is a distance x/2 above the base.
x ˙.
c04.qxd 2/8/12 7:34 PM Page 307 SAMPLE PROBLEM 4/12 A rocket of initial total mass m0 is fired vertically up from the north pole and accelerates until the fuel, which burns at a constant rate, is exhausted. The relative nozzle velocity of the exhaust gas has a constant value u, and the nozzle exhausts at atmospheric pressure throughout the flight. If the residual mass of the rocket structure and machinery is mb when burnout occurs, determine the expression for the maximum velocity reached by the rocket. Neglect atmospheric resistance and the variation of gravity with altitude.
Solution I (F ma Solution).
We adopt the approach illustrated with Fig.
4/7b and treat the thrust as an external force on the rocket. With the neglect of the back pressure p across the nozzle and the atmospheric resistance R, Eq. 4/21 or Newton’s second law gives But the thrust is T mu so that the equation of motion becomes Multiplication by dt, division by m, and rearrangement give which is now in a form which can be integrated. The velocity v corresponding to the time t is given by the integration or Since the fuel is burned at the constant rate m the mass at any time t is m m0 If we let mb stand for the mass of the rocket when burnout oc-curs, then the time at burnout becomes tb (mb m0)/ (m0 mb)/ This time gives the condition for maximum velocity, which is Ans.
The quantity is a negative number since the mass decreases with time.
Solution II (Variable-Mass Solution).
If we use Eq. 4/20, then F mg and the equation becomes But mu T so that the equation of motion becomes which is the same as formulated with Solution I.
T mg mv ˙ m ˙u mg mv ˙ m ˙u [ΣF mv ˙ m ˙u] m ˙ vmax u ln m0 mb g m ˙ (m0 mb) (m ˙).
m ˙ m ˙t.
m ˙, v u ln m0 m gt v 0 dv u m m0 dm m g t 0 dt dv u dm m g dt m ˙u mg mv ˙ m ˙u T mg mv ˙ 308 Chapter 4 Kinetics of Systems of Particles T mg v Vertical launch from the north pole is taken only to eliminate any com-plication due to the earth’s rotation in figuring the absolute trajectory of the rocket.
Helpful Hints The neglect of atmospheric resis-tance is not a bad assumption for a first approximation inasmuch as the velocity of the ascending rocket is smallest in the dense part of the at-mosphere and greatest in the rar-efied region. Also for an altitude of 320 km, the acceleration due to gravity is 91 percent of the value at the surface of the earth.
c04.qxd 2/8/12 7:34 PM Page 308 Article 4/7 Problems 309 4/71 The space shuttle, together with its central fuel tank and two booster rockets, has a total mass of 2.04(106) kg at liftoff. Each of the two booster rockets pro-duces a thrust of 11.80(106) N, and each of the three main engines of the shuttle produces a thrust of 2.00(106) N. The specific impulse (ratio of exhaust velocity to gravitational acceleration) for each of the three main engines of the shuttle is 455 s. Calculate the initial vertical acceleration a of the assembly with all five engines operating and find the rate at which fuel is being consumed by each of the shuttle’s three engines.
Problem 4/71 4/72 A tank truck for washing down streets has a total weight of 20,000 lb when its tank is full. With the spray turned on, 80 lb of water per second issue from the nozzle with a velocity of 60 ft/sec relative to the truck at the 30 angle shown. If the truck is to accel-erate at the rate of 2 ft/sec2 when starting on a level road, determine the required tractive force P be-tween the tires and the road when (a) the spray is turned on and (b) the spray is turned off.
Problem 4/72 a 30° PROBLEMS Introductory Problems 4/69 At the instant of vertical launch the rocket expels exhaust at the rate of 220 kg/s with an exhaust ve-locity of 820 m/s. If the initial vertical acceleration is 6.80 m/s2, calculate the total mass of the rocket and fuel at launch.
Problem 4/69 4/70 When the rocket reaches the position in its trajec-tory shown, it has a mass of 3 Mg and is beyond the effect of the earth’s atmosphere. Gravitational accel-eration is 9.60 m/s2. Fuel is being consumed at the rate of 130 kg/s, and the exhaust velocity relative to the nozzle is 600 m/s. Compute the n- and t-components of acceleration of the rocket.
Problem 4/70 Vert.
Horiz.
30° t n a = 6.8 m/s2 c04.qxd 2/8/12 7:34 PM Page 309 4/73 A tank, which has a mass of 50 kg when empty, is propelled to the left by a force P and scoops up fresh water from a stream flowing in the opposite direc-tion with a velocity of 1.5 m/s. The entrance area of the scoop is 2000 mm2, and water enters the scoop at a rate equal to the velocity of the scoop relative to the stream. Determine the force P at a certain in-stant for which 80 kg of water have been ingested and the velocity and acceleration of the tank are 2 m/s and 0.4 m/s2, respectively. Neglect the small impact pressure at the scoop necessary to elevate the water in the tank.
Problem 4/73 4/74 A small rocket of initial mass m0 is fired vertically upward near the surface of the earth ( g constant). If air resistance is neglected, determine the manner in which the mass m of the rocket must vary as a func-tion of the time t after launching in order that the rocket may have a constant vertical acceleration a, with a constant relative velocity u of the escaping gases with respect to the nozzle.
4/75 The magnetometer boom for a spacecraft consists of a large number of triangular-shaped units which spring into their deployed configuration upon re-lease from the canister in which they were folded and packed prior to release. Write an expression for the force F which the base of the canister must exert on the boom during its deployment in terms of the increasing length x and its time derivatives. The mass of the boom per unit of deployed length is .
Treat the supporting base on the spacecraft as a fixed platform and assume that the deployment takes place outside of any gravitational field. Neglect the dimension b compared with x.
1.5 m/s a P v Problem 4/75 4/76 Fresh water issues from the two 30-mm-diameter holes in the bucket with a velocity of 2.5 m/s in the directions shown. Calculate the force P required to give the bucket an upward acceleration of 0.5 m/s2 from rest if it contains 20 kg of water at that time.
The empty bucket has a mass of 0.6 kg.
Problem 4/76 P 20° 20° x b 310 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:34 PM Page 310 Article 4/7 Problems 311 4/79 A railroad coal car weighs 54,600 lb empty and car-ries a total load of 180,000 lb of coal. The bins are equipped with bottom doors which permit discharg-ing coal through an opening between the rails. If the car dumps coal at the rate of 20,000 lb/sec in a down-ward direction relative to the car, and if frictional resistance to motion is 4 lb per ton of total remain-ing weight, determine the coupler force P required to give the car an acceleration of 0.15 ft/sec2 in the direction of P at the instant when half the coal has been dumped.
Problem 4/79 4/80 The figure represents an idealized one-dimensional structure of uniform mass per unit length moving horizontally with a velocity v0 when its front end col-lides with an immovable barrier and crushes. The force F required to initiate and maintain an accor-dionlike deformation is constant. Neglect the length b of the collapsed portion of the structure compared with the movement of s of the undeformed portion following the impact. The undeformed part may be viewed as a body of decreasing mass. Derive the dif-ferential equation which relates F to s, and by using Eq. 4/20 carefully. Check your expression by applying Eq. 4/6 to both parts together as a system of constant mass.
Problem 4/80 L F L – s s b v0 s ¨ s ˙, P Representative Problems 4/77 The upper end of the open-link chain of length L and mass per unit length is lowered at a constant speed v by the force P. Determine the reading R of the platform scale in terms of x.
Problem 4/77 4/78 At a bulk loading station, gravel leaves the hopper at the rate of 220 lb/sec with a velocity of 10 ft/sec in the direction shown and is deposited on the moving flatbed truck. The tractive force between the driving wheels and the road is 380 lb, which overcomes the 200 lb of frictional road resistance. Determine the acceleration a of the truck 4 seconds after the hop-per is opened over the truck bed, at which instant the truck has a forward speed of 1.5 mi/hr. The empty weight of the truck is 12,000 lb.
Problem 4/78 v 60° 10 ft/sec x L P v c04.qxd 2/8/12 7:34 PM Page 311 4/81 A coil of heavy flexible cable with a total length of 100 m and a mass of 1.2 kg/m is to be laid along a straight horizontal line. The end is secured to a post at A, and the cable peels off the coil and emerges through the horizontal opening in the cart as shown.
The cart and drum together have a mass of 40 kg.
If the cart is moving to the right with a velocity of 2 m/s when 30 m of cable remain in the drum and the tension in the rope at the post is 2.4 N, deter-mine the force P required to give the cart and drum an acceleration of 0.3 m/s2. Neglect all friction.
Problem 4/81 4/82 By lowering a scoop as it skims the surface of a body of water, the aircraft (nicknamed the “Super Scooper”) is able to ingest 4.5 m3 of fresh water during a 12-second run. The plane then flies to a fire area and makes a massive water drop with the ability to re-peat the procedure as many times as necessary. The plane approaches its run with a velocity of 280 km/h and an initial mass of 16.4 Mg. As the scoop enters the water, the pilot advances the throttle to provide an additional 300 hp (223.8 kW) needed to prevent undue deceleration. Determine the initial decelera-tion when the scooping action starts. (Neglect the difference between the average and the initial rates of water intake.) Problem 4/82 v Scoop P x A 4/83 An open-link chain of length L 8 m with a mass of 48 kg is resting on a smooth horizontal surface when end A is doubled back on itself by a force P applied to end A. (a) Calculate the required value of P to give A a constant velocity of 1.5 m/s. (b) Calculate the accel-eration a of end A if P 20 N and if v 1.5 m/s when x 4 m.
Problem 4/83 4/84 A small rocket-propelled vehicle weighs 125 lb, in-cluding 20 lb of fuel. Fuel is burned at the constant rate of 2 lb/sec with an exhaust velocity relative to the nozzle of 400 ft/sec. Upon ignition the vehicle is released from rest on the 10 incline. Calculate the maximum velocity v reached by the vehicle. Neglect all friction.
Problem 4/84 10° L x x – 2 P A v 312 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:34 PM Page 312 Article 4/7 Problems 313 4/87 The cart carries a pile of open-link chain of mass per unit length. The chain passes freely through the hole in the cart and is brought to rest, link by link, by the tension T in the portion of the chain resting on the ground and secured at its end A. The cart and the chain on it move under the action of the con-stant force P and have a velocity v0 and mass m0 when x 0. Determine expressions for the accelera-tion a and velocity v of the cart in terms of x if all friction is neglected. Also find T. Observe that the transition link 2 is decelerated from the velocity v to zero velocity by the tension T transmitted by the last horizontal link 1. Also note that link 2 exerts no force on the following link 3 during the transition.
Explain why the term is absent if Eq. 4/20 is ap-plied to this problem.
Problem 4/87 4/88 The open-link chain of length L and mass per unit length is released from rest in the position shown, where the bottom link is almost touching the platform and the horizontal section is supported on a smooth surface. Friction at the corner guide is negligible. De-termine (a) the velocity v1 of end A as it reaches the corner and (b) its velocity v2 as it strikes the platform.
(c) Also specify the total loss Q of energy.
Problem 4/88 h L – h A m ˙u 4/85 Determine the force P required to give the open-link chain of total length L a constant velocity v The chain has a mass per unit length. Also, by applying the impulse-momentum equation to the left-hand portion of the system, verify that the force R sup-porting the pile of chain equals the weight of the pile. Neglect the small size and mass of the pulley and any friction in the pulley.
Problem 4/85 4/86 A coal car with an empty mass of 25 Mg is moving freely with a speed of 1.2 m/s under a hopper which opens and releases coal into the moving car at the constant rate of 4 Mg per second. Determine the dis-tance x moved by the car during the time that 32 Mg of coal are deposited in the car. Neglect any frictional resistance to rolling along the horizontal track.
Problem 4/86 v0 y h P y ˙.
x A P v 1 2 3 Transition link 2 c04.qxd 2/8/12 7:34 PM Page 313 4/89 In the figure is shown a system used to arrest the motion of an airplane landing on a field of restricted length. The plane of mass m rolling freely with a ve-locity v0 engages a hook which pulls the ends of two heavy chains, each of length L and mass per unit length, in the manner shown. A conservative calcu-lation of the effectiveness of the device neglects the retardation of chain friction on the ground and any other resistance to the motion of the airplane. With these assumptions, compute the velocity v of the air-plane at the instant when the last link of each chain is put in motion. Also determine the relation be-tween the displacement x and the time t after con-tact with the chain. Assume each link of the chain acquires its velocity v suddenly upon contact with the moving links.
Problem 4/89 4/90 The free end of the open-link chain of total length L and mass per unit length is released from rest at x 0. Determine the force R on the fixed end and the tension T1 in the chain at the lower end of the nonmoving part in terms of x. Also find the total loss Q of energy when x L.
Problem 4/90 x x x – 2 L v0 v 4/91 Replace the chain of Prob. 4/90 by a flexible rope or bicycle chain of mass per unit length and total length L. The free end is released from rest at x 0 and falls under the influence of gravity. Determine the acceleration a of the free end, the force R at the fixed end, and the tension T1 in the rope at the loop, all in terms of x. (Note that a is greater than g. What happens to the energy of the system when x L?) 4/92 One end of the pile of chain falls through a hole in its support and pulls the remaining links after it in a steady flow. If the links which are initially at rest acquire the velocity of the chain suddenly and with-out frictional resistance or interference from the support or from adjacent links, find the velocity v of the chain as a function of x if v 0 when x 0.
Also find the acceleration a of the falling chain and the energy Q lost from the system as the last link leaves the platform. (Hint: Apply Eq. 4/20 and treat the product xv as the variable when solving the dif-ferential equation. Also note at the appropriate step that dx v dt.) The total length of the chain is L, and its mass per unit length is .
Problem 4/92 x 314 Chapter 4 Kinetics of Systems of Particles c04.qxd 2/8/12 7:34 PM Page 314 4/8 CHAPTER REVIEW In this chapter we have extended the principles of dynamics for the motion of a single mass particle to the motion of a general system of particles. Such a system can form a rigid body, a nonrigid (elastic) solid body, or a group of separate and unconnected particles, such as those in a defined mass of liquid or gaseous particles. The following summarizes the principal results of Chapter 4.
1. We derived the generalized form of Newton’s second law, which is expressed as the principle of motion of the mass center, Eq. 4/1 in Art. 4/2. This principle states that the vector sum of the external forces acting on any system of mass particles equals the total sys-tem mass times the acceleration of the center of mass.
2. In Art. 4/3, we established a work-energy principle for a system of particles, Eq. 4/3a, and showed that the total kinetic energy of the system equals the energy of the mass-center translation plus the en-ergy due to motion of the particles relative to the mass center.
3. The resultant of the external forces acting on any system equals the time rate of change of the linear momentum of the system, Eq. 4/6 in Art. 4/4.
4. For a fixed point O and the mass center G, the resultant vector mo-ment of all external forces about the point equals the time rate of change of angular momentum about the point, Eq. 4/7 and Eq. 4/9 in Art. 4/4. The principle for an arbitrary point P, Eqs. 4/11 and 4/13, has an additional term and thus does not follow the form of the equations for O and G.
5. In Art. 4/5 we developed the law of conservation of dynamical en-ergy, which applies to a system in which the internal kinetic friction is negligible.
6. Conservation of linear momentum applies to a system in the absence of an external linear impulse. Similarly, conservation of angular mo-mentum applies when there is no external angular impulse.
7. For applications involving steady mass flow, we developed a rela-tion, Eq. 4/18 in Art. 4/6, between the resultant force on a system, the corresponding mass flow rate, and the change in fluid velocity from entrance to exit.
8. Analysis of angular momentum in steady mass flow resulted in Eq.
4/19a in Art. 4/6, which is a relation between the resultant moment of all external forces about a fixed point O on or off the system, the mass flow rate, and the incoming and outgoing velocities.
9. Finally, in Art. 4/7 we developed the equation of linear motion for variable-mass systems, Eq. 4/20. Common examples of such systems are rockets and flexible chains and ropes.
The principles developed in this chapter enable us to treat the mo-tion of both rigid and nonrigid bodies in a unified manner. In addition, the developments in Arts. 4/2–4/5 will serve to place on a rigorous basis the treatment of rigid-body kinetics in Chapters 6 and 7.
Article 4/8 Chapter Review 315 c04.qxd 2/8/12 7:34 PM Page 315 4/95 In an operational design test of the equipment of the fire truck, the water cannon is delivering fresh water through its 2-in.-diameter nozzle at the rate of 1400 gal/min at the 20 angle. Calculate the total friction force F exerted by the pavement on the tires of the truck, which remains in a fixed position with its brakes locked. (There are 231 in.3 in 1 gal.) Problem 4/95 4/96 A small rocket of initial mass m0 is fired vertically up near the surface of the earth ( g constant), and the mass rate of exhaust m and the relative exhaust velocity u are constant. Determine the velocity v as a function of the time t of flight if the air resistance is neglected and if the mass of the rocket case and ma-chinery is negligible compared with the mass of the fuel carried.
4/97 The two balls are attached to the light rigid rod, which is suspended by a cord from the support above it. If the balls and rod, initially at rest, are struck with the force F 12 lb, calculate the corresponding acceleration of the mass center and the rate at which the angular velocity of the bar is changing.
Problem 4/97 2 lb 4 lb F 7″ 3″ 6″ 10″ ¨ a 20° v 316 Chapter 4 Kinetics of Systems of Particles REVIEW PROBLEMS 4/93 Each of the identical steel balls weighs 4 lb and is fastened to the other two by connecting bars of neg-ligible weight and unequal length. In the absence of friction at the supporting horizontal surface, deter-mine the initial acceleration of the mass center of the assembly when it is subjected to the horizontal force F 20 lb applied to the supporting ball. The assembly is initially at rest in the vertical plane. Can you show that is initially horizontal?
Problem 4/93 4/94 A 2-oz bullet is fired horizontally with a velocity v 1000 ft/sec into the slender bar of a 3-lb pendu-lum initially at rest. If the bullet embeds itself in the bar, compute the resulting angular velocity of the pen-dulum immediately after the impact. Treat the sphere as a particle and neglect the mass of the rod. Why is the linear momentum of the system not conserved?
Problem 4/94 10″ 10″ O 2 oz 3 lb Before After v ω F a a c04.qxd 2/8/12 7:34 PM Page 316 4/98 The rocket shown is designed to test the operation of a new guidance system. When it has reached a cer-tain altitude beyond the effective influence of the earth’s atmosphere, its mass has decreased to 2.80 Mg, and its trajectory is 30 from the vertical. Rocket fuel is being consumed at the rate of 120 kg/s with an exhaust velocity of 640 m/s relative to the nozzle.
Gravitational acceleration is 9.34 m/s2 at its altitude.
Calculate the n- and t-components of the accelera-tion of the rocket.
Problem 4/98 4/99 A two-stage rocket is fired vertically up and is above the atmosphere when the first stage burns out and the second stage separates and ignites. The second stage carries 1200 kg of fuel and has an empty mass of 200 kg. Upon ignition the second stage burns fuel at the rate of 5.2 kg/s and has a constant exhaust velocity of 3000 m/s relative to its nozzle. Determine the acceleration of the second stage 60 seconds after ignition and find the maximum acceleration and the time t after ignition at which it occurs. Neglect the variation of g and take it to be 8.70 m/s2 for the range of altitude averaging about 400 km.
Horiz.
Vert.
n t 30° Article 4/8 Review Problems 317 4/100 The three identical spheres, each of mass m, are supported in the vertical plane on the 30 incline.
The spheres are welded to the two connecting rods of negligible mass. The upper rod, also of negligible mass, is pivoted freely to the upper sphere and to the bracket at A. If the stop at B is suddenly re-moved, determine the velocity v with which the upper sphere hits the incline. (Note that the corre-sponding velocity of the middle sphere is v/2.) Ex-plain the loss of energy which has occurred after all motion has ceased.
Problem 4/100 4/101 A jet of fresh water under pressure issues from the 3/4-in.-diameter fixed nozzle with a velocity v 120 ft/sec and is diverted into the two equal streams. Neglect any energy loss in the streams and compute the force F required to hold the vane in place.
Problem 4/101 v A F 30° 30° 10″ 10″ 30° 60° 60° A B c04.qxd 2/8/12 7:34 PM Page 317 Problem 4/103 4/104 The upper end of the open-link chain of length L and mass per unit length is released from rest with the lower end just touching the platform of the scale. Determine the expression for the force F read on the scale as a function of the distance x through which the upper end has fallen. (Comment: The chain acquires a free-fall velocity of be-cause the links on the scale exert no force on those above, which are still falling freely. Work the prob-lem in two ways: first, by evaluating the time rate of change of momentum for the entire chain and second, by considering the force F to be composed of the weight of the links at rest on the scale plus the force necessary to divert an equivalent stream of fluid.) Problem 4/104 x L 2gx A B T 60° 60° C 318 Chapter 4 Kinetics of Systems of Particles 4/102 An ideal rope or bicycle-type chain of length L and mass per unit length is resting on a smooth hori-zontal surface when end A is doubled back on itself by a force P applied to end A. End B of the rope is secured to a fixed support. Determine the force P required to give A a constant velocity v. (Hint: The action of the loop can be modeled by inserting a cir-cular disk of negligible mass as shown in the sepa-rate sketch and then taking the disk radius as zero.
It is easily shown that the tensions in the rope at C, D, and B are all equal to P under the ideal condi-tions imposed and with constant velocity.) Problem 4/102 4/103 In the static test of a jet engine and exhaust nozzle assembly, air is sucked into the engine at the rate of 30 kg/s and fuel is burned at the rate of 1.6 kg/s.
The flow area, static pressure, and axial-flow veloc-ity for the three sections shown are as follows: Sec. A Sec. B Sec. C Flow area, m2 0.15 0.16 0.06 Static pressure, kPa 14 140 14 Axial-flow velocity, m/s 120 315 600 Determine the tension T in the diagonal member of the supporting test stand and calculate the force F exerted on the nozzle flange at B by the bolts and gasket to hold the nozzle to the engine housing.
L B x x – 2 P C D A r c04.qxd 2/8/12 7:34 PM Page 318 4/105 The open-link chain of total length L and of mass per unit length is released from rest at x 0 at the same instant that the platform starts from rest at y 0 and moves vertically up with a constant ac-celeration a. Determine the expression for the total force R exerted on the platform by the chain t sec-onds after the motion starts.
Problem 4/105 4/106 The three identical 2-kg spheres are welded to the connecting rods of negligible mass and are hanging by a cord from point A. The spheres are initially at rest when a horizontal force F 16 N is applied to the upper sphere. Calculate the initial acceleration of the mass center of the spheres, the rate at which the angular velocity is increasing, and the initial acceleration a of the top sphere.
Problem 4/106 F A 300 mm 2 kg 2 kg 2 kg 60° 60° ¨ a x y L Article 4/8 Review Problems 319 4/107 The diverter section of pipe between A and B is de-signed to allow the parallel pipes to clear an ob-struction. The flange of the diverter is secured at C by a heavy bolt. The pipe carries fresh water at the steady rate of 5000 gal/min under a static pressure of 130 lb/in.2 entering the diverter. The inside di-ameter of the pipe at A and at B is 4 in. The ten-sions in the pipe at A and B are balanced by the pressure in the pipe acting over the flow area.
There is no shear or bending of the pipes at A or B.
Calculate the moment M supported by the bolt at C. (Recall that 1 gallon contains 231 in.3) Problem 4/107 4/108 The chain of length L and mass per unit length is released from rest on the smooth horizontal sur-face with a negligibly small overhang x to initiate motion. Determine (a) the acceleration a as a func-tion of x, (b) the tension T in the chain at the smooth corner as a function of x, and (c) the veloc-ity v of the last link A as it reaches the corner.
Problem 4/108 x L – x A x A B C v v 8″ c04.qxd 2/8/12 7:34 PM Page 319 Problem 4/110 4/111 Replace the pile of chain in Prob. 4/92 by a coil of rope of mass per unit length and total length L as shown and determine the velocity of the falling sec-tion in terms of x if it starts from rest at x 0.
Show that the acceleration is constant at g/2. The rope is considered to be perfectly flexible in bend-ing but inextensible and constitutes a conservative system (no energy loss). Rope elements acquire their velocity in a continuous manner from zero to v in a small transition section of the rope at the top of the coil. For comparison with the chain of Prob. 4/92, this transition section may be consid-ered to have negligible length without violating the requirement that there be no energy loss in the present problem. Also determine the force R exerted by the platform on the coil in terms of x and explain why R becomes zero when x 2L/3. Neglect the dimensions of the coil compared with x.
Problem 4/111 x v C A B D 150 mm 150 mm 200 mm 75 mm 320 Chapter 4 Kinetics of Systems of Particles 4/109 A rope or hinged-link bicycle-type chain of length L and mass per unit length is released from rest with x 0. Determine the expression for the total force R exerted on the fixed platform by the chain as a function of x. Note that the hinged-link chain is a conservative system during all but the last in-crement of motion. Compare the result with that of Prob. 4/105 if the upward motion of the platform in that problem is taken to be zero.
Problem 4/109 4/110 The centrifugal pump handles 20 m3 of fresh water per minute with inlet and outlet velocities of 18 m/s.
The impeller is turned clockwise through the shaft at O by a motor which delivers 40 kW at a pump speed of 900 rev/min. With the pump filled but not turning, the vertical reactions at C and D are each 250 N. Calculate the forces exerted by the founda-tion on the pump at C and D while the pump is running. The tensions in the connecting pipes at A and B are exactly balanced by the respective forces due to the static pressure in the water. (Sugges-tion: Isolate the entire pump and water within it between sections A and B and apply the momen-tum principle to the entire system.) x L c04.qxd 2/8/12 7:34 PM Page 320 4/112 The chain of mass per unit length passes over the small freely turning pulley and is released from rest with only a small imbalance h to initiate mo-tion. Determine the acceleration a and velocity v of the chain and the force R supported by the hook at A, all in terms of h as it varies from essentially zero to H. Neglect the weight of the pulley and its sup-porting frame and the weight of the small amount of chain in contact with the pulley. (Hint: The force R does not equal two times the equal tensions T in the chain tangent to the pulley.) Article 4/8 Review Problems 321 Problem 4/112 H A h h c04.qxd 2/8/12 7:34 PM Page 321 c04.qxd 2/8/12 7:34 PM Page 322 PART II Dynamics of Rigid Bodies c05.qxd 2/10/12 10:07 AM Page 323 Rigid-body kinematics describes the relationships between the linear and angular motions of bodies without re-gard to the forces and moments associated with such motions. The designs of gears, cams, connecting links, and many other moving machine parts are largely kinematic problems.
R. Ian Lloyd/Masterile c05.qxd 2/10/12 10:07 AM Page 324 325 5/1 Introduction In Chapter 2 on particle kinematics, we developed the relationships governing the displacement, velocity, and acceleration of points as they moved along straight or curved paths. In rigid-body kinematics we use these same relationships but must also account for the rotational mo-tion of the body. Thus rigid-body kinematics involves both linear and angular displacements, velocities, and accelerations.
We need to describe the motion of rigid bodies for two important reasons. First, we frequently need to generate, transmit, or control cer-tain motions by the use of cams, gears, and linkages of various types.
Here we must analyze the displacement, velocity, and acceleration of the motion to determine the design geometry of the mechanical parts.
Furthermore, as a result of the motion generated, forces may be devel-oped which must be accounted for in the design of the parts.
Second, we must often determine the motion of a rigid body caused by the forces applied to it. Calculation of the motion of a rocket under the influence of its thrust and gravitational attraction is an example of such a problem.
We need to apply the principles of rigid-body kinematics in both sit-uations. This chapter covers the kinematics of rigid-body motion which may be analyzed as occurring in a single plane. In Chapter 7 we will pre-sent an introduction to the kinematics of motion in three dimensions.
5/1 Introduction 5/2 Rotation 5/3 Absolute Motion 5/4 Relative Velocity 5/5 Instantaneous Center of Zero Velocity 5/6 Relative Acceleration 5/7 Motion Relative to Rotating Axes 5/8 Chapter Review CHAPTER OUTLINE 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:07 AM Page 325 Rigid-Body Assumption In the previous chapter we defined a rigid body as a system of parti-cles for which the distances between the particles remain unchanged.
Thus, if each particle of such a body is located by a position vector from reference axes attached to and rotating with the body, there will be no change in any position vector as measured from these axes. This is, of course, an ideal case since all solid materials change shape to some ex-tent when forces are applied to them.
Nevertheless, if the movements associated with the changes in shape are very small compared with the movements of the body as a whole, then the assumption of rigidity is usually acceptable. The dis-placements due to the flutter of an aircraft wing, for instance, do not af-fect the description of the flight path of the aircraft as a whole, and thus the rigid-body assumption is clearly acceptable. On the other hand, if the problem is one of describing, as a function of time, the internal wing stress due to wing flutter, then the relative motions of portions of the wing cannot be neglected, and the wing may not be considered a rigid body. In this and the next two chapters, almost all of the material is based on the assumption of rigidity.
Plane Motion A rigid body executes plane motion when all parts of the body move in parallel planes. For convenience, we generally consider the plane of motion to be the plane which contains the mass center, and we treat the body as a thin slab whose motion is confined to the plane of the slab.
This idealization adequately describes a very large category of rigid-body motions encountered in engineering.
The plane motion of a rigid body may be divided into several cate-gories, as represented in Fig. 5/1.
Translation is defined as any motion in which every line in the body remains parallel to its original position at all times. In translation there is no rotation of any line in the body. In rectilinear translation, part a of Fig. 5/1, all points in the body move in parallel straight lines.
In curvilinear translation, part b, all points move on congruent curves.
We note that in each of the two cases of translation, the motion of the body is completely specified by the motion of any point in the body, since all points have the same motion. Thus, our earlier study of the motion of a point (particle) in Chapter 2 enables us to describe completely the translation of a rigid body.
Rotation about a fixed axis, part c of Fig. 5/1, is the angular motion about the axis. It follows that all particles in a rigid body move in circu-lar paths about the axis of rotation, and all lines in the body which are perpendicular to the axis of rotation (including those which do not pass through the axis) rotate through the same angle in the same time.
Again, our discussion in Chapter 2 on the circular motion of a point en-ables us to describe the motion of a rotating rigid body, which is treated in the next article.
General plane motion of a rigid body, part d of Fig. 5/1, is a com-bination of translation and rotation. We will utilize the principles of rel-ative motion covered in Art. 2/8 to describe general plane motion.
326 Chapter 5 Plane Kinematics of Rigid Bodies These nickel microgears are only 150 microns (150(106) m) thick and have potential application in microscopic robots.
David Parker/Photo Researchers, Inc.
c05.qxd 2/10/12 10:07 AM Page 326 Note that in each of the examples cited, the actual paths of all parti-cles in the body are projected onto the single plane of motion as repre-sented in each figure.
Analysis of the plane motion of rigid bodies is accomplished either by directly calculating the absolute displacements and their time deriva-tives from the geometry involved or by utilizing the principles of relative motion. Each method is important and useful and will be covered in turn in the articles which follow.
5/2 Rotation The rotation of a rigid body is described by its angular motion.
Figure 5/2 shows a rigid body which is rotating as it undergoes plane motion in the plane of the figure. The angular positions of any two lines 1 and 2 attached to the body are specified by 1 and 2 measured from any convenient fixed reference direction. Because the angle is invariant, the relation 2 1 upon differentiation with respect to time gives and or, during a finite interval, 2 1.
Thus, all lines on a rigid body in its plane of motion have the same an-gular displacement, the same angular velocity, and the same angular acceleration.
¨1 ¨2 ˙1 ˙2 Article 5/2 Rotation 327 B A B′ A′ B A B′ A′ B A B A B′ B′ A′ (a) Rectilinear translation Type of Rigid-Body Plane Motion Example Rocket test sled Parallel-link swinging plate Compound pendulum (b) Curvilinear translation (c) Fixed-axis rotation (d) General plane motion θ Connecting rod in a reciprocating engine Figure 5/1 1 2 β 1 θ 2 θ Figure 5/2 c05.qxd 2/10/12 10:07 AM Page 327 Note that the angular motion of a line depends only on its angular position with respect to any arbitrary fixed reference and on the time derivatives of the displacement. Angular motion does not require the presence of a fixed axis, normal to the plane of motion, about which the line and the body rotate.
328 Chapter 5 Plane Kinematics of Rigid Bodies Angular-Motion Relations The angular velocity and angular acceleration of a rigid body in plane rotation are, respectively, the first and second time derivatives of the angular position coordinate of any line in the plane of motion of the body. These definitions give (5/1) The third relation is obtained by eliminating dt from the first two. In each of these relations, the positive direction for and , clockwise or counterclockwise, is the same as that chosen for . Equations 5/1 should be recognized as analogous to the defining equations for the rectilinear motion of a particle, expressed by Eqs. 2/1, 2/2, and 2/3. In fact, all rela-tions which were described for rectilinear motion in Art. 2/2 apply to the case of rotation in a plane if the linear quantities s, v, and a are replaced by their respective equivalent angular quantities , , and . As we pro-ceed further with rigid-body dynamics, we will find that the analogies between the relationships for linear and angular motion are almost complete throughout kinematics and kinetics. These relations are im-portant to recognize, as they help to demonstrate the symmetry and unity found throughout mechanics.
For rotation with constant angular acceleration, the integrals of Eqs. 5/1 becomes Here 0 and 0 are the values of the angular position coordinate and an-gular velocity, respectively, at t 0, and t is the duration of the motion considered. You should be able to carry out these integrations easily, as they are completely analogous to the corresponding equations for recti-linear motion with constant acceleration covered in Art. 2/2.
The graphical relationships described for s, v, a, and t in Figs. 2/3 and 2/4 may be used for , , and merely by substituting the corre-sponding symbols. You should sketch these graphical relations for plane 0 0t 1 2 t2 2 0 2 2( 0) 0 t d d or ˙ d ˙ ¨ d d dt ˙ or d2 dt2 ¨ d dt ˙ KEY CONCEPTS c05.qxd 2/10/12 10:07 AM Page 328 rotation. The mathematical procedures for obtaining rectilinear velocity and displacement from rectilinear acceleration may be applied to rota-tion by merely replacing the linear quantities by their corresponding an-gular quantities.
Rotation about a Fixed Axis When a rigid body rotates about a fixed axis, all points other than those on the axis move in concentric circles about the fixed axis. Thus, for the rigid body in Fig. 5/3 rotating about a fixed axis normal to the plane of the figure through O, any point such as A moves in a circle of radius r. From the previous discussion in Art. 2/5, you should already be familiar with the relationships between the linear motion of A and the angular motion of the line normal to its path, which is also the angular motion of the rigid body. With the notation and for the angular velocity and angular acceleration, respectively, of the body we have Eqs. 2/11, rewritten as (5/2) These quantities may be expressed alternatively using the cross-prod-uct relationship of vector notation. The vector formulation is especially important in the analysis of three-dimensional motion. The angular veloc-ity of the rotating body may be expressed by the vector normal to the plane of rotation and having a sense governed by the right-hand rule, as shown in Fig. 5/4a. From the definition of the vector cross product, we see that the vector v is obtained by crossing into r. This cross product gives the correct magnitude and direction for v and we write The order of the vectors to be crossed must be retained. The reverse order gives r v.
v r ˙ r at r an r2 v2/r v v r ¨ ˙ ˙ A O v ω r · θ (a) A O v = × r ω ω α (b) · = × r) × ( an = ω ω ω × r at = α α ω Figure 5/4 O A α ω t v = rω at = rα 2 ω an = r n r Figure 5/3 Article 5/2 Rotation 329 c05.qxd 2/10/12 10:07 AM Page 329 The acceleration of point A is obtained by differentiating the cross-product expression for v, which gives Here stands for the angular acceleration of the body. Thus, the vector equivalents to Eqs. 5/2 are (5/3) and are shown in Fig. 5/4b.
For three-dimensional motion of a rigid body, the angular-velocity vector may change direction as well as magnitude, and in this case, the angular acceleration, which is the time derivative of angular veloc-ity, will no longer be in the same direction as .
˙, at r an ( r) v r ˙ v r ( r) ˙ r a v ˙ r ˙ ˙ r 330 Chapter 5 Plane Kinematics of Rigid Bodies This pulley-cable system is part of an elevator mechanism.
These pulleys and cables are part of the San Francisco cable-car system.
© Steven Haggard/Alamy © Nomad/SUPERSTOCK c05.qxd 2/10/12 10:07 AM Page 330 SAMPLE PROBLEM 5/1 A flywheel rotating freely at 1800 rev/min clockwise is subjected to a vari-able counterclockwise torque which is first applied at time t 0. The torque pro-duces a counterclockwise angular acceleration 4t rad/s2, where t is the time in seconds during which the torque is applied. Determine (a) the time required for the flywheel to reduce its clockwise angular speed to 900 rev/min, (b) the time required for the flywheel to reverse its direction of rotation, and (c) the total number of revolutions, clockwise plus counterclockwise, turned by the flywheel during the first 14 seconds of torque application.
Solution.
The counterclockwise direction will be taken arbitrarily as positive.
(a) Since is a known function of the time, we may integrate it to obtain angular velocity. With the initial angular velocity of 1800(2)/60 60 rad/s, we have Substituting the clockwise angular speed of 900 rev/min or 900(2)/60 30 rad/s gives Ans.
(b) The flywheel changes direction when its angular velocity is momentarily zero. Thus, Ans.
(c) The total number of revolutions through which the flywheel turns during 14 seconds is the number of clockwise turns N1 during the first 9.71 seconds, plus the number of counterclockwise turns N2 during the remainder of the inter-val. Integrating the expression for in terms of t gives us the angular displace-ment in radians. Thus, for the first interval or N1 1220/2 194.2 revolutions clockwise.
For the second interval or N2 410/2 65.3 revolutions counterclockwise. Thus, the total number of revolutions turned during the 14 seconds is Ans.
We have plotted versus t and we see that 1 is represented by the negative area and 2 by the positive area. If we had integrated over the entire interval in one step, we would have obtained 2 1.
N N1 N2 194.2 65.3 259 rev 2 [60t 2 3 t3]14 9.71 410 rad 2 0 d 14 9.71 (60 2t2) dt 1 [60t 2 3 t3]9.71 0 1220 rad 1 0 d 9.71 0 (60 2t2) dt [d dt] 0 60 2t2 t2 30 t 9.71 s 30 60 2t2 t2 15 t 6.86 s [d dt] 60 d t 0 4t dt 60 2t2 Article 5/2 Rotation 331 Angular velocity , rad/s CCW ω –60π 64.8π 0 0 14 2 10 12 6 8 4 6.86 9.71 –30π θ1 θ2 Time t, s We could have converted the origi-nal expression for into the units of rev/s2, in which case our inte-grals would have come out directly in revolutions.
Again note that the minus sign sig-nifies clockwise in this problem.
Helpful Hints We must be very careful to be consis-tent with our algebraic signs. The lower limit is the negative (clockwise) value of the initial angular velocity.
Also we must convert revolutions to radians since is in radian units.
c05.qxd 2/10/12 10:07 AM Page 331 SAMPLE PROBLEM 5/2 The pinion A of the hoist motor drives gear B, which is attached to the hoisting drum. The load L is lifted from its rest position and acquires an upward velocity of 3 ft/sec in a vertical rise of 4 ft with constant acceleration. As the load passes this position, compute (a) the acceleration of point C on the cable in con-tact with the drum and (b) the angular velocity and angular acceleration of the pinion A.
Solution. (a) If the cable does not slip on the drum, the vertical velocity and acceleration of the load L are, of necessity, the same as the tangential velocity v and tangential acceleration at of point C. For the rectilinear motion of L with constant acceleration, the n- and t-components of the acceleration of C become Ans.
(b) The angular motion of gear A is determined from the angular motion of gear B by the velocity v1 and tangential acceleration a1 of their common point of contact. First, the angular motion of gear B is determined from the motion of point C on the attached drum. Thus, Then from v1 rAA rBB and a1 rAA rBB, we have Ans.
Ans.
SAMPLE PROBLEM 5/3 The right-angle bar rotates clockwise with an angular velocity which is de-creasing at the rate of 4 rad/s2. Write the vector expressions for the velocity and acceleration of point A when 2 rad/s.
Solution.
Using the right-hand rule gives The velocity and acceleration of A become Ans.
Ans.
The magnitudes of v and a are v 0.62 0.82 1 m/s and a 2.82 0.42 2.83 m/s2 a 2.8i 0.4j m/s2 [a an at] at 4k (0.4i 0.3j) 1.2i 1.6j m/s2 [at r] an 2k (0.6i 0.8j) 1.6i 1.2j m/s2 [an ( r)] v 2k (0.4i 0.3j) 0.6i 0.8j m/s [v r] 2k rad/s and 4k rad/s2 A rB rA B 18/12 6/12 0.562 1.688 rad/sec2 CW A rB rA B 18/12 6/12 1.5 4.5 rad/sec CW B at /r 1.125/(24/12) 0.562 rad/sec2 [at r] B v/r 3/(24/12) 1.5 rad/sec [v r] aC (4.5)2 (1.125)2 4.64 ft/sec2 [a an 2 at 2] an 32/(24/12) 4.5 ft/sec2 [an v2/r] a at v2/2s 32/[2(4)] 1.125 ft/sec2 [v2 2as] 332 Chapter 5 Plane Kinematics of Rigid Bodies C B A 12″ 36″ 48″ L 4′ 3 ft /sec x y A 0.4 m 0.3 m ω C B A 6″ 18″ v = 3 ft /sec a = 1.125 ft /sec2 at = 1.125 ft /sec2 A α B α A v1 a1 aC ω B ω an = 4.5 ft /sec2 Helpful Hint Recognize that a point on the cable changes the direction of its velocity after it contacts the drum and acquires a normal component of acceleration.
c05.qxd 2/10/12 10:07 AM Page 332 PROBLEMS Introductory Problems 5/1 The circular disk of radius rotates about a fixed axis through point O with the angular properties rad/s and rad/s2 with directions as shown in the figure. Determine the instantaneous values of the velocity and acceleration of point A.
Problem 5/1 5/2 The triangular plate rotates about a fixed axis through point O with the angular properties indi-cated. Determine the instantaneous velocity and acceleration of point A. Take all given variables to be positive.
Problem 5/2 ω α h b O x y A ω α A r O x y r – – 4 3 2 r 0.16 m Article 5/2 Problems 333 5/3 The body is formed of slender rod and rotates about a fixed axis through point O with the indicated angu-lar properties. If rad/s and rad/s2, deter-mine the instantaneous velocity and acceleration of point A.
Problem 5/3 5/4 A torque applied to a flywheel causes it to accelerate uniformly from a speed of 200 rev/min to a speed of 800 rev/min in 4 seconds. Determine the number of revolutions N through which the wheel turns during this interval. (Suggestion: Use revolutions and min-utes for units in your calculations.) 5/5 The drive mechanism imparts to the semicircular plate simple harmonic motion of the form , where is the amplitude of the oscillation and is its circular frequency. Determine the amplitudes of the angular velocity and angular acceleration and state where in the motion cycle these maxima occur. Note that this motion is not that of a freely pivoted and undriven body undergoing arbitrarily large-amplitude angular motion.
Problem 5/5 θ O 0 0 0t 0 sin 0.2 m 0.5 m ω 20° O x y A α 7 4 c05.qxd 2/10/12 10:07 AM Page 333 5/10 The bent flat bar rotates about a fixed axis through point O. At the instant depicted, its angular proper-ties are rad/s and with directions as indicated in the figure. Determine the instanta-neous velocity and acceleration of point A.
Problem 5/10 Representative Problems 5/11 The angular acceleration of a body which is rotating about a fixed axis is given by , where the constant (no units). Determine the angular displacement and time elapsed when the angular ve-locity has been reduced to one-third its initial value rad/s.
5/12 The angular position of a radial line in a rotating disk is given by the clockwise angle where is in radians and t is in seconds. Calculate the angular displacement of the disk during the interval in which its angular acceleration increases from 42 rad/s2 to 66 rad/s2.
5/13 In order to test an intentionally weak adhesive, the bottom of the small 0.3-kg block is coated with adhe-sive and then the block is pressed onto the turntable with a known force. The turntable starts from rest at time and uniformly accelerates with . If the adhesive fails at exactly determine the ultimate shear force which the adhe-sive supports. What is the angular displacement of the turntable at the time of failure?
t 3 s, 2 rad/s2 t 0 2t3 3t2 4, 0 12 k 0.1 k2 0.3 m 0.5 m O A x y ω 30° 105° α 8 rad/s2 5 334 Chapter 5 Plane Kinematics of Rigid Bodies 5/6 The mass center G of the car has a velocity of 40 mi/hr at position A and 1.52 seconds later at B has a velocity of 50 mi/hr. The radius of curvature of the road at B is 180 ft. Calculate the angular velocity of the car at B and the average angular velocity of the car between A and B.
Problem 5/6 5/7 The rectangular plate is rotating about its corner axis through O with a constant angular velocity rad/s. Determine the magnitudes of the velocity v and acceleration a of the corner A by (a) using the scalar relations and (b) using the vector relations.
Problem 5/7 5/8 If the rectangular plate of Prob. 5/7 starts from rest and point B has an initial acceleration of 5.5 m/s2, determine the distance b if the plate reaches an angu-lar speed of 300 rev/min in 2 seconds with a constant angular acceleration.
5/9 A shaft is accelerated from rest at a constant rate to a speed of 3600 rev/min and then is immediately decel-erated to rest at a constant rate within a total time of 10 seconds. How many revolutions N has the shaft turned during this interval?
z y x b 300 mm 400 mm O B A ω 10 30° 18″ A B G G av c05.qxd 2/10/12 10:07 AM Page 334 Problem 5/13 5/14 The plate OAB forms an equilateral triangle which rotates counterclockwise with increasing speed about point O. If the normal and tangential compo-nents of acceleration of the centroid C at a certain instant are 80 m/s2 and 30 m/s2, respectively, deter-mine the values of and at this same instant. The angle is the angle between line AB and the fixed horizontal axis.
Problem 5/14 A B C O θ 150 mm 150 mm 150 mm ¨ ˙ 0.4 m O P ω Article 5/2 Problems 335 5/15 Experimental data for a rotating control element reveal the plotted relation between angular velocity and the angular coordinate as shown. Approximate the angular acceleration of the element when Problem 5/15 5/16 The rotating arm starts from rest and acquires a rotational speed rev/min in 2 seconds with constant angular acceleration. Find the time t after starting before the acceleration vector of end P makes an angle of with the arm OP.
Problem 5/16 P N O 6″ 45 N 600 0 0 2 4 6 8 10 2 4 6 8 10 , rad θ , rad/s ω 6 rad.
c05.qxd 2/10/12 10:07 AM Page 335 Problem 5/19 5/20 Point A of the circular disk is at the angular position at time The disk has angular velocity at and subsequently experiences a constant angular acceleration Deter-mine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time Problem 5/20 5/21 Repeat Prob. 5/20, except now the angular accelera-tion of the disk is given by where t is in sec-onds and is in radians per second squared.
Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time s.
5/22 Repeat Prob. 5/20, except now the angular accelera-tion of the disk is given by , where is in ra-dians per second and is in radians per second squared. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time s.
5/23 The disk of Prob. 5/20 is at the angular position at time . Its angular velocity at is rad/s, and then it experiences an angular acceleration given by , where is in radians and is in radians per second squared. Determine the angular position of point A at time s.
t 2 2 0 0.1 t 0 t 0 0 t 1 2 t 2 2t, α θ A O 200 mm x y t 1 s.
2 rad/s2.
t 0 0 0.1 rad/s t 0.
0 A B C O y x 4″ 45° 336 Chapter 5 Plane Kinematics of Rigid Bodies 5/17 The belt-driven pulley and attached disk are rotat-ing with increasing angular velocity. At a certain in-stant the speed v of the belt is 1.5 m/s, and the total acceleration of point A is 75 m/s2. For this instant determine (a) the angular acceleration of the pul-ley and disk, (b) the total acceleration of point B, and (c) the acceleration of point C on the belt.
Problem 5/17 5/18 Magnetic tape is being fed over and around the light pulleys mounted in a computer. If the speed v of the tape is constant and if the magnitude of the accelera-tion of point A on the tape is 4/3 times that of point B, calculate the radius r of the smaller pulley.
Problem 5/18 5/19 The circular disk rotates about its center O. For the instant represented, the velocity of A is in./sec and the tangential acceleration of B is in./sec2. Write the vector expressions for the angular velocity and angular acceleration of the disk. Use these results to write the vector ex-pression for the acceleration of point C.
(aB)t 6i vA 8j v v A B 4″ r 150 mm 150 mm A C v v B c05.qxd 2/10/12 10:07 AM Page 336 5/24 During its final spin cycle, a front-loading washing machine has a spin rate of 1200 rev/min. Once power is removed, the drum is observed to uniformly decel-erate to rest in 25 s. Determine the number of revolu-tions made during this period as well as the number of revolutions made during the first half of it.
Problem 5/24 5/25 The solid cylinder rotates about its z-axis. At the instant represented, point P on the rim has a veloc-ity whose x-component is ft/sec, and .
Determine the angular velocity of line AB on the face of the cylinder. Does the element line BC have an angular velocity?
Problem 5/25 5/26 The two V-belt pulleys form an integral unit and ro-tate about the fixed axis at O. At a certain instant, point A on the belt of the smaller pulley has a veloc-ity , and point B on the belt of the larger pulley has an acceleration as shown.
For this instant determine the magnitude of the acceleration of point C and sketch the vector in your solution.
aC aB 45 m/s2 vA 1.5 m/s C y z x P B A 6″ θ 20 4.2 ω O Article 5/2 Problems 337 Problem 5/26 5/27 A clockwise variable torque is applied to a flywheel at time causing its clockwise angular accelera-tion to decrease linearly with angular displacement during 20 revolutions of the wheel as shown. If the clockwise speed of the flywheel was 300 rev/min at , determine its speed N after turning the 20 revolutions. (Suggestion: Use units of revolutions instead of radians.) Problem 5/27 5/28 The design characteristics of a gear-reduction unit are under review. Gear B is rotating clockwise with a speed of 300 rev/min when a torque is applied to gear A at time s to give gear A a counterclock-wise acceleration which varies with time for a duration of 4 seconds as shown. Determine the speed of gear B when Problem 5/28 t 6 s.
NB t 2 , rev/s2 α , rev θ 0 0 0.6 1.8 20 t 0 t 0 800 mm 360 mm B aB A C vA 150 mm O 2 6 t, s 0 0 4 8 A, CCW rad — — s2 α 2b b A B c05.qxd 2/10/12 10:07 AM Page 337 5/3 Absolute Motion We now develop the approach of absolute-motion analysis to de-scribe the plane kinematics of rigid bodies. In this approach, we make use of the geometric relations which define the configuration of the body involved and then proceed to take the time derivatives of the defining geometric relations to obtain velocities and accelerations.
In Art. 2/9 of Chapter 2 on particle kinematics, we introduced the application of absolute-motion analysis for the constrained motion of connected particles. For the pulley configurations treated, the relevant velocities and accelerations were determined by successive differentia-tion of the lengths of the connecting cables. In this earlier treatment, the geometric relations were quite simple, and no angular quantities had to be considered. Now that we will be dealing with rigid-body mo-tion, however, we find that our defining geometric relations include both linear and angular variables and, therefore, the time derivatives of these quantities will involve both linear and angular velocities and linear and angular accelerations.
In absolute-motion analysis, it is essential that we be consistent with the mathematics of the description. For example, if the angular po-sition of a moving line in the plane of motion is specified by its counter-clockwise angle measured from some convenient fixed reference axis, then the positive sense for both angular velocity and angular acceler-ation will also be counterclockwise. A negative sign for either quantity will, of course, indicate a clockwise angular motion. The defining rela-tions for linear motion, Eqs. 2/1, 2/2, and 2/3, and the relations involving angular motion, Eqs. 5/1 and 5/2 or 5/3, will find repeated use in the mo-tion analysis and should be mastered.
The absolute-motion approach to rigid-body kinematics is quite straightforward, provided the configuration lends itself to a geometric description which is not overly complex. If the geometric configuration is awkward or complex, analysis by the principles of relative motion may be preferable. Relative-motion analysis is treated in this chapter beginning with Art. 5/4. The choice between absolute- and relative-motion analyses is best made after experience has been gained with both approaches.
The next three sample problems illustrate the application of absolute-motion analysis to three commonly encountered situations. The kine-matics of a rolling wheel, treated in Sample Problem 5/4, is especially important and will be useful in much of the problem work because the rolling wheel in various forms is such a common element in mechanical systems.
¨ ˙ 338 Chapter 5 Plane Kinematics of Rigid Bodies Ski-lift pulley tower near the Matter-horn in Switzerland.
Tim Macpherson/Stone/Getty Images c05.qxd 2/10/12 10:07 AM Page 338 SAMPLE PROBLEM 5/4 A wheel of radius r rolls on a flat surface without slipping. Determine the angular motion of the wheel in terms of the linear motion of its center O. Also determine the acceleration of a point on the rim of the wheel as the point comes into contact with the surface on which the wheel rolls.
Solution.
The figure shows the wheel rolling to the right from the dashed to the full position without slipping. The linear displacement of the center O is s, which is also the arc length CA along the rim on which the wheel rolls. The ra-dial line CO rotates to the new position CO through the angle , where is measured from the vertical direction. If the wheel does not slip, the arc CA must equal the distance s. Thus, the displacement relationship and its two time deriv-atives give Ans.
where vO aO and The angle , of course, must be in radians. The acceleration aO will be directed in the sense opposite to that of vO if the wheel is slowing down. In this event, the angular acceleration will have the sense opposite to that of .
The origin of fixed coordinates is taken arbitrarily but conveniently at the point of contact between C on the rim of the wheel and the ground. When point C has moved along its cycloidal path to C, its new coordinates and their time de-rivatives become For the desired instant of contact, 0 and Ans.
Thus, the acceleration of the point C on the rim at the instant of contact with the ground depends only on r and and is directed toward the center of the wheel. If desired, the velocity and acceleration of C at any position may be ob-tained by writing the expressions v and a Application of the kinematic relationships for a wheel which rolls without slipping should be recognized for various configurations of rolling wheels such as those illustrated on the right. If a wheel slips as it rolls, the foregoing relations are no longer valid.
y ¨j.
x ¨i y ˙j x ˙i x ¨ 0 and y ¨ r2 aO(1 cos ) r2 sin aO sin r2 cos x ¨ v ˙O(1 cos ) vO ˙ sin y ¨ v ˙O sin vO ˙ cos x ˙ r ˙(1 cos ) vO(1 cos ) y ˙ r ˙ sin vO sin x s r sin r( sin ) y r r cos r(1 cos ) ¨.
˙ ˙, s ¨, v ˙O s ˙, aO r vO r s r Article 5/3 Absolute Motion 339 C s x y vO aO α ω θ r s A s C′ O′ O C O C C O O C O Helpful Hints These three relations are not entirely unfamiliar at this point, and their ap-plication to the rolling wheel should be mastered thoroughly.
Clearly, when 0, the point of contact has zero velocity so that 0. The acceleration of the con-tact point on the wheel will also be obtained by the principles of relative motion in Art. 5/6.
y ˙ x ˙ c05.qxd 2/10/12 10:07 AM Page 339 SAMPLE PROBLEM 5/5 The load L is being hoisted by the pulley and cable arrangement shown.
Each cable is wrapped securely around its respective pulley so it does not slip.
The two pulleys to which L is attached are fastened together to form a single rigid body. Calculate the velocity and acceleration of the load L and the corre-sponding angular velocity and angular acceleration of the double pulley under the following conditions: Case (a) Pulley 1: 1 0 (pulley at rest) Pulley 2: 2 2 rad/sec, 2 3 rad/sec2 Case (b) Pulley 1: 1 1 rad/sec, 1 4 rad/sec2 Pulley 2: 2 2 rad/sec, 2 2 rad/sec2 Solution.
The tangential displacement, velocity, and acceleration of a point on the rim of pulley 1 or 2 equal the corresponding vertical motions of point A or B since the cables are assumed to be inextensible.
Case (a). With A momentarily at rest, line AB rotates to AB through the angle d during time dt. From the diagram we see that the displacements and their time derivatives give With vD r22 4(2) 8 in./sec and aD r22 4(3) 12 in./sec2, we have for the angular motion of the double pulley Ans.
Ans.
The corresponding motion of O and the load L is Ans.
Ans.
Case (b). With point C, and hence point A, in motion, line AB moves to AB during time dt. From the diagram for this case, we see that the displacements and their time derivatives give we have for the angular motion of the double pulley Ans.
Ans.
The corresponding motion of O and the load L is Ans.
Ans. aO (aA)t AO aC AO 16 4(2) 8 in./sec2 vO vA AO vC AO 4 4(1/3) 16/3 in./sec (aB)t (aA)t AB aD aC AB 8 16 12 2 rad/sec2 (CW) vB vA AB vD vC AB 8 4 12 1/3 rad/sec (CCW) aC r11 4(4) 16 in./sec2 aD r22 4(2) 8 in./sec2 vC r11 4(1) 4 in./sec vD r22 4(2) 8 in./sec With dsO dsA AO d vO vA AO aO (aA)t AO dsB dsA AB d vB vA AB (aB)t (aA)t AB aO AO 4(1) 4 in./sec2 vO AO 4(2/3) 8/3 in./sec (aB)t /AB aD /AB 12/12 1 rad/sec2 (CW) vB /AB vD /AB 8/12 2/3 rad/sec (CCW) dsO AO d vO AO aO AO dsB AB d vB AB (aB)t AB ˙2 ˙1 ˙2 ˙1 340 Chapter 5 Plane Kinematics of Rigid Bodies L A O C D 1 + + r1 = 4″ r2 = 4″ 2 B ω 8″ 4″ 2 ω1 A O B Case (a) dsO dsB vO aO (aB)t vB B′ dθ A O B Case (b) dsA dsO dsB dθ vA vO vB aO (aB)t (aA)t A′ B′ Helpful Hints Recognize that the inner pulley is a wheel rolling along the fixed line of the left-hand cable. Thus, the expres-sions of Sample Problem 5/4 hold.
Since B moves along a curved path, in addition to its tangential component of acceleration (aB)t, it will also have a normal component of acceleration to-ward O which does not affect the an-gular acceleration of the pulley.
The diagrams show these quantities and the simplicity of their linear re-lationships. The visual picture of the motion of O and B as AB rotates through the angle d should clarify the analysis.
Again, as in case (a), the differential rotation of line AB as seen from the figure establishes the relation be-tween the angular velocity of the pul-ley and the linear velocities of points A, O, and B. The negative sign for (aB)t aD produces the acceleration diagram shown but does not destroy the linearity of the relationships.
c05.qxd 2/10/12 10:07 AM Page 340 SAMPLE PROBLEM 5/6 Motion of the equilateral triangular plate ABC in its plane is controlled by the hydraulic cylinder D. If the piston rod in the cylinder is moving upward at the constant rate of 0.3 m/s during an interval of its motion, calculate for the instant when 30 the velocity and acceleration of the center of the roller B in the hor-izontal guide and the angular velocity and angular acceleration of edge CB.
Solution.
With the x-y coordinates chosen as shown, the given motion of A is vA 0.3 m/s and aA 0. The accompanying motion of B is given by x and its time derivatives, which may be obtained from x2 y2 b2. Differentiating gives With y b sin , x b cos , and 0, the expressions become Substituting the numerical values vA 0.3 m/s and 30 gives Ans.
Ans.
The negative signs indicate that the velocity and acceleration of B are both to the right since x and its derivatives are positive to the left.
The angular motion of CB is the same as that of every line on the plate, in-cluding AB. Differentiating y b sin gives The angular acceleration is Substitution of the numerical values gives Ans.
Ans.
Both and are counterclockwise since their signs are positive in the sense of the positive measurement of . (0.3)2 (0.2)2 2 3 2 1 3 1.732 rad/s2 0.3 0.2 2 3 1.732 rad/s ˙ vA b ˙ sec tan vA 2 b2 sec2 tan y ˙ b ˙ cos ˙ vA b sec aB (0.3)2(2/3)3 0.2 0.693 m/s2 vB 0.3 1 3 0.1732 m/s aB x ¨ vA 2 b sec3 vB x ˙ vA tan y ¨ xx ¨ x ˙2 yy ¨ y ˙2 0 x ¨ x ˙2 y ˙2 x y x y ¨ xx ˙ yy ˙ 0 x ˙ y x y ˙ y ¨ y ˙ Article 5/3 Absolute Motion 341 x y B C A b b b = 0.2 m D x y θ Helpful Hint Observe that it is simpler to differ-entiate a product than a quotient.
Thus, differentiate 0 rather than yy ˙/x.
x ˙ yy ˙ xx ˙ c05.qxd 2/10/12 10:07 AM Page 341 5/32 The small vehicle rides on rails and is driven by the 400-mm-diameter friction wheel turned by an electric motor. Determine the speed v of the vehicle if the friction-drive wheel is rotating at a speed of 300 rev/min and if no slipping occurs.
Problem 5/32 5/33 The Scotch-yoke mechanism converts rotational mo-tion of the disk to oscillatory translation of the shaft.
For given values of , and d, determine the velocity and acceleration of point P of the shaft.
Problem 5/33 5/34 Slider A moves in the horizontal slot with a constant speed v for a short interval of motion. Determine the angular velocity of bar AB in terms of the displace-ment Problem 5/34 xA B A v L 60° xA.
α ω θ P A r O d , , , r v 800 mm N 400 mm 342 Chapter 5 Plane Kinematics of Rigid Bodies PROBLEMS Introductory Problems 5/29 The fixed hydraulic cylinder C imparts a constant upward velocity v to the collar B, which slips freely on rod OA. Determine the resulting angular velocity in terms of v, the displacement s of point B, and the fixed distance d.
Problem 5/29 5/30 Point A is given a constant acceleration a to the right starting from rest with x essentially zero. Determine the angular velocity of link AB in terms of x and a.
Problem 5/30 5/31 The telephone-cable reel is rolled down the incline by the cable leading from the upper drum and wrapped around the inner hub of the reel. If the upper drum is turned at the constant rate calculate the time required for the center of the reel to move 100 ft along the incline. No slipping occurs.
Problem 5/31 1 ω 48″ 16″ 24″ 1 2 rad/sec, a x A C B b b O B A C d s v θ OA c05.qxd 2/10/12 10:07 AM Page 342 5/35 The cables at A and B are wrapped securely around the rims and the hub of the integral pulley as shown.
If the cables at A and B are given upward velocities of 3 ft/sec and 4 ft/sec, respectively, calculate the ve-locity of the center O and the angular velocity of the pulley.
Problem 5/35 5/36 The wheel of radius r rolls without slipping, and its center O has a constant velocity to the right. De-termine expressions for the magnitudes of the veloc-ity v and acceleration a of point A on the rim by differentiating its x- and y-coordinates. Represent your results graphically as vectors on your sketch and show that v is the vector sum of two vectors, each of which has a magnitude Problem 5/36 y x A r O θ vO xO vO.
vO 3 ft/sec 4 ft/sec 4″ 6″ A A B B O B Article 5/3 Problems 343 5/37 Determine the acceleration of the shaft B for if the crank OA has an angular acceleration and an angular velocity at this position. The spring maintains contact between the roller and the surface of the plunger.
Problem 5/37 5/38 The collar C moves to the left on the fixed guide with speed v. Determine the angular velocity as a function of v, the collar position s, and the height h.
Problem 5/38 v s h O A B C θ OA θ ω O B 20 mm 80 mm A ˙ 4 rad/s ¨ 8 rad/s2 60 c05.qxd 2/10/12 10:07 AM Page 343 Representative Problems 5/42 Calculate the angular velocity of the slender bar AB as a function of the distance x and the constant angular velocity of the drum.
Problem 5/42 5/43 The circular cam is mounted eccentrically about its fixed bearing at O and turns counterclockwise at the constant angular velocity The cam causes the fork A and attached control rod to oscillate in the hori-zontal x-direction. Write expressions for the velocity and acceleration of the control rod in terms of the angle measured from the vertical. The contact surfaces of the fork are vertical.
Problem 5/43 5/44 Rotation of the lever OA is controlled by the motion of the contacting circular disk whose center is given a horizontal velocity v. Determine the expression for the angular velocity of the lever OA in terms of x.
Problem 5/44 v B A x r O ax vx .
B A h x r 0 ω 0 344 Chapter 5 Plane Kinematics of Rigid Bodies 5/39 The linear actuator is designed for rapid horizontal velocity v of jaw C for a slow change in the distance between A and B. If the hydraulic cylinder decreases this distance at the rate u, determine the horizontal velocity of jaw C in terms of the angle Problem 5/39 5/40 The telephone-cable reel rolls without slipping on the horizontal surface. If point A on the cable has a velocity to the right, compute the veloc-ity of the center O and the angular velocity of the reel. (Be careful not to make the mistake of assum-ing that the reel rolls to the left.) Problem 5/40 5/41 As end A of the slender bar is pulled to the right with the velocity v, the bar slides on the surface of the fixed half-cylinder. Determine the angular velocity of the bar in terms of x.
Problem 5/41 A r x v B θ ˙ 1.8 m 0.6 m vA A O vA 0.8 m/s A B C L — 2 L — 2 L — 2 L — 2 θ .
A O b x θ ω c05.qxd 2/10/12 10:07 AM Page 344 5/45 Motion of the sliders B and C in the horizontal guide is controlled by the vertical motion of the slider A.
If A is given an upward velocity determine as a function of the magnitude v of the equal and oppo-site velocities which B and C have as they move toward one another.
Problem 5/45 5/46 Derive an expression for the upward velocity v of the car hoist in terms of . The piston rod of the hydraulic cylinder is extending at the rate .
Problem 5/46 5/47 The cable from drum A turns the double wheel B, which rolls on its hubs without slipping. Determine the angular velocity and angular acceleration of drum C for the instant when the angular velocity and angular acceleration of A are and respectively, both in the counterclockwise direction.
3 rad/sec2, 4 rad/sec θ θ L b 2b b s ˙ 2b 2b b b b b B C A vA θ θ vA, Article 5/3 Problems 345 Problem 5/47 5/48 The flywheel turns clockwise with a constant speed of The connecting link AB slides through the pivoted collar at C. Calculate the angu-lar velocity of AB for the instant when Problem 5/48 5/49 For the instant represented when the piston rod of the hydraulic cylinder C imparts a ver-tical motion to the pin B consisting of and For this instant determine the angular velocity and the angular acceleration of link OA. Members OA and AB make equal angles with the horizontal at this instant.
Problem 5/49 y O B C A 200 mm 300 mm y ¨ 100 mm/s2.
y ˙ 400 mm/s y 160 mm, 16″ 8″ A θ B C O 60.
600 rev/min.
10″ 6″ 4″ A C B 4″ c05.qxd 2/10/12 10:07 AM Page 345 Problem 5/52 5/53 Angular oscillation of the slotted link is achieved by the crank OA, which rotates clockwise at the steady speed Determine an expression for the angular velocity of the slotted link in terms of .
Problem 5/53 5/54 Link OA revolves counterclockwise with an angular velocity of Link AB slides through the piv-oted collar at C. Determine the angular velocity of AB when Problem 5/54 40.
3 rad/sec.
A B O N 2.5″ 9″ θ β ˙ N 120 rev/min.
v s r ω 346 Chapter 5 Plane Kinematics of Rigid Bodies 5/50 Link OA has an angular velocity as it passes the position shown. Determine the corre-sponding angular velocity of the slotted link CB.
Solve by considering the relation between the infini-tesimal displacements involved.
Problem 5/50 5/51 Show that the expressions and hold for the motion of the center O of the wheel which rolls on the concave or convex circular arc, where and are the absolute angular velocity and accelera-tion, respectively, of the wheel. (Hint: Follow the ex-ample of Sample Problem 5/4 and allow the wheel to roll a small distance. Be very careful to identify the correct absolute angle through which the wheel turns in each case in determining its angular veloc-ity and angular acceleration.) Problem 5/51 5/52 Film passes through the guide rollers shown and is being wound onto the reel, which is turned at a con-stant angular velocity . Determine the acceleration of the film as it enters the rollers. The thick-ness of the film is t, and s is sufficiently large so that the change in the angle made by the film with the horizontal is negligible.
a v ˙ at r v r O C A B 320 mm 120 mm 80 mm OA ω CB OA 8 rad/s t R R O r t O r 4″ 8″ A O C B θ 3 rad/sec c05.qxd 2/10/12 10:07 AM Page 346 5/55 The piston rod of the hydraulic cylinder gives point B a velocity as shown. Determine the magnitude of the velocity of point C in terms of .
Problem 5/55 5/56 The Geneva wheel is a mechanism for producing intermittent rotation. Pin P in the integral unit of wheel A and locking plate B engages the radial slots in wheel C, thus turning wheel C one-fourth of a rev-olution for each revolution of the pin. At the engage-ment position shown, For a constant clockwise angular velocity of wheel A, determine the corresponding counterclockwise angu-lar velocity of wheel C for (Note that the motion during engagement is governed by the geom-etry of triangle with changing .) Problem 5/56 200 mm P A C B O1 O2 200/ 2 mm 200/ 2 mm θ 1 ω 2 ω O1O2P 20.
2 1 2 rad/s 45.
vB θ b b b O A B C vC vB Article 5/3 Problems 347 5/57 One of the most common mechanisms is the slider-crank. Express the angular velocity and angular acceleration of the connecting rod AB in terms of the crank angle for a given constant crank speed Take and to be positive counterclock-wise.
Problem 5/57 5/58 The rod AB slides through the pivoted collar as end A moves along the slot. If A starts from rest at and moves to the right with a constant acceleration of , calculate the angular acceleration of AB at the instant when Problem 5/58 x 8″ A B C x 6 in.
4 in./sec2 x 0 0 ω θ A B l r O AB AB 0.
AB AB c05.qxd 2/10/12 10:07 AM Page 347 5/4 Relative Velocity The second approach to rigid-body kinematics is to use the principles of relative motion. In Art. 2/8 we developed these principles for motion relative to translating axes and applied the relative-velocity equation [2/20] to the motions of two particles A and B.
Relative Velocity Due to Rotation We now choose two points on the same rigid body for our two parti-cles. The consequence of this choice is that the motion of one point as seen by an observer translating with the other point must be circular since the radial distance to the observed point from the reference point does not change. This observation is the key to the successful under-standing of a large majority of problems in the plane motion of rigid bodies.
This concept is illustrated in Fig. 5/5a, which shows a rigid body moving in the plane of the figure from position AB to AB during time t. This movement may be visualized as occurring in two parts. First, the body translates to the parallel position AB with the displacement rB. Second, the body rotates about B through the angle . From the nonrotating reference axes x-y attached to the reference point B, you can see that this remaining motion of the body is one of simple rotation about B, giving rise to the displacement rA/B of A with respect to B. To the nonrotating observer attached to B, the body appears to undergo fixed-axis rotation about B with A executing circular motion as empha-sized in Fig. 5/5b. Therefore, the relationships developed for circular motion in Arts. 2/5 and 5/2 and cited as Eqs. 2/11 and 5/2 (or 5/3) de-scribe the relative portion of the motion of point A.
Point B was arbitrarily chosen as the reference point for attachment of our nonrotating reference axes x-y. Point A could have been used just as well, in which case we would observe B to have circular motion about A considered fixed as shown in Fig. 5/5c. We see that the sense of the vA vB vA/B 348 Chapter 5 Plane Kinematics of Rigid Bodies Y X B (a) (b) Motion relative to B Motion relative to A (c) B ΔrB Δθ Δθ ΔrA ΔrB Δ rA / B Δ rA / B Δ rB / A A A x r r r y B′ A′ A′ y′ x′ A″ B Δθ A r B′ Figure 5/5 c05.qxd 2/10/12 10:07 AM Page 348 rotation, counterclockwise in this example, is the same whether we choose A or B as the reference, and we see that rB/A rA/B.
With B as the reference point, we see from Fig. 5/5a that the total displacement of A is where rA/B has the magnitude r as approaches zero. We note that the relative linear motion rA/B is accompanied by the absolute angular motion , as seen from the translating axes x-y. Dividing the expres-sion for rA by the corresponding time interval t and passing to the limit, we obtain the relative-velocity equation (5/4) This expression is the same as Eq. 2/20, with the one restriction that the distance r between A and B remains constant. The magnitude of the rel-ative velocity is thus seen to be vA/B (rA/B/t) which, with becomes (5/5) Using r to represent the vector rA/B from the first of Eqs. 5/3, we may write the relative velocity as the vector (5/6) where is the angular-velocity vector normal to the plane of the motion in the sense determined by the right-hand rule. A critical observation seen from Figs. 5/5b and c is that the relative linear velocity is always perpendicular to the line joining the two points in question.
Interpretation of the Relative-Velocity Equation We can better understand the application of Eq. 5/4 by visualizing the separate translation and rotation components of the equation.
These components are emphasized in Fig. 5/6, which shows a rigid body vA/B r vA/B r ˙, lim tl0 (r/t) lim tl0 vA vB vA/B rA rB rA/B Article 5/4 Relative Velocity 349 Path of A Path of B B A = + r vB v A B A vB vB vB v A v A / B v A / B B A ω r Figure 5/6 c05.qxd 2/10/12 10:07 AM Page 349 in plane motion. With B chosen as the reference point, the velocity of A is the vector sum of the translational portion vB, plus the rotational por-tion vA/B r, which has the magnitude vA/B r, where the absolute angular velocity of AB. The fact that the relative linear ve-locity is always perpendicular to the line joining the two points in ques-tion is an important key to the solution of many problems. To reinforce your understanding of this concept, you should draw the equivalent dia-gram where point A is used as the reference point rather than B.
Equation 5/4 may also be used to analyze constrained sliding con-tact between two links in a mechanism. In this case, we choose points A and B as coincident points, one on each link, for the instant under con-sideration. In contrast to the previous example, in this case, the two points are on different bodies so they are not a fixed distance apart. This second use of the relative-velocity equation is illustrated in Sample Problem 5/10.
Solution of the Relative-Velocity Equation Solution of the relative-velocity equation may be carried out by scalar or vector algebra, or a graphical analysis may be employed. A sketch of the vector polygon which represents the vector equation should always be made to reveal the physical relationships involved.
From this sketch, you can write scalar component equations by project-ing the vectors along convenient directions. You can usually avoid solv-ing simultaneous equations by a careful choice of the projections.
Alternatively, each term in the relative-motion equation may be written in terms of its i- and j-components, from which you will obtain two scalar equations when the equality is applied, separately, to the coeffi-cients of the i- and j-terms.
Many problems lend themselves to a graphical solution, particularly when the given geometry results in an awkward mathematical expres-sion. In this case, we first construct the known vectors in their correct positions using a convenient scale. Then we construct the unknown vec-tors which complete the polygon and satisfy the vector equation. Fi-nally, we measure the unknown vectors directly from the drawing.
The choice of method to be used depends on the particular problem at hand, the accuracy required, and individual preference and experi-ence. All three approaches are illustrated in the sample problems which follow.
Regardless of which method of solution we employ, we note that the single vector equation in two dimensions is equivalent to two scalar equations, so that at most two scalar unknowns can be determined. The unknowns, for instance, might be the magnitude of one vector and the direction of another. We should make a systematic identification of the knowns and unknowns before attempting a solution.
˙, 350 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:07 AM Page 350 SAMPLE PROBLEM 5/7 The wheel of radius r 300 mm rolls to the right without slipping and has a velocity vO 3 m/s of its center O. Calculate the velocity of point A on the wheel for the instant represented.
Solution I (Scalar-Geometric). The center O is chosen as the reference point for the relative-velocity equation since its motion is given. We therefore write where the relative-velocity term is observed from the translating axes x-y at-tached to O. The angular velocity of AO is the same as that of the wheel which, from Sample Problem 5/4, is vO/r 3/0.3 10 rad/s. Thus, from Eq. 5/5 we have which is normal to AO as shown. The vector sum vA is shown on the diagram and may be calculated from the law of cosines. Thus, Ans.
The contact point C momentarily has zero velocity and can be used alterna-tively as the reference point, in which case, the relative-velocity equation be-comes vA vC vA/C vA/C where The distance 436 mm is calculated separately. We see that vA is normal to AC since A is momentarily rotating about point C.
Solution II (Vector). We will now use Eq. 5/6 and write where We now solve the vector equation Ans.
The magnitude vA 4.36 m/s and direction agree with the previous solution.
19 42 (1.732)2 4i 1.732j m/s vO 3i m/s r0 0.2(i cos 30 j sin 30) 0.1732i 0.1j m 10k rad/s vA vO vA/O vO r0 AC vA/C AC AC OC vO 0.436 0.300 (3) 4.36 m/s vA vA/C 4.36 m/s vA 2 32 22 2(3)(2) cos 60 19 (m/s)2 vA 4.36 m/s vA/O 0.2(10) 2 m/s [vA/O r0 ˙] vA vO vA/O Article 5/4 Relative Velocity 351 = 30° r = 300 mm r0 = 200 mm vO = 3 m/s θ A O O C A r r0 vO vA vA/O y x 60° Helpful Hints Be sure to visualize vA/O as the veloc-ity which A appears to have in its circular motion relative to O.
The vectors may also be laid off to scale graphically and the magnitude and direction of vA measured di-rectly from the diagram.
The velocity of any point on the wheel is easily determined by using the contact point C as the reference point. You should construct the ve-locity vectors for a number of points on the wheel for practice.
The vector is directed into the paper by the right-hand rule, whereas the positive z-direction is out from the paper; hence, the minus sign.
c05.qxd 2/10/12 10:07 AM Page 351 SAMPLE PROBLEM 5/8 Crank CB oscillates about C through a limited arc, causing crank OA to os-cillate about O. When the linkage passes the position shown with CB horizontal and OA vertical, the angular velocity of CB is 2 rad/s counterclockwise. For this instant, determine the angular velocities of OA and AB.
Solution I (Vector). The relative-velocity equation vA vB vA/B is rewritten as where Substitution gives Matching coefficients of the respective i- and j-terms gives the solutions of which are Ans.
Solution II (Scalar-Geometric). Solution by the scalar geometry of the vec-tor triangle is particularly simple here since vA and vB are at right angles for this special position of the linkages. First, we compute vB, which is and represent it in its correct direction as shown. The vector vA/B must be per-pendicular to AB, and the angle between vA/B and vB is also the angle made by AB with the horizontal direction. This angle is given by The horizontal vector vA completes the triangle for which we have The angular velocities become Ans.
Ans. OA vA OA 0.30 7 1 0.100 3/7 rad/s CW [ v/r] 6/7 rad/s CW AB vA/B AB 0.150 cos cos 0.250 0.075 [ v/r] vA vB tan 0.150(2/7) 0.30/7 m/s vA/B vB/cos 0.150/cos tan 100 50 250 75 2 7 vB 0.075(2) 0.150 m/s [v r] AB 6/7 rad/s and OA 3/7 rad/s 100OA 50AB 0 25(6 7AB) 0 100OAi 150j 175AB j 50ABi OAk 100j 2k (75i) ABk (175i 50j) rA 100j mm rB 75i mm rA/B 175i 50j mm OA OAk CB 2k rad/s AB ABk OA rA CB rB AB rA/B 352 Chapter 5 Plane Kinematics of Rigid Bodies 250 mm 50 mm 75 mm B x rA rA/B rB CB O A C y 100 mm ω vB = 150 mm/s vA vA/B θ Helpful Hints We are using here the first of Eqs.
5/3 and Eq. 5/6.
The minus signs in the answers indi-cate that the vectors AB and OA are in the negative k-direction.
Hence, the angular velocities are clockwise.
Always make certain that the se-quence of vectors in the vector polygon agrees with the equality of vectors specified by the vector equation.
c05.qxd 2/10/12 10:07 AM Page 352 SAMPLE PROBLEM 5/9 The common configuration of a reciprocating engine is that of the slider-crank mechanism shown. If the crank OB has a clockwise rotational speed of 1500 rev/min, determine for the position where 60 the velocity of the piston A, the velocity of point G on the connecting rod, and the angular velocity of the connecting rod.
Solution.
The velocity of the crank pin B as a point on AB is easily found, so that B will be used as the reference point for determining the velocity of A. The relative-velocity equation may now be written The crank-pin velocity is and is normal to OB. The direction of vA is, of course, along the horizontal cylin-der axis. The direction of vA/B must be perpendicular to the line AB as explained in the present article and as indicated in the lower diagram, where the reference point B is shown as fixed. We obtain this direction by computing angle from the law of sines, which gives We now complete the sketch of the velocity triangle, where the angle between vA/B and vA is 90 18.02 72.0 and the third angle is 180 30 72.0 78.0.
Vectors vA and vA/B are shown with their proper sense such that the head-to-tail sum of vB and vA/B equals vA. The magnitudes of the unknowns are now calculated from the trigonometry of the vector triangle or are scaled from the diagram if a graphical solution is used. Solving for vA and vA/B by the law of sines gives Ans.
The angular velocity of AB is counterclockwise, as revealed by the sense of vA/B, and is Ans.
We now determine the velocity of G by writing As seen from the diagram, vG/B has the same direction as vA/B. The vector sum is shown on the last diagram. We can calculate vG with some geometric labor or simply measure its magnitude and direction from the velocity diagram drawn to scale. For simplicity we adopt the latter procedure here and obtain Ans.
As seen, the diagram may be superposed directly on the first velocity diagram.
vG 64.1 ft/sec vG/B GBAB GB AB vA/B 4 14 (34.4) 9.83 ft/sec where vG vB vG/B AB vA/B AB 34.4 14/12 29.5 rad/sec [ v/r] vA/B sin 30 65.4 sin 72.0 vA/B 34.4 ft/sec vA sin 78.0 65.4 sin 72.0 vA 67.3 ft/sec 5 sin 14 sin 60 sin1 0.309 18.02 vB 5 12 1500 (2) 60 65.4 ft/sec [v r] vA vB vA/B Article 5/4 Relative Velocity 353 A O B β θ ω G r = 5″ 4″ 10″ Helpful Hints Remember always to convert to radians per unit time when using v r.
30° = 60° θ ω vA/B vA vB = 65.4 ft/sec 78.0° 18.02° r = 5″ A O B vB vA 72.0° 14″ 10″ 4″ A P G ω B vA/B vG/B vG/B vB = 65.4 ft/sec vG AB A graphical solution to this problem is the quickest to achieve, although its accuracy is limited. Solution by vector algebra can, of course, be used but would involve somewhat more labor in this problem.
c05.qxd 2/10/12 10:07 AM Page 353 SAMPLE PROBLEM 5/10 The power screw turns at a speed which gives the threaded collar C a veloc-ity of 0.8 ft/sec vertically down. Determine the angular velocity of the slotted arm when 30.
Solution.
The angular velocity of the arm can be found if the velocity of a point on the arm is known. We choose a point A on the arm coincident with the pin B of the collar for this purpose. If we use B as our reference point and write vA vB vA/B, we see from the diagram, which shows the arm and points A and B an instant before and an instant after coincidence, that vA/B has a direction along the slot away from O.
The magnitudes of vA and vA/B are the only unknowns in the vector equa-tion, so that it may now be solved. We draw the known vector vB and then obtain the intersection P of the known directions of vA/B and vA. The solution gives Ans.
We note the difference between this problem of constrained sliding contact be-tween two links and the three preceding sample problems of relative velocity, where no sliding contact occurred and where the points A and B were located on the same rigid body in each case. 0.400 rad/sec CCW vA OA 0.693 (18 12 )/cos 30 [ v/r] vA vB cos 0.8 cos 30 0.693 ft/sec 354 Chapter 5 Plane Kinematics of Rigid Bodies 18″ O C B = 30° θ Helpful Hints Physically, of course, this point does not exist, but we can imagine such a point in the middle of the slot and attached to the arm.
A B A O B ω P vA vB = 0.8 ft/sec 30° vA/B Always identify the knowns and un-knowns before attempting the solu-tion of a vector equation.
c05.qxd 2/10/12 10:07 AM Page 354 PROBLEMS Introductory Problems 5/59 Bar AB moves on the horizontal surface. Its mass center has a velocity directed parallel to the y-axis and the bar has a counterclockwise (as seen from above) angular velocity De-termine the velocity of point B.
Problem 5/59 5/60 The cart has a velocity of 4 ft/sec to the right. Deter-mine the angular speed N of the wheel so that point A on the top of the rim has a velocity (a) equal to 4 ft/sec to the left, (b) equal to zero, and (c) equal to 8 ft/sec to the right.
Problem 5/60 10″ A B O C C = 4 ft/sec v vG 30° 0.4 m 0.4 m A G B x y z ω 4 rad/s.
vG 2 m/s Article 5/4 Problems 355 5/61 The speed of the center of the earth as it orbits the sun is and the absolute angular velocity of the earth about its north-south spin axis is Use the value for the radius of the earth and determine the veloci-ties of points A, B, C, and D, all of which are on the equator. The inclination of the axis of the earth is neglected.
Problem 5/61 5/62 A control element in a special-purpose mechanism un-dergoes motion in the plane of the figure. If the velocity of B with respect to A has a magnitude of 0.926 m/s at a certain instant, what is the correspond-ing magnitude of the velocity of C with respect to D?
Problem 5/62 80 mm 70 mm 50 mm A B C D x A v B Sunlight C D y N ω R 6371 km 7.292(105) rad/s.
v 107 257 km/h, c05.qxd 2/10/12 10:07 AM Page 355 5/66 Determine the angular velocity of the telescoping link AB for the position shown where the driving links have the angular velocities indicated.
Problem 5/66 5/67 The ends of bar AB are confined to the circular slot.
By the method of this article, determine the angular velocity of the bar if the velocity of end A is as shown.
Problem 5/67 0.8 m 45° 0.3 m/s A B O 0.3 m/s 165 Dimensions in millimeters 60 150 45 2 rad/s 2 rad/s A B C O 356 Chapter 5 Plane Kinematics of Rigid Bodies 5/63 End A of the 24-in. link has a velocity of 10 ft/sec in the direction shown. At the same instant end B has a velocity whose magnitude is 12 ft/sec as indicated.
Find the angular velocity of the link in two ways.
Problem 5/63 5/64 The circular disk of radius 0.2 m is released very near the horizontal surface with a velocity of its cen-ter to the right and a clockwise angular velocity Determine the velocities of points A and P of the disk. Describe the motion upon contact with the ground.
Problem 5/64 5/65 For the instant represented the curved link has a counterclockwise angular velocity of 4 rad/s, and the roller at B has a velocity of 40 mm/s along the con-straining surface as shown. Determine the magni-tude of the velocity of A.
Problem 5/65 vB 20 mm 45° A B vA ω A P O 0.2 m y x vO 2 rad/s.
vO 0.7 m/s 30° 24″ B A β vA = 10 ft/sec vB = 12 ft/sec c05.qxd 2/10/12 10:07 AM Page 356 5/68 The two pulleys are riveted together to form a single rigid unit, and each of the two cables is securely wrapped around its respective pulley. If point A on the hoisting cable has a velocity deter-mine the magnitudes of the velocity of point O and the velocity of point B on the larger pulley for the position shown.
Problem 5/68 5/69 The right-angle link AB has a clockwise angular velocity of at the instant when Express the velocity of A with respect to B in vector notation for this instant.
Problem 5/69 θ 12″ A B 5″ 3 rad/sec y x 60.
3 rad/sec A B O v 180 mm 90 mm L v 0.9 m/s, Article 5/4 Problems 357 5/70 The magnitude of the absolute velocity of point A on the automobile tire is when A is in the posi-tion shown. What are the corresponding velocity of the car and the angular velocity of the wheel?
(The wheel rolls without slipping.) Problem 5/70 5/71 For the instant represented point B crosses the hori-zontal axis through point O with a downward veloc-ity Determine the corresponding value of the angular velocity of link OA.
Problem 5/71 5/72 The circular disk rolls without slipping with a clock-wise angular velocity For the instant represented, write the vector expressions for the ve-locity of A with respect to B and for the velocity of P.
Problem 5/72 B A C O P 300 mm 200 mm = 4 rad/s ω y x 4 rad/s.
v 180 mm 130 mm 90 mm O A B OA v 0.6 m/s.
v0 650 mm A O vO 12 m/s c05.qxd 2/10/12 10:07 AM Page 357 5/76 Determine the angular velocity of link AB and the velocity of collar B for the instant repre-sented. Assume the quantities and r to be known.
Problem 5/76 5/77 Determine the angular velocity of link AB and the velocity of collar B for the instant represented.
Assume the quantities and r to be known.
Problem 5/77 5/78 Motion of the threaded collars A and B is controlled by the rotation of their respective lead screws. If A has a velocity to the right of and B has a velocity to the left of when in., deter-mine the angular velocity of ACD at this instant.
Problem 5/78 x 6 2 in./sec 3 in./sec r 2r ω B O A 45° 30° 0 0 vB AB r 2r ω B O A 45° 0 0 vB AB 358 Chapter 5 Plane Kinematics of Rigid Bodies Representative Problems 5/73 At the instant represented, the velocity of point A of the 1.2-m bar is to the right. Determine the speed of point B and the angular velocity of the bar. The diameter of the small end wheels may be neglected.
Problem 5/73 5/74 For an interval of its motion the piston rod of the hydraulic cylinder has a velocity as shown. At a certain instant For this in-stant determine the angular velocity of link BC.
Problem 5/74 5/75 Each of the sliding bars A and B engages its respec-tive rim of the two riveted wheels without slipping.
Determine the magnitude of the velocity of point P for the position shown.
Problem 5/75 100 mm 160 mm O P A vA = 0.8 m/s vB = 0.6 m/s B 10″ 20″ A D B C vA θ β BC 60.
vA 4 ft/sec 1.2 m A B 60° 0.5 m vA vB 3 m/s 6″ 3″ 4″ C D B A x c05.qxd 2/10/12 10:07 AM Page 358 5/79 At the instant represented the triangular plate ABD has a clockwise angular velocity of For this instant determine the angular velocity of link BC.
Problem 5/79 5/80 For the instant represented, crank OB has a clock-wise angular velocity and is passing the horizontal position. Determine the correspond-ing velocity of the guide roller A in the 20° slot and the velocity of point C midway between A and B.
Problem 5/80 C O 20″ 20° 10″ ω B A 0.8 rad/sec 3″ 7″ 3″ 5″ 5″ 5″ O D A C B BC 3 rad/sec.
Article 5/4 Problems 359 5/81 The ends of the 0.4-m slender bar remain in contact with their respective support surfaces. If end B has a velocity in the direction shown, deter-mine the angular velocity of the bar and the velocity of end A.
Problem 5/81 5/82 End A of the link has a downward velocity of during an interval of its motion. For the posi-tion where determine the angular velocity of AB and the velocity of the midpoint G of the link. Solve the relative-velocity equations, first, using the geometry of the vector polygon and, sec-ond, using vector algebra.
Problem 5/82 x y B G A vA θ 200 mm vG 30 2 m/s vA A B 105° 30° 0.4 m vB = 0.5 m/s vB 0.5 m/s c05.qxd 2/10/12 10:07 AM Page 359 5/86 The elements of a switching device are shown. If the vertical control rod has a downward velocity v of when and if roller A is in continuous contact with the horizontal surface, determine the magnitude of the velocity of C for this instant.
Problem 5/86 5/87 The Geneva mechanism of Prob. 5/56 is shown again here. By relative-motion principles determine the an-gular velocity of wheel C for Wheel A has a constant clockwise angular velocity Problem 5/87 200 mm P A C B O1 O2 200/ 2 mm 200/ 2 mm θ 1 ω 2 ω 1 2 rad/s.
20.
2 3″ 3″ B C θ A v 60 3 ft/sec 360 Chapter 5 Plane Kinematics of Rigid Bodies 5/83 Horizontal motion of the piston rod of the hydraulic cylinder controls the rotation of link OB about O.
For the instant represented, and OB is horizontal. Determine the angular velocity of OB for this instant.
Problem 5/83 5/84 The flywheel turns clockwise with a constant speed of and the connecting rod AB slides through the pivoted collar at C. For the position determine the angular velocity of AB by using the relative-velocity relations. (Suggestion: Choose a point D on AB coincident with C as a refer-ence point whose direction of velocity is known.) Problem 5/84 5/85 Determine the velocity of point D which will produce a counterclockwise angular velocity of for link AB in the position shown for the four-bar linkage.
Problem 5/85 75 mm A O C B D 150 mm 100 mm 100 mm AB ω 40 rad/s 8″ 16″ θ A C B O AB 45, 600 rev/min, 160 mm 120 mm 180 mm A B O vA = 2 m/s θ vA 2 m/s c05.qxd 2/10/12 5:43 PM Page 360 5/88 A four-bar linkage is shown in the figure (the ground “link” OC is considered the fourth bar). If the drive link OA has a counterclockwise angular velocity determine the angular velocities of links AB and BC.
Problem 5/88 5/89 The elements of the mechanism for deployment of a spacecraft magnetometer boom are shown. Deter-mine the angular velocity of the boom when the driving link OB crosses the y-axis with an angular velocity if at this instant.
Problem 5/89 120 mm 120 mm 200 mm θ A C B O y x ωOB ω tan 4/3 OB 0.5 rad/s 15° 60° 200 mm 80 mm A B O C 240 mm ω 0 0 10 rad/s, Article 5/4 Problems 361 5/90 Ends A and C of the connected links are controlled by the vertical motion of the piston rods of the hy-draulic cylinders. For a short interval of motion, A has an upward velocity of and C has a down-ward velocity of Determine the velocity of B for the instant when Problem 5/90 y C B A 200 mm 250 mm 250 mm y 150 mm.
2 m/s.
3 m/s, c05.qxd 2/10/12 10:07 AM Page 361 362 Chapter 5 Plane Kinematics of Rigid Bodies 5/5 Instantaneous Center of Zero Velocity In the previous article, we determined the velocity of a point on a rigid body in plane motion by adding the relative velocity due to rotation about a convenient reference point to the velocity of the reference point.
We now solve the problem by choosing a unique reference point which momentarily has zero velocity. As far as velocities are concerned, the body may be considered to be in pure rotation about an axis, normal to the plane of motion, passing through this point. This axis is called the instantaneous axis of zero velocity, and the intersection of this axis with the plane of motion is known as the instantaneous center of zero velocity.
This approach provides us with a valuable means for visualizing and an-alyzing velocities in plane motion.
Locating the Instantaneous Center The existence of the instantaneous center is easily shown. For the body in Fig. 5/7, assume that the directions of the absolute velocities of any two points A and B on the body are known and are not parallel. If there is a point about which A has absolute circular motion at the in-stant considered, this point must lie on the normal to vA through A.
Similar reasoning applies to B, and the intersection of the two perpen-diculars fulfills the requirement for an absolute center of rotation at the instant considered. Point C is the instantaneous center of zero velocity and may lie on or off the body. If it lies off the body, it may be visualized as lying on an imaginary extension of the body. The instantaneous cen-ter need not be a fixed point in the body or a fixed point in the plane.
If we also know the magnitude of the velocity of one of the points, say, vA, we may easily obtain the angular velocity of the body and the linear velocity of every point in the body. Thus, the angular velocity of the body, Fig. 5/7a, is which, of course, is also the angular velocity of every line in the body.
Therefore, the velocity of B is vB rB (rB/rA)vA. Once the instanta-neous center is located, the direction of the instantaneous velocity of vA rA C C C A A A B B B rA vA vA vA vB vB vB rB (a) (b) (c) Figure 5/7 c05.qxd 2/10/12 10:07 AM Page 362 Article 5/5 Instantaneous Center of Zero Velocity 363 every point in the body is readily found since it must be perpendicular to the radial line joining the point in question with C.
If the velocities of two points in a body having plane motion are par-allel, Fig. 5/7b or 5/7c, and the line joining the points is perpendicular to the direction of the velocities, the instantaneous center is located by di-rect proportion as shown. We can readily see from Fig. 5/7b that as the parallel velocities become equal in magnitude, the instantaneous center moves farther away from the body and approaches infinity in the limit as the body stops rotating and translates only.
Motion of the Instantaneous Center As the body changes its position, the instantaneous center C also changes its position both in space and on the body. The locus of the in-stantaneous centers in space is known as the space centrode, and the locus of the positions of the instantaneous centers on the body is known as the body centrode. At the instant considered, the two curves are tan-gent at the position of point C. It can be shown that the body-centrode curve rolls on the space-centrode curve during the motion of the body, as indicated schematically in Fig. 5/8.
Although the instantaneous center of zero velocity is momentarily at rest, its acceleration generally is not zero. Thus, this point may not be used as an instantaneous center of zero acceleration in a manner analo-gous to its use for finding velocity. An instantaneous center of zero ac-celeration does exist for bodies in general plane motion, but its location and use represent a specialized topic in mechanism kinematics and will not be treated here.
C Body centrode Space centrode Figure 5/8 This valve gear of a steam locomotive provides an interesting (albeit not cutting-edge) study in rigid-body kinematics.
© Peter Jordan_NE/Alamy c05.qxd 2/10/12 10:07 AM Page 363 364 Chapter 5 Plane Kinematics of Rigid Bodies SAMPLE PROBLEM 5/11 The wheel of Sample Problem 5/7, shown again here, rolls to the right with-out slipping, with its center O having a velocity vO 3 m/s. Locate the instanta-neous center of zero velocity and use it to find the velocity of point A for the position indicated.
Solution.
The point on the rim of the wheel in contact with the ground has no velocity if the wheel is not slipping; it is, therefore, the instantaneous center C of zero velocity. The angular velocity of the wheel becomes The distance from A to C is The velocity of A becomes Ans.
The direction of vA is perpendicular to AC as shown.
SAMPLE PROBLEM 5/12 Arm OB of the linkage has a clockwise angular velocity of 10 rad/sec in the position shown where 45. Determine the velocity of A, the velocity of D, and the angular velocity of link AB for the position shown.
Solution.
The directions of the velocities of A and B are tangent to their circu-lar paths about the fixed centers O and O as shown. The intersection of the two perpendiculars to the velocities from A and B locates the instantaneous center C for the link AB. The distances and shown on the diagram are com-puted or scaled from the drawing. The angular velocity of BC, considered a line on the body extended, is equal to the angular velocity of AC, DC, and AB and is Ans.
Thus, the velocities of A and D are Ans.
Ans.
in the directions shown. vD 15.23 12 (4.29) 5.44 ft/sec vA 14 12 (4.29) 5.00 ft/sec [v r] 4.29 rad/sec CCW BC vB BC OBOB BC 62(10) 142 [ v/r] DC BC, AC, vA AC 0.436(10) 4.36 m/s [v r] AC (0.300)2 (0.200)2 2(0.300)(0.200) cos 120 0.436 m vO / OC 3/0.300 10 rad/s [ v/r] = 30° r = 300 mm r0 = 200 mm vO = 3 m/s θ A O O 0.200 m 0.300 m vA A C 120° θ = 45° A D O B 6″ 6″ 8″ 2 ″ 6 O′ Body extended C A O′ O D B 14″ 15.23″ 2″ 14 vA vD vB 45° Helpful Hints Be sure to recognize that the cosine of 120 is itself negative.
From the results of this problem, you should be able to visualize and sketch the velocities of all points on the wheel.
Helpful Hint For the instant depicted, we should visualize link AB and its body ex-tended to be rotating as a single unit about point C.
c05.qxd 2/10/12 10:07 AM Page 364 Article 5/5 Instantaneous Center of Zero Velocity 365 5/94 The circular disk of Prob. 5/64 is repeated here. If the disk is released very near the horizontal surface with and locate the instan-taneous center of rotation and determine the veloci-ties of points A and P of the disk.
Problem 5/94 5/95 For the instant represented, when crank OA passes the horizontal position, determine the velocity of the center G of link AB by the method of this article.
Problem 5/95 5/96 The bar AB has a clockwise angular velocity of Construct and determine the vector velocity of each end if the instantaneous center of zero veloc-ity is (a) at and (b) at .
Problem 5/96 8″ 4″ 3″ B C1 C2 A y x 5 rad/s C2 C1 5 rad/sec.
G A O 8 rad/s B 90 mm 90 mm 60 mm 90 mm ω A P O 0.2 m y x vO 2 rad/s, vO 0.7 m/s PROBLEMS Introductory Problems 5/91 The slender bar is moving in general plane motion with the indicated linear and angular properties.
Locate the instantaneous center of zero velocity and determine the velocities of points A and B.
Problem 5/91 5/92 The slender bar is moving in general plane motion with the indicated linear and angular properties.
Locate the instantaneous center of zero velocity and determine the velocities of points A and B.
Problem 5/92 5/93 The figure for Prob. 5/83 is repeated here. Solve for the angular velocity of OB by the method of this article.
Problem 5/93 160 mm 120 mm 180 mm A B O vA = 2 m/s θ B A G 0.3 m 0.3 m 2 m/s 20° 4 rad/s B A G 0.3 m 0.3 m 2 m/s 4 rad/s c05.qxd 2/10/12 10:08 AM Page 365 5/97 The bar of Prob. 5/81 is repeated here. By the method of this article, determine the velocity of end A. Both ends remain in contact with their respective support surfaces.
Problem 5/97 5/98 A car mechanic “walks” two wheel/tire units across a horizontal floor as shown. He walks with constant speed v and keeps the tires in the configuration shown with the same position relative to his body.
If there is no slipping at any interface, determine (a) the angular velocity of the lower tire, (b) the an-gular velocity of the upper tire, and (c) the velocities of points A, B, C, and D. The radius of both tires is r.
Problem 5/98 v A B C r P D r A B 105° 30° 0.4 m vB = 0.5 m/s 5/99 The linkage of Prob. 5/80 is repeated here. At the instant represented, crank OB has a clockwise angular velocity and is passing the horizontal position. By the method of this article, determine the corresponding velocity of the guide roller A in the 20° slot and the velocity of point C midway between A and B.
Problem 5/99 5/100 Motion of the bar is controlled by the constrained paths of A and B. If the angular velocity of the bar is counterclockwise as the position is passed, determine the speeds of points A and P.
Problem 5/100 B P A 500 mm 500 mm θ 2 rad/s 45 2 rad/s C O 20″ 20° 10″ ω B A 0.8 rad/sec 366 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 366 Article 5/5 Problems 367 Representative Problems 5/103 The mechanism of Prob. 5/74 is repeated here. For an interval of its motion the piston rod of the hydraulic cylinder has a velocity as shown. At a certain instant By the method of this article, determine the angular veloc-ity of link BC.
Problem 5/103 5/104 The mechanism of Prob. 5/76 is repeated here. By the method of this article, determine the angular velocity of link AB and the velocity of collar B for the instant shown. Assume the quantities and r to be known.
Problem 5/104 r 2r ω B O A 45° 0 0 10″ 20″ A D B C vA θ β BC 60.
vA 4 ft/sec 5/101 Motion of the rectangular plate P is controlled by the two links which cross without touching. For the instant represented where the links are perpendic-ular to each other, the plate has a counterclockwise angular velocity Determine the corre-sponding angular velocities of the two links.
Problem 5/101 5/102 The mechanism of Prob. 5/34 is repeated here. At the instant when the velocity of the slider at A is to the right. Determine the corresponding velocity of slider B and the angular velocity of bar AB if Problem 5/102 xA B A v L 60° L 0.8 m.
v 2 m/s xA 0.85L, y x B D O A P 0.2 m 0.2 m P = 2 rad/s ω A — O – = 0.6 m B — D – = 0.5 m P 2 rad/s.
c05.qxd 2/10/12 10:08 AM Page 367 5/105 The mechanism of Prob. 5/77 is repeated here. By the method of this article, determine the angular velocity of link AB and the velocity of collar B for the instant depicted. Assume the quantities and r to be known.
Problem 5/105 5/106 The rectangular body B is pivoted to the crank OA at A and is supported by the wheel at D. If OA has a counterclockwise angular velocity of de-termine the velocity of point E and the angular ve-locity of body B when the crank OA passes the vertical position shown.
Problem 5/106 120 mm 40 mm 40 mm 80 mm A B O D E 320 mm 2 rad/s, r 2r ω B O A 45° 30° 0 0 5/107 The sliding rails A and B engage the rims of the double wheel without slipping. For the specified ve-locities of A and B, determine the angular velocity of the wheel and the magnitude of the velocity of point P.
Problem 5/107 5/108 Horizontal oscillation of the spring-loaded plunger E is controlled by varying the air pressure in the horizontal pneumatic cylinder F. If the plunger has a velocity of to the right when deter-mine the downward velocity of roller D in the vertical guide and find the angular velocity of ABD for this position.
Problem 5/108 θ 100 mm 200 mm A B E F D vD 30, 2 m/s O P B A 80 mm 120 mm vB = 1.8 m/s vA = 1.2 m/s 368 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 368 Article 5/5 Problems 369 5/111 The mechanism of Prob. 5/55 is repeated here. By the method of this article determine the expression for the magnitude of the velocity of point C in terms of the velocity of the piston rod and the angle .
Problem 5/111 5/112 Link OA has a counterclockwise angular velocity during an interval of its motion.
Determine the angular velocity of link AB and of sector BD for at which instant AB is hori-zontal and BD is vertical.
Problem 5/112 5/113 The mechanism of Prob. 5/84 is repeated here. The flywheel turns clockwise with a constant speed of and the connecting rod AB slides through the pivoted collar at C. For the position determine the angular velocity of AB by the method of this article.
Problem 5/113 8″ 16″ θ A C B O AB 45, 600 rev/min, D B A O 6″ 16″ 8″ θ B 45 4 rad/sec ˙ vB θ b b b O A B C vB 5/109 The rear driving wheel of a car has a diameter of 26 in. and has an angular speed N of on an icy road. If the instantaneous center of zero velocity is 4 in. above the point of contact of the tire with the road, determine the velocity v of the car and the slipping velocity of the tire on the ice.
Problem 5/109 5/110 The elements of the mechanism for deployment of a spacecraft magnetometer boom are repeated here from Prob. 5/89. By the method of this article, determine the angular velocity of the boom when the driving link OB crosses the y-axis with an an-gular velocity if at this instant Problem 5/110 120 mm 120 mm 200 mm θ A C B O y x ωOB ω tan 4/3.
OB 0.5 rad/s 26″ N vs 200 rev/min c05.qxd 2/10/12 10:08 AM Page 369 5/114 The hydraulic cylinder produces a limited horizon-tal motion of point A. If when determine the magnitude of the velocity of D and the angular velocity of ABD for this position.
Problem 5/114 5/115 The flexible band F is attached at E to the rotating sector and leads over the guide pulley. Determine the angular velocities of AD and BD for the posi-tion shown if the band has a velocity of Problem 5/115 5/116 Motion of the roller A against its restraining spring is controlled by the downward motion of the plunger E. For an interval of motion the velocity of E is Determine the velocity of A when becomes 90°.
v 0.2 m/s.
4 m/s O A D B E F 200 mm 75 mm 125 mm 250 mm 100 mm 4 m/s.
A D B O vA θ 400 mm 200 mm 250 mm 45, vA 4 m/s Problem 5/116 5/117 In the design of this mechanism, upward motion of the plunger G controls the motion of a control rod attached at A. Point B of link AH is confined to move with the sliding collar on the fixed vertical shaft ED. If G has a velocity for a short interval, determine the velocity of A for the posi-tion .
Problem 5/117 vG 240 mm 80 mm A B D H G F E θ 200 mm 160 mm 45 v G 2 m/s A O B v D E θ 90 mm 120 mm 60 mm 370 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 370 Article 5/5 Problems 371 Problem 5/119 5/120 The shaft at O drives the arm OA at a clockwise speed of about the fixed bearing at O.
Use the method of the instantaneous center of zero velocity to determine the rotational speed of gear B (gear teeth not shown) if (a) ring gear D is fixed and (b) ring gear D rotates counterclockwise about O with an angular speed of Problem 5/120 A B D O a a 80 rev/min.
90 rev/min 100 mm 200 mm 250 mm O 5/118 Determine the angular velocity of the ram head AE of the rock crusher in the position for which . The crank OB has an angular speed of When B is at the bottom of its circle, D and E are on a horizontal line through F, and lines BD and AE are vertical. The dimensions are , and Carefully construct the configuration graphically, and use the method of this article.
Problem 5/118 5/119 The large roller bearing rolls to the left on its outer race with a velocity of its center O of At the same time the central shaft and inner race rotate counterclockwise with an angular speed of Determine the angular velocity of each of the rollers.
240 rev/min.
0.9 m/s.
θ A B D E F O AE ED DF 15 in.
BD 30 in., OB 4 in.
60 rev/min.
60 c05.qxd 2/10/12 10:08 AM Page 371 372 Chapter 5 Plane Kinematics of Rigid Bodies 5/6 Relative Acceleration Consider the equation vA vB vA/B, which describes the relative velocities of two points A and B in plane motion in terms of nonrotating reference axes. By differentiating the equation with respect to time, we may obtain the relative-acceleration equation, which is or (5/7) In words, Eq. 5/7 states that the acceleration of point A equals the vec-tor sum of the acceleration of point B and the acceleration which A ap-pears to have to a nonrotating observer moving with B.
Relative Acceleration Due to Rotation If points A and B are located on the same rigid body and in the plane of motion, the distance r between them remains constant so that the observer moving with B perceives A to have circular motion about B, as we saw in Art. 5/4 with the relative-velocity relationship. Because the relative motion is circular, it follows that the relative-acceleration term will have both a normal component directed from A toward B due to the change of direction of vA/B and a tangential component perpendicular to AB due to the change in magnitude of vA/B. These acceleration compo-nents for circular motion, cited in Eqs. 5/2, were covered earlier in Art.
2/5 and should be thoroughly familiar by now.
Thus we may write (5/8) where the magnitudes of the relative-acceleration components are (5/9) In vector notation the acceleration components are (5/9a) In these relationships, is the angular velocity and is the angular ac-celeration of the body. The vector locating A from B is r. It is important to observe that the relative acceleration terms depend on the respective absolute angular velocity and absolute angular acceleration.
Interpretation of the Relative-Acceleration Equation The meaning of Eqs. 5/8 and 5/9 is illustrated in Fig. 5/9, which shows a rigid body in plane motion with points A and B moving along separate curved paths with absolute accelerations aA and aB. Con-trary to the case with velocities, the accelerations aA and aB are, in general, not tangent to the paths described by A and B when these (aA/B)t r (aA/B)n ( r) (aA/B)t v ˙A/B r (aA/B)n vA/B 2/r r2 aA aB (aA/B)n (aA/B)t aA aB aA/B v ˙A v ˙B v ˙A/B A A A B Path of A Path of B = + r aA aB B B aB aB aA/B aA/B aA aB (aA/B)t (aA/B)n (aA/B)n (aA/B)t t n ω α Figure 5/9 c05.qxd 2/10/12 10:08 AM Page 372 Article 5/6 Relative Acceleration 373 paths are curvilinear. The figure shows the acceleration of A to be composed of two parts: the acceleration of B and the acceleration of A with respect to B. A sketch showing the reference point as fixed is useful in disclosing the correct sense of each of the two components of the relative-acceleration term.
Alternatively, we may express the acceleration of B in terms of the acceleration of A, which puts the nonrotating reference axes on A rather than B. This order gives Here aB/A and its n- and t-components are the negatives of aA/B and its n- and t-components. To understand this analysis better, you should make a sketch corresponding to Fig. 5/9 for this choice of terms.
Solution of the Relative-Acceleration Equation As in the case of the relative-velocity equation, we can handle the solution to Eq. 5/8 in three different ways, namely, by scalar algebra and geometry, by vector algebra, or by graphical construction. It is helpful to be familiar with all three techniques. You should make a sketch of the vector polygon representing the vector equation and pay close attention to the head-to-tail combination of vectors so that it agrees with the equation. Known vectors should be added first, and the unknown vec-tors will become the closing legs of the vector polygon. It is vital that you visualize the vectors in their geometrical sense, as only then can you understand the full significance of the acceleration equation.
Before attempting a solution, identify the knowns and unknowns, keeping in mind that a solution to a vector equation in two dimensions can be carried out when the unknowns have been reduced to two scalar quantities. These quantities may be the magnitude or direction of any of the terms of the equation. When both points move on curved paths, there will, in general, be six scalar quantities to account for in Eq. 5/8.
Because the normal acceleration components depend on velocities, it is generally necessary to solve for the velocities before the acceleration calcu-lations can be made. Choose the reference point in the relative-acceleration equation as some point on the body in question whose acceleration is ei-ther known or can be easily found. Be careful not to use the instantaneous center of zero velocity as the reference point unless its acceleration is known and accounted for.
An instantaneous center of zero acceleration exists for a rigid body in general plane motion, but will not be discussed here since its use is somewhat specialized.
aB aA aB/A c05.qxd 2/10/12 10:08 AM Page 373 374 Chapter 5 Plane Kinematics of Rigid Bodies SAMPLE PROBLEM 5/13 The wheel of radius r rolls to the left without slipping and, at the instant con-sidered, the center O has a velocity vO and an acceleration aO to the left. Deter-mine the acceleration of points A and C on the wheel for the instant considered.
Solution.
From our previous analysis of Sample Problem 5/4, we know that the angular velocity and angular acceleration of the wheel are The acceleration of A is written in terms of the given acceleration of O. Thus, The relative-acceleration terms are viewed as though O were fixed, and for this relative circular motion they have the magnitudes and the directions shown.
Adding the vectors head-to-tail gives aA as shown. In a numerical problem, we may obtain the combination algebraically or graphically. The algebraic ex-pression for the magnitude of aA is found from the square root of the sum of the squares of its components. If we use n- and t-directions, we have Ans.
The direction of aA can be computed if desired.
The acceleration of the instantaneous center C of zero velocity, considered a point on the wheel, is obtained from the expression where the components of the relative-acceleration term are (aC/O)n r2 di-rected from C to O and (aC/O)t r directed to the right because of the counter-clockwise angular acceleration of line CO about O. The terms are added together in the lower diagram and it is seen that Ans.
aC r2 aC aO aC/O (r cos r02)2 (r sin r0)2 [aO cos (aA/O)n]2 [aO sin (aA/O)t]2 aA (aA)n 2 (aA)t 2 (aA/O)t r0 r0 aO r (aA/O)n r02 r0 vO r 2 aA aO aA/O aO (aA/O)n (aA/O)t vO/r and aO/r O C A θ r r0 vO aO ω α O O C A t t n n n t θ θ (a A/O)t = r0 (aA/O)t aA α α (aC/O)t = r α (aC/O)t = r α aO = r (aA/O)n aO ω 2 (aC/O)n = rω 2 (a C/O)n = rω 2 a C = rω 2 (a A/O)n = r0 Helpful Hints The counterclockwise angular accel-eration of OA determines the posi-tive direction of (aA/O)t. The normal component (aA/O)n is, of course, di-rected toward the reference center O.
If the wheel were rolling to the right with the same velocity vO but still had an acceleration aO to the left, note that the solution for aA would be unchanged.
We note that the acceleration of the instantaneous center of zero veloc-ity is independent of and is di-rected toward the center of the wheel. This conclusion is a useful re-sult to remember.
c05.qxd 2/10/12 10:08 AM Page 374 Article 5/6 Relative Acceleration 375 SAMPLE PROBLEM 5/14 The linkage of Sample Problem 5/8 is repeated here. Crank CB has a con-stant counterclockwise angular velocity of 2 rad/s in the position shown during a short interval of its motion. Determine the angular acceleration of links AB and OA for this position. Solve by using vector algebra.
Solution.
We first solve for the velocities which were obtained in Sample Problem 5/8. They are where the counterclockwise direction (k-direction) is taken as positive. The ac-celeration equation is where, from Eqs. 5/3 and 5/9a, we may write We now substitute these results into the relative-acceleration equation and equate separately the coefficients of the i-terms and the coefficients of the j-terms to give The solutions are Ans.
Since the unit vector k points out from the paper in the positive z-direction, we see that the angular accelerations of AB and OA are both clockwise (negative).
It is recommended that the student sketch each of the acceleration vectors in its proper geometric relationship according to the relative-acceleration equa-tion to help clarify the meaning of the solution.
AB 0.1050 rad/s2 and OA 4.34 rad/s2 18.37 36.7 175AB 100OA 429 50AB 50ABi 175AB j mm/s2 ABk (175i 50j) (aA/B)t AB rA/B (6 7) 2(175i 50j) mm/s2 6 7 k [(6 7 k) (175i 50j)] (aA/B)n AB (AB rA/B) 300i mm/s2 0 2k (2k [75i]) aB CB rB CB (CB rB) 100OAi 100(3 7) 2j mm/s2 OAk 100j (3 7 k) (3 7 k 100j) aA OA rA OA (OA rA) aA aB (aA/B)n (aA/B)t AB 6/7 rad/s and OA 3/7 rad/s 250 mm 50 mm 75 mm B x rA rA/B rB CB O A C y 100 mm ω Helpful Hints Remember to preserve the order of the factors in the cross products.
In expressing the term aA/B be cer-tain that rA/B is written as the vector from B to A and not the reverse.
c05.qxd 2/10/12 10:08 AM Page 375 376 Chapter 5 Plane Kinematics of Rigid Bodies SAMPLE PROBLEM 5/15 The slider-crank mechanism of Sample Problem 5/9 is repeated here. The crank OB has a constant clockwise angular speed of 1500 rev/min. For the in-stant when the crank angle is 60, determine the acceleration of the piston A and the angular acceleration of the connecting rod AB.
Solution.
The acceleration of A may be expressed in terms of the acceleration of the crank pin B. Thus, Point B moves in a circle of 5-in. radius with a constant speed so that it has only a normal component of acceleration directed from B to O.
The relative-acceleration terms are visualized with A rotating in a circle rel-ative to B, which is considered fixed, as shown. From Sample Problem 5/9, the angular velocity of AB for these same conditions is AB 29.5 rad/sec so that directed from A to B. The tangential component (aA/B)t is known in direction only since its magnitude depends on the unknown angular acceleration of AB.
We also know the direction of aA since the piston is confined to move along the horizontal axis of the cylinder. There are now only two scalar unknowns left in the equation, namely, the magnitudes of aA and (aA/B)t, so the solution can be carried out.
If we adopt an algebraic solution using the geometry of the acceleration polygon, we first compute the angle between AB and the horizontal. With the law of sines, this angle becomes 18.02. Equating separately the horizontal com-ponents and the vertical components of the terms in the acceleration equation, as seen from the acceleration polygon, gives The solution to these equations gives the magnitudes Ans.
With the sense of (aA/B)t also determined from the diagram, the angular accelera-tion of AB is seen from the figure representing rotation relative to B to be Ans.
If we adopt a graphical solution, we begin with the known vectors aB and (aA/B)n and add them head-to-tail using a convenient scale. Next we construct the direction of (aA/B)t through the head of the last vector. The solution of the equa-tion is obtained by the intersection P of this last line with a horizontal line through the starting point representing the known direction of the vector sum aA. Scaling the magnitudes from the diagram gives values which agree with the calculated results.
Ans.
aA 3310 ft/sec2 and (aA/B)t 9030 ft/sec2 AB 9030/(14/12) 7740 rad/sec2 clockwise [ at /r] (aA/B)t 9030 ft/sec2 and aA 3310 ft/sec2 0 10,280 sin 60 1015 sin 18.02 (aA/B)t cos 18.02 aA 10,280 cos 60 1015 cos 18.02 (aA/B)t sin 18.02 (aA/B)n 14 12 (29.5)2 1015 ft/sec2 [an r2] aB 5 12 1500[2] 60 2 10,280 ft/sec2 [an r2] aA aB (aA/B)n (aA/B)t A O B θ ω G r = 5″ 4″ 10″ t n A P aA (aA/B)t (aA/B)n AB α (aA/B)t (aA/B)n = 1015 ft/sec2 B 18.02° 18.02° 60° aB = 10,280 ft/sec2 AB ω = 29.5 rad/sec Except where extreme accuracy is required, do not hesitate to use a graphical solution, as it is quick and reveals the physical relationships among the vectors. The known vec-tors, of course, may be added in any order as long as the governing equa-tion is satisfied.
Helpful Hints If the crank OB had an angular ac-celeration, aB would also have a tan-gential component of acceleration.
Alternatively, the relation an v2/r may be used for calculating (aA/B)n, provided the relative velocity vA/B is used for v. The equivalence is easily seen when it is recalled that vA/B r.
c05.qxd 2/10/12 10:08 AM Page 376 Article 5/6 Problems 377 Problem 5/123 5/124 Refer to the rotor blades and sliding bearing block of Prob. 5/123 where . If and when , find the acceleration of point A for this instant.
5/125 The wheel of radius R rolls without slipping, and its center O has an acceleration . A point P on the wheel is a distance r from O. For given values of , R, and r, determine the angle and the veloc-ity of the wheel for which P has no acceleration in this position.
Problem 5/125 5/126 The circular disk rolls to the left without slipping.
If , determine the velocity and acceleration of the center O of the disk.
Problem 5/126 B A 150 mm 200 mm 150 mm O y x aA/B 2.7j m/s2 P O R r θ aO vO vO aO aO 0 ˙ 0 ¨ 5 rad/s2 aO 3 m/s2 θ 800 mm A y aO O x PROBLEMS Introductory Problems 5/121 The center O of the wheel is mounted on the sliding block, which has an acceleration to the right. At the instant when , and . For this instant determine the magni-tudes of the accelerations of points A and B.
Problem 5/121 5/122 The 9-ft steel beam is being hoisted from its hori-zontal position by the two cables attached at A and B. If the initial angular accelerations are and , determine the initial values of (a) the angular acceleration of the beam, (b) the acceleration of point C, and (c) the distance d from A to the point on the center-line of the beam which has zero acceleration.
Problem 5/122 5/123 The two rotor blades of 800-mm radius rotate counterclockwise with a constant angular velocity about the shaft at O mounted in the sliding block. The acceleration of the block is . Determine the magnitude of the ac-celeration of the tip A of the blade when (a) (b) , and (c) . Does the velocity of O or the sense of enter into the calculation?
180 90 0, aO 3 m/s2 ˙ 2 rad/s 3′ 3′ 3′ C A B 15″ 15″ α2 α1 2 0.6 rad/sec2 1 0.2 rad/sec2 θ A B aO O 400 mm ¨ 8 rad/s2 ˙ 3 rad/s 45 aO 8 m/s2 c05.qxd 2/10/12 10:08 AM Page 377 5/127 The bar of Prob. 5/81 is repeated here. The ends of the 0.4-m bar remain in contact with their respec-tive support surfaces. End B has a velocity of and an acceleration of in the direc-tions shown. Determine the angular acceleration of the bar and the acceleration of end A.
Problem 5/127 5/128 Determine the acceleration of point B on the equa-tor of the earth, repeated here from Prob. 5/61. Use the data given with that problem and assume that the earth’s orbital path is circular, consulting Table D/2 as necessary. Consider the center of the sun fixed and neglect the tilt of the axis of the earth.
Problem 5/128 x A v B Sunlight C D y N ω A B 105° 30° 0.4 m vB = 0.5 m/s aB = 0.3 m/s2 0.3 m/s2 0.5 m/s 5/129 A car with tires of 600-mm diameter accelerates at a constant rate from rest to a velocity of 60 km/h in a distance of 40 m. Determine the magnitude of the acceleration of a point A on the top of the wheel as the car reaches a speed of 10 km/h.
5/130 A car has a forward acceleration without slipping its 24-in.-diameter tires. Deter-mine the velocity v of the car when a point P on the tire in the position shown will have zero horizontal component of acceleration.
Problem 5/130 5/131 Determine the angular acceleration of AB for the position shown if link OB has a constant angu-lar velocity Problem 5/131 r r A O B ω 2 r .
AB 24″ a v P O 45° a 12 ft/sec2 378 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 378 Article 5/6 Problems 379 5/134 The load L is lowered by the two pulleys which are fastened together and rotate as a single unit. For the instant represented, drum A has a counter-clockwise angular velocity of which is de-creasing by each second. Simultaneously, drum B has a clockwise angular velocity of which is increasing by each sec-ond. Calculate the accelerations of points C and D and the load L.
Problem 5/134 5/135 The mechanism of Prob. 5/76 is repeated here. The angular velocity of the disk is constant. For the instant represented, determine the angular acceler-ation of link AB and the acceleration of col-lar B. Assume the quantities and r to be known.
Problem 5/135 r 2r ω B O A 45° 0 0 aB AB 0 y x 6″ 6″ B A D C O 20″ 10″ L 2 rad/sec 6 rad/sec, 4 rad/sec 4 rad/sec, 5/132 Determine the angular acceleration of link AB and the linear acceleration of A for if and at that position. Carry out your solu-tion using vector notation.
Problem 5/132 Representative Problems 5/133 The end rollers of bar AB are constrained to the slot shown. If roller A has a downward velocity of 1.2 m/s and this speed is constant over a small mo-tion interval, determine the tangential acceleration of roller B as it passes the topmost position. The value of R is 0.5 m.
Problem 5/133 vA R 1.5R A B y x θ O B A 500 mm 400 mm 400 mm ¨ 3 rad/s2 ˙ 0 90 c05.qxd 2/10/12 10:08 AM Page 379 5/136 Crank OA oscillates between the dashed positions shown and causes small angular motion of crank BC through the connecting link AB. When OA crosses the horizontal position with AB horizontal and BC vertical, it has an angular velocity and zero angular acceleration. Determine the angular acceleration of BC for this position.
Problem 5/136 5/137 The shaft of the wheel unit rolls without slipping on the fixed horizontal surface. If the velocity and acceleration of point O are to the right and to the left, respectively, determine the accelerations of points A and D.
Problem 5/137 5/138 The hydraulic cylinder imparts motion to point B which causes link OA to rotate. For the instant shown where OA is vertical and AB is horizontal, the velocity of pin B is and is increasing at the rate of For this position determine the angular acceleration of OA.
20 m/s2.
4 m/s vB y x 10″ 2″ vO = 3 ft/sec O A D C B aO = 4 ft/sec2 4 ft/sec2 3 ft/sec O l l r A B C ω Problem 5/138 5/139 The velocity of roller A is to the right as shown, and this velocity is momentarily decreas-ing at a rate of Determine the correspond-ing value of the angular acceleration of bar AB as well as the tangential acceleration of roller B along the circular guide. The value of R is 0.6 m.
Problem 5/139 5/140 The bar AB from Prob. 5/73 is repeated here. If the velocity of point A is to the right and is constant for an interval including the position shown, deter-mine the tangential acceleration of point B along its path and the angular acceleration of the bar.
Problem 5/140 1.2 m A B 60° 0.5 m vA 3 m/s 15° B vA R 2R A R/2 2 m/s2.
vA 0.5 m/s 240 mm A B O 120 mm 45° vB 380 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 380 Article 5/6 Problems 381 Problem 5/143 5/144 The sliding collar moves up and down the shaft, causing an oscillation of crank OB. If the velocity of A is not changing as it passes the null position where AB is horizontal and OB is vertical, determine the angular acceleration of OB in that position.
Problem 5/144 5/145 For the linkage shown, if and is con-stant when the two links become perpendicular to one another, determine the angular acceleration of CB for this position.
Problem 5/145 B A C vA 7″ 5″ 5″ vA 20 in./sec O B l A vA r 10″ 20″ A D B C vA θ β 5/141 The center O of the wooden spool is moving verti-cally downward with a speed and this speed is increasing at the rate of Determine the accelerations of points A, P, and B.
Problem 5/141 5/142 Link OA has a constant counterclockwise angular velocity during a short interval of its motion. For the position shown determine the angular accelera-tions of AB and BC.
Problem 5/142 5/143 The linkage of Prob. 5/74 is shown again here. For the instant when the hydraulic cylin-der gives A a velocity which is increas-ing by each second. For this instant determine the angular acceleration of link BC.
3 ft/sec vA 4 ft/sec 60, O A B r r ω r r 2 C vO O B A P 0.48 m 0.8 m x y 5 m/s2.
vO 2 m/s, c05.qxd 2/10/12 10:08 AM Page 381 5/146 The mechanism of Prob. 5/75 is repeated here.
Each of the sliding bars A and B engages its respec-tive rim of the two riveted wheels without slipping.
If, in addition to the information shown, bar A has an acceleration of to the right and there is no acceleration of bar B, calculate the magnitude of the acceleration of P for the instant depicted.
Problem 5/146 5/147 The four-bar linkage of Prob. 5/88 is repeated here.
If the angular velocity and angular acceleration of drive link OA are and respectively, both counterclockwise, determine the angular accelerations of bars AB and BC for the instant represented.
Problem 5/147 5/148 The elements of a simplified clam-shell bucket for a dredge are shown. With the block at O considered fixed and with the constant velocity of the control cable at C equal to determine the angular acceleration of the right-hand bucket jaw when as the bucket jaws are closing.
45 0.5 m/s, v 15° 60° , 200 mm 80 mm A B O C 240 mm ω 0 α 0 5 rad/s2, 10 rad/s O P 100 mm 160 mm A B vB = 0.6 m/s vA = 0.8 m/s 2 m/s2 Problem 5/148 5/149 The revolving crank ED and connecting link CD cause the rigid frame ABO to oscillate about O. For the instant represented ED and CD are both per-pendicular to FO, and the crank ED has an angular velocity of and an angular acceleration of both counterclockwise. For this in-stant determine the acceleration of point A with re-spect to point B.
Problem 5/149 E D A C O B F 12″ 3′ 4′ 6′ 4′ 3′ 0.06 rad/sec2, 0.4 rad/sec θ 500 mm 600 mm B C A v O 90° 382 Chapter 5 Plane Kinematics of Rigid Bodies c05.qxd 2/10/12 10:08 AM Page 382 Article 5/6 Problems 383 5/152 For a short interval of motion, link OA has a con-stant angular velocity Determine the angular acceleration of link AB for the in-stant when OA is parallel to the horizontal axis through B.
Problem 5/152 5/153 The elements of a power hacksaw are shown in the figure. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counterclockwise speed of determine the acceleration of the blade for the position where and find the corre-sponding angular acceleration of the link AB.
Problem 5/153 θ B O A 450 mm 100 mm 100 mm 90, 60 rev/min, 200 mm O B A ω 60 mm 120 mm AB 4 rad/s.
5/150 If link AB of the four-bar linkage has a constant counterclockwise angular velocity of dur-ing an interval which includes the instant repre-sented, determine the angular acceleration of AO and the acceleration of point D. Express your re-sults in vector notation.
Problem 5/150 5/151 The crank OA of the offset slider-crank mechanism rotates with a constant clockwise angular velocity Determine the angular acceleration of link AB and the acceleration of B for the depicted position.
Problem 5/151 O B 60° 45° A 15° ω 0 OA = 75 mm AB = 225 mm 0 10 rad/s.
75 mm A O C B D 150 mm 100 mm 100 mm AB ω y x 40 rad/s c05.qxd 2/10/12 10:08 AM Page 383 5/154 The mechanism of Prob. 5/115 is repeated here where the flexible band F attached to the sector at E is given a constant velocity of as shown.
For the instant when BD is perpendicular to OA, determine the angular acceleration of BD.
Problem 5/154 5/155 An oil pumping rig is shown in the figure. The flexi-ble pump rod D is fastened to the sector at E and is always vertical as it enters the fitting below D. The link AB causes the beam BCE to oscillate as the weighted crank OA revolves. If OA has a constant clockwise speed of 1 rev every 3 s, determine the acceleration of the pump rod D when the beam and the crank OA are both in the horizontal position shown.
4 m/s O A D B E F 200 mm 75 mm 125 mm 250 mm 100 mm 4 m/s Problem 5/155 5/156 A mechanism for pushing small boxes from an as-sembly line onto a conveyor belt is shown with arm OD and crank CB in their vertical positions. For the configuration shown, crank CB has a constant clockwise angular velocity of Determine the acceleration of E.
Problem 5/156 rad/s.
3.3 m 3 m 0.9 m 1.95 m 0.6 m A O B C D E 384 Chapter 5 Plane Kinematics of Rigid Bodies 600 mm 200 mm 100 mm 100 mm 200 mm 50 mm 200 mm 400 mm A B C O D E c05.qxd 2/10/12 10:08 AM Page 384 5/7 Motion Relative to Rotating Axes In our discussion of the relative motion of particles in Art. 2/8 and in our use of the relative-motion equations for the plane motion of rigid bodies in this present chapter, we have used nonrotating reference axes to describe relative velocity and relative acceleration. Use of rotating reference axes greatly facilitates the solution of many problems in kine-matics where motion is generated within a system or observed from a system which itself is rotating. An example of such a motion is the movement of a fluid particle along the curved vane of a centrifugal pump, where the path relative to the vanes of the impeller becomes an important design consideration.
We begin the description of motion using rotating axes by consider-ing the plane motion of two particles A and B in the fixed X-Y plane, Fig.
5/10a. For the time being, we will consider A and B to be moving inde-pendently of one another for the sake of generality. We observe the mo-tion of A from a moving reference frame x-y which has its origin attached to B and which rotates with an angular velocity . We may write this angular velocity as the vector = k k, where the vector is normal to the plane of motion and where its positive sense is in the positive z-direction (out from the paper), as established by the right-hand rule. The absolute position vector of A is given by (5/10) where i and j are unit vectors attached to the x-y frame and r xi yj stands for rA/B, the position vector of A with respect to B.
Time Derivatives of Unit Vectors To obtain the velocity and acceleration equations we must succes-sively differentiate the position-vector equation with respect to time. In contrast to the case of translating axes treated in Art. 2/8, the unit vec-tors i and j are now rotating with the x-y axes and, therefore, have time derivatives which must be evaluated. These derivatives may be seen from Fig. 5/10b, which shows the infinitesimal change in each unit vec-tor during time dt as the reference axes rotate through an angle d dt. The differential change in i is di, and it has the direction of j and a magnitude equal to the angle d times the length of the vector i, which is unity. Thus, di d j.
Similarly, the unit vector j has an infinitesimal change dj which points in the negative x-direction, so that dj d i. Dividing by dt and replacing di/dt by , dj/dt by , and d/dt by result in By using the cross product, we can see from Fig. 5/10c that i = j and j i. Thus, the time derivatives of the unit vectors may be written as (5/11) i ˙ i and j ˙ j i ˙ j and j ˙ i ˙ j ˙ i ˙ rA rB r rB (xi yj) ˙ ˙ Article 5/7 Motion Relative to Rotating Axes 385 r = rA/B rA rB Y X O B A x y y j i i k × j ω × i ω j x θ · θ ω = x y z dθ dθ θ di = d j θ dj = –d i ω (a) (b) (c) Figure 5/10 c05.qxd 2/10/12 10:08 AM Page 385 Relative Velocity We now use the expressions of Eqs. 5/11 when taking the time de-rivative of the position-vector equation for A and B to obtain the rela-tive-velocity relation. Differentiation of Eq. 5/10 gives But r. Also, since the observer in x-y measures velocity components and , we see that vrel, which is the velocity relative to the x-y frame of refer-ence. Thus, the relative-velocity equation becomes (5/12) Comparison of Eq. 5/12 with Eq. 2/20 for nonrotating reference axes shows that vA/B r vrel, from which we conclude that the term r is the difference between the relative velocities as measured from nonrotating and rotating axes.
To illustrate further the meaning of the last two terms in Eq. 5/12, the motion of particle A relative to the rotating x-y plane is shown in Fig. 5/11 as taking place in a curved slot in a plate which represents the rotating x-y reference system. The velocity of A as measured relative to the plate, vrel, would be tangent to the path fixed in the x-y plate and would have a magnitude , where s is measured along the path. This rel-ative velocity may also be viewed as the velocity vA/P relative to a point P attached to the plate and coincident with A at the instant under consid-eration. The term r has a magnitude and a direction normal to r and is the velocity relative to B of point P as seen from nonrotating axes attached to B.
The following comparison will help establish the equivalence of, and clarify the differences between, the relative-velocity equations written for rotating and nonrotating reference axes: (5/12a) In the second equation, the term vP/B is measured from a nonrotating position—otherwise, it would be zero. The term vA/P is the same as vrel and is the velocity of A as measured in the x-y frame. In the third equa-tion, vP is the absolute velocity of P and represents the effect of the moving coordinate system, both translational and rotational. The fourth equation is the same as that developed for nonrotating axes, Eq. 2/20, and it is seen that vA/B vP/B + vA/P = r vrel.
r ˙ s ˙ vA vB r vrel x ˙i y ˙j y ˙ x ˙ xi ˙ yj ˙ xi yj (xi yj) r ˙B (xi ˙ yj ˙) (x ˙i y ˙j) r ˙A r ˙B d dt (xi yj) 386 Chapter 5 Plane Kinematics of Rigid Bodies r s rB Y X O B Path of A x y vA/B ω P (fixed to path and coincident with A) A vrel = vA/P × r = vP/B ω Figure 5/11 c05.qxd 2/10/12 10:08 AM Page 386 Transformation of a Time Derivative Equation 5/12 represents a transformation of the time derivative of the position vector between rotating and nonrotating axes. We may eas-ily generalize this result to apply to the time derivative of any vector quantity V Vxi Vyj. Accordingly, the total time derivative with re-spect to the X-Y system is The first two terms in the expression represent that part of the total de-rivative of V which is measured relative to the x-y reference system, and the second two terms represent that part of the derivative due to the ro-tation of the reference system.
With the expressions for and from Eqs. 5/11, we may now write (5/13) Here V represents the difference between the time derivative of the vector as measured in a fixed reference system and its time derivative as measured in the rotating reference system. As we will see in Art. 7/2, where three-dimensional motion is introduced, Eq. 5/13 is valid in three dimensions, as well as in two dimensions.
The physical significance of Eq. 5/13 is illustrated in Fig. 5/12, which shows the vector V at time t as observed both in the fixed axes X-Y and in the rotating axes x-y. Because we are dealing with the effects of rotation only, we may draw the vector through the coordinate origin without loss of generality. During time dt, the vector swings to position V, and the observer in x-y measures the two components (a) dV due to its change in magnitude and (b) V d due to its rotation d relative to x-y. To the rotating observer, then, the derivative (dV/dt)xy which the observer measures has the components dV/dt and V d/dt . The re-maining part of the total time derivative not measured by the rotating observer has the magnitude V d/dt and, expressed as a vector, is V.
Thus, we see from the diagram that which is Eq. 5/13.
Relative Acceleration The relative-acceleration equation may be obtained by differentiat-ing the relative-velocity relation, Eq. 5/12. Thus, aA aB ˙ r r ˙ v ˙rel (V ˙)XY (V ˙)xy V V ˙ dV dtXY dV dtxy V j ˙ i ˙ dV dtXY (V ˙xi V ˙y j) (Vxi ˙ Vy j ˙) Article 5/7 Motion Relative to Rotating Axes 387 Y X x y (dV)XY (dV)xy d⏐V⏐ = dV dγ V′ V Vdβ Vdθ β θ γ · θ ω = Figure 5/12 c05.qxd 2/10/12 10:08 AM Page 387 In the derivation of Eq. 5/12 we saw that Therefore, the third term on the right side of the acceleration equation becomes With the aid of Eqs. 5/11, the last term on the right side of the equation for aA becomes Substituting this into the expression for aA and collecting terms, we obtain (5/14) Equation 5/14 is the general vector expression for the absolute accel-eration of a particle A in terms of its acceleration arel measured relative to a moving coordinate system which rotates with an angular velocity and an angular acceleration . The terms and ( r) are shown in Fig. 5/13. They represent, respectively, the tangential and nor-mal components of the acceleration aP/B of the coincident point P in its circular motion with respect to B. This motion would be observed from a set of nonrotating axes moving with B. The magnitude of is and its direction is tangent to the circle. The magnitude of ( r) is r2 and its direction is from P to B along the normal to the circle.
The acceleration of A relative to the plate along the path, arel, may be expressed in rectangular, normal and tangential, or polar coordinates in the rotating system. Frequently, n- and t-components are used, and these components are depicted in Fig. 5/13. The tangential component has the magnitude (arel)t , where s is the distance measured along the path to A. The normal component has the magnitude (arel)n , where is the radius of curvature of the path as measured in x-y. The sense of this vector is always toward the center of curvature.
Coriolis Acceleration The term 2 vrel, shown in Fig. 5/13, is called the Coriolis acceler-ation. It represents the difference between the acceleration of A rela-tive to P as measured from nonrotating axes and from rotating axes.
vrel 2/ s ¨ r ¨ ˙ r ˙ r ˙ aA aB ˙ r ( r) 2 vrel arel vrel arel (x ˙i y ˙j) (x ¨i y ¨j) v ˙rel d dt (x ˙i y ˙j) (x ˙i ˙ y ˙j ˙) (x ¨i y ¨j) r ˙ ( r vrel) ( r) vrel r vrel r ˙ d dt (xi yj) (xi ˙ yj ˙) (x ˙i y ˙j) 388 Chapter 5 Plane Kinematics of Rigid Bodies Named after the French military engineer G. Coriolis (1792–1843), who was the first to call attention to this term.
r s rB Y n X O B Path of A x y P t A (arel)t (arel)n 2 × vrel ω × ( × r) ω ω × r ω · ,ω ω · Figure 5/13 c05.qxd 2/10/12 10:08 AM Page 388 The direction is always normal to the vector vrel, and the sense is estab-lished by the right-hand rule for the cross product.
The Coriolis acceleration aCor 2 vrel is difficult to visualize be-cause it is composed of two separate physical effects. To help with this visualization, we will consider the simplest possible motion in which this term appears. In Fig. 5/14a we have a rotating disk with a radial slot in which a small particle A is confined to slide. Let the disk turn with a constant angular velocity and let the particle move along the slot with a constant speed vrel relative to the slot. The velocity of A has the two components (a) due to motion along the slot and (b) x due to the rotation of the slot. The changes in these two velocity compo-nents due to the rotation of the disk are shown in part b of the figure for the interval dt, during which the x-y axes rotate with the disk through the angle d to x-y.
The velocity increment due to the change in direction of vrel is d and that due to the change in magnitude of x is dx, both being in the y-direction normal to the slot. Dividing each increment by dt and adding give the sum , which is the magnitude of the Coriolis ac-celeration 2 vrel.
Dividing the remaining velocity increment x d due to the change in direction of x by dt gives x or x2, which is the acceleration of a point P fixed to the slot and momentarily coincident with the particle A.
We now see how Eq. 5/14 fits these results. With the origin B in that equation taken at the fixed center O, aB 0. With constant angular ve-locity, . With vrel constant in magnitude and no curvature to the slot, arel 0. We are left with Replacing r by xi, by k, and vrel by gives which checks our analysis from Fig. 5/14.
We also note that this same result is contained in our polar-coordinate analysis of plane curvilinear motion in Eq. 2/14 when we let 0 and 0 and replace r by x and by . If the slot in the disk of Fig. 5/14 had been curved, we would have had a normal component of accelera-tion relative to the slot so that arel would not be zero.
Rotating versus Nonrotating Systems The following comparison will help to establish the equivalence of, and clarify the differences between, the relative-acceleration equations written for rotating and nonrotating reference axes: (5/14a) ˙ ¨ r ¨ aA x2i 2x ˙j x ˙i aA ( r) 2 vrel ˙ r 0 ˙ x ˙ x ˙ 2x ˙ x ˙ x ˙ x ˙ ˙ Article 5/7 Motion Relative to Rotating Axes 389 (a) (b) Y y x x X O A = constant ω xω xω xω dθ dθ dθ x · dθ vrel = x · = constant vrel = x · y y′ x x′ d(x ) = dx + x d = dx + 0 ω ω ω ω Figure 5/14 c05.qxd 2/10/12 10:08 AM Page 389 The equivalence of aP/B and r ( r), as shown in the sec-ond equation, has already been described. From the third equation where aB aP/B has been combined to give aP, it is seen that the rela-tive-acceleration term aA/P, unlike the corresponding relative-velocity term, is not equal to the relative acceleration arel measured from the ro-tating x-y frame of reference.
The Coriolis term is, therefore, the difference between the accelera-tion aA/P of A relative to P as measured in a nonrotating system and the acceleration arel of A relative to P as measured in a rotating system.
From the fourth equation, it is seen that the acceleration aA/B of A with respect to B as measured in a nonrotating system, Eq. 2/21, is a combina-tion of the last four terms in the first equation for the rotating system.
The results expressed by Eq. 5/14 may be visualized somewhat more simply by writing the acceleration of A in terms of the acceleration of the coincident point P. Because the acceleration of P is aP aB r ( r), we may rewrite Eq. 5/14 as (5/14b) When the equation is written in this form, point P may not be picked at random because it is the one point attached to the rotating reference frame coincident with A at the instant of analysis. Again, reference to Fig. 5/13 should be made to clarify the meaning of each of the terms in Eq. 5/14 and its equivalent, Eq. 5/14b.
aA aP 2 vrel arel ˙ ˙ 390 Chapter 5 Plane Kinematics of Rigid Bodies In summary, once we have chosen our rotating reference system, we must recognize the following quantities in Eqs. 5/12 and 5/14: arel acceleration of A measured relative to the rotating axes vrel velocity of A measured relative to the rotating axes ˙ angular acceleration of the rotating axes angular velocity of the rotating axes r position vector of the coincident point P measured from B aB absolute acceleration of the origin B of the rotating axes vB absolute velocity of the origin B of the rotating axes KEY CONCEPTS Also, keep in mind that our vector analysis depends on the consis-tent use of a right-handed set of coordinate axes. Finally, note that Eqs.
5/12 and 5/14, developed here for plane motion, hold equally well for space motion. The extension to space motion will be covered in Art. 7/6.
c05.qxd 2/10/12 10:08 AM Page 390 SAMPLE PROBLEM 5/16 At the instant represented, the disk with the radial slot is rotating about O with a counterclockwise angular velocity of 4 rad/sec which is decreasing at the rate of 10 rad/sec2. The motion of slider A is separately controlled, and at this in-stant, r 6 in., 5 in./sec, and 81 in./sec2. Determine the absolute veloc-ity and acceleration of A for this position.
Solution.
We have motion relative to a rotating path, so that a rotating coor-dinate system with origin at O is indicated. We attach x-y axes to the disk and use the unit vectors i and j.
Velocity.
With the origin at O, the term vB of Eq. 5/12 disappears and we have The angular velocity as a vector is 4k rad/sec, where k is the unit vector nor-mal to the x-y plane in the z-direction. Our relative-velocity equation becomes Ans.
in the direction indicated and has the magnitude Ans.
Acceleration.
Equation 5/14 written for zero acceleration of the origin of the rotating coordinate system is The terms become The total acceleration is, therefore, Ans.
in the direction indicated and has the magnitude Ans.
Vector notation is certainly not essential to the solution of this problem.
The student should be able to work out the steps with scalar notation just as eas-ily. The correct direction of the Coriolis-acceleration term can always be found by the direction in which the head of the vrel vector would move if rotated about its tail in the sense of as shown.
aA (15)2 (20)2 25 in./sec2 aA (81 96)i (40 60)j 15i 20j in./sec2 arel 81i in./sec2 2 vrel 2(4k) 5i 40j in./sec2 ˙ r 10k 6i 60j in./sec2 ( r) 4k (4k 6i) 4k 24j 96i in./sec2 aA ( r) ˙ r 2 vrel arel vA (24)2 (5)2 24.5 in./sec vA 4k 6i 5i 24j 5i in./sec vA r vrel r ¨ r ˙ Article 5/7 Motion Relative to Rotating Axes 391 r O A = 4 rad/sec ω · = 10 rad/sec2 ω y x j vA vrel i O ë r v A y x aA arel O A ë vrel 2v · ë r v ë ( ë r) v v vrel aCor v Helpful Hints This equation is the same as vA vP vA/P, where P is a point at-tached to the disk coincident with A at this instant.
Note that the x-y-z axes chosen con-stitute a right-handed system.
Be sure to recognize that ( r) and represent the normal and tangential components of accelera-tion of a point P on the disk coinci-dent with A. This description becomes that of Eq. 5/14b.
˙ r c05.qxd 2/10/12 10:08 AM Page 391 SAMPLE PROBLEM 5/17 The pin A of the hinged link AC is confined to move in the rotating slot of link OD. The angular velocity of OD is 2 rad/s clockwise and is constant for the interval of motion concerned. For the position where 45 with AC hori-zontal, determine the velocity of pin A and the velocity of A relative to the rotat-ing slot in OD.
Solution.
Motion of a point (pin A) along a rotating path (the slot) suggests the use of rotating coordinate axes x-y attached to arm OD. With the origin at the fixed point O, the term vB of Eq. 5/12 vanishes, and we have vA r vrel.
The velocity of A in its circular motion about C is where the angular velocity CA is arbitrarily assigned in a clockwise sense in the positive z-direction (k).
The angular velocity of the rotating axes is that of the arm OD and, by the right-hand rule, is k 2k rad/s. The vector from the origin to the point P on OD coincident with A is r mm. Thus, Finally, the relative-velocity term vrel is the velocity measured by an ob-server attached to the rotating reference frame and is vrel . Substitution into the relative-velocity equation gives Equating separately the coefficients of the i and j terms yields giving Ans.
With a negative value for CA, the actual angular velocity of CA is counterclock-wise, so the velocity of A is up with a magnitude of Ans.
Geometric clarification of the terms is helpful and is easily shown. Using the equivalence between the third and the first of Eqs. 5/12a with vB 0 enables us to write vA vP vA/P, where P is the point on the rotating arm OD coinci-dent with A. Clearly, vP 225 (2) = mm/s and its direction is normal to OD. The relative velocity vA/P, which is the same as vrel, is seen from the figure to be along the slot toward O. This conclusion becomes clear when it is ob-served that A is approaching P along the slot from below before coincidence and is receding from P upward along the slot following coincidence. The velocity of A is tangent to its circular arc about C. The vector equation can now be satisfied since there are only two remaining scalar unknowns, namely, the magnitude of vA/P and the magnitude of vA. For the 45 position, the figure requires vA/P mm/s and vA 900 mm/s, each in its direction shown. The angular velocity of AC is AC vA / AC 900/225 4 rad/s counterclockwise [ v/r] 4502 4502 2 OP vA 225(4) 900 mm/s CA 4 rad/s and x ˙ vrel 4502 mm/s (225/2)CA x ˙ and (225/2)CA 4502 (225/2)CA (i j) 4502j x ˙i x ˙i r 2k 2252i 4502j mm/s OPi (450 225)2 (225)2 i 2252i vA CA rCA CAk (225/2)(i j) (225/2)CA (i j) 392 Chapter 5 Plane Kinematics of Rigid Bodies A direct conversion between the two reference systems is obtained from the geometry of the unit circle and gives and j I sin J cos i I cos J sin 450 mm O C A D y x 225 mm 225 mm = 45° θ = 2 rad/s ω C O A A A 45° P P P vA vA/P = vrel vP = ë r v Y y x O i j I J θ θ θ X Helpful Hints It is clear enough physically that CA will have a counterclockwise angular velocity for the conditions specified, so we anticipate a negative value for CA.
Solution of the problem is not re-stricted to the reference axes used. Al-ternatively, the origin of the x-y axes, still attached to OD, could be chosen at the coincident point P on OD. This choice would merely replace the r term by its equal, vP. As a further se-lection, all vector quantities could be expressed in terms of X-Y compo-nents using unit vectors I and J.
c05.qxd 2/10/12 10:08 AM Page 392 SAMPLE PROBLEM 5/18 For the conditions of Sample Problem 5/17, determine the angular accelera-tion of AC and the acceleration of A relative to the rotating slot in arm OD.
Solution.
We attach the rotating coordinate system x-y to arm OD and use Eq.
5/14. With the origin at the fixed point O, the term aB becomes zero so that From the solution to Sample Problem 5/17, we make use of the values 2k rad/s, CA 4k rad/s, and vrel 450 i mm/s and write Substitution into the relative-acceleration equation yields Equating separately the i and j terms gives and Solving for the two unknowns gives Ans.
If desired, the acceleration of A may also be written as We make use here of the geometric representation of the relative-acceleration equation to further clarify the problem. The geometric approach may be used as an alternative solution. Again, we introduce point P on OD coincident with A.
The equivalent scalar terms are We start with the known vectors and add them head-to-tail for each side of the equation beginning at R and ending at S, where the intersection of the known directions of (aA)t and arel establishes the solution. Closure of the polygon deter-mines the sense of each of the two unknown vectors, and their magnitudes are easily calculated from the figure geometry. arel x ¨ along OD, sense unknown 2 vrel 2vrel directed as shown (aP)t ˙ r r ˙ 0 since constant (aP)n ( r) OP2 from P to O (aA)n CA (CA rCA) rCA 2 from A to C (aA)t ˙CA rCA r ˙CA rCA normal to CA, sense unknown aA (225/2)(32)(i j) (3600/2)(i j) 7640i 2550j mm/s2 ˙CA 32 rad/s2 and x ¨ arel 8910 mm/s2 (225 ˙CA 3600)/2 18002 (225 ˙CA 3600)/2 9002 x ¨ 1 2 (225 ˙CA 3600)i 1 2 (225 ˙CA 3600)j 9002i 18002j x ¨i arel x ¨i 2 vrel 2(2k) (4502i) 18002j mm/s2 ( r) 2k (2k 2252i) 9002i mm/s2 ˙ r 0 since constant ˙CAk 225 2 (i j) 4k 4k 225 2 [i j] aA ˙CA rCA CA (CA rCA) 2 aA ˙ r ( r) 2 vrel arel Article 5/7 Motion Relative to Rotating Axes 393 450 mm O C A D y x 225 mm 225 mm = 45° θ = 2 rad/s ω R S aA (aA)n (aA)t (aP)n arel 2 ë vrel v 2 ë vrel v ω vrel Helpful Hints If the slot had been curved with a radius of curvature , the term arel would have had a component vrel 2/ normal to the slot and directed to-ward the center of curvature in addi-tion to its component along the slot.
It is always possible to avoid a si-multaneous solution by projecting the vectors onto the perpendicular to one of the unknowns.
c05.qxd 2/10/12 10:08 AM Page 393 SAMPLE PROBLEM 5/19 Aircraft B has a constant speed of 150 m/s as it passes the bottom of a circu-lar loop of 400-m radius. Aircraft A flying horizontally in the plane of the loop passes 100 m directly below B at a constant speed of 100 m/s. (a) Determine the instantaneous velocity and acceleration which A appears to have to the pilot of B, who is fixed to his rotating aircraft. (b) Compare your results for part (a) with the case of erroneously treating the pilot of aircraft B as nonrotating.
Solution (a). We begin by clearly defining the rotating coordinate system x-y-z which best helps us to answer the questions. With x-y-z attached to aircraft B as shown, the terms vrel and arel in Eqs. 5/12 and 5/14 will be the desired results.
The terms in Eq. 5/12 are Ans.
The terms in Eq. 5/14, in addition to those listed above, are Ans.
(b) For motion relative to translating frames, we use Eqs. 2/20 and 2/21 of Chapter 2: Again, we see that vrel vA/B and arel aA/B. The rotation of pilot B makes a dif-ference in what he observes!
The scalar result can be obtained by considering a complete circular motion of aircraft B, during which it rotates 2 radians in a time t : Because the speed of aircraft B is constant, there is no tangential acceleration and thus the angular acceleration of this aircraft is zero.
˙ 2 2/vB vB 2 vB vB aA/B aA aB 0 56.2j 56.2j m/s2 vA/B vA vB 100i 150i 50i m/s arel 4.69k m/s2 Solving for arel gives 2[0.375k (87.5i)] arel 0 56.2j 0 (100j) 0.375k [0.375k (100j)] aA aB ˙ r ( r) 2 vrel arel Eq. 5/14: ˙ 0 aB vB 2 j 1502 400 j 56.2j m/s2 aA 0 vrel 87.5i m/s Solving for vrel gives 100i 150i 0.375k (100j) vrel vA vB r vrel Eq. 5/12: r rA/B 100j m vB k 150 400 k 0.375k rad/s vB 150i m/s vA 100i m/s 394 Chapter 5 Plane Kinematics of Rigid Bodies 100 m B A y z x = 400 m ρ 100 m (z out) x y B r A vB aB vA Helpful Hint Because we choose the rotating frame x-y-z to be fixed to aircraft B, the an-gular velocity of the aircraft and the term in Eqs. 5/12 and 5/14 are identical.
c05.qxd 2/10/12 10:08 AM Page 394 PROBLEMS Introductory Problems 5/157 The disk rotates with angular speed The small ball A is moving along the radial slot with speed relative to the disk. De-termine the absolute velocity of the ball and state the angle between this velocity vector and the positive x-axis.
Problem 5/157 5/158 In addition to the conditions stated in the previous problem, the ball speed u (relative to the disk) is in-creasing at the rate of and the angular rate of the disk is decreasing at the rate of Determine the Coriolis acceleration rela-tive to the disk-fixed Bxy coordinate system. Also determine the absolute acceleration of ball A and the angle between this acceleration vector and the positive x-axis.
5/159 The disk rotates about a fixed axis through O with angular velocity and angular acceler-ation at the instant represented, in the directions shown. The slider A moves in the straight slot. Determine the absolute velocity and acceleration of A for the same instant, when in., and y ¨ 30 in./sec2.
y ˙ 24 in./sec, y 8 3 rad/sec2 5 rad/sec 0.8 rad/s2.
150 mm/s2 y B A u x ω 125 mm u 100 mm/s 2 rad/s.
Article 5/7 Problems 395 Problem 5/159 5/160 The disk rotates about a fixed axis through O with angular velocity and angular acceleration in the directions shown at a certain instant. The small sphere A moves in the circular slot, and at the same instant, and Deter-mine the absolute velocity and acceleration of A at this instant.
Problem 5/160 y x 15″ O A α ω β ¨ 4 rad/sec2.
30, ˙ 2 rad/sec, 3 rad/sec2 5 rad/sec y x 6″ O A α ω y c05.qxd 2/10/12 10:08 AM Page 395 Problem 5/163 5/164 A stationary pole A is viewed by an observer P who is sitting on a small merry-go-round which rotates about a fixed vertical axis at B with a constant an-gular velocity as shown. Determine the apparent velocity of A as seen by the observer P. Does this velocity depend on the location of the observer on the merry-go-round?
Problem 5/164 y Ω B P A x d S y x R B A N θ z Ω v 396 Chapter 5 Plane Kinematics of Rigid Bodies 5/161 The slotted wheel rolls to the right without slip-ping, with a constant speed of its center O. Simultaneously, motion of the sliding block A is controlled by a mechanism not shown so that with Determine the magnitude of the acceleration of A for the instant when and .
Problem 5/161 5/162 The disk rolls without slipping on the horizontal surface, and at the instant represented, the center O has the velocity and acceleration shown in the figure. For this instant, the particle A has the indi-cated speed u and time rate of change of speed , both relative to the disk. Determine the absolute velocity and acceleration of particle A.
Problem 5/162 5/163 An experimental vehicle A travels with constant speed v relative to the earth along a north–south track. Determine the Coriolis acceleration as a function of the latitude . Assume an earth-fixed rotating frame Bxyz and a spherical earth. If the vehicle speed is , determine the mag-nitude of the Coriolis acceleration at (a) the equa-tor and (b) the north pole.
v 500 km/h aCor y x 0.24 m vO = 3 m/s aO = 5 m/s2 u · = 7 m/s2 u = 2 m/s O A 0.30 m u ˙ v 8″ O A θ x 30 x 6 in.
x ¨ 0.
x ˙ 1.5 ft/sec v 2 ft/sec c05.qxd 2/10/12 10:08 AM Page 396 5/165 The small collar A is sliding on the bent bar with speed u relative to the bar as shown. Simultane-ously, the bar is rotating with angular velocity about the fixed pivot B. Take the x-y axes to be fixed to the bar and determine the Coriolis acceler-ation of the slider for the instant represented. In-terpret your result.
Problem 5/165 Representative Problems 5/166 The fire truck is moving forward at a speed of 35 mi/hr and is decelerating at the rate of Simultaneously, the ladder is being raised and ex-tended. At the instant considered the angle is and is increasing at the constant rate of 10 deg/sec.
Also at this instant the extension b of the ladder is 5 ft, with and For this instant determine the acceleration of the end A of the ladder (a) with respect to the truck and (b) with respect to the ground.
Problem 5/166 B 20 b A θ b ¨ 1 ft/sec2.
b ˙ 2 ft/sec 30 10 ft/sec2.
y L x d A B u ω Article 5/7 Problems 397 5/167 For an alternative solution to Prob. assign r-coordinates with origin at B as shown. Then make use of the polar-coordinate relations for the acceleration of A relative to B. The r- and com-ponents of the absolute acceleration should coin-cide with the components along and normal to the ladder which would be found in Prob. 5/166.
Problem 5/167 5/168 Aircraft B has a constant speed of at the bottom of a circular loop of 400-m radius. Aircraft A flying horizontally in the plane of the loop passes 100 m directly under B at a constant speed of . With coordinate axes attached to B as shown, determine the acceleration which A appears to have to the pilot of B for this instant.
Problem 5/168 100 m x B A z y = 400 m ρ 360 km/h 540 km/h = 30° θ 20′ θ r A B b - 5/166 c05.qxd 2/10/12 10:08 AM Page 397 5/171 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity of its mass center B and angular velocity about a vertical axis. The velocity of B is constant, but the angular rate is decreasing at Person A is stationary on the dock. What velocity and acceleration of A are observed by a passenger fixed to and rotating with the ship?
Treat the problem as two-dimensional.
Problem 5/171 5/172 All conditions of the previous problem remain, except now person A is running to the right with a constant speed (with his instanta-neous location still as indicated in the figure).
Determine the velocity and acceleration which A appears to have relative to a passenger fixed to and rotating with the ship.
5/173 Two boys A and B are sitting on opposite sides of a horizontal turntable which rotates at a constant counterclockwise angular velocity as seen from above. Boy A throws a ball toward B by giving it a horizontal velocity u relative to the turntable to-ward B. Assume that the ball has no horizontal acceleration once released and write an expression for the magnitude of the acceleration which B would observe the ball to have in the plane of the turntable just after it is thrown. Sketch the path of the ball on the turntable as observed by B.
Problem 5/173 O A u B r ω arel vA 1.6 m/s y x 10° 100 m 15° B A ω vB 0.5 deg/s2.
1 deg/s vB 1 m/s 398 Chapter 5 Plane Kinematics of Rigid Bodies 5/169 Bar OA has a counterclockwise angular velocity Rod BC slides freely through the piv-oted collar attached to OA. Determine the angular velocity of rod BC and the velocity of collar A relative to rod BC.
Problem 5/169 5/170 A smooth bowling alley is oriented north–south as shown. A ball A is released with speed v along the lane as shown. Because of the Coriolis effect, it will deflect a distance as shown. Develop a general ex-pression for . The bowling alley is located at a lati-tude in the northern hemisphere. Evaluate your expression for the conditions and Should bowlers prefer east–west alleys?
State any assumptions.
Problem 5/170 Not to scale L A δ v N 40.
L 60 ft, v 15 ft/sec, 45° 60° 250 mm A C B O ω 0 BC 0 2 rad/s.
c05.qxd 2/10/12 10:08 AM Page 398 5/174 Car B turns onto the circular off-ramp with a speed v.
Car A, traveling with the same speed v, continues in a straight line. Prove that the velocity which A appears to have to an observer riding in and turn-ing with car B is zero when car A passes the posi-tion shown regardless of the angle Problem 5/174 5/175 For the conditions and conclusion of Prob. 5/174, show that the acceleration which car A appears to have to an observer in and turning with car B is equal to in the direction normal to the true velocity of A.
5/176 For the instant represented, link CB is rotating counterclockwise at a constant rate and its pin A causes a clockwise rotation of the slotted member ODE. Determine the angular velocity and angular acceleration of ODE for this instant.
Problem 5/176 A D B O C N E 120 mm 120 mm 45° N 4 rad/s, v2/R θ v v R A B .
Article 5/7 Problems 399 5/177 Cars A and B are both traveling on the curved inter-secting roads with equal constant speeds of 30 mi/hr. For the positions shown, obtain the vector expressions for the velocity and acceleration which A appears to have to an observer in B who rotates with the car. The x-y axes are attached to car B.
Problem 5/177 5/178 The disk rotates about a fixed axis through point O with a clockwise angular velocity and a counterclockwise angular acceleration at the instant under consideration. The value of r is 200 mm. Pin A is fixed to the disk but slides freely within the slotted member BC. Determine the veloc-ity and acceleration of A relative to slotted member BC and the angular velocity and angular acceleration of BC.
Problem 5/178 3r ω O B A C r 60° 0 0 5 rad/s2 0 20 rad/s 180′ A B 45° 180′ 20′ 45° y x c05.qxd 2/10/12 10:08 AM Page 399 5/181 The figure shows the vanes of a centrifugal-pump impeller which turns with a constant clockwise speed of . The fluid particles are ob-served to have an absolute velocity whose compo-nent in the r-direction is at discharge from the vane. Furthermore, the magnitude of the velocity of the particles measured relative to the vane is increasing at the rate of just be-fore they leave the vane. Determine the magnitude of the total acceleration of a fluid particle an in-stant before it leaves the impeller. The radius of curvature of the vane at its end is 8 in.
Problem 5/181 5/182 The crank OA revolves clockwise with a constant angular velocity of 10 rad/s within a limited arc of its motion. For the position determine the angular velocity of the slotted link CB and the accel-eration of A as measured relative to the slot in CB.
Problem 5/182 θ 200 mm θ 2 A B O C 30 45°+r ω 6″ ρ = 8″ 80 ft/sec2 10 ft/sec 200 rev/min 400 Chapter 5 Plane Kinematics of Rigid Bodies 5/179 All conditions of the previous problem remain the same, except now, rather than rotating about a fixed center, the disk rolls without slipping on the horizontal surface. If the disk has a clockwise angu-lar velocity of and a counterclockwise angular acceleration of determine the ve-locity and acceleration of pin A relative to the slot-ted member BC and the angular velocity and angular acceleration of BC. The value of r is 200 mm. Neglect the distance from the center of pin A to the edge of the disk.
Problem 5/179 5/180 Two satellites are in circular equatorial orbits of different altitudes. Satellite A is in a geosynchro-nous orbit (one with the same period as the earth’s rotation so that it “hovers” over the same spot on the equator). Satellite B has an orbit of radius Calculate the velocity which A ap-pears to have to an observer fixed in B when the elevation angle is (a) 0 and (b) The x-y axes are attached to B, whose antenna always points toward the center of the earth ( -direction).
Consult and Appendix D for the neces-sary orbital information.
Problem 5/180 A θ B N rB rA y x Art. 3/13 y 90.
rB 30 000 km.
3r ω O B A C 60° 0 r 5 rad/s2, 20 rad/s c05.qxd 2/10/12 10:08 AM Page 400 Article 5/7 Problems 401 5/183 The Geneva wheel of Prob. 5/56 is shown again here. Determine the angular acceleration of wheel C for the instant when Wheel A has a constant clockwise angular velocity of 2 rad/s.
Problem 5/183 200 mm P A C B O1 O2 200/ 2 mm 200/ 2 mm θ 1 ω 2 ω 20.
2 5/184 The space shuttle A is in an equatorial circular orbit of 240-km altitude and is moving from west to east.
Determine the velocity and acceleration which it ap-pears to have to an observer B fixed to and rotating with the earth at the equator as the shuttle passes overhead. Use for the radius of the earth. Also use Fig. 1/1 for the appropriate value of g and carry out your calculations to 4-figure accuracy.
Problem 5/184 y x B A 240 km R 6378 km c05.qxd 2/10/12 10:08 AM Page 401 402 Chapter 5 Plane Kinematics of Rigid Bodies 5/8 Chapter Review In Chapter 5 we have applied our knowledge of basic kinematics from Chapter 2 to the plane motion of rigid bodies. We approached the problem in two ways.
1. Absolute-Motion Analysis First, we wrote an equation which describes the general geometric configuration of a given problem in terms of knowns and unknowns.
Then we differentiated this equation with respect to time to obtain ve-locities and accelerations, both linear and angular.
2. Relative-Motion Analysis We applied the principles of relative motion to rigid bodies and found that this approach enables us to solve many problems which are too awkward to handle by mathematical differentiation. The relative-velocity equation, the instantaneous center of zero velocity, and the relative-acceleration equation all require that we visualize clearly and analyze correctly the case of circular motion of one point around an-other point, as viewed from nonrotating axes.
Solution of the Velocity and Acceleration Equations The relative-velocity and relative-acceleration relationships are vec-tor equations which we may solve in any one of three ways: 1. by a scalar-geometric analysis of the vector polygon, 2. by vector algebra, or 3. by a graphical construction of the vector polygon.
Rotating Coordinate Systems Finally, in Chapter 5 we introduced rotating coordinate systems which enable us to solve problems where the motion is observed relative to a rotating frame of reference. Whenever a point moves along a path which itself is turning, analysis by rotating axes is indicated if a relative-motion approach is used. In deriving Eq. 5/12 for velocity and Eq. 5/14 for acceleration, where the relative terms are measured from a rotating reference system, it was necessary for us to account for the time deriva-tives of the unit vectors i and j fixed to the rotating frame. Equations 5/12 and 5/14 also apply to spatial motion, as will be shown in Chapter 7.
An important result of the analysis of rotating coordinate systems is the identification of the Coriolis acceleration. This acceleration repre-sents the fact that the absolute velocity vector may have changes in both direction and magnitude due to rotation of the relative-velocity vector and change in position of the particle along the rotating path.
In Chapter 6 we will study the kinetics of rigid bodies in plane mo-tion. There we will find that the ability to analyze the linear and angular accelerations of rigid bodies is necessary in order to apply the force and moment equations which relate the applied forces to the associated mo-tions. Thus, the material of Chapter 5 is essential to that in Chapter 6.
c05.qxd 2/10/12 10:08 AM Page 402 Article 5/8 Review Problems 403 5/188 The flywheel is rotating with an angular velocity at time when a torque is applied to increase its angular velocity. If the torque is controlled so that the angle between the total acceleration of point A on the rim and the radial line to A remains constant, determine the angular velocity and the angular acceleration as functions of the time t.
Problem 5/188 5/189 The equilateral triangular plate is guided by the two vertex rollers A and B, which are confined to move in the perpendicular slots. The control rod gives A a constant velocity to the left for an in-terval of its motion. Determine the value of for which the horizontal component of the velocity of C is zero.
Problem 5/189 A v A B C b b b θ vA r A ω α t 0 0 REVIEW PROBLEMS 5/185 The frictional resistance to the rotation of a fly-wheel consists of a retardation due to air friction which varies as the square of the angular velocity and a constant frictional retardation in the bear-ing. As a result the angular acceleration of the fly-wheel while it is allowed to coast is given by , where K and k are constants. De-termine an expression for the time required for the flywheel to come to rest from an initial angular velocity 5/186 The wheel slips as it rolls. If and if the velocity of A with respect to B is locate the instantaneous center C of zero velocity and find the velocity of point P.
Problem 5/186 5/187 The bar of is repeated here. If the velocity and tangential acceleration of end A are as indicated in the figure, determine the angular acceler-ation of the bar.
Problem 5/187 vA = 0.3 m/s (aA)t = 0.6 m/s2 0.8 m 45° A B O Prob. 5/67 ω 3″ P O A B D 6″ vO 32 ft/sec, vO 4 ft/sec 0.
K k2 c05.qxd 2/10/12 10:08 AM Page 403 404 Chapter 5 Plane Kinematics of Rigid Bodies 5/190 Roller B of the linkage has a velocity of to the right as the angle passes and bar AB also makes an angle of with the horizontal. Locate the instantaneous center of zero velocity for bar AB and determine its angular velocity Problem 5/190 5/191 The pin A in the bell crank AOD is guided by the flanges of the collar B, which slides with a constant velocity of along the fixed shaft for an interval of motion. For the position deter-mine the acceleration of the plunger CE, whose upper end is positioned by the radial slot in the bell crank.
Problem 5/191 6″ 6″ 90° θ 9″ A E C O B vB D 30 3 ft/sec vB θ A B O 540 mm 360 mm 0.75 m/s AB.
60 60 0.75 m/s 5/192 The helicopter is flying in the horizontal x-direction with a velocity and the plane of rota-tion of the 26-ft-diameter rotor is tilted from the horizontal x-y plane. The rotor blades rotate with an angular velocity For the in-stant represented write the vector expressions for the absolute velocities of rotor tip A and rotor tip B.
Problem 5/192 5/193 The wheel rolls without slipping, and its position is controlled by the motion of the slider B. If B has a constant velocity of to the left, determine the angular velocity of AB and the velocity of the center O of the wheel when Problem 5/193 5/194 If the center O of the wheel of Prob. 5/193 has a constant velocity of to the left, calculate the acceleration of the slider B for the position 0.
6 in./sec O A θ 6″ 4″ 16″ B 0.
10 in./sec x v x′ y, y′ C A B z z′ 10° Ω 10° 800 rev/min.
10 v 120 mi/hr, c05.qxd 2/10/12 10:08 AM Page 404 Article 5/8 Review Problems 405 5/197 The hydraulic cylinder C imparts a velocity v to pin B in the direction shown. The collar slips freely on rod OA. Determine the resulting angular velocity of rod OA in terms of v, the displacement s of pin B, and the fixed distance d, for the angle Problem 5/197 5/198 The figure illustrates a commonly used quick-return mechanism which produces a slow cutting stroke of the tool (attached to D) and a rapid return stroke. If the driving crank OA is turning at the constant rate determine the magnitude of the velocity of point B for the instant when Problem 5/198 500 mm 300 mm D B A 100 mm O C θ 30.
˙ 3 rad/s, β O A B C d s v θ 15.
5/195 In the linkage shown OC has a constant clockwise angular velocity during an interval of motion, while the hydraulic cylinder gives pin A a constant velocity of to the right. For the po-sition shown where OC is vertical and BC is hori-zontal, calculate the angular velocity of BC. Solve by drawing the necessary velocity polygon.
Problem 5/195 5/196 To speed up the unrolling of a telephone cable the trailer with the reel of cable starts from rest and is given an initial acceleration of Simultane-ously, the tow truck pulls the free end of the cable horizontally in the opposite direction with an ini-tial acceleration of If both vehicles start from rest at the same instant, determine the mag-nitude of the total acceleration of point A on the forward end of the horizontal reel diameter (a) just as the motion starts and (b) one second after the start of the motion.
Problem 5/196 2 ft/sec2 A 3 ft/sec2 2 ft/sec2.
3 ft/sec2.
500 mm 400 mm O A B C 300 mm ω vA 1.2 m/s 2 rad/s c05.qxd 2/10/12 10:08 AM Page 405 406 Chapter 5 Plane Kinematics of Rigid Bodies 5/199 The hydraulic cylinder moves pin A to the right with a constant velocity v. Use the fact that the dis-tance from A to B is invariant, where B is the point on AC momentarily in contact with the gear, and write expressions for the angular velocity of the gear and the angular velocity of the rack AC.
Problem 5/199 5/200 For the position shown where point A on the sliding collar has a constant velocity with corresponding lengthening of the hydraulic cylinder AC. For this same position BD is horizon-tal and DE is vertical. Determine the angular accel-eration of DE at this instant.
Problem 5/200 200 mm C B v A θ D 90 mm E 200 mm DE v 0.3 m/s 30, θ ω v D B A r C 5/201 The tilting device maintains a sloshing water bath for washing vegetable produce. If the crank OA os-cillates about the vertical and has a clockwise angu-lar velocity of when OA is vertical, determine the angular velocity of the basket in the position shown where Problem 5/201 5/202 Determine the angular acceleration of the basket of the vegetable washer of for the posi-tion where OA is vertical. In this position OA has an angular velocity of and no angular acceleration.
5/203 A radar station B situated at the equator observes a satellite A in a circular equatorial orbit of altitude and moving from west to east. For the in-stant when the satellite is above the horizon, determine the difference between the velocity of the satellite relative to the radar station, as mea-sured from a nonrotating frame of reference, and the velocity as measured relative to the reference frame of the radar system.
Problem 5/203 30° x y 200 km B A 30 200-km 4 rad/s Prob. 5/201 θ 100 mm 80 mm 240 mm 360 mm A B D O 30.
4 rad/s c05.qxd 2/10/12 10:08 AM Page 406 Article 5/8 Review Problems 407 Problem 5/206 5/207 The crank OA of the four-bar linkage is driven at a constant counterclockwise angular velocity Determine and plot as functions of the crank angle the angular velocities of bars AB and BC over the range State the maximum absolute value of each angular velocity and the value of at which it occurs.
Problem 5/207 5/208 If all conditions in the previous problem remain the same, determine and plot as functions of the crank angle the angular accelerations of bars AB and BC over the range State the maximum absolute value of each angular accelera-tion and the value of at which it occurs.
0 360.
200 mm 70 mm OA = 80 mm A B O C 240 mm ω 0 θ 190 mm 0 360.
0 10 rad/s.
r B O A r θ ω 2 r 5/204 The crank OB revolves clockwise at the constant rate of For the instant when de-termine the angular acceleration of the rod BD, which slides through the pivoted collar at C.
Problem 5/204 Computer-Oriented Problems 5/205 The disk rotates about a fixed axis with a constant angular velocity Pin A is fixed to the disk. Determine and plot the magnitudes of the ve-locity and acceleration of pin A relative to the slot-ted member BC as functions of the disk angle over the range State the maximum and minimum values and also the values of at which they occur. The value of r is 200 mm.
Problem 5/205 5/206 Link OA is given a constant counterclockwise angular velocity Determine the angular velocity of link AB as a function of Compute and plot the ratio for the range Indicate the value of for which the angular velocity of AB is half that of OA.
0 90.
AB/ .
AB .
3r θ ω O B A C r 0 0 360.
0 10 rad/s.
θ ω0 250 mm 600 mm D C B O 90 5 rad/s.
0 c05.qxd 2/10/12 10:08 AM Page 407 408 Chapter 5 Plane Kinematics of Rigid Bodies 5/209 All conditions of Prob. 5/207 remain the same, except the counterclockwise angular velocity of crank OA is when and the constant counterclockwise angular acceleration of the crank is Determine and plot as func-tions of the crank angle the angular velocities of bars AB and BC over the range State the maximum absolute value of each angu-lar velocity and the value of at which it occurs.
5/210 For the Geneva wheel of Prob. 5/56, shown again here, write the expression for the angular velocity of the slotted wheel C during engagement of pin P and plot for the range The driving wheel A has a constant angular veloc-ity Problem 5/210 5/211 The double crank is pivoted at O and permits complete rotation without interference with the pivoted rod CB as it slides through the collar A. If the crank has a constant angular velocity de-termine and plot the ratio as a function of between and By inspection deter-mine the angle for which Problem 5/211 ˙ 0.
180.
0 ˙/ ˙ ˙ , 200 mm P A C B O1 O2 200/ 2 mm 200/ 2 mm θ 1 ω 2 ω 1 2 rad/s.
45 45.
2 2 0 360.
20 rad/s2.
0 10 rad/s 5/212 For the slider-crank configuration shown, derive the expression for the velocity of the piston (taken positive to the right) as a function of Sub-stitute the numerical data of Sample Problem 5/15 and calculate as a function of for Plot versus and find its maximum magnitude and the corresponding value of . (By symmetry anticipate the results for Problem 5/212 5/213 For the slider-crank of derive the expression for the acceleration of the piston (taken positive to the right) as a function of for constant. Substitute the numerical data of Sample Problem and calculate as a function of for Plot versus and find the value of for which (By sym-metry anticipate the results for 180 360).
aA 0.
aA 0 180.
aA 5/15 ˙ aA Prob. 5/212, θ ω vA x l y r B O A 180 360).
vA 0 180.
vA .
vA 2r r C O O B B O A θ β A c05.qxd 2/10/12 10:08 AM Page 408 c05.qxd 2/10/12 10:08 AM Page 409 By changing between a fully outstretched and a tucked or pike position, a diver can cause large changes in her angular speed about an axis perpendicular to the plane of the trajectory. Conservation of angular momentum is the key issue here. The rigid-body principles of this chapter apply here, even though the human body is of course not rigid.
© Belinda Images/SUPERSTOCK c06.qxd 2/10/12 2:13 PM Page 410 411 6/1 Introduction The kinetics of rigid bodies treats the relationships between the ex-ternal forces acting on a body and the corresponding translational and rotational motions of the body. In Chapter 5 we developed the kinematic relationships for the plane motion of rigid bodies, and we will use these relationships extensively in this present chapter, where the effects of forces on the two-dimensional motion of rigid bodies are examined.
For our purpose in this chapter, a body which can be approximated as a thin slab with its motion confined to the plane of the slab will be considered to be in plane motion. The plane of motion will contain the mass center, and all forces which act on the body will be projected onto the plane of motion. A body which has appreciable dimensions normal to the plane of motion but is symmetrical about that plane of motion through the mass center may be treated as having plane motion. These idealizations clearly fit a very large category of rigid-body motions.
6/1 Introduction Section A Force, Mass, and Acceleration 6/2 General Equations of Motion 6/3 Translation 6/4 Fixed-Axis Rotation 6/5 General Plane Motion Section B Work and Energy 6/6 Work-Energy Relations 6/7 Acceleration from Work-Energy; Virtual Work Section C Impulse and Momentum 6/8 Impulse-Momentum Equations 6/9 Chapter Review CHAPTER OUTLINE 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 411 Background for the Study of Kinetics In Chapter 3 we found that two force equations of motion were re-quired to define the motion of a particle whose motion is confined to a plane. For the plane motion of a rigid body, an additional equation is needed to specify the state of rotation of the body. Thus, two force equa-tions and one moment equation or their equivalent are required to de-termine the state of rigid-body plane motion.
The kinetic relationships which form the basis for most of the analysis of rigid-body motion were developed in Chapter 4 for a general system of particles. Frequent reference will be made to these equations as they are further developed in Chapter 6 and applied specifically to the plane motion of rigid bodies. You should refer to Chapter 4 frequently as you study Chapter 6. Also, before proceeding make sure that you have a firm grasp of the calculation of velocities and accelerations as developed in Chapter 5 for rigid-body plane motion. Unless you can determine ac-celerations correctly from the principles of kinematics, you frequently will be unable to apply the force and moment principles of kinetics. Con-sequently, you should master the necessary kinematics, including the calculation of relative accelerations, before proceeding.
Successful application of kinetics requires that you isolate the body or system to be analyzed. The isolation technique was illustrated and used in Chapter 3 for particle kinetics and will be employed consis-tently in the present chapter. For problems involving the instanta-neous relationships among force, mass, and acceleration, the body or system should be explicitly defined by isolating it with its free-body dia-gram. When the principles of work and energy are employed, an active-force diagram which shows only those external forces which do work on the system may be used in lieu of the free-body diagram. The impulse-momentum diagram should be constructed when impulse-momentum methods are used. No solution of a problem should be attempted with-out first defining the complete external boundary of the body or system and identifying all external forces which act on it.
In the kinetics of rigid bodies which have angular motion, we must introduce a property of the body which accounts for the radial distribu-tion of its mass with respect to a particular axis of rotation normal to the plane of motion. This property is known as the mass moment of iner-tia of the body, and it is essential that we be able to calculate this prop-erty in order to solve rotational problems. We assume that you are familiar with the calculation of mass moments of inertia. Appendix B treats this topic for those who need instruction or review.
Organization of the Chapter Chapter 6 is organized in the same three sections in which we treated the kinetics of particles in Chapter 3. Section A relates the forces and moments to the instantaneous linear and angular accelera-tions. Section B treats the solution of problems by the method of work and energy. Section C covers the methods of impulse and momentum.
Virtually all of the basic concepts and approaches covered in these three sections were treated in Chapter 3 on particle kinetics. This repe-tition will help you with the topics of Chapter 6, provided you understand 412 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 412 the kinematics of rigid-body plane motion. In each of the three sections, we will treat three types of motion: translation, fixed-axis rotation, and general plane motion.
Article 6/2 General Equations of Motion 413 6/2 General Equations of Motion In Arts. 4/2 and 4/4 we derived the force and moment vector equa-tions of motion for a general system of mass. We now apply these results by starting, first, with a general rigid body in three dimensions. The force equation, Eq. 4/1, [4/1] tells us that the resultant ΣF of the external forces acting on the body equals the mass m of the body times the acceleration of its mass cen-ter G. The moment equation taken about the mass center, Eq. 4/9, [4/9] shows that the resultant moment about the mass center of the external forces on the body equals the time rate of change of the angular momen-tum of the body about the mass center.
Recall from our study of statics that a general system of forces act-ing on a rigid body may be replaced by a resultant force applied at a cho-sen point and a corresponding couple. By replacing the external forces by their equivalent force-couple system in which the resultant force acts through the mass center, we may visualize the action of the forces and the corresponding dynamic response of the body with the aid of Fig. 6/1.
ΣMG H ˙ G a ΣF ma SECTION A FORCE, MASS, AND ACCELERATION Equivalent Force-Couple System (b) ≡ ≡ Free-Body Diagram (a) Kinetic Diagram (c) ma – F1 F2 F3 F4 ΣF ΣMG H · G G G G Figure 6/1 c06.qxd 2/10/12 2:13 PM Page 413 Part a of the figure shows the relevant free-body diagram. Part b of the figure shows the equivalent force-couple system with the resultant force applied through G. Part c of the figure is a kinetic diagram, which repre-sents the resulting dynamic effects as specified by Eqs. 4/1 and 4/9. The equivalence between the free-body diagram and the kinetic diagram en-ables us to clearly visualize and easily remember the separate transla-tional and rotational effects of the forces applied to a rigid body. We will express this equivalence mathematically as we apply these results to the treatment of rigid-body plane motion.
Plane-Motion Equations We now apply the foregoing relationships to the case of plane mo-tion. Figure 6/2 represents a rigid body moving with plane motion in the x-y plane. The mass center G has an acceleration and the body has an angular velocity k and an angular acceleration k, both taken positive in the z-direction. Because the z-direction of both and remains perpendicular to the plane of motion, we may use scalar notation and to represent the angular velocity and an-gular acceleration.
The angular momentum about the mass center for the general sys-tem was expressed in Eq. 4/8a as HG Σi where i is the posi-tion vector relative to G of the representative particle of mass mi. For our rigid body, the velocity of mi relative to G is i, which has a magnitude i and lies in the plane of motion normal to i. The product i is then a vector normal to the x-y plane in the sense of , and its magnitude is Thus, the magnitude of HG becomes HG The summation, which may also be written as 2 dm, is de-fined as the mass moment of inertia of the body about the z-axis through G. (See Appendix B for a discussion of the calculation of mass moments of inertia.) We may now write where is a constant property of the body. This property is a measure of the rotational inertia, which is the resistance to change in rotational ve-locity due to the radial distribution of mass around the z-axis through G. With this substitution, our moment equation, Eq. 4/9, becomes where is the angular acceleration of the body.
We may now express the moment equation and the vector form of the generalized Newton’s second law of motion, Eq. 4/1, as (6/1) Equations 6/1 are the general equations of motion for a rigid body in plane motion. In applying Eqs. 6/1, we express the vector force equation ΣMG I ΣF ma ˙ ΣMG H ˙G I ˙ I I HG I I Σi 2mi.
Σi 2mi i 2.
˙i ˙i mi ˙i ˙ a, 414 Chapter 6 Plane Kinetics of Rigid Bodies y x mi G a – F4 F3 F2 F1 β α ω i ρ Figure 6/2 c06.qxd 2/10/12 2:13 PM Page 414 in terms of its two scalar components using x-y, n-t, or r- coordinates, whichever is most convenient for the problem at hand.
Alternative Derivation It is instructive to use an alternative approach to derive the mo-ment equation by referring directly to the forces which act on the repre-sentative particle of mass mi, as shown in Fig. 6/3. The acceleration of mi equals the vector sum of and the relative terms i2 and i, where the mass center G is used as the reference point. It follows that the re-sultant of all forces on mi has the components mii2, and mii in the directions shown. The sum of the moments of these force compo-nents about G in the sense of becomes Similar moment expressions exist for all particles in the body, and the sum of these moments about G for the resultant forces acting on all particles may be written as But the origin of coordinates is taken at the mass center, so that Σmixi 0 and Σmi yi 0. Thus, the moment sum becomes as before. The contribution to ΣMG of the forces internal to the body is, of course, zero since they occur in pairs of equal and opposite forces of action and reaction between interacting particles. Thus, ΣMG, as before, represents the sum of moments about the mass center G of only the ex-ternal forces acting on the body, as disclosed by the free-body diagram.
We note that the force component mii2 has no moment about G and conclude, therefore, that the angular velocity has no influence on the moment equation about the mass center.
The results embodied in our basic equations of motion for a rigid body in plane motion, Eqs. 6/1, are represented diagrammatically in Fig. 6/4, ΣMG Σmi i 2 I my mx ΣMG Σmi i 2 a sin Σmi xi a cos Σmi yi MGi mi i 2 (mi a sin )xi (mi a cos )yi mi a, a Article 6/2 General Equations of Motion 415 y x mi mia – xi G or β α ω ω mi i ρ α mi 2 i ρ ω yi Figure 6/3 a – G ma – G α ≡ F1 F2 F3 I – α Free-Body Diagram Kinetic Diagram Figure 6/4 c06.qxd 2/10/12 2:13 PM Page 415 which is the two-dimensional counterpart of parts a and c of Fig. 6/1 for a general three-dimensional body. The free-body diagram discloses the forces and moments appearing on the left-hand side of our equations of motion.
The kinetic diagram discloses the resulting dynamic response in terms of the translational term and the rotational term which appear on the right-hand side of Eqs. 6/1.
As previously mentioned, the translational term will be ex-pressed by its x-y, n-t, or r- components once the appropriate inertial reference system is designated. The equivalence depicted in Fig. 6/4 is basic to our understanding of the kinetics of plane motion and will be employed frequently in the solution of problems.
Representation of the resultants and will help ensure that the force and moment sums determined from the free-body diagram are equated to their proper resultants.
Alternative Moment Equations In Art. 4/4 of Chapter 4 on systems of particles, we developed a gen-eral equation for moments about an arbitrary point P, Eq. 4/11, which is [4/11] where is the vector from P to the mass center G and is the mass-cen-ter acceleration. As we have shown earlier in this article, for a rigid body in plane motion becomes Also, the cross product is sim-ply the moment of magnitude of about P. Therefore, for the two-dimensional body illustrated in Fig. 6/5 with its free-body diagram and kinetic diagram, we may rewrite Eq. 4/11 simply as (6/2) Clearly, all three terms are positive in the counterclockwise sense for the example shown, and the choice of P eliminates reference to F1 and F3.
If we had wished to eliminate reference to F2 and F3, for example, by choosing their intersection as the reference point, then P would lie on the opposite side of the vector, and the clockwise moment of ma ma ΣMP I mad ma mad ma I.
H ˙ G a ΣMP H ˙ G ma I ma ma I ma 416 Chapter 6 Plane Kinetics of Rigid Bodies a – G P d ma – P G α I – α – ρ ≡ F1 F2 F3 Free-Body Diagram Kinetic Diagram Figure 6/5 c06.qxd 2/10/12 2:13 PM Page 416 about P would be a negative term in the equation. Equation 6/2 is easily remembered as it is merely an expression of the familiar principle of mo-ments, where the sum of the moments about P equals the combined mo-ment about P of their sum, expressed by the resultant couple ΣMG and the resultant force ΣF In Art. 4/4 we also developed an alternative moment equation about P, Eq. 4/13, which is [4/13] For rigid-body plane motion, if P is chosen as a point fixed to the body, then in scalar form becomes IP, where IP is the mass moment of inertia about an axis through P and is the angular acceleration of the body. So we may write the equation as (6/3) where the acceleration of P is aP and the position vector from P to G is When 0, point P becomes the mass center G, and Eq. 6/3 re-duces to the scalar form ΣMG previously derived. When point P becomes a point O fixed in an inertial reference system and attached to the body (or body extended), then aP 0, and Eq. 6/3 in scalar form re-duces to (6/4) Equation 6/4 then applies to the rotation of a rigid body about a nonac-celerating point O fixed to the body and is the two-dimensional simplifi-cation of Eq. 4/7.
Unconstrained and Constrained Motion The motion of a rigid body may be unconstrained or constrained.
The rocket moving in a vertical plane, Fig. 6/6a, is an example of uncon-strained motion as there are no physical confinements to its motion.
ΣMO IO I, .
ΣMP I P maP (H ˙ P)rel ΣMP (H ˙ P)rel maP ma.
I Article 6/2 General Equations of Motion 417 B G A F mg y x G (a) Unconstrained Motion (b) Constrained Motion T a – x a – x a – y a – y α α y x Figure 6/6 c06.qxd 2/10/12 2:13 PM Page 417 The two components and of the mass-center acceleration and the angular acceleration may be determined independently of one another by direct application of Eqs. 6/1.
The bar in Fig. 6/6b, on the other hand, undergoes a constrained motion, where the vertical and horizontal guides for the ends of the bar impose a kinematic relationship between the acceleration com-ponents of the mass center and the angular acceleration of the bar.
Thus, it is necessary to determine this kinematic relationship from the principles established in Chapter 5 and to combine it with the force and moment equations of motion before a solution can be car-ried out.
In general, dynamics problems which involve physical constraints to motion require a kinematic analysis relating linear to angular accelera-tion before the force and moment equations of motion can be solved. It is for this reason that an understanding of the principles and methods of Chapter 5 is so vital to the work of Chapter 6.
Systems of Interconnected Bodies Upon occasion, in problems dealing with two or more connected rigid bodies whose motions are related kinematically, it is convenient to analyze the bodies as an entire system.
Figure 6/7 illustrates two rigid bodies hinged at A and subjected to the external forces shown. The forces in the connection at A are internal to the system and are not disclosed. The resultant of all external forces must equal the vector sum of the two resultants and and the sum of the moments about some arbitrary point such as P of all external forces must equal the moment of the resultants, Thus, we may state (6/5) ΣMP ΣI Σmad ΣF Σma m2a2d2.
m1a1d1 I22 I11 m2a2, m1a1 ay ax 418 Chapter 6 Plane Kinetics of Rigid Bodies P Kinetic Diagram of System Free-Body Diagram of System A A G1 G1 G2 m1a – 1 a – 2 a – 1 d1 d2 m2a – 2 I – 2 2 α 2 α 1 α I – 1 1 α G2 ≡ ≡ Figure 6/7 c06.qxd 2/10/12 2:13 PM Page 418 where the summations on the right-hand side of the equations repre-sent as many terms as there are separate bodies.
If there are more than three remaining unknowns in a system, how-ever, the three independent scalar equations of motion, when applied to the system, are not sufficient to solve the problem. In this case, more ad-vanced methods such as virtual work (Art. 6/7) or Lagrange’s equations (not discussed in this book) could be employed, or else the system could be dismembered and each part analyzed separately with the resulting equations solved simultaneously.
Article 6/2 General Equations of Motion 419 When an interconnected system has more than one degree of freedom, that is, requires more than one coordinate to specify completely the configuration of the system, the more advanced equations of Lagrange are generally used. See the first author’s Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, for a treatment of Lagrange’s equations.
Analysis Procedure In the solution of force-mass-acceleration problems for the plane motion of rigid bodies, the following steps should be taken once you un-derstand the conditions and requirements of the problem: 1. Kinematics. First, identify the class of motion and then solve for any needed linear and angular accelerations which can be deter-mined solely from given kinematic information. In the case of con-strained plane motion, it is usually necessary to establish the relation between the linear acceleration of the mass center and the angular ac-celeration of the body by first solving the appropriate relative-velocity and relative-acceleration equations. Again, we emphasize that success in working force-mass-acceleration problems in this chapter is contingent on the ability to describe the necessary kinematics, so that frequent re-view of Chapter 5 is recommended.
2. Diagrams. Always draw the complete free-body diagram of the body to be analyzed. Assign a convenient inertial coordinate system and label all known and unknown quantities. The kinetic diagram should also be constructed so as to clarify the equivalence between the applied forces and the resulting dynamic response.
3. Equations of Motion. Apply the three equations of motion from Eqs. 6/1, being consistent with the algebraic signs in relation to the choice of reference axes. Equation 6/2 or 6/3 may be employed as an alternative to the second of Eqs. 6/1. Combine these relations with the results from any needed kinematic analysis. Count the number of unknowns and be certain that there are an equal number of indepen-dent equations available. For a solvable rigid-body problem in plane motion, there can be no more than the five scalar unknowns which can be determined from the three scalar equations of motion, obtained from Eqs. 6/1, and the two scalar component relations which come from the relative-acceleration equation.
KEY CONCEPTS c06.qxd 2/10/12 2:13 PM Page 419 In the following three articles the foregoing developments will be applied to three cases of motion in a plane: translation, fixed-axis rota-tion, and general plane motion.
6/3 Translation Rigid-body translation in plane motion was described in Art. 5/1 and illustrated in Figs. 5/1a and 5/1b, where we saw that every line in a trans-lating body remains parallel to its original position at all times. In recti-linear translation all points move in straight lines, whereas in curvilinear translation all points move on congruent curved paths. In either case, there is no angular motion of the translating body, so that both and are zero. Therefore, from the moment relation of Eqs. 6/1, we see that all reference to the moment of inertia is eliminated for a translating body.
420 Chapter 6 Plane Kinetics of Rigid Bodies ≡ ≡ G G A B dB dA t t n n ma – x y G G Path of G F1 F2 F3 F3 F2 F1 d A P Free-Body Diagram (a) Rectilinear Translation ( = 0, = 0) α ω (b) Curvilinear Translation ( = 0, = 0) α ω Kinetic Diagram Free-Body Diagram Kinetic Diagram ma – n ma – t Path of G Figure 6/8 c06.qxd 2/10/12 2:13 PM Page 420 For a translating body, then, our general equations for plane mo-tion, Eqs. 6/1, may be written (6/6) For rectilinear translation, illustrated in Fig. 6/8a, if the x-axis is chosen in the direction of the acceleration, then the two scalar force equations become ΣFx and ΣFy 0. For curvilinear translation, Fig.
6/8b, if we use n-t coordinates, the two scalar force equations become ΣFn and ΣFt In both cases, ΣMG 0.
We may also employ the alternative moment equation, Eq. 6/2, with the aid of the kinetic diagram. For rectilinear translation we see that ΣMP and ΣMA 0. For curvilinear translation the kinetic diagram permits us to write ΣMA in the clockwise sense and ΣMB in the counterclockwise sense. Thus, we have complete freedom to choose a convenient moment center.
mat dB mandA mad mat.
man may max ΣMG I 0 ΣF ma Article 6/3 Translation 421 The methods of this article apply to this motorcycle if its roll (lean) angle is constant for an interval of time.
© Howard Sayer/Alamy c06.qxd 2/10/12 2:13 PM Page 421 SAMPLE PROBLEM 6/1 The pickup truck weighs 3220 lb and reaches a speed of 30 mi/hr from rest in a distance of 200 ft up the 10-percent incline with constant acceleration. Cal-culate the normal force under each pair of wheels and the friction force under the rear driving wheels. The effective coefficient of friction between the tires and the road is known to be at least 0.8.
Solution.
We will assume that the mass of the wheels is negligible compared with the total mass of the truck. The truck may now be simulated by a single rigid body in rectilinear translation with an acceleration of The free-body diagram of the complete truck shows the normal forces N1 and N2, the friction force F in the direction to oppose the slipping of the dri-ving wheels, and the weight W represented by its two components. With tan1 1/10 5.71, these components are W cos 3220 cos 5.71 3200 lb and W sin 3220 sin 5.71 320 lb. The kinetic diagram shows the resul-tant, which passes through the mass center and is in the direction of its accel-eration. Its magnitude is Applying the three equations of motion, Eqs. 6/1, for the three unknowns gives Ans.
(a) (b) Solving (a) and (b) simultaneously gives Ans.
In order to support a friction force of 804 lb, a coefficient of friction of at least F/N2 804/1763 0.46 is required. Since our coefficient of friction is at least 0.8, the surfaces are rough enough to support the calculated value of F so that our result is correct.
Alternative Solution.
From the kinetic diagram we see that N1 and N2 can be obtained independently of one another by writing separate moment equations about A and B.
Ans.
Ans. N1 1441 lb 3200(60) 320(24) 120N1 484(24) [ΣMB mad] N2 1763 lb 120N2 60(3200) 24(320) 484(24) [ΣMA mad] N1 1441 lb N2 1763 lb [ΣMG I 0] 60N1 804(24) N2(60) 0 [ΣFy may 0] N1 N2 3200 0 [ΣFx max] F 320 484 F 804 lb ma 3220 32.2 (4.84) 484 lb a (44)2 2(200) 4.84 ft/sec2 [v2 2as] 422 Chapter 6 Plane Kinetics of Rigid Bodies 60″ 10 1 24″ 60″ G A θ B A F N2 θ W sin θ W cos N1 y x B 10 1 ma ≡ Helpful Hints Without this assumption, we would be obliged to account for the rela-tively small additional forces which produce moments to give the wheels their angular acceleration.
Recall that 30 mi/hr is 44 ft/sec.
We must be careful not to use the friction equation F N here since we do not have a case of slipping or impending slipping. If the given co-efficient of friction were less than 0.46, the friction force would be N2, and the car would be unable to attain the acceleration of 4.84 ft/sec2. In this case, the unknowns would be N1, N2, and a.
The left-hand side of the equation is evaluated from the free-body dia-gram, and the right-hand side from the kinetic diagram. The positive sense for the moment sum is arbi-trary but must be the same for both sides of the equation. In this prob-lem, we have taken the clockwise sense as positive for the moment of the resultant force about B.
c06.qxd 2/10/12 2:13 PM Page 422 SAMPLE PROBLEM 6/2 The vertical bar AB has a mass of 150 kg with center of mass G midway be-tween the ends. The bar is elevated from rest at 0 by means of the parallel links of negligible mass, with a constant couple M 5 applied to the lower link at C. Determine the angular acceleration of the links as a function of and find the force B in the link DB at the instant when 30.
Solution.
The motion of the bar is seen to be curvilinear translation since the bar itself does not rotate during the motion. With the circular motion of the mass center G, we choose n- and t-coordinates as the most convenient descrip-tion. With negligible mass of the links, the tangential component At of the force at A is obtained from the free-body diagram of AC, where ΣMC 0 and At 5/1.5 3.33 kN. The force at B is along the link. All applied forces are shown on the free-body diagram of the bar, and the kinetic diagram is also indi-cated, where the resultant is shown in terms of its two components.
The sequence of solution is established by noting that An and B depend on the n-summation of forces and, hence, on at 30. The value of depends on the variation of with . This dependency is established from a force summation in the t-direction for a general value of , where Thus, we begin with Ans.
With a known function of , the angular velocity of the links is obtained from Substitution of 30 gives and The force B may be obtained by a moment summation about A, which elimi-nates An and At and the weight. Or a moment summation may be taken about the intersection of An and the line of action of which eliminates An and Using A as a moment center gives Ans.
The component An could be obtained from a force summation in the n-direction or from a moment summation about G or about the intersection of B and the line of action of mr. B 2.14 kN 1.8 cos 30 B 2.02(1.2) cos 30 2.06(0.6) [ΣMA mad] mr.
mr, mr 0.15(1.5)(9.15) 2.06 kN mr2 0.15(1.5)(8.97) 2.02 kN (2)30 8.97 (rad/s)2 30 9.15 rad/s2 2 29.6 13.08 sin 0 d 0 (14.81 6.54 cos ) d [ d d] 14.81 6.54 cos rad/s2 3.33 0.15(9.81) cos 0.15(1.5a) [ΣFt mat] AC.
(at)A at ¨ mr2 ma M/AC kNm Article 6/3 Translation 423 0.6 m 1.8 m 1.5 m 1.5 m G D C A θ B M n t θ G ≡ mr –α mr – 2 ω 0.15(9.81) kN r – = 1.5 m B An An At At Ct Cn t n G M Helpful Hints Generally speaking, the best choice of reference axes is to make them co-incide with the directions in which the components of the mass-center acceleration are expressed. Examine the consequences of choosing hori-zontal and vertical axes.
The force and moment equations for a body of negligible mass become the same as the equations of equilib-rium. Link BD, therefore, acts as a two-force member in equilibrium.
c06.qxd 2/10/12 2:13 PM Page 423 6/4 The uniform slender bar of mass m is freely pivoted at point O of the frame of mass M. Determine the force P required to maintain the bar perpendicular to the incline of angle as the system accelerates in transla-tion down the incline. The coefficient of kinetic fric-tion between the frame and the incline is Problem 6/4 6/5 What acceleration a of the collar along the horizontal guide will result in a steady-state deflection of the pendulum from the vertical? The slender rod of length l and the particle each have mass m. Friction at the pivot P is negligible.
Problem 6/5 l m 15° P a m 15 M O m P k μ θ k.
424 Chapter 6 Plane Kinetics of Rigid Bodies PROBLEMS Introductory Problems 6/1 For what acceleration a of the frame will the uniform slender rod maintain the orientation shown in the fig-ure? Neglect the friction and mass of the small rollers at A and B.
Problem 6/1 6/2 The right-angle bar with equal legs weighs 6 lb and is freely hinged to the vertical plate at C. The bar is pre-vented from rotating by the two pegs A and B fixed to the plate. Determine the acceleration a of the plate for which no force is exerted on the bar by either peg A or B.
Problem 6/2 6/3 In Prob. if the plate is given a horizontal accelera-tion calculate the force exerted on the bar by either peg A or B.
a 2g, 6/2, a C B A 8″ 8″ 30° a A B c06.qxd 2/10/12 2:13 PM Page 424 6/6 The uniform box of mass m slides down the rough in-cline. Determine the location d of the effective normal force N. The effective normal force is located at the centroid of the nonuniform pressure distribution which the incline exerts on the bottom surface of the block.
Problem 6/6 6/7 The homogeneous create of mass m is mounted on small wheels as shown. Determine the maximum force P which can be applied without overturning the crate about its lower front edge with and its lower back edge with Problem 6/7 6/8 Determine the value of P which will cause the homo-geneous cylinder to begin to roll up out of its rectan-gular recess. The mass of the cylinder is m and that of the cart is M. The cart wheels have negligible mass and friction.
Problem 6/8 P m M G r/2 r/2 P h b c B A h 0.
(b) h b (a) b v h N d m G k μ θ Article 6/3 Problems 425 6/9 Determine the acceleration of the initially stationary 20-kg body when the 50-N force P is applied as shown.
The small wheels at B are ideal, and the feet at A are small.
Problem 6/9 6/10 Repeat the previous problem for the case when the wheels and feet have been reversed as shown in the figure for this problem. Compare your answer to the stated result for the previous problem.
Problem 6/10 6/11 The uniform 30-kg bar OB is secured to the acceler-ating frame in the position from the horizontal by the hinge at O and roller at A. If the horizontal acceleration of the frame is compute the force on the roller and the x- and y-components of the force supported by the pin at O.
Problem 6/11 30° 3000 mm 1000 mm B y x A a O FA a 20 m/s2, 30 P = 50 N G A B 20 kg μs = 0.40 μk = 0.30 μ μ 0.4 m 0.8 m P = 50 N G A B 20 kg μs = 0.40 μk = 0.30 μ μ 0.4 m 0.8 m c06.qxd 2/10/12 2:13 PM Page 425 Problem 6/14 6/15 Repeat the questions of the previous problem for the 3200-lb front-engine car shown, and compare your answers with those listed for the previous problem.
Problem 6/15 Representative Problems 6/16 The uniform 4-m boom has a mass of 60 kg and is pivoted to the back of a truck at A and secured by a cable at C. Calculate the magnitude of the total force supported by the connection at A if the truck starts from rest with an acceleration of Problem 6/16 A B a C 60° 4 m 2 m 5 m/s2.
v G 24″ A B 66″ 44″ 24″ v A B 66″ 44″ G 426 Chapter 6 Plane Kinetics of Rigid Bodies 6/12 The rear-wheel-drive lawn mower, when placed into gear while at rest, is observed to momentarily spin its rear tires as it accelerates. If the coefficients of friction between the rear tires and the ground are and determine the forward acceleration a of the mower. The mass of the mower and attached bag is 50 kg with center of mass at G. Assume that the operator does not push on the handle, so that Problem 6/12 6/13 The 6-kg frame AC and 4-kg uniform slender bar AB of length l slide with negligible friction along the fixed horizontal rod under the action of the 80-N force. Calculate the tension T in wire BC and the x- and y-components of the force exerted on the bar by the pin at A. The x-y plane is vertical.
Problem 6/13 6/14 The mass center of the rear-engine 3200-lb car is at G. Determine the normal forces and exerted by the road on the front and rear pairs of tires for the conditions of (a) being stationary and (b) braking from a forward velocity v with all wheels locked. The coefficient of kinetic friction is 0.90 at all tire/road interfaces. Express all answers in terms of pounds and as percentages of the vehicle weight.
NB NA 80 N 60° 60° C A B l x y 900 mm 1000 mm 200 mm 500 mm 215 mm P A G C B P 0.
k 0.50, s 0.70 c06.qxd 2/10/12 2:13 PM Page 426 6/17 The loaded trailer has a mass of 900 kg with center of mass at G and is attached at A to a rear-bumper hitch. If the car and trailer reach a velocity of on a level road in a distance of from rest with constant acceleration, compute the vertical component of the force supported by the hitch at A.
Neglect the small friction force exerted on the rela-tively light wheels.
Problem 6/17 6/18 Arm AB of a classifying accelerometer has a weight of 0.25 lb with mass center at G and is pivoted freely to the frame F at A. The torsional spring at A is set to preload the arm with an applied clockwise moment of 2 lb-in. Determine the downward acceleration a of the frame at which the contact at B will separate and break the electrical circuit.
Problem 6/18 6/19 The uniform 60-lb log is supported by the two cables and used as a battering ram. If the log is released from rest in the position shown, calculate the initial tension induced in each cable immediately after re-lease and the corresponding angular acceleration of the cables.
2″ A G a B F 1.5″ 1.2 m 0.9 m 0.5 m A G x y 30 m 60 km/h Article 6/3 Problems 427 Problem 6/19 6/20 Determine the magnitude P and direction of the force required to impart a rearward acceleration to the loaded wheelbarrow with no ro-tation from the position shown. The combined weight of the wheelbarrow and its load is 500 lb with center of gravity at G. Compare the normal force at B under acceleration with that for static equilibrium in the position shown. Neglect the friction and mass of the wheel.
Problem 6/20 6/21 Solid homogeneous cylinders 400 mm high and 250 mm in diameter are supported by a flat conveyor belt which moves horizontally. If the speed of the belt increases according to where t is the time in seconds measured from the instant the increase begins, calculate the value of t for which the cylinders begin to tip over. Cleats on the belt prevent the cylinders from slipping.
Problem 6/21 v 1.2 0.9t2 m/s, 24′′ A a B G P 40′′ 8′′ 20′′ θ a 5 ft/sec2 2′ A B C 2′ 2′ 1′ 60° 60° 60 lb c06.qxd 2/10/12 2:13 PM Page 427 6/24 The riding power mower has a mass of 140 kg with center of mass at The operator has a mass of 90 kg with center of mass at Calculate the mini-mum effective coefficient of friction which will per-mit the front wheels of the mower to lift off the ground as the mower starts to move forward.
Problem 6/24 6/25 The 25-kg bar BD is attached to the two light links AB and CD and moves in the vertical plane. The lower link is subjected to a clockwise torque applied through its shaft at A. If each link has an angular velocity as it passes the horizontal position, calculate the force which the upper link exerts on the bar at D at this instant.
Also find the angular acceleration of the links at this position.
Problem 6/25 300 mm 500 mm 600 mm C A M B D G ω 5 rad/s M 200 N m G1 G2 450 mm 100 mm 750 mm 300 mm 900 mm A B G2.
G1.
428 Chapter 6 Plane Kinetics of Rigid Bodies 6/22 The block A and attached rod have a combined mass of 60 kg and are confined to move along the guide under the action of the 800-N applied force.
The uniform horizontal rod has a mass of 20 kg and is welded to the block at B. Friction in the guide is negligible. Compute the bending moment M exerted by the weld on the rod at B.
Problem 6/22 6/23 The parallelogram linkage shown moves in the verti-cal plane with the uniform 8-kg bar EF attached to the plate at E by a pin which is welded both to the plate and to the bar. A torque (not shown) is applied to link AB through its lower pin to drive the links in a clockwise direction. When reaches the links have an angular acceleration and an angular velocity of and respectively. For this instant calculate the magnitudes of the force F and torque M supported by the pin at E.
Problem 6/23 Horizontal D B E F θ θ A 800 mm 800 mm 1200 mm Welded pin C 3 rad/s, 6 rad/s2 60, 800 N 1.4 m 60 A B 60 c06.qxd 2/10/12 2:13 PM Page 428 6/26 A jet transport with a landing speed of re-duces its speed to with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At the lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Problem 6/26 6/27 The uniform L-shaped bar pivots freely at point P of the slider, which moves along the horizontal rod.
Determine the steady-state value of the angle if (a) and (b) For what value of a would the steady-state value of be zero?
Problem 6/27 6/28 The van seen from the rear is traveling at a speed v around a turn of mean radius r banked inward at an angle . The effective coefficient of friction between the tires and the road is . Determine (a) the proper bank angle for a given v to eliminate any tendency to slip or tip, and (b) the maximum speed v before the van tips or slips for a given . Note that the forces and the acceleration lie in the plane of the figure so that the problem may be treated as one of plane mo-tion even though the velocity is normal to this plane.
θ P a l 2l a g/2.
a 0 v 15 m 2.4 m R A G B 1.8 m 3 m 60 km/h 200 km/h Article 6/3 Problems 429 Problem 6/28 6/29 The parallelogram linkage is used to transfer crates from platform A to platform B and is hydraulically operated. The oil pressure in the cylinder is pro-grammed to provide a smooth transition of motion from to rad given by where t is in seconds. Determine the force at D on the pin (a) just after the start of the motion with and t essentially zero and (b) when The crate and platform have a combined mass of 200 kg with mass center at G. The mass of each link is small and may be neglected.
Problem 6/29 600 mm 1200 mm A B D C E F θ θ0 G 480 mm t 1 s.
6 1 cos t 2 0 /3 0 h b/2 b/2 r G θ c06.qxd 2/10/12 2:13 PM Page 429 Problem 6/31 6/32 The uniform 200-kg bar AB is raised in the vertical plane by the application of a constant couple applied to the link at C. The mass of the links is small and may be neglected. If the bar starts from rest at determine the magnitude of the force supported by the pin at A as the position is passed.
Problem 6/32 1.5 m 1.5 m D C M 1.75 m 0.5 m 0.75 m B A θ θ 60 0, M 3 kNm v 0.6 m 1.0 m 0.8 m 0.4 m B A O G 0.6 m 0.4 m θ θ 430 Chapter 6 Plane Kinetics of Rigid Bodies 6/30 The 1800-kg rear-wheel-drive car accelerates forward at a rate of If the modulus of each of the rear and front springs is estimate the resulting momentary nose-up pitch angle . (This upward pitch angle during acceleration is called squat, while the downward pitch during braking is called dive!) Neglect the unsprung mass of the wheels and tires.
(Hint: Begin by assuming a rigid vehicle.) Problem 6/30 6/31 The two wheels of the vehicle are connected by a 20-kg link AB with center of mass at G. The link is pinned to the wheel at B, and the pin at A fits into a smooth horizontal slot in the link. If the vehicle has a constant speed of determine the magnitude of the force supported by the pin at B for the position 30.
4 m/s, A 1500 mm 1500 mm B a G 600 mm 35 kN/m, g/2.
c06.qxd 2/10/12 2:13 PM Page 430 6/4 Fixed-Axis Rotation Rotation of a rigid body about a fixed axis O was described in Art.
5/2 and illustrated in Fig. 5/1c. For this motion, we saw that all points in the body describe circles about the rotation axis, and all lines of the body in the plane of motion have the same angular velocity and angu-lar acceleration .
The acceleration components of the mass center for circular mo-tion are most easily expressed in n-t coordinates, so we have an and at as shown in Fig. 6/9a for rotation of the rigid body about the fixed axis through O. Part b of the figure represents the free-body diagram, and the equivalent kinetic diagram in part c of the figure shows the force resultant in terms of its n- and t-components and the resultant couple Our general equations for plane motion, Eqs. 6/1, are directly ap-plicable and are repeated here.
[6/1] Thus, the two scalar components of the force equation become ΣFn and ΣFt In applying the moment equation about G, we must ac-count for the moment of the force applied to the body at O, so this force must not be omitted from the free-body diagram.
For fixed-axis rotation, it is generally useful to apply a moment equation directly about the rotation axis O. We derived this equation previously as Eq. 6/4, which is repeated here.
[6/4] From the kinetic diagram in Fig. 6/9c, we may obtain Eq. 6/4 very easily by evaluating the moment of the resultants about O, which becomes ΣMO Application of the parallel-axis theorem for mass mo-ments of inertia, IO gives ΣMO (IO IO.
For the common case of rotation of a rigid body about a fixed axis through its mass center G, clearly, 0, and therefore ΣF 0. The re-sultant of the applied forces then is the couple We may combine the resultant-force component and resultant couple by moving to a parallel position through point Q on line OG, Fig. 6/10, located by Using the parallel-axis theorem and IO gives q Point Q is called the center of percussion and has the unique prop-erty that the resultant of all forces applied to the body must pass through it. It follows that the sum of the moments of all forces about the center of percussion is always zero, ΣMQ 0.
kO 2/ r.
kO 2m mr(r).
I mrq mat I mat I.
a mr 2 mr 2) mr 2, I matr.
I ΣMO IO mr.
mr2 ΣMG I ΣF ma I.
ma r, r2 Article 6/4 Fixed-Axis Rotation 431 G O G O r – ma – n ma – t I –α ≡ Free-Body Diagram (b) G n t or α ω ω O Fixed-Axis Rotation (a) Kinetic Diagram (c) a – n = r – 2 ω a – t = r –α Figure 6/9 G Q O r – mr – 2 ω mr –α kO 2 ——– r – q = α Figure 6/10 c06.qxd 2/10/12 2:13 PM Page 431 SAMPLE PROBLEM 6/3 The concrete block weighing 644 lb is elevated by the hoisting mechanism shown, where the cables are securely wrapped around the respective drums. The drums, which are fastened together and turn as a single unit about their mass cen-ter at O, have a combined weight of 322 lb and a radius of gyration about O of 18 in.
If a constant tension P of 400 lb is maintained by the power unit at A, determine the vertical acceleration of the block and the resultant force on the bearing at O.
Solution I.
The free-body and kinetic diagrams of the drums and concrete block are drawn showing all forces which act, including the components Ox and Oy of the bearing reaction. The resultant of the force system on the drums for centroidal rotation is the couple IO, where Taking moments about the mass center O for the pulley in the sense of the angular acceleration gives (a) The acceleration of the block is described by (b) From at r, we have a (12/12). With this substitution, Eqs. (a) and (b) are combined to give Ans.
The bearing reaction is computed from its components. Since 0, we use the equilibrium equations Ans.
Solution II.
We may use a more condensed approach by drawing the free-body diagram of the entire system, thus eliminating reference to T, which becomes in-ternal to the new system. From the kinetic diagram for this system, we see that the moment sum about O must equal the resultant couple for the drums, plus the moment of the resultant ma for the block. Thus, from the principle of Eq. 6/5 we have With a (12/12), the solution gives, as before, a 3.67 ft/sec2.
We may equate the force sums on the entire system to the sums of the resul-tants. Thus, Ox 400 cos 45 0 Ox 283 lb [ΣFx Σmax] Oy 1322 lb Oy 322 644 400 sin 45 322 32.2 (0) 644 32.2 (3.67) [ΣFy Σmay] [ΣMO I mad] 400 24 12 644 12 12 22.5 644 32.2 a 12 12 I O (283)2 (1322)2 1352 lb Oy 322 717 400 sin 45 0 Oy 1322 lb [ΣFy 0] Ox 400 cos 45 0 Ox 283 lb [ΣFx 0] a T 717 lb 3.67 rad/sec2 a 3.67 ft/sec2 T 644 644 32.2 a [ΣFy may] 400 24 12 T 12 12 22.5 [ΣMG I] I IO 18 12 2 322 32.2 22.5 lb-ft-sec2 [I k2m] I 432 Chapter 6 Plane Kinetics of Rigid Bodies 45° 24″ 12″ A O P = 400 lb W = 322 lb kO = 18″ 644 lb T ma Ox Oy O α O x a y 45° 322 lb 644 lb 400 lb I _ α ≡ ≡ ma Ox Oy O α O x a y 45° 322 lb 644 lb 400 lb I _ α ≡ Helpful Hints Be alert to the fact that the tension T is not 644 lb. If it were, the block would not accelerate.
Do not overlook the need to express kO in feet when using g in ft/sec2.
c06.qxd 2/10/12 2:13 PM Page 432 SAMPLE PROBLEM 6/4 The pendulum has a mass of 7.5 kg with center of mass at G and has a ra-dius of gyration about the pivot O of 295 mm. If the pendulum is released from rest at 0, determine the total force supported by the bearing at the instant when 60. Friction in the bearing is negligible.
Solution.
The free-body diagram of the pendulum in a general position is shown along with the corresponding kinetic diagram, where the components of the resultant force have been drawn through G.
The normal component On is found from a force equation in the n-direction, which involves the normal acceleration Since the angular velocity of the pendulum is found from the integral of the angular acceleration and since Ot de-pends on the tangential acceleration it follows that should be obtained first.
To this end with IO the moment equation about O gives and for 60 The remaining two equations of motion applied to the 60 position yield Ans.
The proper sense for Ot may be observed at the outset by applying the moment equation ΣMG where the moment about G due to Ot must be clockwise to agree with . The force Ot may also be obtained initially by a moment equation about the center of percussion Q, shown in the lower figure, which avoids the ne-cessity of computing . First, we must obtain the distance q, which is Ans.
Ot 10.37 N Ot(0.348) 7.5(9.81)(cos 60)(0.348 0.250) 0 [ΣMQ 0] q (0.295)2 0.250 0.348 m [q kO 2/ r ] I, O (155.2)2 (10.37)2 155.6 N Ot 10.37 N Ot 7.5(9.81) cos 60 7.5(0.25)(28.2) cos 60 [ΣFt mr] On 155.2 N On 7.5(9.81) sin 60 7.5(0.25)(48.8) [ΣFn mr2] 2 48.8 (rad/s)2 0 d /3 0 28.2 cos d [ d d] 28.2 cos rad/s2 7.5(9.81)(0.25) cos (0.295)2(7.5) [ΣMO IO] kO 2m, r, r2.
Article 6/4 Fixed-Axis Rotation 433 θ r _ = 250 mm O G n α ω t 7.5(9.81) N G Ot O G On I _ α mr _ α mr _ ω 2 ≡ O q G Q mr _ α mr _ ω 2 Helpful Hints The acceleration components of G are, of course, and r.
at r2 an Review the theory again and satisfy yourself that ΣMO IO Note especially here that the force summations are taken in the posi-tive direction of the acceleration components of the mass center G.
mrq.
mr 2 I c06.qxd 2/10/12 6:34 PM Page 433 Problem 6/35 6/36 The automotive dynamometer is able to simulate road conditions for an acceleration of 0.5g for the loaded pickup truck with a gross weight of 5200 lb.
Calculate the required moment of inertia of the dy-namometer drum about its center O assuming that the drum turns freely during the acceleration phase of the test.
Problem 6/36 A 36′′ 15′′ O 3 m 3 m 1 m P C B A θ 434 Chapter 6 Plane Kinetics of Rigid Bodies PROBLEMS Introductory Problems 6/33 The uniform 20-kg slender bar is pivoted at O and swings freely in the vertical plane. If the bar is re-leased from rest in the horizontal position, calculate the initial value of the force R exerted by the bearing on the bar an instant after release.
Problem 6/33 6/34 The 20-kg uniform steel plate is freely hinged about the z-axis as shown. Calculate the force supported by each of the bearings at A and B an instant after the plate is released from rest in the horizontal y-z plane.
Problem 6/34 6/35 The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with Determine the initial angular acceleration of the beam and the magnitude of the force supported by the pin at A due to the appli-cation of a force on the attached cable.
P 300 N FA 0.
80 mm 250 mm 80 mm A B z y x 400 mm O 1.6 m c06.qxd 2/10/12 2:13 PM Page 434 6/37 A momentum wheel for dynamics-class demonstra-tions is shown. It is basically a bicycle wheel modi-fied with rim band-weighting, handles, and a pulley for cord startup. The heavy rim band causes the radius of gyration of the 7-lb wheel to be 11 in. If a steady 10-lb pull T is applied to the cord, determine the angular acceleration of the wheel. Neglect bear-ing friction.
Problem 6/37 6/38 Determine the angular acceleration and the force on the bearing at O for (a) the narrow ring of mass m and (b) the flat circular disk of mass m immediately after each is released from rest in the vertical plane with OC horizontal.
Problem 6/38 6/39 The 30-in. slender bar weighs 20 lb and is mounted on a vertical shaft at O. If a torque lb-in. is applied to the bar through its shaft, calculate the horizontal force R on the bearing as the bar starts to rotate.
M 100 O C r (a) O C r (b) 24″ T = 10 lb 30° 4″ Article 6/4 Problems 435 Problem 6/39 6/40 The uniform slender bar AB has a mass of 8 kg and swings in a vertical plane about the pivot at A. If when compute the force sup-ported by the pin at A at that instant.
Problem 6/40 6/41 The uniform quarter-circular sector of mass m is re-leased from rest with one straight edge vertical as shown. Determine the initial angular acceleration and the horizontal and vertical components of the reaction at the ideal pivot at O.
Problem 6/41 O b m Vertical Horizontal 900 mm θ A B 30, ˙ 2 rad/s M O 18″ 12″ c06.qxd 2/10/12 2:13 PM Page 435 Problem 6/44 6/45 If the slender-bar assembly is released from rest while in the horizontal position shown, determine its angular acceleration. The mass per unit length of the bar is . Neglect friction at the bearing O.
Problem 6/45 6/46 An air table is used to study the elastic motion of flexible spacecraft models. Pressurized air escaping from numerous small holes in the horizontal surface provides a supporting air cushion which largely eliminates friction. The model shown consists of a cylindrical hub of radius r and four appendages of length l and small thickness t. The hub and the four appendages all have the same depth d and are con-structed of the same material of density . Assume that the spacecraft is rigid and determine the mo-ment M which must be applied to the hub to spin the model from rest to an angular velocity in a time period of seconds. (Note that for a spacecraft with highly flexible appendages, the moment must be judiciously applied to the rigid hub to avoid undesir-able large elastic deflections of the appendages.) Problem 6/46 2b 3b b b O 2r 3r m r O 436 Chapter 6 Plane Kinetics of Rigid Bodies 6/42 The circular sector of uniform thickness and mass m is released from rest when one of its straight edges is vertical as shown. Determine the initial angular ac-celeration about the ideal pivot at O. Evaluate your general expression for and Compare your results to the stated answer for the previous problem.
Problem 6/42 6/43 The square frame is composed of four equal lengths of uniform slender rod, and the ball attachment at O is suspended in a socket (not shown). Beginning from the position shown, the assembly is rotated about axis A-A and released. Determine the initial angular acceleration of the frame. Repeat for a rotation about axis B-B. Neglect the small mass, offset, and friction of the ball.
Problem 6/43 6/44 If the system is released from rest while in the hori-zontal position shown, determine the angular accel-eration of the lightweight right-angle shaft. The sphere of radius r has mass m. Neglect friction at the bearing O.
A A B B O b b b/2 b/2 b 45 45 O b m β Vertical .
/2 z r M t l d c06.qxd 2/10/12 2:13 PM Page 436 6/47 The narrow ring of mass m is free to rotate in the vertical plane about O. If the ring is released from rest at determine expressions for the n- and t-components of the force at O in terms of .
Problem 6/47 Representative Problems 6/48 Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia of the disk is considered. The system is released from rest, the cord does not slip on the disk, and bearing friction at O may be neglected.
Problem 6/48 6/49 The solid homogeneous cylinder weighs 300 lb and is free to rotate about the horizontal axis O-O. If the cylinder, initially at rest, is acted upon by the 100-lb force shown, calculate the horizontal component R of the force supported by each of the two symmetri-cally placed bearings when the 100-lb force is first applied.
4 kg 6 kg 5 kg 0.25 m A B O O C n t r θ 0, Article 6/4 Problems 437 Problem 6/49 6/50 The solid cylindrical rotor B has a mass of 43 kg and is mounted on its central axis C-C. The frame A ro-tates about the fixed vertical axis O-O under the ap-plied torque The rotor may be unlocked from the frame by withdrawing the locking pin P.
Calculate the angular acceleration of the frame A if the locking pin is (a) in place and (b) withdrawn.
Neglect all friction and the mass of the frame.
Problem 6/50 O O M A P B 200 mm 250 mm C C M 30 Nm.
12″ 100 lb O O 4″ 6″ c06.qxd 2/10/12 2:13 PM Page 437 6/53 The bar A of mass m is formed into a circular arc of radius r and is attached to the hub by the light rods.
The curved bar oscillates about the vertical axis under the action of a torsional spring B. At the instant under consideration, the angular velocity is and the angu-lar acceleration is . Write expressions for the moment M exerted by the spring on the hub and the horizontal force R exerted by the shaft on the hub.
Problem 6/53 6/54 The uniform slender bar is released from rest in the horizontal position shown. Determine the value of x for which the angular acceleration is a maximum, and determine the corresponding angular accelera-tion .
Problem 6/54 l x G O ω α 90° r r A B O O 90 438 Chapter 6 Plane Kinetics of Rigid Bodies 6/51 The uniform 40-lb bar is released from rest in the horizontal position shown and strikes the fixed cor-ner B at the center of percussion of the bar. Deter-mine the t-component of the force exerted by the bearing O on the bar just prior to impact, during im-pact, and just after impact.
Problem 6/51 6/52 Each of the two grinding wheels has a diameter of 6 in., a thickness of in., and a specific weight of When switched on, the machine acceler-ates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec.
Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.
Problem 6/52 ω 425 lb/ft3.
3/4 30° 6′ t n O A B c06.qxd 2/10/12 2:13 PM Page 438 6/55 The uniform rectangular slab is released from rest in the position shown. Determine the value of x for which the angular acceleration is a maximum, and determine the corresponding angular acceleration. Compare your answers with those listed for Prob. 6/54.
Problem 6/55 6/56 The spring is uncompressed when the uniform slen-der bar is in the vertical position shown. Determine the initial angular acceleration of the bar when it is released from rest in a position where the bar has been rotated clockwise from the position shown. Neglect any sag of the spring, whose mass is negligible.
Problem 6/56 6/57 A gimbal pedestal supports a payload in the space shuttle and deploys it when the doors of the cargo bay are opened in orbit. The payload is modeled as a homogeneous rectangular block with a mass of 6000 kg.
The torque on the gimbal axis O-O is 30 sup-plied by a d-c brushless motor. With the shuttle or-biting in a “weightless” condition, determine the time t required to bring the payload from its stowed position at to its deployed position at if the torque is applied for the first of travel and then reversed for the remaining to bring the pay-load to a stop ( ˙ 0).
45 45 90 0 Nm l k A G O B m l — 4 l — 4 l — 4 l — 4 30 x G b O b — 2 Article 6/4 Problems 439 Problem 6/57 6/58 A uniform slender bar of mass m and length 2b is mounted in a right-angle frame of negligible mass.
The bar and frame rotate in the vertical plane about a fixed axis at O. If the bar is released from rest in the vertical position derive an expression for the magnitude of the force exerted by the bearing at O on the frame as a function of .
Problem 6/58 45° 45° O G b b b θ ( 0), 2.5 m 0.8 m 1.5 m 1.5 m θ O O c06.qxd 2/10/12 2:13 PM Page 439 6/61 The 12-kg cylinder supported by the bearing brack-ets at A and B has a moment of inertia about the vertical through its mass center G equal to The disk and brackets have a moment of inertia about the vertical z-axis of rotation equal to If a torque is applied to the disk through its shaft with the disk initially at rest, calculate the horizontal x-components of force supported by the bearings at A and B.
Problem 6/61 6/62 The 24-kg uniform slender bar AB is mounted on end rollers of negligible mass and rotates about the fixed point O as it follows the circular path in the vertical plane. The bar is released from a position which gives it an angular velocity as it passes the position Calculate the forces and ex-erted by the guide on the rollers for this instant.
Problem 6/62 A O G B 300 mm 300 mm 600 mm 600 mm θ FB FA 45.
2 rad/s 100 mm 100 mm 100 mm A B M G x y z z0 M 16 Nm 0.60 kg m2.
0.080 kg m2.
z0-axis 440 Chapter 6 Plane Kinetics of Rigid Bodies 6/59 The uniform semicircular bar of mass m and radius r is hinged freely about a horizontal axis through A. If the bar is released from rest in the position shown, where AB is horizontal, determine the initial angu-lar acceleration of the bar and the expression for the force exerted on the bar by the pin at A. (Note carefully that the initial tangential acceleration of the mass center is not vertical.) Problem 6/59 6/60 A device for impact testing consists of a 34-kg pendu-lum with mass center at G and with radius of gyration about O of 620 mm. The distance b for the pendulum is selected so that the force on the bearing at O has the least possible value during impact with the specimen at the bottom of the swing. Determine b and calculate the magnitude of the total force R on the bearing O an instant after release from rest at Problem 6/60 600 mm Specimen b O G θ 60.
A B r c06.qxd 2/10/12 2:13 PM Page 440 6/63 The mass of gear A is 20 kg and its centroidal radius of gyration is 150 mm. The mass of gear B is 10 kg and its centroidal radius of gyration is 100 mm. Cal-culate the angular acceleration of gear B when a torque of 12 is applied to the shaft of gear A.
Neglect friction.
Problem 6/63 6/64 Prior to deployment of its two instrument arms AB, the spacecraft shown in the upper view is spinning at the constant rate of 1 revolution per second. Each instrument arm, shown in the lower view, has a mass of 10 kg with mass center at G. Calculate the tension T in the deployment cable prior to release. Also find the magnitude of the force on the pin at A. Neglect any acceleration of the center O of the spacecraft.
Problem 6/64 500 mm 400 mm 100 mm A A B B A T O G B ω O ω A B M rB rA rA = 240 mm rB = 180 mm Nm Article 6/4 Problems 441 6/65 Disk B weighs 50 lb and has a centroidal radius of gyration of 8 in. The power unit C consists of a motor M and a disk A, which is driven at a constant angular speed of 1600 rev/min. The coefficients of static and kinetic friction between the two disks are and respectively. Disk B is ini-tially stationary when contact with disk A is estab-lished by application of the constant force Determine the angular acceleration of B and the time t required for B to reach its steady-state speed.
Problem 6/65 6/66 Two slender bars AB, each of mass m and length l, are pivoted at A to the plate. The plate rotates in the horizontal plane about a fixed vertical axis through its center O and is given a constant angular accelera-tion . (a) Determine the force F exerted on each of the two rollers as the assembly starts to rotate.
(b) Find the total force on the pin at A and show that it remains constant as long as (c) Determine the angular velocity at which contact with the rollers ceases.
Problem 6/66 l — 2 l — 2 l — 2 l — 2 A B B A O α F 0.
30° B A A ω M P C rB rA rA = 8 in.
rB = 10 in.
P 3 lb.
k 0.60, s 0.80 c06.qxd 2/10/12 2:13 PM Page 441 6/69 The uniform slender bar of mass m and length l is released from rest in the vertical position and pivots on its square end about the corner at O. (a) If the bar is observed to slip when find the coeffi-cient of static friction between the bar and the corner. (b) If the end of the bar is notched so that it cannot slip, find the angle at which contact be-tween the bar and the corner ceases.
Problem 6/69 6/70 The uniform rectangular block is released from rest with essentially zero and pivots in the vertical plane about the center A of its lower face on the fixed corner. (a) If the block is observed to slip when find the coefficient of static friction between the block and the corner. (b) If the bottom face of the block is notched so that it cannot slip, find the angle at which contact between the block and the corner ceases.
Problem 6/70 A θ 20″ 12″ α 30, θ O l s 30, 442 Chapter 6 Plane Kinetics of Rigid Bodies 6/67 The robotic device consists of the stationary pedestal OA, arm AB pivoted at A, and arm BC pivoted at B.
The rotation axes are normal to the plane of the fig-ure. Estimate (a) the moment applied to arm AB required to rotate it about joint A at counter-clockwise from the position shown with joint B locked and (b) the moment applied to arm BC required to rotate it about joint B at the same rate with joint A locked. The mass of arm AB is 25 kg and that of BC is 4 kg, with the stationary portion of joint A excluded entirely and the mass of joint B divided equally between the two arms. Assume that the cen-ters of mass and are in the geometric centers of the arms and model the arms as slender rods.
Problem 6/67 6/68 Each of the two uniform slender bars OA and BC has a mass of 8 kg. The bars are welded at A to form a T-shaped member and are rotating freely about a horizontal axis through O. If the bars have an angu-lar velocity of 4 rad/s as OA passes the horizontal position shown, calculate the total force R supported by the bearing at O.
Problem 6/68 0.5 m 0.25 m 0.25 m A B C O A G1 G2 45° 700 mm 350 mm 90° B C O G2 G1 MB 4 rad/s2 MA c06.qxd 2/10/12 2:13 PM Page 442 Article 6/5 General Plane Motion 443 KEY CONCEPTS 6/5 General Plane Motion The dynamics of a rigid body in general plane motion combines translation and rotation. In Art. 6/2 we represented such a body in Fig.
6/4 with its free-body diagram and its kinetic diagram, which discloses the dynamic resultants of the applied forces. Figure 6/4 and Eqs. 6/1, which apply to general plane motion, are repeated here for convenient reference.
[6/1] Direct application of these equations expresses the equivalence between the externally applied forces, as disclosed by the free-body diagram, and their force and moment resultants, as represented by the kinetic diagram. ΣMG I ΣF ma a – G ma – G α ≡ F1 F2 F3 I – α Free-Body Diagram Kinetic Diagram Figure 6/4, repeated Solving Plane-Motion Problems Keep in mind the following considerations when solving plane-motion problems.
Choice of Coordinate System. The force equation of Eq. 6/1 should be expressed in whatever coordinate system most readily describes the acceleration of the mass center. You should consider rectangular, normal-tangential, and polar coordinates.
Choice of Moment Equation. In Art. 6/2 we also showed, with the aid of Fig. 6/5, the application of the alternative relation for moments about any point P, Eq. 6/2. This figure and this equation are also re-peated here for easy reference.
[6/2] In some instances, it may be more convenient to use the alternative mo-ment relation of Eq. 6/3 when moments are taken about a point P whose acceleration is known. Note also that the equation for moments about a ΣMP I mad c06.qxd 2/10/12 2:13 PM Page 443 nonaccelerating point O on the body, Eq. 6/4, constitutes still another alternative moment relation and at times may be used to advantage.
Constrained versus Unconstrained Motion. In working a problem in general plane motion, we first observe whether the motion is uncon-strained or constrained, as illustrated in the examples of Fig. 6/6. If the motion is constrained, we must account for the kinematic relationship between the linear and the angular accelerations and incorporate it into our force and moment equations of motion. If the motion is uncon-strained, the accelerations can be determined independently of one an-other by direct application of the three motion equations, Eqs. 6/1.
Number of Unknowns. In order for a rigid-body problem to be solvable, the number of unknowns cannot exceed the number of inde-pendent equations available to describe them, and a check on the suffi-ciency of the relationships should always be made. At the most, for plane motion we have three scalar equations of motion and two scalar components of the vector relative-acceleration equation for constrained motion. Thus, we can handle as many as five unknowns for each rigid body.
Identification of the Body or System. We emphasize the impor-tance of clearly choosing the body to be isolated and representing this isolation by a correct free-body diagram. Only after this vital step has been completed can we properly evaluate the equivalence between the external forces and their resultants.
Kinematics. Of equal importance in the analysis of plane motion is a clear understanding of the kinematics involved. Very often, the diffi-culties experienced at this point have to do with kinematics, and a thor-ough review of the relative-acceleration relations for plane motion will be most helpful.
Consistency of Assumptions. In formulating the solution to a problem, we recognize that the directions of certain forces or accelera-tions may not be known at the outset, so that it may be necessary to make initial assumptions whose validity will be proved or disproved when the solution is carried out. It is essential, however, that all as-sumptions made be consistent with the principle of action and reaction 444 Chapter 6 Plane Kinetics of Rigid Bodies a – G P d ma – P G α I – α – ρ ≡ F1 F2 F3 Free-Body Diagram Kinetic Diagram Figure 6/5, repeated c06.qxd 2/10/12 2:13 PM Page 444 Article 6/5 General Plane Motion 445 and with any kinematic requirements, which are also called conditions of constraint.
Thus, for example, if a wheel is rolling on a horizontal surface, its cen-ter is constrained to move on a horizontal line. Furthermore, if the un-known linear acceleration a of the center of the wheel is assumed positive to the right, then the unknown angular acceleration will be positive in a clockwise sense in order that a r, if we assume the wheel does not slip. Also, we note that, for a wheel which rolls without slipping, the static friction force between the wheel and its supporting surface is generally less than its maximum value, so that F sN. But if the wheel slips as it rolls, a r, and a kinetic friction force is generated which is given by F kN.
It may be necessary to test the validity of either assumption, slipping or no slipping, in a given problem. The difference between the coefficients of sta-tic and kinetic friction, s and k, is sometimes ignored, in which case, is used for either or both coefficients.
Look ahead to Prob. 6/103 to see a special-case problem involving a crash-test dummy such as the one shown here.
Media Bakery c06.qxd 2/10/12 2:13 PM Page 445 446 Chapter 6 Plane Kinetics of Rigid Bodies SAMPLE PROBLEM 6/5 A metal hoop with a radius r 6 in. is released from rest on the 20 incline.
If the coefficients of static and kinetic friction are s 0.15 and k 0.12, deter-mine the angular acceleration of the hoop and the time t for the hoop to move a distance of 10 ft down the incline.
Solution.
The free-body diagram shows the unspecified weight mg, the normal force N, and the friction force F acting on the hoop at the contact point C with the incline. The kinetic diagram shows the resultant force through G in the direc-tion of its acceleration and the couple The counterclockwise angular accelera-tion requires a counterclockwise moment about G, so F must be up the incline.
Assume that the hoop rolls without slipping, so that r. Application of the components of Eqs. 6/1 with x- and y-axes assigned gives Elimination of F between the first and third equations and substitution of the kinematic assumption r give Alternatively, with our assumption of r for pure rolling, a moment sum about C by Eq. 6/2 gives directly. Thus, To check our assumption of no slipping, we calculate F and N and compare F with its limiting value. From the above equations, But the maximum possible friction force is Because our calculated value of 0.1710mg exceeds the limiting value of 0.1410mg, we conclude that our assumption of pure rolling was wrong. Therefore, the hoop slips as it rolls and r. The friction force then becomes the kinetic value The motion equations now give Ans.
The time required for the center G of the hoop to move 10 ft from rest with con-stant acceleration is Ans.
[x 1 2 at2] 0.1128(32.2) 6/12 7.26 rad/sec2 0.1128mg(r) mr2 [ΣMG I] a 0.229(32.2) 7.38 ft/sec2 mg sin 20 0.1128mg ma [ΣFx max] F 0.12(0.940mg) 0.1128mg [F kN] a Fmax 0.15(0.940mg) 0.1410mg [Fmax sN] N mg cos 20 0.940mg F mg sin 20 m g 2 sin 20 0.1710mg [ΣMC I mad] mgr sin 20 mr2 a r mar a g 2 sin 20 a a a g 2 sin 20 32.2 2 (0.342) 5.51 ft/sec2 a Fr mr2 [ΣMG I] N mg cos 20 0 [ΣFy may 0] mg sin 20 F ma [ΣFx max] a I.
ma 20° 6″ G s = 0.15 μ k = 0.12 μ 20° F a _ x G mg y r = 6″ N C G C I _ α ma _ α ≡ Helpful Hints Because all of the mass of a hoop is a distance r from its center G, its mo-ment of inertia about G must be mr2.
Note that is independent of both m and r.
a Note that is independent of m but dependent on r.
c06.qxd 2/10/12 2:13 PM Page 446 Article 6/5 General Plane Motion 447 SAMPLE PROBLEM 6/6 The drum A is given a constant angular acceleration 0 of 3 rad/s2 and causes the 70-kg spool B to roll on the horizontal surface by means of the con-necting cable, which wraps around the inner hub of the spool. The radius of gyration of the spool about its mass center G is 250 mm, and the coefficient of static friction between the spool and the horizontal surface is 0.25. Deter-mine the tension T in the cable and the friction force F exerted by the hori-zontal surface on the spool.
Solution.
The free-body diagram and the kinetic diagram of the spool are drawn as shown. The correct direction of the friction force may be assigned in this problem by observing from both diagrams that with counterclockwise angu-lar acceleration, a moment sum about point G (and also about point D) must be counterclockwise. A point on the connecting cable has an acceleration at r 0.25(3) 0.75 m/s2, which is also the horizontal component of the acceleration of point D on the spool. It will be assumed initially that the spool rolls without slipping, in which case it has a counterclockwise angular acceleration 0.75/0.30 2.5 rad/s2. The acceleration of the mass center G is, therefore, r 0.45(2.5) 1.125 m/s2.
With the kinematics determined, we now apply the three equations of mo-tion, Eqs. 6/1, (a) (b) Solving (a) and (b) simultaneously gives Ans.
To establish the validity of our assumption of no slipping, we see that the sur-faces are capable of supporting a maximum friction force Fmax sN 0.25(687) 171.7 N. Since only 75.8 N of friction force is required, we conclude that our assumption of rolling without slipping is valid.
If the coefficient of static friction had been 0.1, for example, then the fric-tion force would have been limited to 0.1(687) 68.7 N, which is less than 75.8 N, and the spool would slip. In this event, the kinematic relation r would no longer hold. With (aD)x known, the angular acceleration would be Using this relation along with F kN 68.7 N, we would then re-solve the three equations of motion for the unknowns T, and .
Alternatively, with point C as a moment center in the case of pure rolling, we may use Eq. 6/2 and obtain T directly. Thus, Ans.
where the previous kinematic results for no slipping have been incorporated. We could also write a moment equation about point D to obtain F directly. T 154.6 N 0.3T 70(0.25)2(2.5) 70(1.125)(0.45) [ΣMC I mar] a, (aD)x]/GD.
[a a F 75.8 N and T 154.6 N F(0.450) T(0.150) 70(0.250)2(2.5) [ΣMG I] N 70(9.81) 0 N 687 N [ΣFy may] F T 70(1.125) [ΣFx max] a (aD)x/DC k 250 mm 150 mm 450 mm G B A 0 α 70(9.81) N k _ = 250 mm I _ α ma _ G G x F T D C C y N ≡ α a _ Helpful Hints The relation between and is the kinematic constraint which accom-panies the assumption that the spool rolls without slipping.
Be careful not to make the mistake of using for of the spool, which is not a uniform circular disk.
I 1 2 mr2 a Our principles of relative accelera-tion are a necessity here. Hence, the relation (aG/D)t should be recognized.
GD The flexibility in the choice of mo-ment centers provided by the kinetic diagram can usually be employed to simplify the analysis.
c06.qxd 2/10/12 2:13 PM Page 447 448 Chapter 6 Plane Kinetics of Rigid Bodies SAMPLE PROBLEM 6/7 The slender bar AB weighs 60 lb and moves in the vertical plane, with its ends constrained to follow the smooth horizontal and vertical guides. If the 30-lb force is applied at A with the bar initially at rest in the position for which 30, calculate the resulting angular acceleration of the bar and the forces on the small end rollers at A and B.
Solution.
The bar undergoes constrained motion, so that we must establish the relationship between the mass-center acceleration and the angular accelera-tion. The relative-acceleration equation aA aB aA/B must be solved first, and then the equation aG aB aG/B is next solved to obtain expressions relat-ing and . With assigned in its clockwise physical sense, the acceleration polygons which represent these equations are shown, and their solution gives Next we construct the free-body diagram and the kinetic diagram as shown.
With and now known in terms of , the remaining unknowns are and the forces A and B. We now apply Eqs. 6/1, which give Solving the three equations simultaneously gives us the results Ans.
As an alternative solution, we can use Eq. 6/2 with point C as the moment center and avoid the necessity of solving three equations simultaneously. This choice eliminates reference to forces A and B and gives directly. Thus, Ans.
With determined, we can now apply the force equations independently and get Ans.
Ans. 30 B 60 32.2 (1.732)(4.42) B 15.74 lb [ΣFx max] A 60 60 32.2 (1.0)(4.42) A 68.2 lb [ΣFy may] 43.9 9.94 4.42 rad/sec2 60 32.2 (1.732)(2 cos 30) 60 32.2 (1.0)(2 sin 30) 30(4 cos 30) 60(2 sin 30) 1 12 60 32.2 (42) [ΣMC I Σmad] A 68.2 lb B 15.74 lb 4.42 rad/sec2 A 60 60 32.2 (1.0) [ΣFy may] 30 B 60 32.2 (1.732) [ΣFx max] 30(2 cos 30) A(2 sin 30) B(2 cos 30) 1 12 60 32.2 (42) [ΣMG I] ay ax ay a sin 30 2 sin 30 1.0 ft/sec2 ax a cos 30 2 cos 30 1.732 ft/sec2 a a y G B A 30 lb 2′ 2′ x θ = 30° θ 30° aA/B = 4 α (aG/B)t = 2 α a _ a _ x aA aB aB a _ y 30° 30 lb 60 lb A B C C G G ≡ I _ α ma _ y ma _ x dx dy Helpful Hints If the application of the relative-ac-celeration equations is not perfectly clear at this point, then review Art.
5/6. Note that the relative normal ac-celeration term is absent since there is no angular velocity of the bar.
Recall that the moment of inertia of a slender rod about its center is 1 12 ml2.
From the kinetic diagram, Since both terms of the sum are clockwise, in the same sense as they are positive.
I, maydx.
maxdy Σmad c06.qxd 2/10/12 2:13 PM Page 448 Article 6/5 General Plane Motion 449 SAMPLE PROBLEM 6/8 A car door is inadvertently left slightly open when the brakes are applied to give the car a constant rearward acceleration a. Derive expressions for the angu-lar velocity of the door as it swings past the 90 position and the components of the hinge reactions for any value of . The mass of the door is m, its mass center is a distance from the hinge axis O, and the radius of gyration about O is kO.
Solution.
Because the angular velocity increases with , we need to find how the angular acceleration varies with so that we may integrate it over the in-terval to obtain . We obtain from a moment equation about O. First, we draw the free-body diagram of the door in the horizontal plane for a general position .
The only forces in this plane are the components of the hinge reaction shown here in the x- and y-directions. On the kinetic diagram, in addition to the resul-tant couple shown in the sense of , we represent the resultant force in terms of its components by using an equation of relative acceleration with re-spect to O. This equation becomes the kinematic equation of constraint and is The magnitudes of the components are then where and For a given angle , the three unknowns are , Ox, and Oy. We can eliminate Ox and Oy by a moment equation about O, which gives Now we integrate first to a general position and get Ans.
To find Ox and Oy for any given value of , the force equations give Ans.
Ans. mar 2 kO 2 (3 cos 2) sin mr ar kO 2 sin cos mr 2ar kO 2 (1 cos ) sin Oy mr cos mr2 sin [ΣFy may] ma1 r2 kO 2 (1 2 cos 3 cos2) ma 2ar 2 kO 2 (1 cos ) cos ar 2 kO 2 sin2 Ox ma mr2 cos mr sin [ΣFx max] 1 kO 2ar For /2, 2 2ar kO 2 (1 cos ) 0 d 0 ar kO 2 sin d [ d d] Solving for gives ar kO 2 sin [ΣMO I Σmad] 0 m(kO 2 r2) mr(r) ma(r sin ) ¨.
˙ maO ma m(aG/O)n mr2 m(aG/O)t mr ma a aG aO (aG/O)n (aG/O)t ma I r a θ ω O – r Ox Oy θ G x y – Iα – mrα – mr 2 ω G ≡ O mao = ma Helpful Hints Point O is chosen because it is the only point on the door whose accel-eration is known.
Be careful to place in the sense of positive with respect to rotation about O.
The free-body diagram shows that there is zero moment about O. We use the transfer-of-axis theorem here and substitute If this relation is not totally familiar, review Art. B/1 in Appendix B.
We may also use Eq. 6/3 with O as a moment center where the scalar values of the terms are IO and maO be-comes sin .
The kinetic diagram shows clearly the terms which make up and may.
max rma mkO 2 ΣMO IO maO r 2.
k2 kO 2 mr c06.qxd 2/10/12 2:13 PM Page 449 450 Chapter 6 Plane Kinetics of Rigid Bodies PROBLEMS Introductory Problems 6/71 The uniform square plate of mass m is lying motion-less on the horizontal surface when the force P is applied at A as shown. Determine the resulting initial acceleration of point B. Friction is negligible.
Problem 6/71 6/72 The L-shaped bar of mass m is lying motionless on the horizontal surface when the force P is applied at A as shown. Determine the initial acceleration of point A. Neglect friction and the thickness of the bar.
Problem 6/72 6/73 The body consists of a uniform slender bar and a uniform disk, each of mass It rests on a smooth surface. Determine the angular acceleration and the acceleration of the mass center of the body when the force is applied as shown. The value of the mass m of the entire body is 1.2 kg.
P 6 N m/2.
y x P A B l l y x P A B b b Problem 6/73 6/74 Repeat Prob. 6/73, except now the location of force P has been changed. The value of the mass m of the entire body is 1.2 kg.
Problem 6/74 6/75 Above the earth’s atmosphere at an altitude of 400 km where the acceleration due to gravity is a certain rocket has a total remaining mass of 300 kg and is directed 30 from the vertical. If the thrust T from the rocket motor is 4 kN and if the rocket noz-zle is tilted through an angle of as shown, calcu-late the angular acceleration of the rocket and the x- and y-components of the acceleration of its mass center G. The rocket has a centroidal radius of gyra-tion of 1.5 m.
Problem 6/75 G 3 m 1° T x y 30° 1 8.69 m/s2, P = 6 N y x m/2 m/2 500 mm 200 mm y x m/2 m/2 P = 6 N 500 mm 200 mm c06.qxd 2/10/12 2:13 PM Page 450 Article 6/5 Problems 451 6/80 The uniform disk of mass pivots freely on the cart of mass Determine the acceleration of the assembly and the angular acceleration of the disk under the action of the force applied to a cord wrapped securely around the disk.
Problem 6/80 6/81 The fairing which covers the spacecraft package in the nose of the booster rocket is jettisoned when the rocket is in space where gravitational attraction is negligible. A mechanical actuator moves the two halves slowly from the closed position I to position II at which point the fairings are released to rotate freely about their hinges at O under the influence of the constant acceleration a of the rocket. When posi-tion III is reached, the hinge at O is released and the fairings drift away from the rocket. Determine the angular velocity of the fairing at the position.
The mass of each fairing is m with center of mass at G and radius of gyration about O.
Problem 6/81 90° I II III G a O G r ω kO 90 P = 75 N 0.2 m m2 m1 20° O P 75 N m1 5 kg.
m2 8 kg 6/76 The 10-kg wheel with a radius of gyration of 180 mm about its center O is released from rest on the in-cline and slips as it rolls. If the coefficient of kinetic friction is calculate the acceleration of the center O of the wheel and its angular accel-eration .
Problem 6/76 6/77 How large would the coefficient of static friction have to be in order that the wheel of Prob. 6/76 not slip as it rolls?
6/78 The solid homogeneous cylinder is released from rest on the ramp. If and de-termine the acceleration of the mass center G and the friction force exerted by the ramp on the cylinder.
Problem 6/78 6/79 Repeat Prob. 6/78, except let and k 0.10.
s 0.15, 30, μs μk W = 8 lb G 6″ θ k 0.20, s 0.30, 40, s 60° μk = 0.30 200 mm O aO k 0.30, 60 c06.qxd 2/10/12 2:13 PM Page 451 6/82 Determine the angular acceleration of each of the two wheels as they roll without slipping down the in-clines. For wheel A investigate the case where the mass of the rim and spokes is negligible and the mass of the bar is concentrated along its centerline.
For wheel B assume that the thickness of the rim is negligible compared with its radius so that all of the mass is concentrated in the rim. Also specify the minimum coefficient of static friction required to prevent each wheel from slipping.
Problem 6/82 Representative Problems 6/83 A uniform slender rod of length l and mass m is secured to a circular hoop of radius l as shown. The mass of the hoop is negligible. If the rod and hoop are released from rest on a horizontal surface in the position illustrated, determine the initial values of the friction force F and normal force N under the hoop if friction is sufficient to prevent slipping.
θ θ B A r r s Problem 6/83 6/84 The uniform 12-kg square panel is suspended from point C by the two wires at A and B. If the wire at B suddenly breaks, calculate the tension T in the wire at A an instant after the break occurs.
Problem 6/84 6/85 The uniform steel beam of mass m and length l is suspended by the two cables at A and B. If the cable at B suddenly breaks, determine the tension T in the cable at A immediately after the break occurs. Treat the beam as a slender rod and show that the result is independent of the length of the beam.
Problem 6/85 A B l 60° 60° A B C b b 45° 45° l 452 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 452 Article 6/5 Problems 453 Problem 6/88 6/89 Repeat Prob. 6/88, except let and 6/90 The uniform rectangular panel of mass m is moving to the right when wheel B drops off the horizontal support rail. Determine the resulting angular accel-eration and the force in the strap at A immedi-ately after wheel B rolls off the rail. Neglect friction and the mass of the small straps and wheels.
Problem 6/90 6/91 The uniform slender bar AB has a mass of 0.8 kg and is driven by crank OA and constrained by link CB of negligible mass. If OA has an angular acceleration and an angular velocity when both OA and CB are normal to AB, calculate the force in CB for this instant. (Suggestion: Con-sider the use of Eq. 6/3 with A as a moment center.) Problem 6/91 w0 100 mm 200 mm 300 mm ω w0 α O C B A 0 2 rad/s 0 4 rad/s2 h m A B v b – – 6 b – – 6 2b – – 3 TA 30.
T 50 N μs = 0.10 μk = 0.08 T 200 mm 75 mm G m = 25 kg k = 175 mm θ μ μ 6/86 The circular disk of mass m and radius r is rolling through the bottom of the circular path of radius R.
If the disk has an angular velocity , determine the force N exerted by the path on the disk.
Problem 6/86 6/87 The system is released from rest with the cable taut, and the homogeneous cylinder does not slip on the rough incline. Determine the angular acceleration of the cylinder and the minimum coefficient of fric-tion for which the cylinder will not slip.
Problem 6/87 6/88 The circular disk of 200-mm radius has a mass of 25 kg with centroidal radius of gyration mm and has a concentric circular groove of 75-mm radius cut into it. A steady force T is applied at an angle to a cord wrapped around the groove as shown. If and determine the angular acceleration of the disk, the acceleration a of its mass center G, and the friction force F which the surface exerts on the disk.
k 0.08, s 0.10, 0, T 30 N, k 175 7 kg 10 kg 0.15 m 20° O s R ω r c06.qxd 2/10/12 2:13 PM Page 453 6/92 The crank OA rotates in the vertical plane with a constant clockwise angular velocity of For the position where OA is horizontal, calculate the force under the light roller B of the 10-kg slender bar AB.
Problem 6/92 6/93 Each of the two solid disk wheels weighs 20 lb, and the inner solid cylinder weighs 16 lb. The disk wheels and the inner disk are mounted on the small central shaft O-O and can rotate independently of each other. Friction in the shaft bearings is negligi-ble, whereas friction between the incline and the large disk wheels is sufficient to prevent slippage of the wheels. Determine the acceleration of the center O after the assembly is released on the incline.
The cord wrapped securely around the inner cylin-der is fastened to point A.
Problem 6/93 O O O A 10° 12″ 6″ 10 1.0 m 0.4 m 0.8 m B A O 0 ω 4.5 rad/s.
0 6/94 The robotic device of Prob. 6/67 is repeated here.
Member AB is rotating about joint A with a counter-clockwise angular velocity of and this rate is increasing at Determine the moment exerted by arm AB on arm BC if joint B is held in a locked condition. The mass of arm BC is 4 kg, and the arm may be treated as a uniform slender rod.
Problem 6/94 6/95 The uniform slender rod of mass m and length L is released from rest in the inverted vertical position shown. Neglect friction and the mass of the small end roller and find the initial acceleration of A.
Evaluate your result for Problem 6/95 L m A θ 30.
A G1 G2 45° 700 mm 350 mm 90° B C O MB 4 rad/s2.
2 rad/s, 454 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 454 Article 6/5 Problems 455 6/98 The assembly consisting of a uniform slender bar (mass m/5) and a rigidly attached uniform disk (mass 4m/5) is freely pinned to point O on the collar that in turn slides on the fixed horizontal guide. The assembly is at rest when the collar is given a sudden acceleration a to the left as shown. Determine the initial angular acceleration of the assembly.
Problem 6/98 6/99 The uniform 12-ft pole is hinged to the truck bed and released from the vertical position as the truck starts from rest with an acceleration of If the acceleration remains constant during the motion of the pole, calculate the angular velocity of the pole as it reaches the horizontal position.
Problem 6/99 a 12′ O ω 90° 3 ft/sec2.
P O a L/2 m 5 L/4 4m 5 L/4 6/96 In an investigation of whiplash resulting from rear-end collision, sudden rotation of the head is modeled by using a homogeneous solid sphere of mass m and radius r pivoted about a tangent axis (at the neck) to represent the head. If the axis at O is given a con-stant acceleration a with the head initially at rest, determine expressions for the initial angular acceler-ation of the head and its angular velocity as a function of the angle of rotation. Assume that the neck is relaxed so that no moment is applied to the head at O.
Problem 6/96 6/97 The uniform 15-kg bar is supported on the horizon-tal surface at A by a small roller of negligible mass.
If the coefficient of kinetic friction between end B and the vertical surface is 0.30, calculate the initial acceleration of end A as the bar is released from rest in the position shown.
Problem 6/97 2.4 m 40° = 0.30 A x y B wk μ Vertical G O r G O aO = a c06.qxd 2/10/12 2:13 PM Page 455 6/100 The T-shaped body of mass m is composed of two identical slender bars welded together. If the body is released from rest in the vertical plane in the position shown, determine the initial acceleration of point A.
Neglect the small mass and friction of the roller.
Problem 6/100 6/101 A bowling ball with a circumference of 27 in. weighs 14 lb and has a radius of gyration of 3.28 in. If the ball is released with a velocity of but with no angular velocity as it touches the alley floor, compute the distance traveled by the ball before it begins to roll without slipping. The coefficient of friction between the ball and the floor is 0.20.
Problem 6/101 6/102 The compound pendulum of mass m and radius of gyration about O is freely hinged to the trolley, which is given a constant horizontal acceleration a from rest with the pendulum initially at rest with Determine an expression for the angular ac-celeration and the n- and t-components of the force at O as functions of . Calculate the maxi-mum value reached by if Problem 6/102 θ a G O n r t a 0.5g.
¨ 0.
kO v0 s = 0 v v ω ω = r 20 ft/sec 30° A l l –– 2 l –– 2 6/103 In a study of head injury against the instrument panel of a car during sudden or crash stops where lap belts without shoulder straps or airbags are used, the segmented human model shown in the figure is analyzed. The hip joint O is assumed to remain fixed relative to the car, and the torso above the hip is treated as a rigid body of mass m freely pivoted at O. The center of mass of the torso is at G with the initial position of OG taken as vertical.
The radius of gyration of the torso about O is If the car is brought to a sudden stop with a constant deceleration a, determine the velocity v relative to the car with which the model’s head strikes the in-strument panel. Substitute the values and and compute v.
Problem 6/103 6/104 The uniform slender bar of mass m and length L with small end rollers is released from rest in the position shown with the lower roller in contact with the horizontal plane. Determine the normal force N under the lower roller and the angular acceleration of the bar immediately after release.
Problem 6/104 L θ θ r r G O a 10g 45, kO 550 mm, r 800 mm, r 450 mm, m 50 kg, kO.
456 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 456 Article 6/5 Problems 457 Problem 6/106 6/107 The small rollers at the ends of the uniform slender bar are confined to the circular slot in the vertical surface. If the bar is released from rest in the posi-tion shown, determine the initial angular accelera-tion . Neglect the mass and friction of the rollers.
Problem 6/107 30° B l A M m s O 40° 30° r μ 6/105 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in.
The piston and piston pin A together weigh 1.80 lb.
The engine is running at a constant speed of 3000 so that the angular velocity of the crank is Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Problem 6/105 6/106 A particle of mass m is embedded at the periphery of the otherwise uniform disk of mass M and radius r as shown. The disk starts from rest and does not slip on the rough incline. (a) For the position shown, what condition on m will cause the disk to begin to roll up the incline? (b) If deter-mine the initial angular acceleration of the disk and the minimum value of the coefficient of static friction required for the no-slip condition.
m 4M, θ ω O B A G 1.7′′ 1.3′′ 3′′ 90.
3000(2)/60 100 rad/sec.
rev/min, c06.qxd 2/10/12 2:13 PM Page 457 458 Chapter 6 Plane Kinetics of Rigid Bodies 6/108 The small end rollers of the 8-lb uniform slender bar are constrained to move in the slots, which lie in a vertical plane. At the instant when , the angular velocity of the bar is counter-clockwise. Determine the angular acceleration of the bar, the reactions at A and B, and the accelera-tions of points A and B under the action of the 6-lb force P. Neglect the friction and mass of the small rollers.
2 rad/sec 30 Problem 6/108 105° B P = 6 lb l = 4 ft A θ c06.qxd 2/10/12 2:13 PM Page 458 SECTION B WORK AND ENERGY 6/6 Work-Energy Relations In our study of the kinetics of particles in Arts. 3/6 and 3/7, we de-veloped the principles of work and energy and applied them to the mo-tion of a particle and to selected cases of connected particles. We found that these principles were especially useful in describing motion which resulted from the cumulative effect of forces acting through distances.
Furthermore, when the forces were conservative, we were able to deter-mine velocity changes by analyzing the energy conditions at the begin-ning and end of the motion interval. For finite displacements, the work-energy method eliminates the necessity for determining the accel-eration and integrating it over the interval to obtain the velocity change. These same advantages are realized when we extend the work-energy principles to describe rigid-body motion.
Before carrying out this extension, you should review the defini-tions and concepts of work, kinetic energy, gravitational and elastic potential energy, conservative forces, and power treated in Arts. 3/6 and 3/7 because we will apply them to rigid-body problems. You should also review Arts. 4/3 and 4/4 on the kinetics of systems of par-ticles, in which we extended the principles of Arts. 3/6 and 3/7 to en-compass any general system of mass particles, which includes rigid bodies.
Work of Forces and Couples The work done by a force F has been treated in detail in Art. 3/6 and is given by where dr is the infinitesimal vector displacement of the point of applica-tion of F, as shown in Fig. 3/2a. In the equivalent scalar form of the in-tegral, is the angle between F and the direction of the displacement, and ds is the magnitude of the vector displacement dr.
We frequently need to evaluate the work done by a couple M which acts on a rigid body during its motion. Figure 6/11 shows a couple M Fb acting on a rigid body which moves in the plane of the couple. During time dt the body rotates through an angle d, and line AB moves to AB. We may consider this motion in two parts, first a translation to AB and then a rotation d about A. We see immediately that during the translation the work done by one of the forces cancels that done by the other force, so that the net work done is dU F(b d) M d due to the rotational part of the motion. If the couple acts in the sense opposite to the rotation, the work done is negative. During a finite rota-tion, the work done by a couple M whose plane is parallel to the plane of motion is, therefore, U M d U Fdr or U (F cos ) ds Article 6/6 Work-Energy Relations 459 ≡ M F θ d θ d θ b d F F F A B b B″ B′ A′ Figure 6/11 c06.qxd 2/10/12 2:13 PM Page 459 Kinetic Energy We now use the familiar expression for the kinetic energy of a parti-cle to develop expressions for the kinetic energy of a rigid body for each of the three classes of rigid-body plane motion illustrated in Fig. 6/12.
(a) Translation. The translating rigid body of Fig. 6/12a has a mass m and all of its particles have a common velocity v. The kinetic en-ergy of any particle of mass mi of the body is Ti so for the en-tire body T or (6/7) This expression holds for both rectilinear and curvilinear translation.
(b) Fixed-axis rotation. The rigid body in Fig. 6/12b rotates with an angular velocity about the fixed axis through O. The kinetic energy of a representative particle of mass mi is Ti Thus, for the entire body T But the moment of inertia of the body about O is IO so (6/8) Note the similarity in the forms of the kinetic energy expressions for translation and rotation. You should verify that the dimensions of the two expressions are identical.
(c) General plane motion. The rigid body in Fig. 6/12c exe-cutes plane motion where, at the instant considered, the velocity of its mass center G is and its angular velocity is . The velocity vi of a representative particle of mass mi may be expressed in terms of the mass-center velocity and the velocity i relative to the mass center as shown. With the aid of the law of cosines, we write the kinetic en-ergy of the body as the sum ΣTi of the kinetic energies of all its parti-cles. Thus, Because and are common to all terms in the third summation, we may factor them out. Thus, the third term in the expression for T becomes since Σmi yi 0. The kinetic energy of the body is then T or (6/9) where is the moment of inertia of the body about its mass center. This expression for kinetic energy clearly shows the separate contributions to the total kinetic energy resulting from the translational velocity of the mass center and the rotational velocity about the mass center.
v I T 1 2 mv 2 1 2 I2 1 2 2Σmii 2 1 2 v 2Σmi my v Σmii cos v Σmi yi 0 v T Σ 1 2 mivi 2 Σ 1 2 mi(v 2 i 22 2vi cos ) v v T 1 2 IO 2 Σmiri 2, 1 2 2Σmiri 2.
1 2 mi(ri)2.
T 1 2 mv2 1 2 v2Σmi Σ 1 2 miv2 1 2 miv2, 460 Chapter 6 Plane Kinetics of Rigid Bodies (a) Translation mi v i = v v (c) General Plane Motion mi G yi v i – v – v (b) Fixed-Axis Rotation mi ri v i = ri ω i ω ρ ω i ρ θ θ G O ω Figure 6/12 c06.qxd 2/10/12 2:13 PM Page 460 The kinetic energy of plane motion may also be expressed in terms of the rotational velocity about the instantaneous center C of zero velocity.
Because C momentarily has zero velocity, the proof leading to Eq. 6/8 for the fixed point O holds equally well for point C, so that, alternatively, we may write the kinetic energy of a rigid body in plane motion as (6/10) In Art. 4/3 we derived Eq. 4/4 for the kinetic energy of any system of mass. We now see that this expression is equivalent to Eq. 6/9 when the mass system is rigid. For a rigid body, the quantity in Eq.
4/4 is the velocity of the representative particle relative to the mass center and is the vector i, which has the magnitude i. The summation term in Eq. 4/4 becomes which brings Eq. 4/4 into agreement with Eq. 6/9.
Potential Energy and the Work-Energy Equation Gravitational potential energy Vg and elastic potential energy Ve were covered in detail in Art. 3/7. Recall that the symbol U (rather than U) is used to denote the work done by all forces except the weight and elastic forces, which are accounted for in the potential-energy terms.
The work-energy relation, Eq. 3/15a, was introduced in Art. 3/6 for particle motion and was generalized in Art. 4/3 to include the motion of a general system of particles. This equation [4/2] applies to any mechanical system. For application to the motion of a sin-gle rigid body, the terms T1 and T2 must include the effects of transla-ton and rotation as given by Eqs. 6/7, 6/8, 6/9, or 6/10, and U1-2 is the work done by all external forces. On the other hand, if we choose to ex-press the effects of weight and springs by means of potential energy rather than work, we may rewrite the above equation as [4/3a] where the prime denotes the work done by all forces other than weight and spring forces.
When applied to an interconnected system of rigid bodies, Eq. 4/3a includes the effect of stored elastic energy in the connections, as well as that of gravitational potential energy for the various members. The term includes the work of all forces external to the system (other than gravitational forces), as well as the negative work of internal fric-tion forces, if any. The terms T1 and T2 are the initial and final kinetic energies of all moving parts over the interval of motion in question.
When the work-energy principle is applied to a single rigid body, ei-ther a free-body diagram or an active-force diagram should be used. In the case of an interconnected system of rigid bodies, an active-force dia-gram of the entire system should be drawn in order to isolate the system and disclose all forces which do work on the system. Diagrams should U 1-2 T1 V1 U 1-2 T2 V2 T1 U1-2 T2 1 2 I 2, 1 2 2 Σmi i 2 Σ 1 2 mi(i)2 ˙i T 1 2 IC 2 Article 6/6 Work-Energy Relations 461 c06.qxd 2/10/12 2:13 PM Page 461 also be drawn to disclose the initial and final positions of the system for the given interval of motion.
The work-energy equation provides a direct relationship between the forces which do work and the corresponding changes in the motion of a mechanical system. However, if there is appreciable internal me-chanical friction, then the system must be dismembered in order to dis-close the kinetic-friction forces and account for the negative work which they do. When the system is dismembered, however, one of the primary advantages of the work-energy approach is automatically lost. The work-energy method is most useful for analyzing conservative systems of interconnected bodies, where energy loss due to the negative work of friction forces is negligible.
Power The concept of power was discussed in Art. 3/6, which treated work-energy for particle motion. Recall that power is the time rate at which work is performed. For a force F acting on a rigid body in plane motion, the power developed by that force at a given instant is given by Eq. 3/16 and is the rate at which the force is doing work. The power is given by where dr and v are, respectively, the differential displacement and the velocity of the point of application of the force.
Similarly, for a couple M acting on the body, the power developed by the couple at a given instant is the rate at which it is doing work, and is given by where d and are, respectively, the differential angular displacement and the angular velocity of the body. If the senses of M and are the same, the power is positive and energy is supplied to the body. Con-versely, if M and have opposite senses, the power is negative and en-ergy is removed from the body. If the force F and the couple M act simultaneously, the total instantaneous power is We may also express power by evaluating the rate at which the total me-chanical energy of a rigid body or a system of rigid bodies is changing.
The work-energy relation, Eq. 4/3, for an infinitesimal displacement is where dU is the work of the active forces and couples applied to the body or to the system of bodies. Excluded from dU are the work of dU dT dV P Fv M P dU dt M d dt M P dU dt Fdr dt Fv 462 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 462 gravitational forces and that of spring forces, which are accounted for in the dV term. Dividing by dt gives the total power of the active forces and couples as Thus, we see that the power developed by the active forces and couples equals the rate of change of the total mechanical energy of the body or system of bodies.
We note from Eq. 6/9 that, for a given body, the first term may be written where R is the resultant of all forces acting on the body and is the re-sultant moment about the mass center G of all forces. The dot product accounts for the case of curvilinear motion of the mass center, where and are not in the same direction.
v a M ma v I() R v M 1 2 m(a v v a) I ˙ T ˙ dT dt d dt 1 2 mvv 1 2 I2 P dU dt T ˙ V ˙ d dt (T V) Article 6/6 Work-Energy Relations 463 Power-generating wind turbines near Mojave, California.
Savid Nunuk/PhotoResearchers, Inc.
c06.qxd 2/10/12 2:13 PM Page 463 SAMPLE PROBLEM 6/9 The wheel rolls up the incline on its hubs without slipping and is pulled by the 100-N force applied to the cord wrapped around its outer rim. If the wheel starts from rest, compute its angular velocity after its center has moved a dis-tance of 3 m up the incline. The wheel has a mass of 40 kg with center of mass at O and has a centroidal radius of gyration of 150 mm. Determine the power input from the 100-N force at the end of the 3-m motion interval.
Solution.
Of the four forces shown on the free-body diagram of the wheel, only the 100-N pull and the weight of 40(9.81) 392 N do work. The friction force does no work as long as the wheel does not slip. By use of the concept of the instantaneous center C of zero velocity, we see that a point A on the cord to which the 100-N force is applied has a velocity vA [(200 100)/100]v. Hence, point A on the cord moves a distance of (200 100)/100 3 times as far as the center O. Thus, with the effect of the weight included in the U-term, the work done on the wheel becomes The wheel has general plane motion, so that the initial and final kinetic en-ergies are The work-energy equation gives Alternatively, the kinetic energy of the wheel may be written The power input from the 100-N force when 30.3 rad/s is Ans.
[P Fv] P100 100(0.3)(30.3) 908 W [T 1 2 IC2] T 1 2 40[(0.15)2 (0.10)2]2 0.6502 [T1 U1-2 T2] 0 595 0.6502 30.3 rad/s 0.6502 [T 1 2 mv 2 1 2 I2] T1 0 T2 1 2 40(0.10)2 1 2 40(0.15)22 U1-2 100 200 100 100 (3) (392 sin 15)(3) 595 J 464 Chapter 6 Plane Kinetics of Rigid Bodies 15° 100 mm 200 mm 100 N O 15° 3 m 392 N 100 mm 200 mm 100 N O A C F N ω v v vA Helpful Hints Since the velocity of the instanta-neous center C on the wheel is zero, it follows that the rate at which the friction force does work is continu-ously zero. Hence, F does no work as long as the wheel does not slip. If the wheel were rolling on a moving platform, however, the friction force would do work, even if the wheel were not slipping.
Note that the component of the weight down the plane does negative work.
Be careful to use the correct radius in the expression v r for the ve-locity of the center of the wheel.
Recall that IC where IO The velocity here is that of the appli-cation point of the 100-N force.
mkO 2.
I mOC2, I c06.qxd 2/10/12 2:13 PM Page 464 SAMPLE PROBLEM 6/10 The 4-ft slender bar weighs 40 lb with mass center at B and is released from rest in the position for which is essentially zero. Point B is confined to move in the smooth vertical guide, while end A moves in the smooth horizontal guide and compresses the spring as the bar falls. Determine (a) the angular velocity of the bar as the position 30 is passed and (b) the velocity with which B strikes the horizontal surface if the stiffness of the spring is 30 lb/in.
Solution.
With the friction and mass of the small rollers at A and B neglected, the system may be treated as being conservative.
Part (a). For the first interval of motion from 0 (state 1) to 30 (state 2), the spring is not engaged, so that there is no Ve term in the energy equation. If we adopt the alternative of treating the work of the weight in the Vg term, then there are no other forces which do work, and 0.
Since we have a constrained plane motion, there is a kinematic relation be-tween the velocity vB of the center of mass and the angular velocity of the bar.
This relation is easily obtained by using the instantaneous center C of zero veloc-ity and noting that vB Thus, the kinetic energy of the bar in the 30 posi-tion becomes With a datum established at the initial position of the mass center B, our initial and final gravitational potential energies are We now substitute into the energy equation and obtain Ans.
Part (b). We define state 3 as that for which 90. The initial and final spring potential energies are In the final horizontal position, point A has no velocity, so that the bar is, in effect, rotating about A. Hence, its final kinetic energy is The final gravitational potential energy is Substituting into the energy equation gives Ans.
Alternatively, if the bar alone constitutes the system, the active-force dia-gram shows the weight, which does positive work, and the spring force kx, which does negative work. We would then write which is identical with the previous result.
80 45 0.828vB 2 [T1 U1-3 T3] vB 6.50 ft/sec 0 0 0 0.828vB 2 45 80 [T1 V1 U 1-3 T3 V3] V3 40(2) 80 ft-lb [Vg Wh] T3 1 2 1 3 40 32.2 42 vB 24/12 2 0.828vB 2 [T 1 2 IA2] V1 0 V3 1 2 (30)(24 18)2 1 12 45 ft-lb [Ve 1 2 kx2] 2.72 rad/sec 0 0 0 1.4492 10.72 [T1 V1 U 1-2 T2 V2] V2 40(2 cos 30 2) 10.72 ft-lb V1 0 T 1 2 40 32.2 12 12 2 1 2 1 12 40 32.2 422 1.4492 [T 1 2 mv2 1 2 I2] CB.
U 1-2 Article 6/6 Work-Energy Relations 465 D A 24″ 24″ θ 18″ k B B C A 24 ″ 12″ ω D vA vB 30° 40 lb (Alternative Active-Force Diagram) kx Helpful Hints We recognize that the forces acting on the bar at A and B are normal to the respective directions of motion and, hence, do no work.
If we convert k to lb/ft, we have Always check the consistency of your units. 45 ft-lb Ve 1 2 30 lb in.12 in.
ft 24 18 12 ft 2 c06.qxd 2/10/12 2:13 PM Page 465 SAMPLE PROBLEM 6/11 In the mechanism shown, each of the two wheels has a mass of 30 kg and a centroidal radius of gyration of 100 mm. Each link OB has a mass of 10 kg and may be treated as a slender bar. The 7-kg collar at B slides on the fixed vertical shaft with negligible friction. The spring has a stiffness k 30 kN/m and is con-tacted by the bottom of the collar when the links reach the horizontal position. If the collar is released from rest at the position 45 and if friction is sufficient to prevent the wheels from slipping, determine (a) the velocity vB of the collar as it first strikes the spring and (b) the maximum deformation x of the spring.
Solution.
The mechanism executes plane motion and is conservative with the neglect of kinetic friction losses. We define states 1, 2, and 3 to be at 45, 0, and maximum spring deflection, respectively. The datum for zero gravita-tional potential energy Vg is conveniently taken through O as shown.
(a) For the interval from 45 to 0, we note that the initial and final ki-netic energies of the wheels are zero since each wheel starts from rest and mo-mentarily comes to rest at 0. Also, at position 2, each link is merely rotating about its point O so that The collar at B drops a distance 0.265 m so that Also, 0. Hence, Ans.
(b) At the condition of maximum deformation x of the spring, all parts are mo-mentarily at rest, which makes T3 0. Thus, Solution for the positive value of x gives Ans.
It should be noted that the results of parts (a) and (b) involve a very simple net energy change despite the fact that the mechanism has undergone a fairly complex sequence of motions. Solution of this and similar problems by other than a work-energy approach is not an inviting prospect.
x 60.1 mm 0 2(10)(9.81) x 2 7(9.81)x 1 2 (30)(103)x2 [T1 V1 U 1-3 T3 V3] 0 2(10)(9.81) 0.265 2 7(9.81)(0.265) 0 vB 2.54 m/s [T1 V1 U 1-2 T2 V2] 0 44.2 0 6.83vB 2 0 U 1-2 V1 2(10)(9.81) 0.265 2 7(9.81)(0.265) 44.2 J V2 0 0.375/2 1 3 10(0.375)2 vB 0.375 2 1 2 7vB 2 6.83vB 2 T2 [2( 1 2 IO2 )]links [ 1 2 mv2]collar 466 Chapter 6 Plane Kinetics of Rigid Bodies O O A B A Vg=0 k vB v = vO 375 mm 375 mm θ θ 150 mm 150 mm Helpful Hint With the work of the weight of the collar B included in the potential-energy terms, there are no other forces external to the system which do work. The friction force acting under each wheel does no work since the wheel does not slip, and, of course, the normal force does no work here. Hence, 0.
U 1-2 c06.qxd 2/10/12 2:13 PM Page 466 PROBLEMS (In the following problems neglect any energy loss due to kinetic friction unless otherwise instructed.) Introductory Problems 6/109 The uniform slender bar of mass m and length L is released from rest when in the horizontal position shown. Determine its angular velocity and mass-center speed as it passes the vertical position.
Problem 6/109 6/110 The slender rod (mass m, length L) has a particle (mass 2m) attached to one end. If the body is nudged away from the vertical equilibrium position shown, determine its angular speed after it has rotated 180 .
Problem 6/110 6/111 The 32.2-lb wheel is released from rest and rolls on its hubs without slipping. Calculate the velocity v of the center O of the wheel after it has moved a distance ft down the incline. The radius of gyration of the wheel about O is 5 in.
x 10 L –– 2 L –– 2 O m 2m O m L –– 4 3L –– 4 Article 6/6 Problems 467 Problem 6/111 6/112 The uniform quarter-circular sector is released from rest with one edge vertical as shown. Determine its subsequent maximum angular velocity.
Problem 6/112 6/113 The velocity of the 8-kg cylinder is 0.3 m/s at a certain instant. What is its speed v after dropping an additional 1.5 m? The mass of the grooved drum is 12 kg, its centroidal radius of gyration is and the radius of its groove is The frictional moment at O is a constant 3 .
Problem 6/113 O = 210 mm m = 12 kg ri = 200 mm ro = 300 mm k 8 kg Nm 200 mm.
ri k 210 mm, O b m Vertical 8″ 2″ x O 10 1 c06.qxd 2/10/12 2:13 PM Page 467 Problem 6/116 6/117 The 15-kg slender bar OA is released from rest in the vertical position and compresses the spring of stiffness as the horizontal position is passed. Determine the proper setting of the spring, by specifying the distance h, which will result in the bar having an angular velocity as it crosses the horizontal position. What is the effect of x on the dynamics of the problem?
Problem 6/117 h A O x = 400 mm 600 mm ω 4 rad/s k 20 kN/m θ θ B A r r 468 Chapter 6 Plane Kinetics of Rigid Bodies 6/114 The log is suspended by the two parallel 5-m cables and used as a battering ram. At what angle should the log be released from rest in order to strike the object to be smashed with a velocity of 4 m/s?
Problem 6/114 6/115 The T-shaped body of total mass m is constructed of uniform rod. If it is released from rest while in the position shown, determine the vertical force reac-tion at O as it passes the vertical position (120 after release).
Problem 6/115 6/116 The two wheels of Prob. 6/82, shown again here, represent two extreme conditions of distribution of mass. For case A all of the mass m is assumed to be concentrated in the center of the hoop in the axial bar of negligible diameter. For case B all of the mass m is assumed to be concentrated in the rim.
Determine the velocity of the center of each hoop after it has traveled a distance x down the incline from rest. The hoops roll without slipping.
b –– 4 b –– 4 b O 30° v = 4 m/s 5 m 5 m θ θ c06.qxd 2/10/12 2:13 PM Page 468 6/118 The wheel is composed of a 10-kg hoop stiffened by four thin spokes, each with a mass of 2 kg. A hori-zontal force of 40 N is applied to the wheel initially at rest. Calculate the angular velocity of the wheel after its center has moved 3 m. Friction is sufficient to prevent slipping.
Problem 6/118 6/119 A steady 5-lb force is applied normal to the handle of the hand-operated grinder. The gear inside the housing with its shaft and attached handle together weigh 3.94 lb and have a radius of gyration about their axis of 2.85 in. The grinding wheel with its at-tached shaft and pinion (inside housing) together weigh 1.22 lb and have a radius of gyration of 2.14 in.
If the gear ratio between gear and pinion is 4:1, cal-culate the speed N of the grinding wheel after 6 com-plete revolutions of the handle starting from rest.
Problem 6/119 6″ 5 lb N O O 300 mm 40 N Article 6/6 Problems 469 Representative Problems 6/120 The 1.2-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position where the spring is unstretched. If the bar is ob-served to momentarily stop in the position determine the spring constant k. For your com-puted value of k, what is the angular velocity of the bar when Problem 6/120 6/121 Specify the unstretched length l0 of the spring of stiffness which will result in a velocity of for the contact at A if the toggle is given a slight nudge from its null position at The toggle has a mass of 1.5 kg and a radius of gyration about O of 55 mm. Motion occurs in the horizontal plane.
Problem 6/121 l 50 mm 50 mm O A θ = 30° 60 mm 0.
0.25 m/s k 1400 N/m O 1.2 kg A B k 0.6 m 0.6 m 0.2 m θ 25?
50, 0 c06.qxd 2/10/12 2:13 PM Page 469 6/124 The two identical links, each of length b and mass m, may be treated as uniform slender bars. If they are released from rest in the position shown with end A constrained by the smooth vertical guide, determine the velocity v with which A reaches O with essentially zero.
Problem 6/124 6/125 The torsional spring has a stiffness of 30 m/rad and is undeflected when the 6-kg uniform slender bar is in the upright position. If the bar is released from rest in the horizontal position shown, deter-mine its angular velocity as it passes the vertical position. Friction is negligible.
Problem 6/125 O K 0.8 m ω N A B O b b θ 470 Chapter 6 Plane Kinetics of Rigid Bodies 6/122 The 50-kg flywheel has a radius of gyration about its shaft axis and is subjected to the torque where is in radians. If the flywheel is at rest when determine its angu-lar velocity after 5 revolutions.
Problem 6/122 6/123 The 12-lb lever OA with 10-in. radius of gyration about O is initially at rest in the vertical position where the attached spring of stiffness is unstretched. Calculate the constant moment M applied to the lever through its shaft at O which will give the lever an angular velocity as the lever reaches the horizontal position Problem 6/123 15″ 15″ 8″ θ G A M k = 3 lb/in.
O 0.
4 rad/sec k 3 lb/in.
( 90), M 0, M 2(1 e0.1) Nm, k 0.4 m c06.qxd 2/10/12 2:13 PM Page 470 6/126 The wheel consists of a 4-kg rim of 250-mm radius with hub and spokes of negligible mass. The wheel is mounted on the 3-kg yoke OA with mass center at G and with a radius of gyration about O of 350 mm.
If the assembly is released from rest in the horizon-tal position shown and if the wheel rolls on the cir-cular surface without slipping, compute the velocity of point A when it reaches .
Problem 6/126 6/127 The uniform slender bar ABC weighs 6 lb and is initially at rest with end A bearing against the stop in the horizontal guide. When a constant couple -in. is applied to end C, the bar rotates causing end A to strike the side of the vertical guide with a velocity of 10 ft/sec. Calculate the loss of energy E due to friction in the guides and rollers. The mass of the rollers may be neglected.
Problem 6/127 B A C M 45° 8″ 8″ M 72 lb A′ 300 mm 250 mm O A G 500 mm A Article 6/6 Problems 471 6/128 The center of the 200-lb wheel with centroidal radius of gyration of 4 in. has a velocity of 2 ft/sec down the incline in the position shown. Calculate the normal reaction N under the wheel as it rolls past position A. Assume that no slipping occurs.
Problem 6/128 6/129 The uniform slender bar of mass m pivots freely about a horizontal axis through O. If the bar is released from rest in the horizontal position shown where the spring is unstretched, it is observed to rotate a maximum of 30 clockwise. The spring con-stant and the distance Determine (a) the mass m of the bar and (b) the angular velocity of the bar when the angular displacement is 15 clockwise from the release position.
Problem 6/129 b –– 4 3b –– 4 b –– 4 O A m C B k b 200 mm.
k 200 N/m A 30° 24″ 12″ 6″ c06.qxd 2/20/12 3:23 PM Page 471 6/132 The electric motor shown is delivering 4 kW at 1725 rev/min to a pump which it drives. Calculate the angle through which the motor deflects under load if the stiffness of each of its four spring mounts is 15 kN/m. In what direction does the motor shaft turn?
Problem 6/132 6/133 The two uniform right-angle bars are released from rest when in the position at which the spring of modulus is un-stretched. The bars then rotate in a vertical plane about the fixed centers of the attached light gears, thus maintaining the same angle for both bars.
Determine the angular speed of the bars as the position is passed.
Problem 6/133 θ 60 mm 180 mm 120 mm 40 mm θ k = 450 N/m 20 k 450 N/m 0, 4.2-kg 200 mm δ 472 Chapter 6 Plane Kinetics of Rigid Bodies 6/130 The system is released from rest when the angle Determine the angular velocity of the uni-form slender bar when equals 60 . Use the values Problem 6/130 6/131 The two identical steel frames with the dimensions shown are fabricated from the same bar stock and are hinged at the midpoints A and B of their sides.
If the frame is resting in the position shown on a horizontal surface with negligible friction, deter-mine the velocity v with which each of the upper ends of the frame hits the horizontal surface if the cord at C is cut.
Problem 6/131 θ b b b b c c B A C O C m2 m1 B A 2b b 2b θ m1 1 kg, m2 1.25 kg, and b 0.4 m.
90.
c06.qxd 2/10/12 2:13 PM Page 472 6/134 A lid-support mechanism is being designed for a storage chest to limit the angular velocity of the 10-lb uniform lid to 1.5 rad/sec for when it is released from rest with essentially equal to 90 .
Two identical mechanisms are included as indi-cated on the pictorial sketch. Specify the necessary stiffness k of each of the two springs, which are compressed 2 in. upon closure. Neglect the weight of the links and any friction in the sliding collars C.
Also, the thickness of the lid is small compared with its other dimensions.
Problem 6/134 6/135 Each of the two links has a mass of 2 kg and a cen-troidal radius of gyration of 60 mm. The slider at B has a mass of 3 kg and moves freely in the vertical guide. The spring has a stiffness of 6 kN/m. If a constant torque is applied to link OA through its shaft at O starting from the rest posi-tion at determine the angular velocity of OA when 0.
45, M 20 Nm 2″ C A O 2″ θ 10″ 14″ 6″ B A A B B O C C 0 Article 6/6 Problems 473 Problem 6/135 6/136 The system is at rest with the spring unstretched when The 5-kg uniform slender bar is then given a slight clockwise nudge. The value of b is 0.4 m. (a) If the bar comes to momentary rest when determine the spring constant k. (b) For the value find the angular velocity of the bar when Problem 6/136 k A b O 1.25b θ 25.
k 90 N/m, 40, 0.
θ 400 mm 200 mm 200 mm 50 mm A B M O c06.qxd 2/10/12 2:13 PM Page 473 6/139 The 8-kg crank OA, with mass center at G and ra-dius of gyration about O of 0.22 m, is connected to the 12-kg uniform slender bar AB. If the linkage is released from rest in the position shown, compute the velocity v of end B as OA swings through the vertical.
Problem 6/139 6/140 The figure shows the cross section of a uniform 200-lb ventilator door hinged about its upper hori-zontal edge at O. The door is controlled by the spring-loaded cable which passes over the small pulley at A. The spring has a stiffness of 15 lb per foot of stretch and is undeformed when If the door is released from rest in the horizontal po-sition, determine the maximum angular velocity reached by the door and the corresponding angle .
Problem 6/140 60° k A O θ 0.
1.0 m 0.4 m 0.18 m 0.8 m B A O G 474 Chapter 6 Plane Kinetics of Rigid Bodies 6/137 The body shown is constructed of uniform slender rod and consists of a ring of radius r attached to a straight section of length 2r. The body pivots freely about a ball-and-socket joint at O. If the body is at rest in the vertical position shown and is given a slight nudge, compute its angular velocity after a 90 rotation about (a) axis A-A and (b) axis B-B.
Problem 6/137 6/138 A facility for testing the performance of motorized golf carts consists of an endless belt where the angle can be adjusted. The cart of mass m is slowly brought up to its rated ground speed v with the braking torque M on the upper pulley con-stantly adjusted so that the cart remains in a fixed position A on the test stand. With no cart on the belt, a torque is required on the pulley to over-come friction and turn the pulleys regardless of speed. Friction is sufficient to prevent the wheels from slipping on the belt. Determine an expression for the power P absorbed by the braking torque M.
Do the static friction forces between the wheels and the belt do work?
Problem 6/138 r v v A M θ M0 A 2r A O B B r c06.qxd 2/10/12 2:13 PM Page 474 6/141 Motive power for the experimental 10-Mg bus comes from the energy stored in a rotating flywheel which it carries. The flywheel has a mass of 1500 kg and a radius of gyration of 500 mm and is brought up to a maximum speed of 4000 rev/min. If the bus starts from rest and acquires a speed of 72 km/h at the top of a hill 20 m above the starting position, compute the reduced speed N of the flywheel. Assume that 10 percent of the energy taken from the flywheel is lost. Neglect the rotational energy of the wheels of the bus. The 10-Mg mass includes the flywheel.
Problem 6/141 6/142 The two identical uniform bars are released from rest from the position shown in the vertical plane.
Determine the angular velocity of AB when the bars become collinear.
Problem 6/142 6/143 The figure shows the cross section of a garage door which is a uniform rectangular panel 8 by 8 ft and weighing 200 lb. The door carries two spring as-semblies, one on each side of the door, like the one shown. Each spring has a stiffness of 50 lb/ft and is unstretched when the door is in the open position shown. If the door is released from rest in this posi-tion, calculate the velocity of the edge at A as it strikes the garage floor.
A B C b b 45° 45° 45° Article 6/6 Problems 475 Problem 6/143 6/144 The uniform slender rod of length l is released from rest in the dashed vertical position. With what speed does end A strike the 30 incline? Neglect the small mass and friction of the end rollers.
Problem 6/144 l l A B 30° x vA 10′ 8′ 8′ A B c06.qxd 2/10/12 2:13 PM Page 475 6/146 Motion of the 600-mm slender bar of mass 4 kg is controlled by the constrained movement of its small rollers A and B of negligible mass and friction. The bar starts from rest in the horizontal position with and moves in the vertical plane under the ac-tion of the constant force applied normal to the bar at end C. Calculate the velocity v with which roller A strikes the wall of the vertical guide at Problem 6/146 C B A P 300 mm 300 mm θ 90.
P 50 N 0 476 Chapter 6 Plane Kinetics of Rigid Bodies 6/145 The 10-kg double wheel with radius of gyration of 125 mm about O is connected to the spring of stiffness by a cord which is wrapped securely around the inner hub. If the wheel is re-leased from rest on the incline with the spring stretched 225 mm, calculate the maximum velocity v of its center O during the ensuing motion. The wheel rolls without slipping.
Problem 6/145 5 1 75 mm 200 mm O k k 600 N/m c06.qxd 2/10/12 2:13 PM Page 476 6/7 Acceleration from Work-Energy; Virtual Work In addition to using the work-energy equation to determine the veloc-ities due to the action of forces acting over finite displacements, we may also use the equation to establish the instantaneous accelerations of the members of a system of interconnected bodies as a result of the active forces applied. We may also modify the equation to determine the configu-ration of such a system when it undergoes a constant acceleration.
Work-Energy Equation for Differential Motions For an infinitesimal interval of motion, Eq. 4/3 becomes The term dU represents the total work done by all active nonpotential forces acting on the system under consideration during the infinitesimal displacement of the system. The work of potential forces is included in the dV-term. If we use the subscript i to denote a representative body of the interconnected system, the differential change in kinetic energy T for the entire system becomes where d and di are the respective changes in the magnitudes of the velocities and where the summation is taken over all bodies of the sys-tem. But for each body, d and where represents the infinitesimal linear displacement of the center of mass and where di represents the infinitesimal angular displacement of the body in the plane of motion. We note that d is identical to d where is the component of along the tangent to the curve described by the mass center of the body in question. Also i represents the angular acceleration of the representative body. Consequently, for the entire system d This change may also be written as d where Ri and are the resultant force and resultant couple acting on body i and where di dik. These last two equations merely show us that the differential change in kinetic energy equals the differential work done on the system by the resultant forces and resultant couples acting on all the bodies of the system.
The term dV represents the differential change in the total gravita-tional potential energy Vg and the total elastic potential energy Ve and has the form dV d(Σmighi Σ 1 2 kj xj 2) Σmig dhi Σkj xj dxj MGi si ΣMGi di dT ΣRi si ΣIii di dT Σmiai ¨i, ai (ai)t si, (ai)t si ai d si Iii di, Iii di si miai mivi dvi vi dT d(Σ 1 2mivi 2 Σ 1 2 Iii 2) Σmivi dvi ΣIii di dU dT dV Article 6/7 Acceleration from Work-Energy; Virtual Work 477 c06.qxd 2/10/12 2:13 PM Page 477 where hi represents the vertical distance of the center of mass of the representative body of mass mi above any convenient datum plane and where xj stands for the deformation, tensile or compressive, of a repre-sentative elastic member of the system (spring) whose stiffness is kj.
The complete expression for dU may now be written as d (6/11) When Eq. 6/11 is applied to a system of one degree of freedom, the terms d and will be positive if the accelerations are in the same direction as the respective displacements and negative if they are in the opposite direction. Equation 6/11 has the advantage of relating the accel-erations to the active forces directly, which eliminates the need for dis-membering the system and then eliminating the internal forces and reactive forces by simultaneous solution of the force-mass-acceleration equations for each member.
Virtual Work In Eq. 6/11 the differential motions are differential changes in the real or actual displacements which occur. For a mechanical system which assumes a steady-state configuration during constant accelera-tion, we often find it convenient to introduce the concept of virtual work. The concepts of virtual work and virtual displacement were intro-duced and used to establish equilibrium configurations for static sys-tems of interconnected bodies (see Chapter 7 of Vol. 1 Statics).
A virtual displacement is any assumed and arbitrary displacement, linear or angular, away from the natural or actual position. For a sys-tem of connected bodies, the virtual displacements must be consistent with the constraints of the system. For example, when one end of a link is hinged about a fixed pivot, the virtual displacement of the other end must be normal to the line joining the two ends. Such requirements for displacements consistent with the constraints are purely kinematic and provide what are known as the equations of constraint.
If a set of virtual displacements satisfying the equations of con-straint and therefore consistent with the constraints is assumed for a mechanical system, the proper relationship between the coordinates which specify the configuration of the system will be determined by ap-plying the work-energy relationship of Eq. 6/11, expressed in terms of virtual changes. Thus, (6/11a) It is customary to use the differential symbol d to refer to differential changes in the real displacements, whereas the symbol is used to sig-nify virtual changes, that is, differential changes which are assumed rather than real. U Σmiai si ΣIii i Σmi g hi Σkj xj xj Iii di si miai si ΣIii di Σmi g dhi Σkj xj dxj dU Σmiai 478 Chapter 6 Plane Kinetics of Rigid Bodies c06.qxd 2/10/12 2:13 PM Page 478 SAMPLE PROBLEM 6/12 The movable rack A has a mass of 3 kg, and rack B is fixed. The gear has a mass of 2 kg and a radius of gyration of 60 mm. In the position shown, the spring, which has a stiffness of 1.2 kN/m, is stretched a distance of 40 mm. For the instant represented, determine the acceleration a of rack A under the action of the 80-N force. The plane of the figure is vertical.
Solution.
The given figure represents the active-force diagram for the entire system, which is conservative.
During an infinitesimal upward displacement dx of rack A, the work dU done on the system is 80 dx, where x is in meters, and this work equals the sum of the corresponding changes in the total energy of the system. These changes, which appear in Eq. 6/11, are as follows: The change in potential energies of the system, from Eq. 6/11, becomes Substitution into Eq. 6/11 gives us Canceling dx and solving for a give Ans.
We see that using the work-energy method for an infinitesimal displace-ment has given us the direct relation between the applied force and the resulting acceleration. It was unnecessary to dismember the system, draw two free-body diagrams, apply ΣF twice, apply ΣMG and F kx, eliminate un-wanted terms, and finally solve for a.
I ma a 16.76/3.78 4.43 m/s2 80 dx 3a dx 0.781a dx 29.4 dx 9.81 dx 24 dx dVspring kj xj dxj 1200(0.04) dx/2 24 dx dVgear 2g(dx/2) g dx 9.81 dx dVrack 3g dx 3(9.81) dx 29.4 dx [dV Σmi g dhi Σkj xj dxj] dTgear 2 a 2 dx 2 2(0.06)2 a/2 0.08 dx/2 0.08 0.781a dx dTrack 3a dx [dT Σmiai dsi ΣIii di] Article 6/7 Acceleration from Work-Energy; Virtual Work 479 80 N 80 mm A B Helpful Hints Note that none of the remaining forces external to the system do any work. The work done by the weight and by the spring is accounted for in the potential-energy terms.
Note that for the gear is its mass-center acceleration, which is half that for the rack A. Also, its displace-ment is dx/2. For the rolling gear, the angular acceleration from a r becomes i (a/2)/0.08, and the an-gular displacement from ds r d becomes di (dx/2)/0.08.
Note here that the displacement of the spring is one-half that of the rack. Hence, xi x/2.
ai c06.qxd 2/10/12 2:13 PM Page 479 SAMPLE PROBLEM 6/13 A constant force P is applied to end A of the two identical and uniform links and causes them to move to the right in their vertical plane with a horizontal ac-celeration a. Determine the steady-state angle made by the bars with one an-other.
Solution.
The figure constitutes the active-force diagram for the system. To find the steady-state configuration, consider a virtual displacement of each bar from the natural position assumed during the acceleration. Measurement of the displacement with respect to end A eliminates any work done by force P during the virtual displacement. Thus, The terms involving acceleration in Eq. 6/11a reduce to We choose the horizontal line through A as the datum for zero potential en-ergy. Thus, the potential energy of the links is and the virtual change in potential energy becomes Substitution into the work-energy equation for virtual changes, Eq. 6/11a, gives from which Ans.
Again, in this problem we see that the work-energy approach obviated the necessity for dismembering the system, drawing separate free-body diagrams, applying motion equations, eliminating unwanted terms, and solving for .
2 tan1 2a g 0 mal cos 2 mgl 2 sin 2 Vg 2mg l 2 cos 2 mgl 2 sin 2 Vg 2mg l 2 cos 2 ma l cos 2 ma l 2 sin 2 3l 2 sin 2 ma s ma( s1) ma( s2) U 0 480 Chapter 6 Plane Kinetics of Rigid Bodies s2 V g = 0 s1 A P a l/2 l/2 l/2 l/2 θ Helpful Hints Note that we use the symbol to refer to an assumed or virtual differential change rather than the symbol d, which refers to an infinitesimal change in the real displacement.
Here we are evaluating the work done by the resultant forces and couples in the virtual displacement.
Note that 0 for both bars.
We have chosen to use the angle to describe the configuration of the links, although we could have used the distance between the two ends of the links just as well.
The last two terms in Eq. 6/11a ex-press the virtual changes in gravita-tional and elastic potential energy.
c06.qxd 2/10/12 2:13 PM Page 480 PROBLEMS Introductory Problems 6/147 The position of the horizontal platform of mass is controlled by the parallel slender links of masses m and 2m. Determine the initial angular accelera-tion of the links as they start from their supported position shown under the action of a force P applied normal to AB at its end.
Problem 6/147 6/148 The uniform slender bar of mass m is shown in its equilibrium position in the vertical plane before the couple M is applied to the end of the bar. Determine the initial angular acceleration of the bar upon application of M. The mass of each guide roller is negligible.
Problem 6/148 6/149 The two uniform slender bars are hinged at O and supported on the horizontal surface by their end rollers of negligible mass. If the bars are released from rest in the position shown, determine their initial angular acceleration as they collapse in the vertical plane. (Suggestion: Make use of the instan-taneous center of zero velocity in writing the ex-pression for dT.) b b M k θ A P B O C D E m0 m b b b 2m θ θ m0 Article 6/7 Problems 481 Problem 6/149 6/150 Links A and B each weigh 8 lb, and bar C weighs 12 lb. Calculate the angle assumed by the links if the body to which they are pinned is given a steady horizontal acceleration a of 4 ft/sec2.
Problem 6/150 6/151 The mechanism shown moves in the vertical plane.
The vertical bar AB weighs 10 lb, and each of the two links weighs 6 lb with mass center at G and with a radius of gyration of 10 in. about its bearing (O or C). The spring has a stiffness of 15 lb/ft and an unstretched length of 18 in. If the support at D is suddenly withdrawn, determine the initial angu-lar acceleration of the links.
Problem 6/151 18″ 18″ 18″ G G O A C B D 60° 60° OG = CG = 8″ —– —– A a C B 18″ 18″ θ θ θ θ A O B b b c06.qxd 2/10/12 2:13 PM Page 481 6/154 The box and load of the dump truck have a mass m with mass center at G and a moment of inertia IA about the pivot at A. Determine the angular accel-eration of the box when it is started from rest in the position shown under the application of the couple M to link CD. Neglect the mass of the links.
The figure ABDC is a parallelogram.
Problem 6/154 6/155 Each of the uniform bars OA and OB has a mass of 2 kg and is freely hinged at O to the vertical shaft, which is given an upward acceleration a g/2. The links which connect the light collar C to the bars have negligible mass, and the collar slides freely on the shaft. The spring has a stiffness k 130 N/m and is uncompressed for the position equivalent to 0. Calculate the angle assumed by the bars under conditions of steady acceleration.
Problem 6/155 θ θ 200 C B A a O 200 200 200 200 200 Dimensions in millimeters θ A B C D M G a b 482 Chapter 6 Plane Kinetics of Rigid Bodies Representative Problems 6/152 The load of mass m is given an upward acceleration a from its supported rest position by the application of the forces P. Neglect the mass of the links compared with m and determine the initial acceleration a.
Problem 6/152 6/153 The cargo box of the food-delivery truck for aircraft servicing has a loaded mass m and is elevated by the application of a couple M on the lower end of the link which is hinged to the truck frame. The horizontal slots allow the linkage to unfold as the cargo box is elevated. Determine the upward accel-eration of the box in terms of h for a given value of M. Neglect the mass of the links.
Problem 6/153 b b b b h M m θ θ θ θ b b b b a P P m c06.qxd 2/10/12 2:13 PM Page 482 6/156 The linkage consists of the two slender bars and moves in the horizontal plane under the influence of force P. Link OC has a mass m and link AC has a mass 2m. The sliding block at B has negligible mass. Without dismembering the system, determine the initial angular acceleration of the links as P is applied at A with the links initially at rest. (Sugges-tion: Replace P by its equivalent force-couple system.) Problem 6/156 6/157 The portable work platform is elevated by means of the two hydraulic cylinders articulated at points C.
The pressure in each cylinder produces a force F. The platform, man, and load have a combined mass m, and the mass of the linkage is small and may be ne-glected. Determine the upward acceleration a of the platform and show that it is independent of both b and .
Problem 6/157 b b B B C C b/2 b/2 b/2 b/2 θ θ P A B b b b C O θ θ Article 6/7 Problems 483 6/158 Each of the three identical uniform panels of a seg-mented industrial door has mass m and is guided in the tracks (one shown dashed). Determine the hori-zontal acceleration a of the upper panel under the action of the force P. Neglect any friction in the guide rollers.
Problem 6/158 6/159 The mechanical tachometer measures the rota-tional speed N of the shaft by the horizontal mo-tion of the collar B along the rotating shaft. This movement is caused by the centrifugal action of the two 12-oz weights A, which rotate with the shaft.
Collar C is fixed to the shaft. Determine the rota-tional speed N of the shaft for a reading 15 .
The stiffness of the spring is 5 lb/in., and it is un-compressed when 0 and 0. Neglect the weights of the links.
Problem 6/159 45° Vertical Horizontal P C B θ θ 5/8″ 1.5″ 1.5″ y A A N β 1/4″ 1/4″ c06.qxd 2/10/12 2:13 PM Page 483 6/162 The aerial tower shown is designed to elevate a workman in a vertical direction. An internal mech-anism at B maintains the angle between AB and BC at twice the angle between BC and the ground. If the combined mass of the man and the cab is 200 kg and if all other masses are neglected, determine the torque M applied to BC at C and the torque MB in the joint at B required to give the cab an initial vertical acceleration of 1.2 m/s2 when it is started from rest in the position 30.
Problem 6/162 6/163 The uniform arm OA has a mass of 4 kg, and the gear D has a mass of 5 kg with a radius of gyration about its center of 64 mm. The large gear B is fixed and cannot rotate. If the arm and small gear are re-leased from rest in the position shown in the verti-cal plane, calculate the initial angular acceleration of OA.
Problem 6/163 B O A D 200 mm 100 mm θ θ A B C M 2 6 m 6 m 484 Chapter 6 Plane Kinetics of Rigid Bodies 6/160 A planetary gear system is shown, where the gear teeth are omitted from the figure. Each of the three identical planet gears A, B, and C has a mass of 0.8 kg, a radius r 50 mm, and a radius of gyra-tion of 30 mm about its center. The spider E has a mass of 1.2 kg and a radius of gyration about O of 60 mm. The ring gear D has a radius R 150 mm and is fixed. If a torque M 5 is applied to the shaft of the spider at O, determine the initial angular acceleration of the spider.
Problem 6/160 6/161 The sector and attached wheels are released from rest in the position shown in the vertical plane.
Each wheel is a solid circular disk weighing 12 lb and rolls on the fixed circular path without slip-ping. The sector weighs 18 lb and is closely approxi-mated by one-fourth of a solid circular disk of 16-in.
radius. Determine the initial angular acceleration of the sector.
Problem 6/161 O 16″ 8″ 8″ M E A D C r R B O Nm c06.qxd 2/10/12 2:13 PM Page 484 6/164 The vehicle is used to transport supplies to and from the bottom of the 25-percent grade. Each pair of wheels, one at A and the other at B, has a mass of 140 kg with a radius of gyration of 150 mm. The drum C has a mass of 40 kg and a radius of gyra-tion of 100 mm. The total mass of the vehicle is 520 kg. The vehicle is released from rest with a re-straining force T of 500 N in the control cable which passes around the drum and is secured at D.
Determine the initial acceleration a of the vehicle.
The wheels roll without slipping.
Article 6/7 Problems 485 Problem 6/164 150 mm C B D T A 400 mm 25 100 c06.qxd 2/10/12 2:13 PM Page 485 6/8 Impulse-Momentum Equations The principles of impulse and momentum were developed and used in Articles 3/9 and 3/10 for the description of particle motion. In that treatment, we observed that those principles were of particular impor-tance when the applied forces were expressible as functions of the time and when interactions between particles occurred during short periods of time, such as with impact. Similar advantages result when the im-pulse-momentum principles are applied to the motion of rigid bodies.
In Art. 4/2 the impulse-momentum principles were extended to cover any defined system of mass particles without restriction as to the connections between the particles of the system. These extended rela-tions all apply to the motion of a rigid body, which is merely a special case of a general system of mass. We will now apply these equations di-rectly to rigid-body motion in two dimensions.
Linear Momentum In Art. 4/4 we defined the linear momentum of a mass system as the vector sum of the linear momenta of all its particles and wrote G Σmivi. With ri representing the position vector to mi, we have vi and G which, for a system whose total mass is constant, may be written as G d(Σmiri)/dt. When we substitute the principle of mo-ments Σmiri to locate the mass center, the momentum becomes G , where is the velocity of the mass center. There-fore, as before, we find that the linear momentum of any mass system, rigid or nonrigid, is [4/5] In the derivation of Eq. 4/5, we note that it was unnecessary to employ the kinematic condition for a rigid body, Fig. 6/13, which is vi i. In that case, we obtain the same result by writing G i). The first sum is and the second sum be-comes Σmii 0 since i is measured from the mass center, making zero.
Next in Art. 4/4 we rewrote Newton’s generalized second law as Eq.
4/6. This equation and its integrated form are (6/12) Equation 6/12 may be written in its scalar-component form, which, for plane motion in the x-y plane, gives (6/12a) ΣFx G ˙x ΣFy G ˙y and (Gx)1 t2 t1 ΣFx dt (Gx)2 (Gy)1 t2 t1 ΣFy dt (Gy)2 ΣF G ˙ and G1 t2 t1 ΣF dt G2 m mv, vΣmi Σmi(v v G mv v r ˙ mr ˙ d(mr)/dt mr Σmir ˙i r ˙i 486 Chapter 6 Plane Kinetics of Rigid Bodies SECTION C IMPULSE AND MOMENTUM vi = ri · – – v = r · G – v – r ri i ρ O ω i = × i ρ ω ρ · mi Figure 6/13 c06.qxd 2/10/12 2:13 PM Page 486 In words, the first of Eqs. 6/12 and 6/12a states that the resultant force equals the time rate of change of momentum. The integrated form of Eqs. 6/12 and 6/12a states that the initial linear momentum plus the lin-ear impulse acting on the body equals the final linear momentum.
As in the force-mass-acceleration formulation, the force summa-tions in Eqs. 6/12 and 6/12a must include all forces acting externally on the body considered. We emphasize again, therefore, that in the use of the impulse-momentum equations, it is essential to construct the com-plete impulse-momentum diagrams so as to disclose all external im-pulses. In contrast to the method of work and energy, all forces exert impulses, whether they do work or not.
Angular Momentum Angular momentum is defined as the moment of linear momentum.
In Art. 4/4 we expressed the angular momentum about the mass center of any prescribed system of mass as HG Σi mivi, which is merely the vector sum of the moments about G of the linear momenta of all particles. We showed in Art. 4/4 that this vector sum could also be writ-ten as HG Σi where is the velocity of mi with respect to G.
Although we have simplified this expression in Art. 6/2 in the course of deriving the moment equation of motion, we will pursue this same ex-pression again for sake of emphasis by using the rigid body in plane mo-tion represented in Fig. 6/13. The relative velocity becomes i, where the angular velocity of the body is k. The unit vector k is di-rected into the paper for the sense of shown. Because i, and are at right angles to one another, the magnitude of is i, and the mag-nitude of i is Thus, we may write HG where is the mass moment of inertia of the body about its mass center.
Because the angular-momentum vector is always normal to the plane of motion, vector notation is generally unnecessary, and we may write the angular momentum about the mass center as the scalar (6/13) This angular momentum appears in the moment-angular-momentum relation, Eq. 4/9, which in scalar notation for plane motion, along with its integrated form, is (6/14) In words, the first of Eqs. 6/14 states that the sum of the moments about the mass center of all forces acting on the body equals the time rate of change of angular momentum about the mass center. The inte-grated form of Eq. 6/14 states that the initial angular momentum about the mass center G plus the external angular impulse about G equals the final angular momentum about G.
The sense for positive rotation must be clearly established, and the algebraic signs of ΣMG, and must be consistent with (HG)2 (HG)1, ΣMG H ˙G and (HG)1 t2 t1 ΣMG dt (HG)2 HG I Σmii 2 I Ik, Σi 2mik i 2mi.
mi ˙i ˙i ˙i, ˙i ˙i mi ˙i, This ice skater can effect a large in-crease in angular speed about a ver-tical axis by drawing her arms closer to the center of her body.
Article 6/8 Impulse-Momentum Equations 487 fstockfoto/shutterstock.com c06.qxd 2/10/12 2:13 PM Page 487 this choice. The impulse-momentum diagram (see Art. 3/9) is again essential. See the Sample Problems which accompany this article for examples of these diagrams.
With the moments about G of the linear momenta of all particles ac-counted for by HG it follows that we may represent the linear mo-mentum G as a vector through the mass center G, as shown in Fig. 6/14a. Thus, G and HG have vector properties analogous to those of the resultant force and couple.
With the establishment of the linear- and angular-momentum re-sultants in Fig. 6/14a, which represents the momentum diagram, the angular momentum HO about any point O is easily written as (6/15) This expression holds at any particular instant of time about O, which may be a fixed or moving point on or off the body.
When a body rotates about a fixed point O on the body or body ex-tended, as shown in Fig. 6/14b, the relations and d may be substituted into the expression for HO, giving HO But IO so that (6/16) In Art. 4/2 we derived Eq. 4/7, which is the moment-angular-momentum equation about a fixed point O. This equation, written in scalar notation for plane motion along with its integrated form, is (6/17) Note that you should not add linear momentum and angular momentum for the same reason that force and moment cannot be added directly.
Interconnected Rigid Bodies The equations of impulse and momentum may also be used for a system of interconnected rigid bodies since the momentum principles are applicable to any general system of constant mass. Figure 6/15 shows the combined free-body diagram and momentum diagram for two interconnected bodies a and b. Equations 4/6 and 4/7, which are ΣF and ΣMO O where O is a fixed reference point, may be written for each member of the system and added. The sums are (6/18) In integrated form for a finite time interval, these expressions become (6/19) t2 t1 ΣF dt (G)system t2 t1 ΣMO dt (HO)system ΣMO (H ˙ O)a (H ˙ O)b … ΣF G ˙ a G ˙ b … H ˙ G ˙ ΣMO H ˙O and (HO)1 t2 t1 ΣMO dt (HO)2 HO IO I mr 2 (I mr 2).
r r v HO I mvd mv I, 488 Chapter 6 Plane Kinetics of Rigid Bodies O O G HG = I _ ω HG = I _ ω ω G = mv _ d v _ G (b) (a) ω G = mv _ v _ r _ Figure 6/14 c06.qxd 2/10/12 2:13 PM Page 488 We note that the equal and opposite actions and reactions in the connec-tions are internal to the system and cancel one another so they are not involved in the force and moment summations. Also, point O is one fixed reference point for the entire system.
Conservation of Momentum In Art. 4/5, we expressed the principles of conservation of momen-tum for a general mass system by Eqs. 4/15 and 4/16. These principles are applicable to either a single rigid body or a system of interconnected rigid bodies. Thus, if ΣF 0 for a given interval of time, then [4/15] which says that the linear-momentum vector undergoes no change in the absence of a resultant linear impulse. For the system of interconnected rigid bodies, there may be linear-momentum changes of individual parts of the system during the interval, but there will be no resultant momentum change for the system as a whole if there is no resultant linear impulse.
Similarly, if the resultant moment about a given fixed point O or about the mass center is zero during a particular interval of time for a single rigid body or for a system of interconnected rigid bodies, then [4/16] which says that the angular momentum either about the fixed point or about the mass center undergoes no change in the absence of a corre-sponding resultant angular impulse. Again, in the case of the intercon-nected system, there may be angular-momentum changes of individual components during the interval, but there will be no resultant angular-momentum change for the system as a whole if there is no resultant an-gular impulse about the fixed point or the mass center. Either of Eqs.
4/16 may hold without the other.
In the case of an interconnected system, the system center of mass is generally inconvenient to use.
As was illustrated previously in Articles 3/9 and 3/10 in the chapter on particle motion, the use of momentum principles greatly facilitates (HO)1 (HO)2 or (HG)1 (HG)2 G1 G2 Article 6/8 Impulse-Momentum Equations 489 F1 Ga Gb F6 F4 F5 F3 Ga = mav _ a Gb = mbv _ b F2 a b (HG)a = I _ a a ω (HG)b = I _ b b ω O Figure 6/15 c06.qxd 2/10/12 2:13 PM Page 489 the analysis of situations where forces and couples act for very short pe-riods of time.
Impact of Rigid Bodies Impact phenomena involve a fairly complex interrelationship of en-ergy and momentum transfer, energy dissipation, elastic and plastic de-formation, relative impact velocity, and body geometry. In Art. 3/12 we treated the impact of bodies modeled as particles and considered only the case of central impact, where the contact forces of impact passed through the mass centers of the bodies, as would always happen with colliding smooth spheres, for example. To relate the conditions after im-pact to those before impact required the introduction of the so-called co-efficient of restitution e or impact coefficient, which compares the relative separation velocity with the relative approach velocity mea-sured along the direction of the contact forces. Although in the classical theory of impact, e was considered a constant for given materials, more modern investigations show that e is highly dependent on geometry and impact velocity as well as on materials. At best, even for spheres and rods under direct central and longitudinal impact, the coefficient of restitution is a complex and variable factor of limited use.
Any attempt to extend this simplified theory of impact utilizing a coefficient of restitution for the noncentral impact of rigid bodies of varying shape is a gross oversimplification which has little practical value. For this reason, we do not include such an exercise in this book, even though such a theory is easily developed and appears in certain ref-erences. We can and do, however, make full use of the principles of con-servation of linear and angular momentum when they are applicable in discussing impact and other interactions of rigid bodies.
490 Chapter 6 Plane Kinetics of Rigid Bodies There are small reaction wheels inside the Hubble Space Telescope that make precision attitude control possible. The principles of angular momen-tum are fundamental to the design and operation of such a control system.
NASA, 2002 c06.qxd 2/10/12 2:14 PM Page 490 SAMPLE PROBLEM 6/14 The force P, which is applied to the cable wrapped around the central hub of the symmetrical wheel, is increased slowly according to P 1.5t, where P is in pounds and t is the time in seconds after P is first applied. Determine the angular velocity 2 of the wheel 10 seconds after P is applied if the wheel is rolling to the left with a velocity of its center of 3 ft/sec at time t 0. The wheel weighs 120 lb with a radius of gyration about its center of 10 in. and rolls without slipping.
Solution.
The impulse-momentum diagram of the wheel discloses the initial linear and angular momenta at time t1 0, all external impulses, and the final linear and angular momenta at time t2 10 sec. The correct direction of the fric-tion force F is that to oppose the slipping which would occur without friction.
Application of the linear impulse-momentum equation and the angular impulse-momentum equation over the entire interval gives Since the force F is variable, it must remain under the integral sign. We eliminate F between the two equations by multiplying the second one by and adding to the first one. Integrating and solving for 2 give Ans.
Alternative Solution.
We could avoid the necessity of a simultaneous solution by applying the second of Eqs. 6/17 about a fixed point O on the horizontal surface. The moments of the 120-lb weight and the equal and opposite force N cancel one an-other, and F is eliminated since its moment about O is zero. Thus, the angular mo-mentum about O becomes HO mr2 where is the centroidal radius of gyration and r is the 18-in. rolling radius. Thus, we see that HO HC since r2 and HC IC Equation 6/17 now gives Solution of this one equation is equivalent to the simultaneous solution of the two previous equations.
120 32.2 10 12 2 18 12 2 2 120 32.2 10 12 2 18 12 2 3 18/12 10 0 1.5t 18 9 12 dt (HO)1 t2 t1 ΣMO dt (HO)2 mkC 2.
kC 2 k2 k m(k2 r2), mk2 mvr I 2 3.13 rad/sec clockwise 12 18 120 32.2 10 12 2 3 18/12 10 0 18 12 F 9 12 (1.5t) dt 120 32.2 10 12 22 (HG)1 t2 t1 ΣMG dt (HG)2 (Gx)1 t2 t1 ΣFx dt (Gx)2 120 32.2 (3) 10 0 (1.5t F) dt 120 32.2 18 12 2 Article 6/8 Impulse-Momentum Equations 491 G P v _ 1 = 3 ft/sec 9″ 18″ mg dt N dt F dt P dt + = G C O ω I _ 1 ω I _ 2 mv _ 1 t1 = 0 t2 = 10 sec mv _ 2 y x + G 18″ 9″ G Helpful Hints Also, we note the clockwise imbal-ance of moments about C, which causes a clockwise angular accelera-tion as the wheel rolls without slip-ping. Since the moment sum about G must also be in the clockwise sense of , the friction force must act to the left to provide it.
Note carefully the signs of the mo-mentum terms. The final linear ve-locity is assumed in the positive x-direction, so is positive. The initial linear velocity is negative, so is negative.
Since the wheel rolls without slip-ping, a positive x-velocity requires a clockwise angular velocity, and vice versa.
(Gx)1 (Gx)2 c06.qxd 2/10/12 2:14 PM Page 491 SAMPLE PROBLEM 6/15 The sheave E of the hoisting rig shown has a mass of 30 kg and a centroidal radius of gyration of 250 mm. The 40-kg load D which is carried by the sheave has an initial downward velocity v1 1.2 m/s at the instant when a clockwise torque is applied to the hoisting drum A to maintain essentially a constant force F 380 N in the cable at B. Compute the angular velocity 2 of the sheave 5 sec-onds after the torque is applied to the drum and find the tension T in the cable at O during the interval. Neglect all friction.
Solution.
The load and the sheave taken together constitute the system, and its impulse-momentum diagram is shown. The tension T in the cable at O and the final angular velocity 2 of the sheave are the two unknowns. We eliminate T ini-tially by applying the moment-angular-momentum equation about the fixed point O, taking counterclockwise as positive.
Substituting into the momentum equation gives Ans.
The linear-impulse-momentum equation is now applied to the system to de-termine T. With the positive direction up, we have Ans.
If we had taken our moment equation around the center C of the sheave in-stead of point O, it would contain both unknowns T and , and we would be obliged to solve it simultaneously with the foregoing force equation, which would also contain the same two unknowns.
5T 1841 T 368 N 70(1.2) 5 0 [T 380 70(9.81)] dt 70[0.375(8.53)] G1 t2 t1 ΣF dt G2 2 8.53 rad/s counterclockwise 37.5 137.4 11.722 11.722 (30 40)(0.3752)(0.375) 30(0.250)22 (HO)2 (mE mD)v2d I2 37.5 Nms (30 40)(1.2)(0.375) 30(0.250)2 1.2 0.375 (HO)1 (mE mD)v1d I1 137.4 N ms t2 t1 ΣMO dt 5 0 [380(0.750) (30 40)(9.81)(0.375)] dt (HO)1 t2 t1 ΣMO dt (HO)2 492 Chapter 6 Plane Kinetics of Rigid Bodies O A B E 375 mm v1 = 1.2 m/s C D O C F dt T dt mtot g dt + = mtotv1 mtotv2 375 mm C C ω I _ 1 ω I _ 2 Helpful Hint The units of angular momentum, which are those of angular impulse, may also be written as kgm2/s.
c06.qxd 2/10/12 2:14 PM Page 492 SAMPLE PROBLEM 6/16 The uniform rectangular block of dimensions shown is sliding to the left on the horizontal surface with a velocity v1 when it strikes the small step at O. As-sume negligible rebound at the step and compute the minimum value of v1 which will permit the block to pivot freely about O and just reach the standing position A with no velocity. Compute the percentage energy loss n for b c.
Solution.
We break the overall process into two subevents: the collision (I) and the subsequent rotation (II).
I. Collision.
With the assumption that the weight mg is nonimpulsive, angu-lar momentum about O is conserved. The initial angular momentum of the block about O just before impact is the moment about O of its linear momentum and is mv1(b/2). The angular momentum about O just after impact when the block is starting its rotation about O is Conservation of angular momentum gives II. Rotation about O.
With the assumptions that the rotation is like that about a fixed frictionless pivot and that the location of the effective pivot O is at ground level, mechanical energy is conserved during the rotation according to Ans.
The percentage loss of energy during the impact is Ans.
1 3 41 c2 b2 n 62.5% for b c n E E 1 2 mv1 2 1 2 IO2 2 1 2mv1 2 1 kO 22 2 v1 2 1 b2 c2 3 3b 2(b2 c2) 2 [(HO)1 (HO)2] mv1 b 2 m 3 (b2 c2)2 2 3v1b 2(b2 c2) m 3 (b2 c2)2 [HO IO] (HO)1 Article 6/8 Impulse-Momentum Equations 493 O A b c v1 O x dt mv1 mg dt O y dt + = G O _ r c b G O mv2 G O IO 2 ω Helpful Hints If the corner of the block struck a spring instead of the rigid step, then the time of the interaction during compression of the spring could be-come appreciable, and the angular impulse about the fixed point at the end of the spring due to the moment of the weight would have to be ac-counted for.
Notice the abrupt change in direc-tion and magnitude of the velocity of G during the impact.
Be sure to use the transfer theorem IO correctly here.
The datum is taken at the initial al-titude of the mass center G. State 3 is taken to be the standing position A, at which the diagonal of the block is vertical.
mr 2 I c06.qxd 2/10/12 2:14 PM Page 493 6/167 The 75-kg flywheel has a radius of gyration about its shaft axis of and is subjected to the torque where t is in seconds.
If the flywheel is at rest at time determine its angular velocity at Problem 6/167 6/168 The constant tensions of 200 N and 160 N are ap-plied to the hoisting cable as shown. If the velocity v of the load is 2 m/s down and the angular velocity of the pulley is 8 rad/s counterclockwise at time determine v and after the cable tensions have been applied for 5 s. Note the independence of the results.
Problem 6/168 15 kg k = 250 mm 300 mm 200 N 160 N 20 kg t 0, M t 3 s.
t 0, M 10(1 et) Nm, k 0.50 m 494 Chapter 6 Plane Kinetics of Rigid Bodies PROBLEMS Introductory Problems 6/165 The mass center G of the slender bar of mass 0.8 kg and length 0.4 m is falling vertically with a velocity at the instant depicted. Calculate the an-gular momentum of the bar about point O if the angular velocity of the bar is (a) clockwise and (b) counterclockwise.
Problem 6/165 6/166 The grooved drums in the two systems shown are identical. In both cases, (a) and (b), the system is at rest at time Determine the angular velocity of each grooved drum at time Neglect fric-tion at the pivot O.
Problem 6/166 m m = 14 kg, k = 225 mm ro = 325 mm, ri = 215 mm _ 10 kg ri ro m 10(9.81) N ri ro O (a) (b) t 4 s.
t 0.
θ ω O G v 0.4 m a ωb 0.3 m b 10 rad/s a 10 rad/s HO v 2 m/s c06.qxd 2/10/12 2:14 PM Page 494 6/169 Determine the angular momentum of the earth about the center of the sun. Assume a homoge-neous earth and a circular earth orbit of radius 149.6 consult Table D/2 for other needed information. Comment on the relative contribu-tions of the terms and Problem 6/169 6/170 The constant 9-lb force is applied to the 80-lb stepped cylinder as shown. The centroidal radius of gyration of the cylinder is and it rolls on the incline without slipping. If the cylinder is at rest when the force is first applied, determine its angu-lar velocity eight seconds later.
Problem 6/170 6/171 The frictional moment acting on a rotating tur-bine disk and its shaft is given by where is the angular velocity of the turbine. If the source of power is cut off while the turbine is run-ning with an angular velocity determine the time t for the speed of the turbine to drop to half of its initial value. The moment of inertia of the tur-bine disk and shaft is I.
0, M ƒ k2 Mƒ 6″ 10″ 10° 9 lb k 8 in., z x y N ω Sunlight mv d.
I (106) km; Article 6/8 Problems 495 6/172 The man is walking with speed to the right when he trips over a small floor discontinuity.
Estimate his angular velocity just after the im-pact. His mass is 76 kg with center-of-mass height and his mass moment of inertia about the ankle joint O is 66 kg where all are proper-ties of the portion of his body above O; i.e., both the mass and moment of inertia do not include the foot.
Problem 6/172 6/173 Repeat the previous problem, only now the man carries a 10-kg backpack as shown. Develop a gen-eral expression for the angular velocity of the man just after impact with the small step. Evaluate your expression for the backpack center-of-mass positions (a) and and (b) Case (b) is the condition of a beltpack. The mass conditions for the man remain unchanged from the previous problem. State any assumptions and com-pare your results with those from the previous problem.
Problem 6/173 v1 h d hB O G d hB 0.
hB 0.3 m d 0.2 m v1 h G O m2, h 0.87 m, v1 1.2 m/s c06.qxd 2/10/12 2:14 PM Page 495 6/177 The wad of clay of mass m is initially moving with a horizontal velocity when it strikes and sticks to the initially stationary uniform slender bar of mass M and length L. Determine the final angular veloc-ity of the combined body and the x-component of the linear impulse applied to the body by the pivot O during the impact.
Problem 6/177 Representative Problems 6/178 The uniform rectangular panel is falling vertically with speed when its small peg A engages in the receptacle. Determine the angular velocity of the body as well as the x- and y-components of its mass-center velocity just after the impact.
Problem 6/178 b – – 2 7b – – – 8 b – – 8 G A v1 x y v1 x y O v1 M m L –– 3 L –– 3 L –– 3 v1 496 Chapter 6 Plane Kinetics of Rigid Bodies 6/174 A uniform slender bar of mass M and length L is translating on the smooth horizontal x-y plane with a velocity when a particle of mass m traveling with a velocity as shown strikes and becomes embedded in the bar. Determine the final linear and angular velocities of the bar with its embedded particle.
Problem 6/174 6/175 The initially stationary uniform disk of mass and radius b is allowed to drop onto the moving belt from a very small elevation. Determine the time t required for the disk to acquire its steady-state angular speed. The belt drive pulley rotates with a constant counterclockwise velocity .
Problem 6/175 6/176 Repeat the previous problem if the belt drive pulley rotates clockwise with a constant angular velocity .
3b ω O A r m2 b k m1 μ m1 vm vM M x y 3L — 4 L — 4 m vm vM c06.qxd 2/10/12 2:14 PM Page 496 6/179 Just after leaving the platform, the diver’s fully ex-tended 80-kg body has a rotational speed of 0.3 rev/s about an axis normal to the plane of the trajectory.
Estimate the angular velocity N later in the dive when the diver has assumed the tuck position.
Make reasonable assumptions concerning the mass moment of inertia of the body in each configuration.
Problem 6/179 6/180 The slender rod of mass and length L has a movable slider of mass which can be tightened at any location x along the rod. The assembly is initially falling in translation with speed A small peg on the left end of the rod becomes engaged in the receptacle. Determine the angular velocity of the body just after impact. For the condition determine the maximum value of and the corresponding value of x. Plot versus x/L for this mass condition.
Problem 6/180 6/181 A cylindrical shell of 400-mm diameter and mass m is rotating about its central horizontal axis with an angular velocity when it is released onto a horizontal surface with no velocity of its cen-ter If slipping between the shell and the surface occurs for 1.5 s, calculate the coefficient of kinetic friction and the maximum velocity v reached by the center of the shell.
k (v0 0).
0 30 rad/s A v1 x m1 m2 2 2 m2 m1/2, 2 v1.
m2 m1 0.7 m 0.3 rev/s 2 m N Article 6/8 Problems 497 Problem 6/181 6/182 Two small variable-thrust jets are actuated to keep the spacecraft angular velocity about the z-axis con-stant at as the two telescoping booms are extended from at a constant rate over a 2-min period. Determine the necessary thrust T for each jet as a function of time where is the time when the telescoping action is begun. The small 10-kg experiment mod-ules at the ends of the booms may be treated as par-ticles, and the mass of the rigid booms is negligible.
Problem 6/182 t 0 r1 1.2 m to r2 4.5 m 0 1.25 rad/s 400 mm v0 ω0 G T r r T z 10 kg 10 kg 1.1 m 1.1 m ω0 c06.qxd 2/10/12 2:14 PM Page 497 6/185 The base B has a mass of 5 kg and a radius of gyration of 80 mm about the central vertical axis shown. Each plate P has a mass of 3 kg. If the sys-tem is freely rotating about the vertical axis with an angular speed with the plates in the vertical position, estimate the angular speed when the plates have moved to the horizontal positions indicated. Neglect friction.
Problem 6/185 6/186 In the initial position shown, the disk of axial mass moment of inertia rotates freely with angular velocity relative to the lightweight frame. The turntable of axial mass moment of inertia rotates freely with angular velocity Then the axis of the disk is turned through an angular displacement What is the resulting angular velocity of the turntable? Assume that the thickness of the disk is sufficiently small so that its axial moment of inertia can be approximated by twice its transverse moment of inertia.
2 90.
1.
It Id N2 N1 10 rev/min 498 Chapter 6 Plane Kinetics of Rigid Bodies 6/183 With the gears initially at rest and the couple M equal to zero, the forces exerted by the frame on the shafts of the gears at A and B are 30 and 16 lb, re-spectively, both upward to support the weights of the two gears. A couple is now ap-plied to the larger gear through its shaft at A. After 4 sec the larger gear has a clockwise angular mo-mentum of 12 ft-lb-sec, and the smaller gear has a counterclockwise angular momentum of 4 ft-lb-sec.
Calculate the new values of the forces and ex-erted by the frame on the shafts during the 4-sec in-terval. Isolate the two gears together as the system.
Problem 6/183 6/184 The phenomenon of vehicle “tripping” is investi-gated here. The sport-utility vehicle is sliding side-ways with speed and no angular velocity when it strikes a small curb. Assume no rebound of the right-side tires and estimate the minimum speed which will cause the vehicle to roll completely over to its right side. The mass of the SUV is 2300 kg and its mass moment of inertia about a longitudi-nal axis through the mass center G is Problem 6/184 v1 760 mm G 880 mm 880 mm 900 kg m2.
v1 v1 A B 10″ 16″ M RB RA M 60 lb-in.
P B N 300 mm 100 mm 100 mm 120 mm 120 mm P c06.qxd 2/10/12 2:14 PM Page 498 Problem 6/186 6/187 The system is initially rotating freely with angular velocity rad/s when the inner rod A is cen-tered lengthwise within the hollow cylinder B as shown in the figure. Determine the angular velocity of the system (a) if the inner rod A has moved so that a length b/2 is protruding from the cylinder, (b) just before the rod leaves the cylinder, and (c) just after the rod leaves the cylinder. Ne-glect the moment of inertia of the vertical support shafts and friction in the two bearings. Both bodies are constructed of the same uniform material. Use the values and and refer to the results of Prob. B/37 as needed.
Problem 6/187 B 2r r A b – – 4 b – – 4 b – – 4 b – – 4 ω r 20 mm, b 400 mm 1 10 Ω θ ω Article 6/8 Problems 499 6/188 The homogeneous sphere of mass m and radius r is projected along the incline of angle with an initial speed and no angular velocity If the co-efficient of kinetic friction is determine the time duration t of the period of slipping. In addition, state the velocity v of the mass center G and the an-gular velocity at the end of the period of slipping.
Problem 6/188 6/189 The homogeneous sphere of Prob. 6/188 is placed on the incline with a clockwise angular velocity but no linear velocity of its center Deter-mine the time duration t of the period of slipping.
In addition, state the velocity v and angular veloc-ity at the end of the period of slipping.
6/190 The 165-lb ice skater with arms extended horizon-tally spins about a vertical axis with a rotational speed of 1 rev/sec. Estimate his rotational speed N if he fully retracts his arms, bringing his hands very close to the centerline of his body. As a reasonable approximation, model the extended arms as uni-form slender rods, each of which is 27 in. long and weighs 15 lb. Model the torso as a solid 135-lb cylin-der 13 in. in diameter. Treat the man with arms re-tracted as a solid 165-lb cylinder of 13-in. diameter.
Neglect friction at the skate–ice interface.
Problem 6/190 27′′ 1 rev/sec 13′′ (v0 0).
0 θ ω0 G v0 m r μk k, (0 0).
v0 c06.qxd 2/10/12 2:14 PM Page 499 Problem 6/192 6/193 A 55-kg dynamics instructor is demonstrating the principles of angular momentum to her class. She stands on a freely rotating platform with her body aligned with the vertical platform axis. With the platform not rotating, she holds a modified bicycle wheel so that its axis is vertical. She then turns the wheel axis to a horizontal orientation without changing the 600-mm distance from the centerline of her body to the wheel center, and her students observe a platform rotation rate of 30 rev/min. If the rim-weighted wheel has a mass of 10 kg and a centroidal radius of gyration and is spinning at a fairly constant rate of 250 rev/min, estimate the mass moment of inertia I of the in-structor (in the posture shown) about the vertical platform axis.
Problem 6/193 Ω 600 mm k 300 mm, 500 Chapter 6 Plane Kinetics of Rigid Bodies 6/191 The elements of a spacecraft with axial mass sym-metry and a reaction-wheel control system are shown in the figure. When the motor exerts a torque on the reaction wheel, an equal and opposite torque is exerted on the spacecraft, thereby changing its angular momentum in the z-direction. If all system elements start from rest and the motor exerts a con-stant torque M for a time period t, determine the final angular velocity of (a) the spacecraft and (b) the wheel relative to the spacecraft. The mass moment of inertia about the z-axis of the entire spacecraft, including the wheel, is I and that of the wheel alone is The spin axis of the wheel is coincident with the z-axis of symmetry of the spacecraft.
Problem 6/191 6/192 The body of the spacecraft weighs 322 lb on earth and has a radius of gyration about its z-axis of 1.5 ft.
Each of the two solar panels may be treated as a uniform flat plate weighing 16.1 lb. If the spacecraft is rotating about its z-axis at the angular rate of 1.0 rad/sec with determine the angular rate after the panels are rotated to the position by an internal mechanism. Neglect the small momentum change of the body about the y-axis.
/2 0, I Iw M z Reaction wheel Motor Iw.
8′ 8′ 6′ 2′ 6′ 2′ 2′ 2′ z x y θ ω c06.qxd 2/10/12 2:14 PM Page 500 6/194 If the dynamics instructor of Prob. 6/193 reorients the wheel axis by 180 with respect to its initial ver-tical position, what rotational speed N will her stu-dents observe? All the given information and the result of Prob. 6/193 may be utilized.
6/195 The slotted circular disk whose mass is 6 kg has a radius of gyration about O of 175 mm. The disk car-ries the four steel balls, each of mass 0.15 kg and lo-cated as shown, and rotates freely about a vertical axis through O with an angular speed of 120 rev/min. Each of the small balls is held in place by a latching device not shown. If the balls are released while the disk is rotating and come to rest relative to the disk at the outer ends of the slots, compute the new angular velocity of the disk. Also find the magnitude of the energy loss due to the impact of the balls with the ends of the slots. Neglect the di-ameter of the balls and discuss this approximation.
Problem 6/195 6/196 A uniform pole of length L, inclined at an angle with the vertical, is dropped and both ends have a velocity v as end A hits the ground. If end A pivots about its contact point during the remainder of the motion, determine the velocity with which end B hits the ground.
Problem 6/196 L A B θ v v v′ v O 120 rev/min 100 mm 200 mm E I 3.45 kg m2 Article 6/8 Problems 501 6/197 The 17.5-Mg lunar landing module with center of mass at G has a radius of gyration of 1.8 m about G. The module is designed to contact the lunar sur-face with a vertical free-fall velocity of 8 km/h. If one of the four legs hits the lunar surface on a small incline and suffers no rebound, compute the angular velocity of the module immediately after impact as it pivots about the contact point. The 9-m dimension is the distance across the diagonal of the square formed by the four feet as corners.
Problem 6/197 6/198 A uniform circular disk which rolls with a velocity v without slipping encounters an abrupt change in the direction of its motion as it rolls onto the incline . Determine the new velocity of the center of the disk as it starts up the incline, and find the fraction n of the initial energy which is lost because of impact with the incline if Problem 6/198 r G v θ 10.
v 8 km/h 9 m G 3 m c06.qxd 2/10/12 2:14 PM Page 501 6/200 A frozen-juice can rests on the horizontal rack of a freezer door as shown. With what maximum angu-lar velocity can the door be “slammed” shut against its seal and not dislodge the can? Assume that the can rolls without slipping on the corner of the rack, and neglect the dimension d compared with the 500-mm distance.
Problem 6/200 35 mm 7 mm Ω d 500 mm 502 Chapter 6 Plane Kinetics of Rigid Bodies 6/199 Determine the minimum velocity v which the wheel must have to just roll over the obstruction. The centroidal radius of gyration of the wheel is k, and it is assumed that the wheel does not slip.
Problem 6/199 r h v c06.qxd 2/10/12 2:14 PM Page 502 6/9 Chapter Review In Chapter 6 we have made use of essentially all the elements of dy-namics studied so far. We noted that a knowledge of kinematics, using both absolute- and relative-motion analysis, is an essential part of the so-lution to problems in rigid-body kinetics. Our approach in Chapter 6 par-alleled Chapter 3, where we developed the kinetics of particles using force-mass-acceleration, work-energy, and impulse-momentum methods.
The following is a summary of the important considerations in the solution of rigid-body kinetics problems in plane motion: 1. Identification of the body or system. It is essential to make an unambiguous decision as to which body or system of bodies is to be analyzed and then isolate the selected body or system by drawing the free-body and kinetic diagrams, the active-force diagram, or the impulse-momentum diagram, whichever is appropriate.
2. Type of motion. Next identify the category of motion as rectilin-ear translation, curvilinear translation, fixed-axis rotation, or gen-eral plane motion. Always make sure that the kinematics of the problem is properly described before attempting to solve the kinetic equations.
3. Coordinate system. Choose an appropriate coordinate system.
The geometry of the particular motion involved is usually the decid-ing factor. Designate the positive sense for moment and force sum-mations and be consistent with the choice.
4. Principle and method. If the instantaneous relationship between the applied forces and the acceleration is desired, then the equiva-lence between the forces and their and resultants, as dis-closed by the free-body and kinetic diagrams, will indicate the most direct approach to a solution.
When motion occurs over an interval of displacement, the work-energy approach is indicated, and we relate initial to final velocities without calculating the acceleration. We have seen the advantage of this approach for interconnected mechanical systems with negligi-ble internal friction.
If the interval of motion is specified in terms of time rather than displacement, the impulse-momentum approach is indicated.
When the angular motion of a rigid body is suddenly changed, the principle of conservation of angular momentum may apply.
5. Assumptions and approximations. By now you should have ac-quired a feel for the practical significance of certain assumptions and approximations, such as treating a rod as an ideal slender bar and neglecting friction when it is minimal. These and other ideal-izations are important to the process of obtaining solutions to real problems.
I ma Article 6/9 Chapter Review 503 c06.qxd 2/10/12 2:14 PM Page 503 Problem 6/203 6/204 Each of the solid circular disk wheels has a mass of 2 kg, and the inner solid cylinder has a mass of 3 kg.
The disks and cylinder are mounted on the small central shaft so that each can rotate independently of the other with negligible friction in the bearings.
Calculate the acceleration of the center of the wheels when the 20-N force is applied as shown. The coeffi-cients of friction between the wheels and the hori-zontal surface are Problem 6/204 75 mm 150 mm 20 N 20 N k 0.30 and s 0.40.
A O N B B′ 110 mm 160 mm 504 Chapter 6 Plane Kinetics of Rigid Bodies REVIEW PROBLEMS 6/201 The force P is applied to the homogeneous crate of mass m. If the coefficient of kinetic friction between the crate and the horizontal platform is deter-mine the limiting values of h so that the crate will slide without tipping about either the front edge or the rear edge.
Problem 6/201 6/202 A person who walks through the revolving door ex-erts a 90-N horizontal force on one of the four door panels. If each panel is modeled by a 60-kg uniform rectangular plate which is 1.2 m in length as viewed from above, determine the angular acceleration of the door unit. Neglect friction.
Problem 6/202 6/203 The preliminary design of a unit for automatically reducing the speed of a freely rotating assembly is shown. Initially the unit is rotating freely about a vertical axis through O at a speed of 600 rev/min with the arms secured in the positions shown by AB. When the arms are released, they swing out-ward and become latched in the dashed positions shown. The disk has a mass of 30 kg with a radius of gyration of 90 mm about O. Each arm has a length of 160 mm and a mass of 0.84 kg and may be treated as a uniform slender rod. Determine the new speed N of rotation and calculate the loss of energy of the system. Would the results be affected by either the direction of rotation or the sequence of release of the rods? E O 0.4 m 0.8 m 15° 90 N c b h k P μ k, c06.qxd 2/10/12 2:14 PM Page 504 6/205 A slender rod of mass and length l is welded at its midpoint A to the rim of the solid circular disk of mass m and radius r. The center of the disk, which rolls without slipping, has a velocity v at the instant when A is at the top of the disk with the rod parallel to the ground. For this instant determine the angular momentum of the combined body about O.
Problem 6/205 6/206 The uniform slender rod of mass m and length l is freely hinged about a horizontal axis through its end O and is given an initial angular velocity as it crosses the vertical position where If the rod swings through a maximum angle de-rive an expression in integral form for the time t from release at is reached. (Ex-press in terms of ) Problem 6/206 r θ β ω0 ω = ω = 0 l O · θ .
0 0 until 90, 0.
0 r A O l v m0 Article 6/9 Review Problems 505 6/207 The uniform rectangular block with the given di-mensions is dropped from rest from the position shown. Corner A strikes the ledge at B and becomes latched to it. Determine the angular velocity of the block immediately after it becomes attached to B. Also find the percentage n of energy loss during the corner attachment for the case Problem 6/207 6/208 Four identical slender rods each of mass m are welded at their ends to form a square, and the cor-ners are then welded to a light metal hoop of radius r.
If the rigid assembly of rods and hoop is allowed to roll down the incline, determine the minimum value of the coefficient of static friction which will prevent slipping.
Problem 6/208 r θ h c b B A b c.
c06.qxd 2/10/12 2:14 PM Page 505 6/211 The small block of mass m slides along the radial slot of the disk while the disk rotates in the hori-zontal plane about its center O. The block is re-leased from rest relative to the disk and moves outward with an increasing velocity along the slot as the disk turns. Determine the expression in terms of r and for the torque M that must be ap-plied to the disk to maintain a constant angular ve-locity of the disk.
Problem 6/211 6/212 The forklift truck with center of mass at has a weight of 3200 lb including the vertical mast. The fork and load have a combined weight of 1800 lb with center of mass at The roller guide at B is capable of supporting horizontal force only, whereas the connection at C, in addition to sup-porting horizontal force, also transmits the vertical elevating force. If the fork is given an upward accel-eration which is sufficient to reduce the force under the rear wheels at A to zero, calculate the corre-sponding reaction at B.
Problem 6/212 2′ 4′ 4′ A C B G2 G1 a 3′ 2′ G2.
G1 O r M m ω r ˙ r ˙ 506 Chapter 6 Plane Kinetics of Rigid Bodies 6/209 A couple is applied at C to the spring-toggle mechanism, which is released from rest in the position In this position the spring, which has a stiffness of 140 N/m, is stretched 150 mm. Bar AB has a mass of 3 kg and BC a mass of 6 kg. Calculate the angular velocity of BC as it crosses the position Motion is in the vertical plane, and friction is negligible.
Problem 6/209 6/210 The link OA and pivoted circular disk are released from rest in the position shown and swing in the vertical plane about the fixed bearing at O. The 6-kg link OA has a radius of gyration about O of 375 mm.
The disk has a mass of 8 kg. The two bearings are assumed to be frictionless. Find the force exerted at O on the link (a) just after release and (b) as OA swings through the vertical position Problem 6/210 250 mm G A O A′ 600 mm 300 mm OA.
FO θ A B C M 250 mm 250 mm 250 mm 0.
45.
M 12 N m c06.qxd 2/10/12 2:14 PM Page 506 6/213 A space telescope is shown in the figure. One of the reaction wheels of its attitude-control system is spinning as shown at 10 rad/s, and at this speed the friction in the wheel bearing causes an internal mo-ment of Both the wheel speed and the friction moment may be considered constant over a time span of several hours. If the mass moment of inertia of the entire spacecraft about the x-axis is determine how much time passes before the line of sight of the initially stationary spacecraft drifts by 1 arc-second, which is 1/3600 degree. All other elements are fixed relative to the spacecraft, and no torquing of the reaction wheel shown is performed to correct the attitude drift.
Neglect external torques.
Problem 6/213 6/214 Each of the solid square blocks is allowed to fall by rotating clockwise from the rest positions shown.
The support at in case is a hinge and in case is a small roller. Determine the angular velocity of each block as edge becomes horizontal just before striking the supporting surface.
Problem 6/214 A O O 250 mm 250 mm 45° 45° Hinge Roller C A 250 mm 250 mm C (a) (b) OC (b) (a) O x G 150(103) kgm2, 106 Nm.
Article 6/9 Review Problems 507 6/215 The mechanical flyball governor operates with a vertical shaft As the shaft speed is in-creased, the rotational radius of the two balls tends to increase, and the weight A is lifted up by the collar Determine the steady-state value of for a rotational speed of 150 rev/min.
Neglect the mass of the arms and collar.
Problem 6/215 6/216 In an acrobatic stunt, man of mass drops from a raised platform onto the end of the light but strong beam with a velocity The boy of mass is propelled upward with a velocity For a given ratio determine in terms of to maxi-mize the upward velocity of the boy. Assume that both man and boy act as rigid bodies.
Problem 6/216 L O B A b v0 L b n mB/mA vB.
mB v0.
mA A 3 lb 20 lb 3 lb O O 1″ 1″ 1″ 1″ 4″ 4″ 2″ B N β β A B.
20-lb 3-lb N O-O.
c06.qxd 2/10/12 2:14 PM Page 507 6/219 Before it hits the ground a falling chimney, such as the one shown, will usually crack at the point where the bending moment is greatest. Show that the position of maximum moment occurs at the center of percussion relative to the upper end for a slender chimney of constant cross section. Neglect any restraining moment at the base.
Problem 6/219 6/220 The two slender bars, each having a mass of 4 kg, are hinged at and pivoted at If a horizontal im-pulse is applied to the end of the lower bar during an interval of during which the bars are still essentially in their vertical rest positions, compute the angular velocity of the upper bar immediately after the impulse.
Problem 6/220 C B A F 1.2 m 1.2 m 2 0.1 s A F dt 14 N s C.
B 508 Chapter 6 Plane Kinetics of Rigid Bodies 6/217 The small block of mass slides in the smooth radial slot of the disk, which turns freely in its bear-ing. If the block is displaced slightly from the center position when the angular velocity of the disk is determine its radical velocity as a function of the radical distance The mass moment of inertia of the disk about its axis of rotation is Problem 6/217 6/218 The pendulum with mass center at is piv-oted at A to the fixed support It has a radius of gyration of 17 in. about and swings through an amplitude For the instant when the pendulum is in the extreme position, calculate the moments applied by the base sup-port to the column at Problem 6/218 C 20″ 8″ x y z O A O G θ 16″ C.
Mx, My, and Mz 60.
O-O CA.
G 6-lb 0 rd m IO r ω IO.
r.
vr 0, m c06.qxd 2/10/12 2:14 PM Page 508 Computer-Oriented Problems 6/221 The system of Prob. 6/120 is repeated here. If the uniform slender bar is released from rest in the position where the spring is un-stretched, determine and plot its angular velocity as a function of over the range where is the value of at which the bar mo-mentarily comes to rest. The value of the spring constant k is 100 N/m, and friction can be ne-glected. State the maximum angular speed and the value of at which it occurs.
Problem 6/221 6/222 The crate slides down the incline with velocity and its corner strikes a small obstacle at Deter-mine the minimum required velocity if the crate is to rotate about A so that it travels on the con-veyor belt on its side as indicated in the figure.
Plot the variation of with for Problem 6/222 6/223 The uniform 4-ft slender bar with light end rollers is released from rest in the vertical plane with essentially zero. Determine and graph the velocity of A as a function of and find the maximum ve-locity of A and the corresponding angle .
2′ 2′ α v1 A 0 45.
v1 v1 A.
v1 O 1.2 kg A B k 0.6 m 0.6 m 0.2 m θ max 0 max, 0 1.2-kg Article 6/9 Review Problems 509 Problem 6/223 6/224 The cart moves to the right with acceleration If and determine the steady-state angular de-flection of the uniform slender rod of mass Treat the small end sphere of mass as a particle.
The spring, which exerts a moment of magnitude on the rod, is undeformed when the rod is vertical.
Problem 6/224 l a B K O m 3m θ M K m 3m.
N m/rad, K 75 m 0.5 kg, l 0.6 m, a 2g.
B A B 4′ θ c06.qxd 2/10/12 2:14 PM Page 509 6/227 The uniform slender bar has a par-ticle attached to its end. The spring constant is and the distance If the bar is released from rest in the horizontal position shown where the spring is unstretched, determine the maximum angular deflection of the bar.
Also determine the value of the angular velocity at Neglect friction.
Problem 6/227 6/228 The uniform 100-kg beam is hanging initially at rest with when the constant force is applied to the cable. Determine (a) the maximum angular velocity reached by the beam with the cor-responding angle and the maximum angle reached by the beam.
Problem 6/228 3 m P C A B θ 1 m 3 m max (b) P 300 N 0 AB b –– 4 3b –– 4 b –– 4 O A 1.2 kg C B k 0.6 kg max/2.
max b 200 mm.
k 300 N/m 0.6-kg 1.2-kg 510 Chapter 6 Plane Kinetics of Rigid Bodies 6/225 The steel I-beam is to be transported by the over-head trolley to which it is hinged at If the trol-ley starts from rest with and is given a constant horizontal acceleration find the maximum values of and The magnitude of the initial swing would constitute a shop safety consideration.
Problem 6/225 6/226 The uniform power pole of mass and length is hoisted into a vertical position with its lower end supported by a fixed pivot at The guy wires sup-porting the pole are accidentally released, and the pole falls to the ground. Plot the x- and y-compo-nents of the force exerted on the pole at in terms of from Can you explain why increases again after going to zero?
Problem 6/226 θ L O y x Oy 0 to 90.
O O.
L m 4 m O a θ .
˙ a 2 m/s2, ˙ 0 O.
c06.qxd 2/10/12 2:14 PM Page 510 6/229 The 30-kg slender bar has an initial angular ve-locity in the vertical position, where the spring is unstretched. Determine the mini-mum angular velocity reached by the bar and the corresponding angle Also find the angular velocity of the bar as it strikes the hori-zontal surface.
Problem 6/229 1.2 m k = 3 kN/m 1.2 m 1.2 m A O ω 0 ω θ .
min 0 4 rad/s Article 6/9 Review Problems 511 6/230 The 60-ft telephone pole of essentially uniform diameter is being hoisted into the vertical position by two cables attached at as shown. The end rests on a fixed support and cannot slip. When the pole is nearly vertical, the fitting at suddenly breaks, releasing both cables. When the angle reaches the speed of the upper end A of the pole is 4.5 ft/sec. From this point, calculate the time which the workman would have to get out of the way before the pole hits the ground. With what speed does end A hit the ground?
Problem 6/230 O x y z A 60′ θ B vA t 10, B O B c06.qxd 2/10/12 2:14 PM Page 511 By proper management of the hydraulic cylinders which support and move this flight simulator, a variety of three-dimensional translational and rotational accelerations can be produced.
Angelo Giampiccolo/PhotoResearchers, Inc.
513 7/1 Introduction Although a large percentage of dynamics problems in engineering can be solved by the principles of plane motion, modern developments have focused increasing attention on problems which call for the analy-sis of motion in three dimensions. Inclusion of the third dimension adds considerable complexity to the kinematic and kinetic relationships. Not only does the added dimension introduce a third component to vectors which represent force, linear velocity, linear acceleration, and linear momentum, but the introduction of the third dimension also adds the possibility of two additional components for vectors representing angu-lar quantities including moments of forces, angular velocity, angular ac-celeration, and angular momentum. It is in three-dimensional motion that the full power of vector analysis is utilized.
A good background in the dynamics of plane motion is extremely useful in the study of three-dimensional dynamics, where the approach to problems and many of the terms are the same as or analogous to those in two dimensions. If the study of three-dimensional dynamics is under-taken without the benefit of prior study of plane-motion dynamics, more 7/1 Introduction Section A Kinematics 7/2 Translation 7/3 Fixed-Axis Rotation 7/4 Parallel-Plane Motion 7/5 Rotation about a Fixed Point 7/6 General Motion Section B Kinetics 7/7 Angular Momentum 7/8 Kinetic Energy 7/9 Momentum and Energy Equations of Motion 7/10 Parallel-Plane Motion 7/11 Gyroscopic Motion: Steady Precession 7/12 Chapter Review CHAPTER OUTLINE 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies time will be required to master the principles and to become familiar with the approach to problems.
The treatment presented in Chapter 7 is not intended as a complete development of the three-dimensional motion of rigid bodies but merely as a basic introduction to the subject. This introduction should, how-ever, be sufficient to solve many of the more common problems in three-dimensional motion and also to lay the foundation for more advanced study. We will proceed as we did for particle motion and for rigid-body plane motion by first examining the necessary kinematics and then pro-ceeding to the kinetics.
514 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies SECTION A KINEMATICS 7/2 Translation Figure 7/1 shows a rigid body translating in three-dimensional space. Any two points in the body, such as A and B, will move along par-allel straight lines if the motion is one of rectilinear translation or will move along congruent curves if the motion is one of curvilinear transla-tion. In either case, every line in the body, such as AB, remains parallel to its original position.
The position vectors and their first and second time derivatives are where rA/B remains constant, and therefore its time derivative is zero.
Thus, all points in the body have the same velocity and the same accel-eration. The kinematics of translation presents no special difficulty, and further elaboration is unnecessary.
7/3 Fixed-Axis Rotation Consider now the rotation of a rigid body about a fixed axis n-n in space with an angular velocity , as shown in Fig. 7/2. The angular ve-locity is a vector in the direction of the rotation axis with a sense estab-lished by the familiar right-hand rule. For fixed-axis rotation, does not change its direction since it lies along the axis. We choose the origin O of the fixed coordinate system on the rotation axis for convenience. Any point such as A which is not on the axis moves in a circular arc in a plane normal to the axis and has a velocity (7/1) which may be seen by replacing r by h b and noting that h 0.
The acceleration of A is given by the time derivative of Eq. 7/1.
Thus, (7/2) a ˙ r ( r) v r rA rB rA/B vA vB aA aB z y x B A rA rB rA/B vA vB Figure 7/1 where has been replaced by its equal, v r. The normal and tan-gential components of a for the circular motion have the familiar magni-tudes an ( r) b2 and at b, where Inasmuch as both v and a are perpendicular to and it follows that 0, 0, 0, and 0 for fixed-axis rotation.
7/4 Parallel-Plane Motion When all points in a rigid body move in planes which are parallel to a fixed plane P, Fig. 7/3, we have a general form of plane motion. The reference plane is customarily taken through the mass center G and is called the plane of motion. Because each point in the body, such as A, has a motion identical with the motion of the corresponding point (A) in plane P, it follows that the kinematics of plane motion covered in Chap-ter 5 provides a complete description of the motion when applied to the reference plane.
7/5 Rotation about a Fixed Point When a body rotates about a fixed point, the angular-velocity vector no longer remains fixed in direction, and this change calls for a more general concept of rotation.
Rotation and Proper Vectors We must first examine the conditions under which rotation vectors obey the parallelogram law of addition and may, therefore, be treated as proper vectors. Consider a solid sphere, Fig. 7/4, which is cut from a rigid body confined to rotate about the fixed point O.
The x-y-z axes here are taken as fixed in space and do not rotate with the body. In part a of the figure, two successive 90 rotations of the sphere about, first, the x-axis and, second, the y-axis result in the mo-tion of a point which is initially on the y-axis in position 1, to positions 2 a ˙ a v ˙ v ˙, ˙.
˙ r r ˙ Article 7/5 Rotation about a Fixed Point 515 ω n n O A n n A b h r ω v = × r ω ω at = ⋅ × r a z y x Fixed axis an = × ( × r) ω ω ω ω ⋅ = α Figure 7/2 P A G A′ x z y x′ y′ Figure 7/3 and 3, successively. On the other hand, if the order of the rotations is re-versed, the point undergoes no motion during the y-rotation but moves to point 3 during the 90 rotation about the x-axis. Thus, the two cases do not produce the same final position, and it is evident from this one special example that finite rotations do not generally obey the parallelo-gram law of vector addition and are not commutative. Thus, finite rota-tions may not be treated as proper vectors.
Infinitesimal rotations, however, do obey the parallelogram law of vector addition. This fact is shown in Fig. 7/5, which represents the combined effect of two infinitesimal rotations d1 and d2 of a rigid body about the respective axes through the fixed point O. As a result of d1, point A has a displacement d1 r, and likewise d2 causes a displace-ment d2 r of point A. Either order of addition of these infinitesimal displacements clearly produces the same resultant displacement, which is d1 r d2 r (d1 d2) r. Thus, the two rotations are equivalent to the single rotation d d1 d2. It follows that the an-gular velocities 1 1 and 2 2 may be added vectorially to give 1 2. We conclude, therefore, that at any instant of time a body with one fixed point is rotating instantaneously about a particular axis passing through the fixed point.
Instantaneous Axis of Rotation To aid in visualizing the concept of the instantaneous axis of rota-tion, we will cite a specific example. Figure 7/6 represents a solid cylin-drical rotor made of clear plastic containing many black particles embedded in the plastic. The rotor is spinning about its shaft axis at the steady rate 1, and its shaft, in turn, is rotating about the fixed vertical axis at the steady rate 2, with rotations in the directions indicated. If the rotor is photographed at a certain instant during its motion, the re-sulting picture would show one line of black dots sharply defined, indi-cating that, momentarily, their velocity was zero. This line of points with no velocity establishes the instantaneous position of the axis of ro-tation O-n. Any dot on this line, such as A, would have equal and oppo-site velocity components, v1 due to 1 and v2 due to 2. All other dots, ˙ ˙ ˙ 516 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 1,2 1 3 3 y x z y x z 2 O O x then y θ θ y then x θ θ (a) (b) Figure 7/4 ω = ⋅ θ θ d 1 θ d 1 θ d 2 θ d 2 θ d θ O A r d θ d 1 θ × r d θ × r d 2 θ × × r Figure 7/5 v1 v2 A n P O 1 ω 2 ω Figure 7/6 such as the one at P, would appear blurred, and their movements would show as short streaks in the form of small circular arcs in planes normal to the axis O-n. Thus, all particles of the body, except those on line O-n, are momentarily rotating in circular arcs about the instantaneous axis of rotation.
If a succession of photographs were taken, we would observe in each photograph that the rotation axis would be defined by a new series of sharply-defined dots and that the axis would change position both in space and relative to the body. For rotation of a rigid body about a fixed point, then, it is seen that the rotation axis is, in general, not a line fixed in the body.
Body and Space Cones Relative to the plastic cylinder of Fig. 7/6, the instantaneous axis of rotation O-A-n generates a right-circular cone about the cylinder axis called the body cone. As the two rotations continue and the cylinder swings around the vertical axis, the instantaneous axis of rotation also generates a right-circular cone about the vertical axis called the space cone. These cones are shown in Fig. 7/7 for this particular example.
We see that the body cone rolls on the space cone and that the angu-lar velocity of the body is a vector which lies along the common ele-ment of the two cones. For a more general case where the rotations are not steady, the space and body cones are not right-circular cones, Fig.
7/8, but the body cone still rolls on the space cone.
Angular Acceleration The angular acceleration of a rigid body in three-dimensional mo-tion is the time derivative of its angular velocity, In contrast to the case of rotation in a single plane where the scalar measures only the change in magnitude of the angular velocity, in three-dimensional motion the vector reflects the change in direction of as well as its change in magnitude. Thus in Fig. 7/8 where the tip of the angular ve-locity vector follows the space curve p and changes in both magnitude and direction, the angular acceleration becomes a vector tangent to this curve in the direction of the change in .
When the magnitude of remains constant, the angular accelera-tion is normal to . For this case, if we let stand for the angular ve-locity with which the vector itself rotates (precesses) as it forms the space cone, the angular acceleration may be written (7/3) This relation is easily seen from Fig. 7/9. The upper part of the figure re-lates the velocity of a point A on a rigid body to its position vector from O and the angular velocity of the body. The vectors , , and in the lower figure bear exactly the same relationship to each other as do the vectors v, r, and in the upper figure.
If we use Fig. 7/2 to represent a rigid body rotating about a fixed point O with the instantaneous axis of rotation n-n, we see that the ˙.
Article 7/5 Rotation about a Fixed Point 517 ω ω = r = × r = constant r v ω ω α = = × = constant Ω Ω Space cone Rigid cone O A r O ω ω Ω ω ω ω ω · · ⏐ ⏐ ⏐ ⏐ Figure 7/9 O p Body cone Space cone ω ω α = ⋅ Figure 7/8 A n ω O Space cone Body cone Figure 7/7 velocity v and acceleration a of any point A in the body are given by the same expressions as apply to the case in which the axis is fixed, namely, [7/1] [7/2] The one difference between the case of rotation about a fixed axis and rotation about a fixed point lies in the fact that for rotation about a fixed point, the angular acceleration will have a component nor-mal to due to the change in direction of , as well as a component in the direction of to reflect any change in the magnitude of . Although any point on the rotation axis n-n momentarily will have zero velocity, it will not have zero acceleration as long as is changing its direction. On the other hand, for rotation about a fixed axis, has only the one component along the fixed axis to reflect the change in the magnitude of . Furthermore, points which lie on the fixed rotation axis clearly have no velocity or acceleration.
Although the development in this article is for the case of rotation about a fixed point, we observe that rotation is a function solely of angu-lar change, so that the expressions for and do not depend on the fix-ity of the point around which rotation occurs. Thus, rotation may take place independently of the linear motion of the rotation point. This con-clusion is the three-dimensional counterpart of the concept of rotation of a rigid body in plane motion described in Art. 5/2 and used through-out Chapters 5 and 6.
˙ ˙ a ˙ r ( r) v r v ˙ 518 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies The engine/propeller units at the wingtips of this aircraft can tilt from a vertical takeoff position to a horizontal position for forward flight.
© Michael Wood/StocktrekImages, Inc.
SAMPLE PROBLEM 7/1 The 0.8-m arm OA for a remote-control mechanism is pivoted about the hori-zontal x-axis of the clevis, and the entire assembly rotates about the z-axis with a constant speed N 60 rev/min. Simultaneously, the arm is being raised at the constant rate 4 rad/s. For the position where 30, determine (a) the an-gular velocity of OA, (b) the angular acceleration of OA, (c) the velocity of point A, and (d) the acceleration of point A. If, in addition to the motion described, the ver-tical shaft and point O had a linear motion, say, in the z-direction, would that mo-tion change the angular velocity or angular acceleration of OA?
Solution. (a) Since the arm OA is rotating about both the x- and the z-axes, it has the components x 4 rad/s and z 2N/60 2(60)/60 6.28 rad/s.
The angular velocity is Ans.
(b) The angular acceleration of OA is Since z is not changing in magnitude or direction, 0. But x is changing di-rection and thus has a derivative which, from Eq. 7/3, is Therefore, Ans.
(c) With the position vector of A given by r 0.693j 0.4k m, the velocity of A from Eq. 7/1 becomes Ans.
(d) The acceleration of A from Eq. 7/2 is Ans.
The angular motion of OA depends only on the angular changes N and so any linear motion of O does not affect and .
˙, 20.1i 38.4j 6.40k m/s2 (10.05i) (10.05i 38.4j 6.40k) r v a ˙ r ( r) 25.1j 0 25.1j rad/s2 ˙x z x 6.28k 4i 25.1j rad/s2 ˙z ˙ ˙x ˙z x z 4i 6.28k rad/s ˙ ˙ Article 7/5 Rotation about a Fixed Point 519 Helpful Hints Alternatively, consider axes x-y-z to be attached to the vertical shaft and clevis so that they rotate. The deriv-ative of x becomes But from Eq. 5/11, we have z i 6.28k i 6.28j. Thus, 4(6.28)j 25.1j rad/s2 as before.
˙x i ˙ 4i ˙.
˙x To compare methods, it is suggested that these results for v and a be ob-tained by applying Eqs. 2/18 and 2/19 for particle motion in spherical coordinates, changing symbols as necessary.
O N 0.8 m x y A z = 30° β ω = ⋅ α ω ω x ω z O x y z ω β x = ⋅ ( ) SAMPLE PROBLEM 7/2 The electric motor with an attached disk is running at a constant low speed of 120 rev/min in the direction shown. Its housing and mounting base are ini-tially at rest. The entire assembly is next set in rotation about the vertical Z-axis at the constant rate N 60 rev/min with a fixed angle of 30. Determine (a) the angular velocity and angular acceleration of the disk, (b) the space and body cones, and (c) the velocity and acceleration of point A at the top of the disk for the instant shown.
Solution.
The axes x-y-z with unit vectors i, j, k are attached to the motor frame, with the z-axis coinciding with the rotor axis and the x-axis coinciding with the horizontal axis through O about which the motor tilts. The Z-axis is vertical and carries the unit vector K j cos k sin .
(a) The rotor and disk have two components of angular velocity: 0 120(2)/60 4 rad/sec about the z-axis and 60(2)/60 2 rad/sec about the Z-axis. Thus, the angular velocity becomes Ans.
The angular acceleration of the disk from Eq. 7/3 is Ans.
(b) The angular velocity vector is the common element of the space and body cones which may now be constructed as shown.
(c) The position vector of point A for the instant considered is From Eq. 7/1 the velocity of A is Ans.
From Eq. 7/2 the acceleration of point A is Ans. 1063j 473k in./sec2 68.4i (5j 10k) (3j 5k) (7.68i) a ˙ r ( r) r v r 5j 10k in. (0 cos )i i(2)(4) cos 30 68.4i rad/sec2 (0 cos sin cos )i (2 sin cos )i (j cos k sin ) [( cos )j (0 sin )k] ˙ (2 cos 30)j (4 2 sin 30)k (3j 5.0k) rad/sec 0k (j cos k sin ) ( cos )j (0 sin )k 0 0k K 520 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies x z y Z A C O N ω 0 OC = 10″ CA = 5″ — — — — γ = 30° γ ω ω π = 2 rad/sec ω π z = 5 rad/sec ω π y = 3 rad/sec ω π 0 = 4 rad/sec Z y z z O O Z Ω Space cone Body cone Helpful Hints Note that 0 y z as shown on the vector diagram.
Remember that Eq. 7/3 gives the complete expression for only for steady precession where is con-stant, which applies to this problem.
Since the magnitude of is con-stant, must be tangent to the base circle of the space cone, which puts it in the plus x-direction in agree-ment with our calculated conclusion.
PROBLEMS Introductory Problems 7/1 Place your textbook on your desk, with fixed axes ori-ented as shown. Rotate the book about the x-axis through a 90 angle and then from this new position rotate it 90 about the y-axis. Sketch the final position of the book. Repeat the process but reverse the order of rotation. From your results, state your conclusion concerning the vector addition of finite rotations. Rec-oncile your observations with Fig. 7/4.
Problem 7/1 7/2 Repeat the experiment of Prob. 7/1 but use a small angle of rotation, say, . Note the near-equal final po-sitions for the two different rotation sequences. What does this observation lead you to conclude for the combination of infinitesimal rotations and for the time derivatives of angular quantities? Reconcile your observations with Fig. 7/5.
7/3 The solid cylinder is rotating about the fixed axis OA with a constant speed rev/min in the direc-tion shown. If the x- and y-components of the velocity of point P are 12 ft/sec and ft/sec, determine its z-component of velocity and the radial distance R from P to the rotation axis. Also find the magnitude of the acceleration of P.
6 N 600 5 z y x Article 7/5 Problems 521 Problem 7/3 7/4 A timing mechanism consists of the rotating distribu-tor arm AB and the fixed contact C. If the arm rotates about the fixed axis OA with a constant angular veloc-ity 30(3i 2j 6k) rad/s, and if the coordinates of the contact C expressed in millimeters are (20, 30, 80), determine the magnitude of the acceleration of the tip B of the distributor arm as it passes point C.
Problem 7/4 30 mm z A C B O y x z x y A 8″ 12″ 4″ P N O 7/7 The rotor B spins about its inclined axis OA at the speed rev/min, where . Simultane-ously, the assembly rotates about the vertical z-axis at the rate . If the total angular velocity of the rotor has a magnitude of 40 rad/s, determine .
Problem 7/7 7/8 A slender rod bent into the shape shown rotates about the fixed line CD at a constant angular rate . Deter-mine the velocity and acceleration of point A.
Problem 7/8 x z D O A C θ d h l y ω x z y A B O β N1 N2 N2 N2 30 N1 200 522 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/5 The rotor and shaft are mounted in a clevis which can rotate about the z-axis with an angular velocity .
With 0 and constant, the rotor has an angular velocity 0 4j 3k rad/s. Find the velocity vA of point A on the rim if its position vector at this instant is r 0.5i 1.2j 1.1k m. What is the rim speed vB of any point B?
Problem 7/5 7/6 The disk rotates with a spin velocity of 15 rad/s about its horizontal z-axis first in the direction (a) and sec-ond in the direction (b). The assembly rotates with the velocity N 10 rad/s about the vertical axis. Con-struct the space and body cones for each case.
Problem 7/6 N (b) (a) z z y Ω θ ω x A B 0 v 7/9 The rod is hinged about the axis O-O of the clevis, which is attached to the end of the vertical shaft. The shaft rotates with a constant angular 0 as shown. If is decreasing at the constant rate p, write ex-pressions for the angular velocity and angular accel-eration of the rod.
Problem 7/9 7/10 The panel assembly and attached x-y-z axes rotate with a constant angular velocity 0.6 rad/sec about the vertical z-axis. Simultaneously, the panels rotate about the y-axis as shown with a constant rate 0 2 rad/sec. Determine the angular acceleration of panel A and find the acceleration of point P for the instant when 90.
Problem 7/10 z y x O A P β Ω B 18″ 16″ 20″ ω0 ω0 z O O A y x ω θ 0 ˙ Article 7/5 Problems 523 Representative Problems 7/11 The motor of Sample Problem 7/2 is shown again here. If the motor pivots about the x-axis at the con-stant rate 3 rad/sec with no rotation about the Z-axis (N 0), determine the angular acceleration of the rotor and disk as the position 30 is passed.
The constant speed of the motor is 120 rev/min. Also find the velocity and acceleration of point A, which is on the top of the disk for this position.
Problem 7/11 7/12 If the motor of Sample Problem 7/2, repeated in Prob.
7/11, reaches a speed of 3000 rev/min in 2 seconds from rest with constant acceleration, determine the total an-gular acceleration of the rotor and disk second after it is turned on if the turntable is rotating at a constant rate N 30 rev/min. The angle 30 is constant.
7/13 The spool A rotates about its axis with an angular velocity of 20 rad/s, first in the sense of a and sec-ond in the sense of b. Simultaneously, the assembly rotates about the vertical axis with an angular veloc-ity 1 10 rad/s. Determine the magnitude of the total angular velocity of the spool and construct the body and space cones for the spool for each case.
Problem 7/13 60° A O ω1 ωb ωa 1 3 x z y Z A C O N ω 0 OC = 10″ CA = 5″ — — — — γ ˙ Problem 7/16 7/17 For the robot of Prob. 7/16, determine the angular velocity and angular acceleration of the jaws A if 60 and 30, both constant, and if 1 2 rad/s, 2 3 4 0, and 5 0.8 rad/s, all constant.
7/18 The wheel rolls without slipping in a circular arc of radius R and makes one complete turn about the vertical y-axis with constant speed in time . Deter-mine the vector expression for the angular accelera-tion of the wheel and construct the space and body cones.
Problem 7/18 z A r x O y R θ β z x y A O2 O1 ω1 ω2 ω 3 ω 5 ω4 524 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/14 In manipulating the dumbbell, the jaws of the robotic device have an angular velocity p 2 rad/s about the axis OG with fixed at 60 . The entire assembly rotates about the vertical Z-axis at the constant rate 0.8 rad/s. Determine the angular velocity and angular acceleration of the dumbbell. Express the results in terms of the given orientation of axes x-y-z, where the y-axis is parallel to the Y-axis.
Problem 7/14 7/15 Determine the angular acceleration of the dumb-bell of Prob. 7/14 for the conditions stated, except that is increasing at the rate of 3 rad/s2 for the in-stant under consideration.
7/16 The robot shown has five degrees of rotational free-dom. The x-y-z axes are attached to the base ring, which rotates about the z-axis at the rate 1. The arm O1O2 rotates about the x-axis at the rate 2 The control arm O2A rotates about axis O1-O2 at the rate 3 and about a perpendicular axis through O2 which is momentarily parallel to the x-axis at the rate 4 Finally, the jaws rotate about axis O2-A at the rate 5. The magnitudes of all angular rates are constant. For the configuration shown, determine the magnitude of the total angular velocity of the jaws for 60 and 45 if 1 2 rad/s, 1.5 rad/s, and 3 4 5 0. Also express the angular accel-eration of arm O1O2 as a vector.
˙ ˙.
˙.
m z a G a m x X Y y γ Z O p ω Ω 7/19 Determine expressions for the velocity v and acceler-ation a of point A on the wheel of Prob. 7/18 for the position shown, where A crosses the horizontal line through the center of the wheel.
7/20 The circular disk of 120-mm radius rotates about the z-axis at the constant rate z 20 rad/s, and the en-tire assembly rotates about the fixed x-axis at the constant rate x 10 rad/s. Calculate the magni-tudes of the velocity v and acceleration a of point B for the instant when 30.
Problem 7/20 7/21 The crane has a boom of length 24 m and is revolving about the vertical axis at the constant rate of 2 rev/min in the direction shown. Simultane-ously, the boom is being lowered at the constant rate 0.10 rad/s. Calculate the magnitudes of the velocity and acceleration of the end P of the boom for the instant when it passes the position 30.
Problem 7/21 β O P ω ˙ OP 200 mm 120 mm y B O x y ′ z θ ωz ωx Article 7/5 Problems 525 7/22 The design of the rotating arm OA of a control mechanism requires that it rotate about the vertical Z-axis at the constant rate rad/s. Simul-taneously, OA oscillates according to sin , where radians and t is in seconds measured from the time when . Determine the angular velocity and the angular acceleration of OA for the instant (a) when s and (b) when s.
The x-y reference axes rotate in the X-Y plane with the angular velocity .
Problem 7/22 7/23 For the rotating and oscillating control arm OA of Prob. 7/22, determine the velocity v and acceleration a of the ball tip A for the condition when .
Distance and sin as defined in Prob. 7/22 with rad/s and rad.
7/24 If the angular velocity rad/s of the rotor in Prob. 7/5 is constant in magnitude, deter-mine the angular acceleration of the rotor for (a) and rad/s (both constant) and (b) and rad/s (both constant).
Find the magnitude of the acceleration of point A in each case, where A has the position vector 1.1k m at the instant represented.
r 0.5i 1.2j 2 tan1(3 4) ˙ 2 0 0 4j 3k 0 /6 4t 0 s 100 mm, b 120 mm, t 1/2 s Ω β y X A s b O Y x θ β β z = · Z t 1/8 t 1/2 0 0 /6 4t 0 ˙ 7/27 The pendulum oscillates about the x-axis according to sin 3t radians, where t is the time in seconds. Simultaneously, the shaft OA revolves about the ver-tical z-axis at the constant rate z 2 rad/sec.
Determine the velocity v and acceleration a of the center B of the pendulum as well as its angular ac-celeration for the instant when t 0.
Problem 7/27 7/28 The solid right-circular cone of base radius r and height h rolls on a flat surface without slipping.
The center B of the circular base moves in a circu-lar path around the z-axis with a constant speed v.
Determine the angular velocity and the angular acceleration of the solid cone.
Problem 7/28 x z y A O r h r B ω θ z x B O A 8″ y z 4″ 6 526 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/25 The vertical shaft and attached clevis rotate about the z-axis at the constant rate 4 rad/s. Simulta-neously, the shaft B revolves about its axis OA at the constant rate 0 3 rad/s, and the angle is decreas-ing at the constant rate of /4 rad/s. Determine the angular velocity and the magnitude of the angular acceleration of shaft B when 30. The x-y-z axes are attached to the clevis and rotate with it.
Problem 7/25 7/26 The right-circular cone A rolls on the fixed right-circular cone B at a constant rate and makes one complete trip around B every 4 seconds. Compute the magnitude of the angular acceleration of cone A during its motion.
Problem 7/26 150 mm 150 mm Z z B O 50 mm A x z y A O B γ 0 ω Ω Article 7/6 General Motion 527 7/6 General Motion The kinematic analysis of a rigid body which has general three-dimensional motion is best accomplished with the aid of our principles of relative motion. We have applied these principles to problems in plane motion and now extend them to space motion. We will make use of both translating axes and rotating reference axes.
Translating Reference Axes Figure 7/10 shows a rigid body which has an angular velocity . We may choose any convenient point B as the origin of a translating refer-ence system x-y-z. The velocity v and acceleration a of any other point A in the body are given by the relative-velocity and relative-acceleration expressions [5/4] [5/7] which were developed in Arts. 5/4 and 5/6 for the plane motion of rigid bodies. These expressions also hold in three dimensions, where the three vectors for each of the equations are also coplanar.
In applying these relations to rigid-body motion in space, we note from Fig. 7/10 that the distance remains constant. Thus, from an ob-server’s position on x-y-z, the body appears to rotate about the point B and point A appears to lie on a spherical surface with B as the center.
Consequently, we may view the general motion as a translation of the body with the motion of B plus a rotation of the body about B.
The relative-motion terms represent the effect of the rotation about B and are identical to the velocity and acceleration expressions dis-cussed in the previous article for rotation of a rigid body about a fixed point. Therefore, the relative-velocity and relative-acceleration equa-tions may be written (7/4) where and are the instantaneous angular velocity and angular accel-eration of the body, respectively.
The selection of the reference point B is quite arbitrary in theory. In practice, point B is chosen for convenience as some point in the body whose motion is known in whole or in part. If point A is chosen as the reference point, the relative-motion equations become where rB/A rA/B. It should be clear that and, thus, are the same vectors for either formulation since the absolute angular motion of the body is independent of the choice of reference point. When we come to ˙ aB aA ˙ rB/A ( rB/A) vB vA rB/A ˙ aA aB ˙ rA/B ( rA/B) vA vB rA/B AB aA aB aA/B vA vB vA/B Z ω rA rB rA/B z Y A B O y x X Figure 7/10 This time-lapse photo of a VTOL air-craft shows a three-dimensional com-bination of translation and rotation.
© Jim Sugar/CORBIS 528 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Z rA rB rA/B Y A B O X ω Ω (Axes) (Body) z y x Figure 7/11 the kinetic equations for general motion, we will see that the mass cen-ter of a body is frequently the most convenient reference point to choose.
If points A and B in Fig. 7/10 represent the ends of a rigid control link in a spatial mechanism where the end connections act as ball-and-socket joints (as in Sample Problem 7/3), it is necessary to impose cer-tain kinematic requirements. Clearly, any rotation of the link about its own axis AB does not affect the action of the link. Thus, the angular ve-locity n whose vector is normal to the link describes its action. It is nec-essary, therefore, that n and rA/B be at right angles, and this condition is satisfied if 0.
Similarly, it is only the component n of the angular acceleration of the link normal to AB which affects its action, so that 0 must also hold.
Rotating Reference Axes A more general formulation of the motion of a rigid body in space calls for the use of reference axes which rotate as well as translate. The description of Fig. 7/10 is modified in Fig. 7/11 to show reference axes whose origin is attached to the reference point B as before, but which rotate with an absolute angular velocity which may be different from the absolute angular velocity of the body.
We now make use of Eqs. 5/11, 5/12, 5/13, and 5/14 developed in Art.
5/7 for describing the plane motion of a rigid body with the use of rotat-ing axes. The extension of these relations from two to three dimensions is easily accomplished by merely including the z-component of the vec-tors, and this step is left to the student to carry out. Replacing in these equations by the angular velocity of our rotating x-y-z axes gives us (7/5) for the time derivatives of the rotating unit vectors attached to x-y-z.
The expression for the velocity and acceleration of point A become (7/6) where vrel and arel are, respectively, the velocity and acceleration of point A measured relative to x-y-z by an ob-server attached to x-y-z.
We again note that is the angular velocity of the axes and may be different from the angular velocity of the body. Also we note that rA/B re-mains constant in magnitude for points A and B fixed to a rigid body, but it will change direction with respect to x-y-z when the angular velocity of the axes is different from the angular velocity of the body. We observe z ¨k y ¨j x ¨i z ˙k y ˙j x ˙i aA aB ˙ rA/B ( rA/B) 2 vrel arel vA vB rA/B vrel i ˙ i j ˙ j k ˙ k n rA/B n rA/B It may be shown that n if the angular velocity of the link about its own axis is not changing. See the first author’s Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 37.
˙n Article 7/6 General Motion 529 further that, if x-y-z are rigidly attached to the body, and vrel and arel are both zero, which makes the equations identical to Eqs. 7/4.
In Art. 5/7 we also developed the relationship (Eq. 5/13) between the time derivative of a vector V as measured in the fixed X-Y system and the time derivative of V as measured relative to the rotating x-y sys-tem. For our three-dimensional case, this relation becomes (7/7) When we apply this transformation to the relative-position vector rA/B rA rB for our rigid body of Fig. 7/11, we obtain or which gives us the first of Eqs. 7/6.
Equations 7/6 are particularly useful when the reference axes are attached to a moving body within which relative motion occurs.
Equation 7/7 may be recast as the vector operator (7/7a) where [ ] stands for any vector V expressible both in X-Y-Z and in x-y-z. If we apply the operator to itself, we obtain the second time deriva-tive, which becomes (7/7b) This exercise is left to the student. Note that the form of Eq. 7/7b is the same as that of the second of Eqs. 7/6 expressed for aA/B aA aB. 2 d[ ] dt xyz d2[ ] dt2 XYZ d2[ ] dt2 xyz ˙ [ ] ( [ ]) d[ ] dt XYZ d[ ] dt xyz [ ] vA vB vrel rA/B drA dt XYZ drB dt XYZ drA/B dt xyz rA/B dV dtXYZ dV dtxyz V Robots welding automobile unit-bodies.
© Giles Barnard/AgeFotostock America, Inc.
530 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies SAMPLE PROBLEM 7/3 Crank CB rotates about the horizontal axis with an angular velocity 1 6 rad/s which is constant for a short interval of motion which includes the position shown. The link AB has a ball-and-socket fitting on each end and connects crank DA with CB. For the instant shown, determine the angular velocity 2 of crank DA and the angular velocity n of link AB.
Solution.
The relative-velocity relation, Eq. 7/4, will be solved first using translating reference axes attached to B. The equation is where n is the angular velocity of link AB taken normal to AB. The velocities of A and B are Also rA/B 50i 100j 100k mm. Substitution into the velocity relation gives Expanding the determinant and equating the coefficients of the i, j, k terms give These equations may be solved for 2, which becomes Ans.
As they stand, the three equations incorporate the fact that n is normal to vA/B, but they cannot be solved until the requirement that n be normal to rA/B is in-cluded. Thus, Combination with two of the three previous equations yields the solutions Thus, with Ans.
n 2 322 42 52 25 rad/s n 2 3 (2i 4j 5k) rad/s nx 4 3 rad/s ny 8 3 rad/s nz 10 3 rad/s 50nx 100ny 100nz 0 [n rA/B 0] 2 6 rad/s 0 2nx ny 2 2nx nz 6 ny nz vA 502 j vB 100(6)i 600i mm/s [v r] vA vB n rA/B D A z x y C ω 1 ω 2 50 mm 100 mm 100 mm v A v B 100 mm B Helpful Hints We select B as the reference point since its motion can easily be deter-mined from the given angular veloc-ity 1 of CB.
The angular velocity of AB is taken as a vector n normal to AB since any rotation of the link about its own axis AB has no influence on the behavior of the linkage.
The relative-velocity equation may be written as vA vB vA/B n rA/B, which requires that vA/B be perpendicular to both n and rA/B. This equation alone does not incorporate the additional require-ment that n be perpendicular to rA/B. Thus, we must also satisfy 0.
n rA/B Article 7/6 General Motion 531 SAMPLE PROBLEM 7/4 Determine the angular acceleration of crank AD in Sample Problem 7/3 for the conditions cited. Also find the angular acceleration of link AB.
Solution.
The accelerations of the links may be found from the second of Eqs.
7/4, which may be written where n, as in Sample Problem 7/3, is the angular velocity of AB taken normal to AB. The angular acceleration of AB is written as In terms of their normal and tangential components, the accelerations of A and B are Also Substitution into the relative-acceleration equation and equating respective coef-ficients of i, j, k give Solution of these equations for gives Ans.
The vector is normal to rA/B but is not normal to vA/B, as was the case with n.
which, when combined with the preceding relations for these same quantities, gives Thus, Ans.
and Ans.
˙n 422 42 32 429 rad/s2 ˙n 4(2i 4j 3k) rad/s2 ˙nx 8 rad/s2 ˙ny 16 rad/s2 ˙nz 12 rad/s2 2 ˙nx 4 ˙ny 4 ˙nz 0 [ ˙n rA/B 0] ˙n ˙2 36 rad/s2 ˙2 32 2 ˙nx ˙ny ˙2 40 2 ˙nx ˙nz 28 ˙ny ˙nz (50 ˙nz 100 ˙nx)j (100 ˙nx 50 ˙ny)k ˙n rA/B (100 ˙ny 100 ˙nz)i n (n rA/B) n 2rA/B 20(50i 100j 100k) mm/s2 aB 1001 2k (0)i 3600k mm/s2 aA 502 2i 50 ˙2 j 1800i 50 ˙2 j mm/s2 ˙n.
aA aB ˙n rA/B n (n rA/B) ˙n ˙2 The component of which is not normal to vA/B gives rise to the change in direction of vA/B.
˙n Helpful Hints If the link AB had an angular veloc-ity component along AB, then a change in both magnitude and direc-tion of this component could occur which would contribute to the actual angular acceleration of the link as a rigid body. However, since any rota-tion about its own axis AB has no in-fluence on the motion of the cranks at C and D, we will concern our-selves only with ˙n.
532 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies SAMPLE PROBLEM 7/5 The motor housing and its bracket rotate about the Z-axis at the constant rate 3 rad/s. The motor shaft and disk have a constant angular velocity of spin p 8 rad/s with respect to the motor housing in the direction shown. If is constant at 30, determine the velocity and acceleration of point A at the top of the disk and the angular acceleration of the disk.
Solution.
The rotating reference axes x-y-z are attached to the motor housing, and the rotating base for the motor has the momentary orientation shown with respect to the fixed axes X-Y-Z. We will use both X-Y-Z components with unit vectors I, J, K and x-y-z components with unit vectors i, j, k. The angular veloc-ity of the x-y-z axes becomes K 3K rad/s.
Velocity.
The velocity of A is given by the first of Eqs. 7/6 where Thus, Ans.
Acceleration.
The acceleration of A is given by the second of Eqs. 7/6 where Substituting into the expression for aA and collecting terms give us and Ans.
Angular Acceleration.
Since the precession is steady, we may use Eq. 7/3 to give us Ans. 0 (24 cos 30)i 20.8i rad/s2 ˙ 3K (3K 8j) aA (0.703)2 (8.09)2 8.12 m/s2 aA 0.703j 8.09k m/s2 7.68k m/s2 arel p (p rA/B) 8j [8j (0.300j 0.120k)] 5.76(j cos 30 k sin 30) 4.99j 2.88k m/s2 2 vrel 2(3K) 0.960i 5.76J 3K (0.599i) 1.557j 0.899k m/s2 ( rA/B) 3K [3K (0.300j 0.120k)] ˙ 0 3.15(j cos 30 k sin 30) 2.73j 1.575k m/s2 aB ( rB) 3K (3K 0.350J) 3.15J aA aB ˙ rA/B ( rA/B) 2 vrel arel vA 1.05i 0.599i 0.960i 0.689i m/s vrel p rA/B 8j (0.300j 0.120k) 0.960i m/s (0.9 cos 30)i (0.36 sin 30)i 0.599i m/s rA/B 3K (0.300j 0.120k) vB rB 3K 0.350J 1.05I 1.05i m/s vA vB rA/B vrel A B Z X x y z 150 mm 120 mm Ω 350 mm 300 mm Y γ p Note that K i J j cos k sin , K j i cos , and K k i sin .
Helpful Hints This choice for the reference axes provides a simple description for the motion of the disk relative to these axes.
Article 7/6 Problems 533 Problem 7/31 7/32 If the angular rate p of the disk in Prob. 7/31 is in-creasing at the rate of 6 rad/s per second and if re-mains constant at 4 rad/s, determine the angular acceleration of the disk at the instant when p reaches 10 rad/s.
7/33 For the conditions of Prob. 7/31, determine the ve-locity vA and acceleration aA of point A on the disk as it passes the position shown. Reference axes x-y-z are attached to the collar at O and its shaft OC.
7/34 An unmanned radar-radio controlled aircraft with tilt-rotor propulsion is being designed for reconnais-sance purposes. Vertical rise begins with 0 and is followed by horizontal flight as approaches 90. If the rotors turn at a constant speed N of 360 rev/min, determine the angular acceleration of rotor A for 30 if is constant at 0.2 rad/s.
Problem 7/34 θ θ x z y A N N ˙ Ω = 4 rad/s p = 10 rad/s 400 mm 300 mm A C O x x0 z y PROBLEMS Introductory Problems 7/29 The solid cylinder has a body cone with a semi-vertex angle of 20. Momentarily the angular veloc-ity has a magnitude of 30 rad/s and lies in the y-z plane. Determine the rate p at which the cylinder is spinning about its z-axis and write the vector expres-sion for the velocity of B with respect to A.
Problem 7/29 7/30 The helicopter is nosing over at the constant rate q rad/s. If the rotor blades revolve at the constant speed p rad/s, write the expression for the angular acceleration of the rotor. Take the y-axis to be attached to the fuselage and pointing forward per-pendicular to the rotor axis.
Problem 7/30 7/31 The collar at O and attached shaft OC rotate about the fixed x0-axis at the constant rate 4 rad/s.
Simultaneously, the circular disk rotates about OC at the constant rate p 10 rad/s. Determine the magnitude of the total angular velocity of the disk and find its angular acceleration .
y z p q z y x 20° 0.4 m ω A B 7/35 End A of the rigid link is confined to move in the x-direction while end B is confined to move along the z-axis. Determine the component n normal to AB of the angular velocity of the link as it passes the position shown with vA 3 ft/sec.
Problem 7/35 Representative Problems 7/36 The small motor M is pivoted about the x-axis through O and gives its shaft OA a constant speed p rad/s in the direction shown relative to its housing. The entire unit is then set into rotation about the vertical Z-axis at the constant angular velocity rad/s. Simultane-ously, the motor pivots about the x-axis at the con-stant rate for an interval of motion. Determine the angular acceleration of the shaft OA in terms of .
Express your result in terms of the unit vectors for the rotating x-y-z axes.
˙ 3′ 7′ 2′ B O A x z y vA Problem 7/36 7/37 The flight simulator is mounted on six hydraulic ac-tuators connected in pairs to their attachment points on the underside of the simulator. By programming the actions of the actuators, a variety of flight condi-tions can be simulated with translational and rota-tional displacements through a limited range of motion. Axes x-y-z are attached to the simulator with origin B at the center of the volume. For the instant represented, B has a velocity and an acceleration in the horizontal y-direction of 3.2 ft/sec and 4 ft/sec2, respectively. Simultaneously, the angular velocities and their time rates of change are x 1.4 rad/sec, 2 rad/sec2, y 1.2 rad/sec, 3 rad/sec2, z 0. For this instant determine the magni-tudes of the velocity and acceleration of point A.
Problem 7/37 60″ B z y ω y ω x x A ˙z ˙y ˙x Ω β y A R M Z O b p x z 534 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/6 Problems 535 Problem 7/39 7/40 The spacecraft is revolving about its z-axis, which has a fixed space orientation, at the constant rate p rad/s. Simultaneously, its solar panels are unfolding at the rate which is programmed to vary with as shown in the graph. Determine the angular acceleration of panel A an instant (a) be-fore and an instant (b) after it reaches the position 18.
Problem 7/40 β β β x y z A p 0 0 2 18 ( ° ) ( °/s) 90 β · ˙ 1 10 x Z Y y vB A L B z X Y 0.6 m X 7/38 The robot of Prob. 7/16 is shown again here, where the coordinate system x-y-z with origin at O2 ro-tates about the X-axis at the rate Nonrotating axes X-Y-Z oriented as shown have their origin at O1. If 2 3 rad/s constant, 3 1.5 rad/s con-stant, 1 5 0, 1.2 m, and 0.6 m, determine the velocity of the center A of the jaws for the instant when 60. The angle lies in the y-z plane and is constant at 45.
Problem 7/38 7/39 For the instant represented collar B is moving along the fixed shaft in the X-direction with a con-stant velocity vB 4 m/s. Also at this instant X 0.3 m and Y 0.2 m. Calculate the velocity of collar A, which moves along the fixed shaft parallel to the Y-axis. Solve, first, by differentiating the relation X 2 Y 2 Z2 L2 with respect to time and, sec-ond, by using the first of Eqs. 7/4 with translating axes attached to B. Each clevis is free to rotate about the axis of the rod.
θ β z X x Y Z y A O2 O1 ω1 ω2 ω 3 ω 5 ω4 O2 A O1O2 ˙ ˙ .
7/41 The disk has a constant angular velocity p about its z-axis, and the yoke A has a constant angular veloc-ity 2 about its shaft as shown. Simultaneously, the entire assembly revolves about the fixed X-axis with a constant angular velocity 1. Determine the ex-pression for the angular acceleration of the disk as the yoke brings it into the vertical plane in the position shown. Solve by picturing the vector changes in the angular-velocity components.
Problem 7/41 7/42 The collar and clevis A are given a constant upward velocity of 8 in./sec for an interval of motion and cause the ball end of the bar to slide in the radial slot in the rotating disk. Determine the angular accelera-tion of the bar when the bar passes the position for which z 3 in. The disk turns at the constant rate of 2 rad/sec.
Problem 7/42 z z y x Ω = 2 rad/sec A B l = 5′′ p z Z x Y X A y ω2 ω1 7/43 The circular disk of 100-mm radius rotates about its z-axis at the constant speed p 240 rev/min, and arm OCB rotates about the Y-axis at the constant speed N 30 rev/min. Determine the velocity v and acceleration a of point A on the disk as it passes the position shown. Use reference axes x-y-z attached to the arm OCB.
Problem 7/43 7/44 Solve Prob. 7/43 by attaching the reference axes x-y-z to the rotating disk.
7/45 For the conditions described in Prob. 7/36, deter-mine the velocity v and acceleration a of the center A of the ball tool in terms of .
7/46 The circular disk is spinning about its own axis ( y-axis) at the constant rate p 10 rad/s. Simulta-neously, the frame is rotating about the Z-axis at the constant rate 4 rad/s. Calculate the angular acceleration of the disk and the acceleration of point A at the top of the disk. Axes x-y-z are attached to the frame, which has the momentary orientation shown with respect to the fixed axes X-Y-Z.
Problem 7/46 p z Z Y X A O y B x 300 mm 100 mm Ω 180 mm 100 mm 100 mm x z p y A B C D X Y Z N O 536 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/6 Problems 537 R A x z r B M p O y C θ Problem 7/48 7/49 For the conditions specified with Sample Problem 7/2, except that is increasing at the steady rate of 3 rad/sec, determine the angular velocity and the angular acceleration of the rotor when the position 30 is passed. (Suggestion: Apply Eq. 7/7 to the vector to find . Note that in Sample Problem 7/2 is no longer the complete angular velocity of the axes.) 7/50 The wheel of radius r is free to rotate about the bent axle CO which turns about the vertical axis at the constant rate p rad/s. If the wheel rolls without slip-ping on the horizontal circle of radius R, determine the expressions for the angular velocity and angu-lar acceleration of the wheel. The x-axis is always horizontal.
Problem 7/50 Y y A x X z Z p B O r b 2 ω 1 ω 7/47 The center O of the spacecraft is moving through space with a constant velocity. During the period of motion prior to stabilization, the spacecraft has a constant rotational rate rad/sec about its z-axis. The x-y-z axes are attached to the body of the craft, and the solar panels rotate about the y-axis at the constant rate rad/sec with respect to the spacecraft. If is the absolute angular velocity of the solar panels, determine Also find the accelera-tion of point A when 30.
Problem 7/47 7/48 The thin circular disk of mass m and radius r is ro-tating about its z-axis with a constant angular veloc-ity p, and the yoke in which it is mounted rotates about the X-axis through OB with a constant angu-lar velocity 1. Simultaneously, the entire assembly rotates about the fixed Y-axis through O with a con-stant angular velocity 2. Determine the velocity v and acceleration a of point A on the rim of the disk as it passes the position shown where the x-y plane of the disk coincides with the X-Y plane. The x-y-z axes are attached to the yoke.
8′ 2′ z x y A O θ Ω 2′ ˙.
1 4 ˙ 1 2 7/51 The gyro rotor shown is spinning at the constant rate of 100 rev/min relative to the x-y-z axes in the direction indicated. If the angle between the gim-bal ring and the horizontal X-Y plane is made to in-crease at the constant rate of 4 rad/s and if the unit is forced to precess about the vertical at the constant rate N 20 rev/min, calculate the magnitude of the angular acceleration of the rotor when 30.
Solve by using Eq. 7/7 applied to the angular velocity of the rotor.
Problem 7/51 7/52 For a short interval of motion, collar A moves along its fixed shaft with a velocity vA 2 m/s in the Y-direction. Collar B, in turn, slides along its fixed vertical shaft. Link AB is 700 mm in length and can turn within the clevis at A to allow for the an-gular change between the clevises. For the instant when A passes the position where y 200 mm, de-termine the velocity of collar B using nonrotating axes attached to B and find the component n, nor-mal to AB, of the angular velocity of the link. Also solve for vB by differentiating the appropriate rela-tion x2 y2 z2 l2.
γ N Z O X y z Y x Problem 7/52 X 300 mm 700 mm v A Y O B Z y y x z A 538 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/7 Angular Momentum 539 7/7 Angular Momentum The force equation for a mass system, rigid or nonrigid, Eq. 4/1 or 4/6, is the generalization of Newton’s second law for the motion of a par-ticle and should require no further explanation. The moment equation for three-dimensional motion, however, is not nearly as simple as the third of Eqs. 6/1 for plane motion since the change of angular momen-tum has a number of additional components which are absent in plane motion.
We now consider a rigid body moving with any general motion in space, Fig. 7/12a. Axes x-y-z are attached to the body with origin at the mass center G. Thus, the angular velocity of the body becomes the an-gular velocity of the x-y-z axes as observed from the fixed reference axes X-Y-Z. The absolute angular momentum HG of the body about its mass center G is the sum of the moments about G of the linear momenta of all elements of the body and was expressed in Art. 4/4 as HG Σ(i mivi), where vi is the absolute velocity of the mass element mi.
But for the rigid body, vi i, where i is the relative velocity of mi with respect to G as seen from nonrotating axes. Thus, we may write where we have factored out from the first summation terms by revers-ing the order of the cross product and changing the sign. With the origin at the mass center G, the first term in HG is zero since Σmii 0.
The second term with the substitution of dm for mi and for i gives (7/8) Before expanding the integrand of Eq. 7/8, we consider also the case of a rigid body rotating about a fixed point O, Fig. 7/12b. The x-y-z axes are attached to the body, and both body and axes have an angular veloc-ity . The angular momentum about O was expressed in Art. 4/4 and is HO Σ(ri mivi), where, for the rigid body, vi ri. Thus, with the substitution of dm for mi and r for ri, the angular momentum is (7/9) Moments and Products of Inertia We observe now that for the two cases of Figs. 7/12a and 7/12b, the position vectors i and ri are given by the same expression xi yj zk.
Thus, Eqs. 7/8 and 7/9 are identical in form, and the symbol H will be used here for either case. We now carry out the expansion of the inte-grand in the two expressions for angular momentum, recognizing that the components of are invariant with respect to the integrals over the body and thus become constant multipliers of the integrals. The cross-product HO [r ( r)] dm HG [ ( )] dm m v HG v Σmi i Σ[ i mi( i)] v SECTION B KINETICS x G Y X Z y v – mi z i ρ ω x y mi z ω ri Y X O Z (a) (b) Figure 7/12 expansion applied to the triple vector product gives, upon collection of terms, Now let (7/10) The quantities Ixx, Iyy, Izz are called the moments of inertia of the body about the respective axes, and Ixy, Ixz, Iyz are the products of inertia with respect to the coordinate axes. These quantities describe the man-ner in which the mass of a rigid body is distributed with respect to the chosen axes. The calculation of moments and products of inertia is ex-plained fully in Appendix B. The double subscripts for the moments and products of inertia preserve a symmetry of notation which has special meaning in their description by tensor notation. Observe that Ixy Iyx, Ixz Izx, and Iyz Izy. With the substitutions of Eqs. 7/10, the expression for H becomes (7/11) and the components of H are clearly (7/12) Equation 7/11 is the general expression for the angular momentum ei-ther about the mass center G or about a fixed point O for a rigid body rotating with an instantaneous angular velocity .
Remember that in each of the two cases represented, the reference axes x-y-z are attached to the rigid body. This attachment makes the Hz Izxx Izyy Izzz Hy Iyxx Iyyy Iyzz Hx Ixxx Ixyy Ixzz (Izxx Izyy Izzz)k (Iyxx Iyyy Iyzz)j H ( Ixxx Ixyy Ixzz)i Izz (x2 y2) dm Iyz yz dm Iyy (z2 x2) dm Ixz xz dm Ixx (y2 z2) dm Ixy xy dm k[ zxx zyy (x2 y2)z] dm j[ yxx (z2 x2)y yzz] dm dH i[(y2 z2)x xyy xzz] dm 540 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies See, for example, the first author’s Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 41.
Article 7/7 Angular Momentum 541 moment-of-inertia integrals and the product-of-inertia integrals of Eqs.
7/10 invariant with time. If the x-y-z axes were to rotate with respect to an irregular body, then these inertia integrals would be functions of the time, which would introduce an undesirable complexity into the angular-momentum relations. An important exception occurs when a rigid body is spinning about an axis of symmetry, in which case, the inertia inte-grals are not affected by the angular position of the body about its spin axis. Thus, for a body rotating about an axis of symmetry, it is frequently convenient to choose one axis of the reference system to coincide with the axis of rotation and allow the other two axes not to turn with the body. In addition to the momentum components due to the angular ve-locity of the reference axes, then, an added angular-momentum com-ponent along the spin axis due to the relative spin about the axis would have to be accounted for.
Principal Axes The array of moments and products of inertia which appear in Eq. 7/12 is called the inertia matrix or inertia tensor. As we change the orientation of the axes relative to the body, the moments and products of inertia will also change in value. It can be shown that there is one unique orientation of axes x-y-z for a given origin for which the products of inertia vanish and the moments of inertia Ixx, Iyy, Izz take on stationary values. For this orientation, the inertia matrix takes the form and is said to be diagonalized. The axes x-y-z for which the products of inertia vanish are called the principal axes of inertia, and Ixx, Iyy, and Izz are called the principal moments of inertia. The principal moments of inertia for a given origin represent the maximum, the minimum, and an intermediate value of the moments of inertia.
If the coordinate axes coincide with the principal axes of inertia, Eq.
7/11 for the angular momentum about the mass center or about a fixed point becomes (7/13) It is always possible to locate the principal axes of inertia for a gen-eral three-dimensional rigid body. Thus, we can express its angular mo-mentum by Eq. 7/13, although it may not always be convenient to do so H Ixxxi Iyyy j Izzzk See, for example, the first author’s Dynamics, 2nd Edition, SI Version, 1975, John Wiley & Sons, Art. 41.
for geometric reasons. Except when the body rotates about one of the principal axes of inertia or when Ixx Iyy Izz, the vectors H and have different directions.
Transfer Principle for Angular Momentum The momentum properties of a rigid body may be represented by the resultant linear-momentum vector G through the mass center and the resultant angular-momentum vector HG about the mass center, as shown in Fig. 7/13. Although HG has the properties of a free vector, we represent it through G for convenience.
These vectors have properties analogous to those of a force and a couple. Thus, the angular momentum about any point P equals the free vector HG plus the moment of the linear-momentum vector G about P.
Therefore, we may write (7/14) This relation, which was derived previously in Chapter 4 as Eq. 4/10, also applies to a fixed point O on the body or body extended, where O merely replaces P. Equation 7/14 constitutes a transfer theorem for an-gular momentum.
7/8 Kinetic Energy In Art. 4/3 on the dynamics of systems of particles, we developed the expression for the kinetic energy T of any general system of mass, rigid or nonrigid, and obtained the result [4/4] where is the velocity of the mass center and i is the position vector of a representative element of mass mi with respect to the mass center. We identified the first term as the kinetic energy due to the translation of the system and the second term as the kinetic energy associated with the motion relative to the mass center. The translational term may be written alternatively as where is the velocity of the mass center and G is the linear momen-tum of the body.
For a rigid body, the relative term becomes the kinetic energy due to rotation about the mass center. Because is the velocity of the rep-resentative particle with respect to the mass center, then for the rigid body we may write it as i, where is the angular velocity of the body. With this substitution, the relative term in the kinetic energy expression becomes Σ 1 2 mi ˙i2 Σ 1 2 mi( i) ( i) ˙i ˙i v r ˙ 1 2 mv 2 1 2 mr ˙ r ˙ 1 2 vG v T 1 2 mv 2 Σ 1 2 mi ˙i2 HP HG r G mv 542 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies P G HG G = mv r Figure 7/13 Article 7/8 Kinetic Energy 543 If we use the fact that the dot and the cross may be interchanged in the triple scalar product, that is, P R, we may write Because is the same factor in all terms of the summation, it may be factored out to give where HG is the same as the integral expressed by Eq. 7/8. Thus, the general expression for the kinetic energy of a rigid body moving with mass-center velocity and angular velocity is (7/15) Expansion of this vector equation by substitution of the expression for HG written from Eq. 7/11 yields (7/16) If the axes coincide with the principal axes of inertia, the kinetic energy is merely (7/17) When a rigid body is pivoted about a fixed point O or when there is a point O in the body which momentarily has zero velocity, the kinetic energy is T This expression reduces to (7/18) where HO is the angular momentum about O, as may be seen by replac-ing i in the previous derivation by ri, the position vector from O. Equa-tions 7/15 and 7/18 are the three-dimensional counterparts of Eqs. 6/9 and 6/8 for plane motion.
T 1 2 HO Σ 1 2mir ˙i r ˙i.
T 1 2 mv 2 1 2 (Ixxx 2 Iyyy 2 Izzz 2) (Ixyxy Ixzxz Iyzyz) T 1 2 mv 2 1 2 (Ixxx 2 Iyyy 2 Izzz 2) T 1 2 v G 1 2 HG v Σ 1 2 mi ˙i2 1 2 Σi mi( i) 1 2 HG ( i) ( i) i ( i) QR P Q Portions of the landing gear for a large aircraft undergo three-dimensional motion during retraction and deployment.
Media Bakery 544 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies SAMPLE PROBLEM 7/6 The bent plate has a mass of 70 kg per square meter of surface area and re-volves about the z-axis at the rate 30 rad/s. Determine (a) the angular mo-mentum H of the plate about point O and (b) the kinetic energy T of the plate.
Neglect the mass of the hub and the thickness of the plate compared with its sur-face dimensions.
Solution.
The moments and products of inertia are written with the aid of Eqs. B/3 and B/9 in Appendix B by transfer from the parallel centroidal axes for each part. First, the masses of the parts are mA (0.100)(0.125)(70) 0.875 kg and mB (0.075)(0.150)(70) 0.788 kg.
Part A Part B The sum of the respective inertia terms gives for the two plates together (a) The angular momentum of the body is given by Eq. 7/11, where z 30 rad/s and x and y are zero. Thus, Ans.
(b) The kinetic energy from Eq. 7/18 becomes Ans. 8.25 J T 1 2 HO 1 2(30k) 30(0.002 21i 0.010 12j 0.018 34k) HO 30(0.002 21i 0.010 12j 0.018 34k) Nms Izz 0.018 34 kg m2 Iyz 0.010 12 kg m2 Iyy 0.010 30 kg m2 Ixz 0.002 21 kg m2 Ixx 0.0257 kg m2 Ixy 0.003 69 kg m2 Iyz 0 0.788(0.125)(0.075) 0.007 38 kg m2 [Iyz Iyz mdydz] Ixz 0 0.788(0.0375)(0.075) 0.002 21 kg m2 [Ixz Ixz mdxdz] Ixy 0 0.788(0.0375)(0.125) 0.003 69 kg m2 [Ixy Ixy mdxdy] 0.013 78 kg m2 Izz 0.788 12 (0.075)2 0.788[(0.125)2 (0.0375)2] [Izz Izz md2] 0.788[(0.0375)2 (0.075)2] 0.007 38 kg m2 Iyy 0.788 12 [(0.075)2 (0.150)2] [Iyy Iyy md2] 0.018 21 kg m2 Ixx 0.788 12 (0.150)2 0.788[(0.125)2 (0.075)2] [Ixx Ixx md2] [Iyz Iyz mdydz] Iyz 0 0.875(0.0625)(0.050) 0.002 73 kg m2 Ixy xy dm, Ixz xz dm Ixy 0 Ixz 0 Izz 0.875 3 (0.125)2 0.004 56 kg m2 [Izz 1 3ml2] Iyy 0.875 3 (0.100)2 0.002 92 kg m2 [Iyy 1 3ml2] 0.875[(0.050)2 (0.0625)2] 0.007 47 kg m2 Ixx 0.875 12 [(0.100)2 (0.125)2] [Ixx Ixx md2] 125 mm 150 mm 75 mm x A B y z 100 mm O ω x A HO B y z O ω Helpful Hints The parallel-axis theorems for trans-ferring moments and products of in-ertia from centroidal axes to parallel axes are explained in Appendix B and are most useful relations.
Recall that the units of angular mo-mentum may also be written in the base units as kgm2/s.
Article 7/8 Problems 545 7/55 The aircraft landing gear viewed from the front is being retracted immediately after takeoff, and the wheel is spinning at the rate corresponding to the takeoff speed of 200 km/h. The 45-kg wheel has a ra-dius of gyration about its z-axis of 370 mm. Neglect the thickness of the wheel and calculate the angular momentum of the wheel about G and about A for the position where is increasing at the rate of 30 per second.
Problem 7/55 7/56 The bent rod has a mass per unit length and ro-tates about the z-axis with an angular velocity .
Determine the angular momentum HO of the rod about the fixed origin O of the axes, which are at-tached to the rod. Also find the kinetic energy T of the rod.
Problem 7/56 b ω b y z b x O 215 mm 920 mm x z A G θ PROBLEMS Introductory Problems 7/53 The three small spheres, each of mass m, are rigidly mounted to the horizontal shaft which rotates with the angular velocity as shown. Neglect the radius of each sphere compared with the other dimensions and write expressions for the magnitudes of their linear momentum G and their angular momentum about the origin O of the coordinates.
Problem 7/53 7/54 The spheres of Prob. 7/53 are replaced by three rods, each of mass m and length l, mounted at their cen-ters to the shaft, which rotates with the angular velocity as shown. The axes of the rods are, respec-tively, in the x-, y-, and z-directions, and their diame-ters are negligible compared with the other dimensions. Determine the angular momentum of the three rods with respect to the coordinate origin O.
Problem 7/54 HO HO b b b b b ω x y O z b b b b b ω x y O z 7/57 Use the results of Prob. 7/56 and determine the an-gular momentum HG of the bent rod of that problem about its mass center G using the given reference axes.
7/58 The slender rod of mass m and length l rotates about the y-axis as the element of a right-circular cone. If the angular velocity about the y-axis is , determine the expression for the angular momen-tum of the rod with respect to the x-y-z axes for the particular position shown.
Problem 7/58 Representative Problems 7/59 The solid half-circular cylinder of mass m revolves about the z-axis with an angular velocity as shown.
Determine its angular momentum H with respect to the x-y-z axes.
Problem 7/59 7/60 The solid circular cylinder of mass m, radius r, and length b revolves about its geometric axis at an an-gular rate p rad/s. Simultaneously, the bracket and attached shaft revolve about the x-axis at the rate rad/s. Write the expression for the angular mo-mentun of the cylinder about O with reference axes as shown.
HO ω O y z x b c r ω θ y z l O x Problem 7/60 7/61 The elements of a reaction-wheel attitude-control system for a spacecraft are shown in the figure.
Point G is the center of mass for the system of the spacecraft and wheels, and x, y, z are principal axes for the system. Each wheel has a mass m and a mo-ment of inertia I about its own axis and spins with a relative angular velocity p in the direction indicated.
The center of each wheel, which may be treated as a thin disk, is a distance b from G. If the spacecraft has angular velocity components x, y, and z, determine the angular momentum HG of the three wheels as a unit.
Problem 7/61 G z y x ω h O r b y x z p 546 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/8 Problems 547 7/64 The rectangular plate, with a mass of 3 kg and a uni-form small thickness, is welded at the 45 angle to the vertical shaft, which rotates with the angular velocity of 20 rad/s. Determine the angular mo-mentum H of the plate about O and find the kinetic energy of the plate.
Problem 7/64 7/65 The circular disk of mass m and radius r is mounted on the vertical shaft with an angle between its plane and the plane of rotation of the shaft. Deter-mine an expression for the angular momentum H of the disk about O. Find the angle which the angular momentum H makes with the shaft if 10.
Problem 7/65 ω α x x ′ z y O r x z y O 45° 200 mm 200 mm 100 mm 100 mm ω π = 20 rad/s 7/62 The gyro rotor is spinning at the constant rate p 100 rev/min relative to the x-y-z axes in the direction indicated. If the angle between the gimbal ring and horizontal X-Y plane is made to increase at the rate of 4 rad/sec and if the unit is forced to precess about the vertical at the constant rate N 20 rev/min, calculate the angular momentum HO of the rotor when 30.
The axial and transverse moments of inertia are Izz 5(103) lb-ft-sec2 and Ixx Iyy 2.5(103) lb-ft-sec2.
Problem 7/62 7/63 The slender steel rod AB weighs 6.20 lb and is secured to the rotating shaft by the rod OG and its fittings at O and G. The angle remains constant at 30, and the entire rigid assembly rotates about the z-axis at the steady rate N 600 rev/min. Calculate the angu-lar momentum HO of AB and its kinetic energy T.
Problem 7/63 y z x G A B β 14″ 14″ 16″ O N γ N Z O X y z p Y x 7/66 The right-circular cone of height h and base radius r spins about its axis of symmetry with an angular rate p. Simultaneously, the entire cone revolves about the x-axis with angular rate . Determine the angular momentum HO of the cone about the origin O of the x-y-z axes and the kinetic energy T for the position shown. The mass of the cone is m.
Problem 7/66 7/67 Each of the slender rods of length l and mass m is welded to the circular disk which rotates about the vertical z-axis with an angular velocity . Each rod makes an angle with the vertical and lies in a plane parallel to the y-z plane. Determine an expres-sion for the angular momentum HO of the two rods about the origin O of the axes.
Problem 7/66 b b l l O y x z β β ω z h x y p O r Ω 7/68 The spacecraft shown has a mass m with mass center G. Its radius of gyration about its z-axis of rotational symmetry is k and that about either the x- or y-axis is k. In space, the spacecraft spins within its x-y-z reference frame at the rate p Simultaneously, a point C on the z-axis moves in a circle about the z0-axis with a frequency ƒ (rotations per unit time). The z0-axis has a constant direction in space. Determine the angular momentum HG of the spacecraft relative to the axes designated. Note that the x-axis always lies in the z-z0 plane and that the y-axis is therefore normal to z0.
Problem 7/68 7/69 The uniform circular disk of Prob. 7/48 with the three components of angular velocity is shown again here. Determine the kinetic energy T and the angu-lar momentum HO with respect to O of the disk for the instant represented, when the x-y plane coin-cides with the X-Y plane. The mass of the disk is m.
Problem 7/69 Y y A x X z Z p B O r b 2 ω 1 ω C G x y z z0 φ ˙.
548 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/8 Problems 549 Problem 7/71 7/72 In a test of the solar panels for a spacecraft, the model shown is rotated about the vertical axis at the angular rate . If the mass per unit area of panel is , write the expression for the angular momentum HO of the assembly about the axes shown in terms of .
Also determine the maximum, minimum, and inter-mediate values of the moment of inertia about the axes through O.
Problem 7/72 θ θ ω x a a b b O y b b z c c b b c m y x m O z c β ω ω r 7/70 The 4-in.-radius wheel weighs 6 lb and turns about its y-axis with an angular velocity p 40 rad/sec in the direction shown. Simultaneously, the fork ro-tates about its x-axis shaft with an angular velocity 10 rad/sec as indicated. Calculate the angular momentum of the wheel about its center O. Also compute the kinetic energy of the wheel.
Problem 7/70 7/71 The assembly, consisting of the solid sphere of mass m and the uniform rod of length 2c and equal mass m, revolves about the vertical z-axis with an angular velocity . The rod of length 2c has a diameter which is small compared with its length and is perpendicu-lar to the horizontal rod to which it is welded with the inclination shown. Determine the combined an-gular momentum HO of the sphere and inclined rod.
O O′ y y ′ z′ z x x ′ 10″ 4″ p = 40 rad/sec π = 10 rad/sec π ω 550 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/9 Momentum and Energy Equations of Motion With the description of angular momentum, inertial properties, and kinetic energy of a rigid body established in the previous two articles, we are ready to apply the general momentum and energy equations of motion.
Momentum Equations In Art. 4/4 of Chapter 4, we established the general linear- and angular-momentum equations for a system of constant mass. These equations are [4/6] [4/7] or [4/9] The general moment relation, Eq. 4/7 or 4/9, is expressed here by the single equation ΣM where the terms are taken either about a fixed point O or about the mass center G. In the derivation of the moment principle, the derivative of H was taken with respect to an absolute co-ordinate system. When H is expressed in terms of components measured relative to a moving coordinate system x-y-z which has an angular veloc-ity , then by Eq. 7/7 the moment relation becomes The terms in parentheses represent that part of due to the change in magnitude of the components of H, and the cross-product term repre-sents that part due to the changes in direction of the components of H.
Expansion of the cross product and rearrangement of terms give (7/19) Equation 7/19 is the most general form of the moment equation about a fixed point O or about the mass center G. The ’s are the angu-lar velocity components of rotation of the reference axes, and the H-components in the case of a rigid body are as defined in Eq. 7/12, where the ’s are the components of the angular velocity of the body.
We now apply Eq. 7/19 to a rigid body where the coordinate axes are attached to the body. Under these conditions, when expressed in the x-y-z coordinates, the moments and products of inertia are invariant with time, (H ˙z Hxy Hyx)k (H ˙y Hzx Hxz)j ΣM (H ˙x Hyz Hzy)i H ˙ (H ˙xi H ˙y j H ˙zk) H ΣM dH dtxyz H H ˙, ΣM H ˙ ΣF G ˙ Article 7/9 Momentum and Energy Equations of Motion 551 and . Thus, for axes attached to the body, the three scalar compo-nents of Eq. 7/19 become (7/20) Equations 7/20 are the general moment equations for rigid-body motion with axes attached to the body. They hold with respect to axes through a fixed point O or through the mass center G. ΣMz H ˙z Hxy Hyx ΣMy H ˙y Hzx Hxz ΣMx H ˙x Hyz Hzy Named after Leonhard Euler (1707–1783), a Swiss mathematician.
KEY CONCEPTS In Art. 7/7 it was mentioned that, in general, for any origin fixed to a rigid body, there are three principal axes of inertia with respect to which the products of inertia vanish. If the reference axes coincide with the principal axes of inertia with origin at the mass center G or at a point O fixed to the body and fixed in space, the factors Ixy, Iyz, Ixz will be zero, and Eqs. 7/20 become (7/21) These relations, known as Euler’s equations, are extremely useful in the study of rigid-body motion.
Energy Equations The resultant of all external forces acting on a rigid body may be re-placed by the resultant force ΣF acting through the mass center and a resultant couple ΣMG acting about the mass center. Work is done by the resultant force and the resultant couple at the respective rates and where is the linear velocity of the mass center and is the angular velocity of the body. Integration over the time from condi-tion 1 to condition 2 gives the total work done during the time interval.
Equating the works done to the respective changes in kinetic energy as expressed in Eq. 7/15 gives (7/22) These equations express the change in translational kinetic energy and the change in rotational kinetic energy, respectively, for the interval during which ΣF or ΣMG acts, and the sum of the two expressions equals T.
v ΣMG, ΣFv ΣMz Izz ˙z (Ixx Iyy)xy ΣMy Iyy ˙y (Izz Ixx)zx ΣMx Ixx ˙x (Iyy Izz)yz 552 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies The work-energy relationship, developed in Chapter 4 for a general system of particles and given by [4/3] was used in Chapter 6 for rigid bodies in plane motion. The equation is equally applicable to rigid-body motion in three dimensions. As we have seen previously, the work-energy approach is of great advantage when we analyze the initial and final end-point conditions of motion.
Here the work done during the interval by all active forces exter-nal to the body or system is equated to the sum of the corresponding changes in kinetic energy T and potential energy V. The potential-energy change is determined in the usual way, as described previously in Art. 3/7.
We will limit our application of the equations developed in this arti-cle to two problems of special interest, parallel-plane motion and gyro-scopic motion, discussed in the next two articles.
7/10 Parallel-Plane Motion When all particles of a rigid body move in planes which are parallel to a fixed plane, the body has a general form of plane motion, as de-scribed in Art. 7/4 and pictured in Fig. 7/3. Every line in such a body which is normal to the fixed plane remains parallel to itself at all times.
We take the mass center G as the origin of coordinates x-y-z which are attached to the body, with the x-y plane coinciding with the plane of mo-tion P. The components of the angular velocity of both the body and the attached axes become x y 0, z 0. For this case, the angular-momentum components from Eq. 7/12 become and the moment relations of Eqs. 7/20 reduce to (7/23) We see that the third moment equation is equivalent to the second of Eqs. 6/1, where the z-axis passes through the mass center, or to Eq. 6/4 if the z-axis passes through a fixed point O.
Equations 7/23 hold for an origin of coordinates at the mass center, as shown in Fig. 7/3, or for any origin on a fixed axis of rotation. The three independent force equations of motion which also apply to parallel-plane motion are clearly Equations 7/23 find special use in describing the effect of dynamic imbalance in rotating machinery and in rolling bodies.
ΣFx max ΣFy may ΣFz 0 ΣMz Izz ˙z ΣMy Iyz ˙z Ixzz 2 ΣMx Ixz ˙z Iyzz 2 Hx Ixzz Hy Iyzz Hz Izzz U 1-2 U 1-2 T V Article 7/10 Parallel-Plane Motion 553 SAMPLE PROBLEM 7/7 The two circular disks, each of mass m1, are connected by the curved bar bent into quarter-circular arcs and welded to the disks. The bar has a mass m2.
The total mass of the assembly is m 2m1 m2. If the disks roll without slip-ping on a horizontal plane with a constant velocity v of the disk centers, deter-mine the value of the friction force under each disk at the instant represented when the plane of the curved bar is horizontal.
Solution.
The motion is identified as parallel-plane motion since the planes of motion of all parts of the system are parallel. The free-body diagram shows the normal forces and friction forces at A and B and the total weight mg acting through the mass center G, which we take as the origin of coordinates which ro-tate with the body.
We now apply Eqs. 7/23, where Iyz 0 and 0. The moment equation about the y-axis requires determination of Ixz. From the diagram showing the geometry of the curved rod and with standing for the mass of the rod per unit length, we have Evaluating the integrals gives The second of Eqs. 7/23 with z v/r and 0 gives But with v constant, 0 so that Thus, Ans.
We also note for the given position that with Iyz 0 and 0, the moment equation about the x-axis gives NAr NBr 0 NA NB mg/2 [ΣMx 0] ˙z FA FB m2v2 2r FA FB 0 FA FB [ΣFx 0] ax v FA FB m2v2 r FAr FBr m2r2 v2 r2 [ΣMy Ixzz 2] ˙z Ixz r3/2 r3/2 r3 m2r2 /2 0 (r sin )(r r cos ) r d Ixz /2 0 (r sin )(r r cos ) r d Ixy xz dm ˙z B r v y G x z r A r r θ θ r x z dm G r y B NB FA NA FB z A mg G x Helpful Hints We must be very careful to observe the correct sign for each of the coor-dinates of the mass element dm which make up the product xz.
When the plane of the curved bar is not horizontal, the normal forces under the disks are no longer equal.
554 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/75 The uniform slender bar of length l and mass m is welded to the shaft, which rotates in bearings A and B with a constant angular velocity . Determine the expression for the force supported by the bearing at B as a function of . Consider only the force due to the dynamic imbalance and assume that the bear-ings can support radial forces only.
Problem 7/75 7/76 If a torque M Mk is applied to the shaft in Prob.
7/75, determine the x- and y-components of the force supported by the bearing B as the bar and shaft start from rest in the position shown. Neglect the mass of the shaft and consider dynamic forces only.
7/77 The paint stirrer shown in the figure is made from a rod of length 7b and mass per unit length. Before immersion in the paint, the stirrer is rotating freely at a constant high angular velocity about its z-axis.
Determine the bending moment M in the rod at the base O of the chuck.
Problem 7/77 b b b b b y ω z x O c A O B θ ω b l y x z PROBLEMS Introductory Problems 7/73 Each of the two rods of mass m is welded to the face of the disk, which rotates about the vertical axis with a constant angular velocity . Determine the bending moment M acting on each rod at its base.
Problem 7/73 7/74 The slender shaft carries two offset particles, each of mass m, and rotates about the z-axis with the con-stant angular rate as indicated. Determine the x- and y-components of the bearing reactions at A and B due to the dynamic imbalance of the shaft for the position shown.
Problem 7/74 R R m m B ω L — 3 L — 3 L — 3 z x A y O ω l l b b Article 7/10 Problems 555 7/78 The 6-kg circular disk and attached shaft rotate at a constant speed 10 000 rev/min. If the center of mass of the disk is 0.05 mm off center, determine the magnitudes of the horizontal forces A and B sup-ported by the bearings because of the rotational im-balance.
Problem 7/78 Representative Problems 7/79 Determine the bending moment M at the tangency point A in the semicircular rod of radius r and mass m as it rotates about the tangent axis with a con-stant and large angular velocity . Neglect the mo-ment mgr produced by the weight of the rod.
Problem 7/79 7/80 If the semicircular rod of Prob. 7/79 starts from rest under the action of a torque MO applied through the collar about its z-axis of rotation, determine the ini-tial bending moment M in the rod at A.
ω x z y A r ω A B 150 mm 200 mm 7/81 The large satellite-tracking antenna has a moment of inertia I about its z-axis of symmetry and a mo-ment of inertia IO about each of the x- and y-axes.
Determine the angular acceleration of the antenna about the vertical Z-axis caused by a torque M ap-plied about Z by the drive mechanism for a given ori-entation .
Problem 7/81 7/82 The plate has a mass of 3 kg and is welded to the fixed vertical shaft, which rotates at the constant speed of 20 rad/s. Compute the moment M applied to the shaft by the plate due to dynamic imbalance.
Problem 7/82 x z y O 45° 200 mm 200 mm 100 mm 100 mm ω π = 20 rad/s θ α Z M x z y 556 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/86 The circular disk of mass m and radius r is mounted on the vertical shaft with a small angle between its plane and the plane of rotation of the shaft. Deter-mine the expression for the bending moment M act-ing on the shaft due to the wobble of the disk at a shaft speed of rad/s.
Problem 7/86 7/87 The thin circular disk of mass m and radius R is hinged about its horizontal tangent axis to the end of a shaft rotating about its vertical axis with an angu-lar velocity . Determine the steady-state angle as-sumed by the plane of the disk with the vertical axis.
Observe any limitation on to ensure that 0.
Problem 7/87 7/88 Determine the normal forces under the two disks of Sample Problem 7/7 for the position where the plane of the curved bar is vertical. Take the curved bar to be at the top of disk A and at the bottom of disk B.
β β z z Detail of hinge at A x A R G ω x ω α x x ′ z y O r 7/83 Each of the two semicircular disks has a mass of 1.20 kg and is welded to the shaft supported in bear-ings A and B as shown. Calculate the forces applied to the shaft by the bearings for a constant angular speed N 1200 rev/min. Neglect the forces of static equilibrium.
Problem 7/83 7/84 Solve Prob. 7/83 for the case where the assembly starts from rest with an initial angular acceleration 900 rad/s2 as a result of a starting torque (cou-ple) M applied to the shaft in the same sense as N.
Neglect the moment of inertia of the shaft about its z-axis and calculate M.
7/85 The uniform slender bar of mass per unit length is freely pivoted about the y-axis at the clevis, which rotates about the fixed vertical z-axis with a constant angular velocity . Determine the steady-state angle assumed by the bar. Length b is greater than length c.
Problem 7/85 b z y x c θ ω 80 mm 80 mm 80 mm 80 mm y z B C D 100 mm 100 mm A N G x Article 7/10 Problems 557 7/89 The uniform square plate of mass m is welded at O to the end of the shaft, which rotates about the verti-cal z-axis with a constant angular velocity . Deter-mine the moment applied to the plate by the weld due only to the rotation.
Problem 7/89 7/90 For the plate of mass m in Prob. 7/89, determine the y- and z-components of the moment applied to the plate by the weld at O necessary to give the plate an angular acceleration starting from rest. Ne-glect the moment due to the weight.
7/91 The uniform slender rod of length l is welded to the bracket at A on the underside of the disk B. The disk rotates about a vertical axis with a constant angular velocity . Determine the value of which will re-sult in a zero moment supported by the weld at A for the position 60 with b l/4.
Problem 7/91 θ l ω A B b ˙ b y x z b/2 O b/2 β ω 7/92 The half-cylindrical shell of radius r, length 2b, and mass m revolves about the vertical z-axis with a con-stant angular velocity as indicated. Determine the magnitude M of the bending moment in the shaft at A due to both the weight and the rotational motion of the shell.
Problem 7/92 7/93 The homogeneous thin triangular plate of mass m is welded to the horizontal shaft, which rotates freely in the bearings at A and B. If the plate is released from rest in the horizontal position shown, deter-mine the magnitude of the bearing reaction at A for the instant just after release.
Problem 7/93 7/94 If the homogeneous triangular plate of Prob. 7/93 is released from rest in the position shown, determine the magnitude of the bearing reaction at A after the plate has rotated 90.
b a y x z A B a A y x z ω r r b b 7/11 Gyroscopic Motion: Steady Precession One of the most interesting of all problems in dynamics is that of gyroscopic motion. This motion occurs whenever the axis about which a body is spinning is itself rotating about another axis. Although the com-plete description of this motion involves considerable complexity, the most common and useful examples of gyroscopic motion occur when the axis of a rotor spinning at constant speed turns (precesses) about an-other axis at a steady rate. Our discussion in this article will focus on this special case.
The gyroscope has important engineering applications. With a mounting in gimbal rings (see Fig. 7/19b), the gyro is free from external moments, and its axis will retain a fixed direction in space regardless of the rotation of the structure to which it is attached. In this way, the gyro is used for inertial guidance systems and other directional control devices. With the addition of a pendulous mass to the inner gimbal ring, the earth’s rotation causes the gyro to precess so that the spin axis will always point north, and this action forms the basis of the gyro compass.
The gyroscope has also found important use as a stabilizing device. The controlled precession of a large gyro mounted in a ship is used to pro-duce a gyroscopic moment to counteract the rolling of a ship at sea. The gyroscopic effect is also an extremely important consideration in the de-sign of bearings for the shafts of rotors which are subjected to forced precessions.
We will first describe gyroscopic action with a simple physical ap-proach which relies on our previous experience with the vector changes encountered in particle kinetics. This approach will help us gain a direct physical insight into gyroscopic action. Next, we will make use of the general momentum relation, Eq. 7/19, for a more complete description.
Simplified Approach Figure 7/14 shows a symmetrical rotor spinning about the z-axis with a large angular velocity p, known as the spin velocity. If we apply two forces F to the rotor axle to form a couple M whose vector is di-rected along the x-axis, we will find that the rotor shaft rotates in the x-z plane about the y-axis in the sense indicated, with a relatively slow angular velocity known as the precession velocity. Thus, we iden-tify the spin axis (p), the torque axis (M), and the precession axis (), where the usual right-hand rule identifies the sense of the rotation vec-tors. The rotor shaft does not turn about the x-axis in the sense of M, as it would if the rotor were not spinning. To aid understanding of this phenomenon, a direct analogy may be made between the rotation vec-tors and the familiar vectors which describe the curvilinear motion of a particle.
Figure 7/15a shows a particle of mass m moving in the x-z plane with constant speed v v. The application of a force F normal to its linear momentum G mv causes a change dG d(mv) in its momentum. We see that dG, and thus dv, is a vector in the direction of the normal force F according to Newton’s second law F which may be written as G ˙, ˙ Spin axis Precession axis Torque axis y F F M p z x Ω ψ ψ = · ⏐ ⏐ Figure 7/14 558 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/11 Gyroscopic Motion: Steady Precession 559 F dt dG. From Fig. 7/15b we see that, in the limit, tan d d F dt/mv or F In vector notation with the force becomes which is the vector equivalent of our familiar scalar relation Fn man for the normal force on the particle, as treated extensively in Chapter 3.
With these relations in mind, we now turn to our problem of rota-tion. Recall now the analogous equation M which we developed for any prescribed mass system, rigid or nonrigid, referred to its mass cen-ter (Eq. 4/9) or to a fixed point O (Eq. 4/7). We now apply this relation to our symmetrical rotor, as shown in Fig. 7/15c. For a high rate of spin p and a low precession rate about the y-axis, the angular momentum is represented by the vector H Ip, where I Izz is the moment of inertia of the rotor about the spin axis.
Initially, we neglect the small component of angular momentum about the y-axis which accompanies the slow precession. The application of the couple M normal to H causes a change dH d(Ip) in the angular momentum. We see that dH, and thus dp, is a vector in the direction of the couple M since M which may also be written M dt dH. Just as the change in the linear-momentum vector of the particle is in the di-rection of the applied force, so is the change in the angular-momentum vector of the gyro in the direction of the couple. Thus, we see that the vectors M, H, and dH are analogous to the vectors F, G, and dG. With this insight, it is no longer strange to see the rotation vector undergo a change which is in the direction of M, thereby causing the axis of the rotor to precess about the y-axis.
In Fig. 7/15d we see that during time dt the angular-momentum vector Ip has swung through the angle d , so that in the limit with tan d d , we have Substituting d /dt for the magnitude of the precession velocity gives us (7/24) We note that M, , and p as vectors are mutually perpendicular, and that their vector relationship may be represented by writing the equa-tion in the cross-product form (7/24a) which is completely analogous to the foregoing relation F m v for the curvilinear motion of a particle as developed from Figs. 7/15a and b.
M I p M Ip d M dt Ip or M I d dt p H ˙, H ˙ F m v ˙j, mv ˙.
F (F = G) x y z dG = d(mv) G = mv (a) = j x y z F dt = d(mv) mv (b) M dθ ω x y z M dt = d(Ip) Ip (d) (c) dψ (M = H) y H = Ip dH = d(Ip) z x Ω ψ = · j · · θ · Figure 7/15 Equations 7/24 and 7/24a apply to moments taken about the mass cen-ter or about a fixed point on the axis of rotation.
The correct spatial relationship among the three vectors may be re-membered from the fact that dH, and thus dp, is in the direction of M, which establishes the correct sense for the precession . Therefore, the spin vector p always tends to rotate toward the torque vector M. Figure 7/16 represents three orientations of the three vectors which are consis-tent with their correct order. Unless we establish this order correctly in a given problem, we are likely to arrive at a conclusion directly opposite to the correct one. Remember that Eq. 7/24, like F ma and M I, is an equation of motion, so that the couple M represents the couple due to all forces acting on the rotor, as disclosed by a correct free-body diagram of the rotor. Also note that, when a rotor is forced to precess, as occurs with the turbine in a ship which is executing a turn, the motion will generate a gyroscopic couple M which obeys Eq. 7/24a in both magni-tude and sense.
In the foregoing discussion of gyroscopic motion, it was assumed that the spin was large and the precession was small. Although we can see from Eq. 7/24 that for given values of I and M, the precession must be small if p is large, let us now examine the influence of on the momentum relations. Again, we restrict our attention to steady preces-sion, where has a constant magnitude.
Figure 7/17 shows our same rotor again. Because it has a moment of inertia about the y-axis and an angular velocity of precession about this axis, there will be an additional component of angular momentum about the y-axis. Thus, we have the two components Hz Ip and Hy I0, where I0 stands for Iyy and, again, I stands for Izz. The total angular mo-mentum is H as shown. The change in H remains dH M dt as previ-ously, and the precession during time dt is the angle d M dt/Hz M dt/(Ip) as before. Thus, Eq. 7/24 is still valid and for steady precession is an exact description of the motion as long as the spin axis is perpen-dicular to the axis around which precession occurs.
Consider now the steady precession of a symmetrical top, Fig.
7/18, spinning about its axis with a high angular velocity p and sup-ported at its point O. Here the spin axis makes an angle with the vertical Z-axis around which precession occurs. Again, we will neglect the small angular-momentum component due to the precession and consider H equal to Ip, the angular momentum about the axis of the top associated with the spin only. The moment about O is due to the weight and is sin , where is the distance from O to the mass center G. From the diagram, we see that the angular-momentum vec-tor HO has a change dHO MO dt in the direction of MO during time dt and that is unchanged. The increment in precessional angle around the Z-axis is Substituting the values MO sin and d /dt gives mgr sin Ip sin or mgr Ip mgr d MO dt Ip sin r mgr 560 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Ω Ω Ω p M Figure 7/16 Ω Ω M Ω p Ω p M Ω Ω y dH H Hy H′ Hz M dt = dH z x dψ Ω ψ = · j Figure 7/17 dHO = MO dt HO ≈ Ip θ θ z Z x X MO mg O G y Y dψ r _ Figure 7/18 Article 7/11 Gyroscopic Motion: Steady Precession 561 which is independent of . Introducing the radius of gyration so that I mk2 and solving for the precessional velocity give (7/25) Unlike Eq. 7/24, which is an exact description for the rotor of Fig.
7/17 with precession confined to the x-z plane, Eq. 7/25 is an approxima-tion based on the assumption that the angular momentum associated with is negligible compared with that associated with p. We will see the amount of the error associated with this approximation when we re-consider steady-state precession later in this article. On the basis of our analysis, the top will have a steady precession at the constant angle only if it is set in motion with a value of which satisfies Eq. 7/25.
When these conditions are not met, the precession becomes unsteady, and may oscillate with an amplitude which increases as the spin veloc-ity decreases. The corresponding rise and fall of the rotation axis is called nutation.
More Detailed Analysis We now make direct use of Eq. 7/19, which is the general angular-momentum equation for a rigid body, by applying it to a body spinning about its axis of rotational symmetry. This equation is valid for rotation about a fixed point or for rotation about the mass center. A spinning top, the rotor of a gyroscope, and a spacecraft are examples of bodies whose motions can be described by the equations for rotation about a point. The general moment equations for this class of problems are fairly complex, and their complete solutions involve the use of elliptic in-tegrals and somewhat lengthy computations. However, a large fraction of engineering problems where the motion is one of rotation about a point involves the steady precession of bodies of revolution which are spinning about their axes of symmetry. These conditions greatly sim-plify the equations and thus facilitate their solution.
Consider a body with axial symmetry, Fig. 7/19a, rotating about a fixed point O on its axis, which is taken to be the z-direction. With O as origin, the x- and y-axes automatically become principal axes of inertia along with the z-axis. This same description may be used for the rota-tion of a similar symmetrical body about its center of mass G, which is taken as the origin of coordinates as shown with the gimbaled gyroscope rotor of Fig. 7/19b. Again, the x- and y-axes are principal axes of inertia for point G. The same description may also be used to represent the ro-tation about the mass center of an axially symmetric body in space, such as the spacecraft in Fig. 7/19c. In each case, we note that, regardless of the rotation of the axes or of the body relative to the axes (spin about the z-axis), the moments of inertia about the x- and y-axes remain con-stant with time. The principal moments of inertia are again designated Izz I and Ixx Iyy I0. The products of inertia are, of course, zero.
Before applying Eq. 7/19, we introduce a set of coordinates which provide a natural description for our problem. These coordinates are gr k2p G x y z z y x (a) (b) (c) O θ z y x G Figure 7/19 shown in Fig. 7/20 for the example of rotation about a fixed point O. The axes X-Y-Z are fixed in space, and plane A contains the X-Y axes and the fixed point O on the rotor axis. Plane B contains point O and is always normal to the rotor axis. Angle measures the inclination of the rotor axis from the vertical Z-axis and is also a measure of the angle between planes A and B. The intersection of the two planes is the x-axis, which is located by the angle from the X-axis. The y-axis lies in plane B, and the z-axis coincides with the rotor axis. The angles and completely specify the position of the rotor axis. The angular displacement of the rotor with respect to axes x-y-z is specified by the angle measured from the x-axis to the x-axis, which is attached to the rotor. The spin velocity becomes p ˙.
562 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies ΣMz ΣMy ΣMx Ωz = cos ψ θ z = Ωz + p ω where p = φ Ωy = y = sin ω ψ θ x Y X B A z Z y x′ ψ θ φ O φ Ωx = x = ω θ · · · · · Figure 7/20 The components of the angular velocity of the rotor and the angu-lar velocity of the axes x-y-z from Fig. 7/20 become It is important to note that the axes and the body have identical x- and y-components of angular velocity, but that the z-components differ by the relative angular velocity p.
The angular-momentum components from Eq. 7/12 become Hz Izzz I( ˙ cos p) Hy Iyyy I0 ˙ sin Hx Ixxx I0 ˙ z ˙ cos z ˙ cos p y ˙ sin y ˙ sin x ˙ x ˙ Article 7/11 Gyroscopic Motion: Steady Precession 563 Substitution of the angular-velocity and angular-momentum com-ponents into Eq. 7/19 yields (7/26) Equations 7/26 are the general equations of rotation of a symmetrical body about either a fixed point O or the mass center G. In a given prob-lem, the solution to the equations will depend on the moment sums ap-plied to the body about the three coordinate axes. We will confine our use of these equations to two particular cases of rotation about a point which are described in the following sections.
Steady-State Precession We now examine the conditions under which the rotor precesses at a steady rate at a constant angle and with constant spin velocity p. Thus, and Eqs. 7/26 become (7/27) From these results, we see that the required moment acting on the rotor about O (or about G) must be in the x-direction since the y- and z-components are zero. Furthermore, with the constant values of , and p, the moment is constant in magnitude. It is also important to note that the moment axis is perpendicular to the plane defined by the pre-cession axis (Z-axis) and the spin axis (z-axis).
We may also obtain Eqs. 7/27 by recognizing that the components of H remain constant as observed in x-y-z so that 0. Because in general ΣM H, we have for the case of steady precession (7/28) which reduces to Eqs. 7/27 upon substitution of the values of and H.
By far the most common engineering examples of gyroscopic motion occur when precession takes place about an axis which is normal to the rotor axis, as in Fig. 7/14. Thus with the substitution /2, z p, , and ΣMx M, we have from Eqs. 7/27 [7/24] M Ip ˙ ΣM H (H ˙)xyz (H ˙)xyz ˙, ΣMz 0 ΣMy 0 ΣMx ˙ sin [I( ˙ cos p) I0 ˙ cos ] p constant, p ˙ 0 constant, ˙ ¨ 0 ˙ constant, ¨ 0 ˙ ΣMz I d dt ( ˙ cos p) ΣMy I0( ¨ sin 2 ˙ ˙ cos ) I ˙( ˙ cos p) ΣMx I0( ¨ ˙2 sin cos ) I ˙( ˙ cos p) sin which we derived initially in this article from a direct analysis of this special case.
Now let us examine the steady precession of the rotor (symmetrical top) of Fig. 7/20 for any constant value of other than /2. The moment ΣMx about the x-axis is due to the weight of the rotor and is sin .
Substitution into Eqs. 7/27 and rearrangement of terms give us We see that is small when p is large, so that the second term on the right-hand side of the equation becomes very small compared with If we neglect this smaller term, we have which, upon use of the previous substitution and mk2 I, becomes [7/25] We derived this same relation earlier by assuming that the angular mo-mentum was entirely along the spin axis.
Steady Precession with Zero Moment Consider now the motion of a symmetrical rotor with no external moment about its mass center. Such motion is encountered with space-craft and projectiles which both spin and precess during flight.
Figure 7/21 represents such a body. Here the Z-axis, which has a fixed direction in space, is chosen to coincide with the direction of the angular momentum HG, which is constant since ΣMG 0. The x-y-z axes are at-tached in the manner described in Fig. 7/20. From Fig. 7/21 the three com-ponents of momentum are 0, HG sin , HG cos .
From the defining relations, Eqs. 7/12, with the notation of this article, these components are also given by I0x, I0y, Iz.
Thus, x x 0 so that is constant. This result means that the mo-tion is one of steady precession about the constant HG vector.
With no x-component, the angular velocity of the rotor lies in the y-z plane along with the Z-axis and makes an angle with the z-axis.
The relationship between and is obtained from tan I0y/(Iz), which is (7/29) Thus, the angular velocity makes a constant angle with the spin axis.
The rate of precession is easily obtained from Eq. 7/27 with M 0, which gives (7/30) It is clear from this relation that the direction of the precession depends on the relative magnitudes of the two moments of inertia. ˙ Ip (I0 I) cos tan I0 I tan HGy /HGz HGz HGy HGx HGz HGy HGx gr k2p ˙ mgr/(Ip) ˙ I ˙p. ˙ mgr I ˙p (I0 I) ˙2 cos mgr 564 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies β ω θ ω z ω y φ ⋅ HGz HGy HG Z y z x G Figure 7/21 Article 7/11 Gyroscopic Motion: Steady Precession 565 If I0 I, then , as indicated in Fig. 7/22a, and the precession is said to be direct. Here the body cone rolls on the outside of the space cone.
If I I0, then , as indicated in Fig. 7/22b, and the precession is said to be retrograde. In this instance, the space cone is internal to the body cone, and and p have opposite signs.
If I I0, then from Eq. 7/29, and Fig. 7/22 shows that both angles must be zero to be equal. For this case, the body has no preces-sion and merely rotates with an angular velocity p. This condition oc-curs for a body with point symmetry, such as with a homogeneous sphere. ˙ θ β ω θ β p z G Z Body cone Space cone ψ ⋅ ψ ⋅ z G p Z Space cone Body cone ω Direct precession I0 > I (a) Retrograde precession I0 < I (b) Figure 7/22 Toy gyroscopes are useful in demon-strating the principles of this article.
Media Bakery 566 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies SAMPLE PROBLEM 7/8 The turbine rotor in a ship’s power plant has a mass of 1000 kg, with center of mass at G and a radius of gyration of 200 mm. The rotor shaft is mounted in bearings A and B with its axis in the horizontal fore-and-aft direction and turns counterclockwise at a speed of 500 rev/min when viewed from the stern. Deter-mine the vertical components of the bearing reactions at A and B if the ship is making a turn to port (left) of 400-m radius at a speed of 25 knots (1 knot 0.514 m/s). Does the bow of the ship tend to rise or fall because of the gyroscopic action?
Solution.
The vertical component of the bearing reactions will equal the static reactions R1 and R2 due to the weight of the rotor, plus or minus the increment R due to the gyroscopic effect. The moment principle from statics easily gives R1 5890 N and R2 3920 N. The given directions of the spin velocity p and the precession velocity are shown with the free-body diagram of the rotor. Be-cause the spin axis always tends to rotate toward the torque axis, we see that the torque axis M points in the starboard direction as shown. The sense of the R’s is, therefore, up at B and down at A to produce the couple M. Thus, the bearing reactions at A and B are The precession velocity is the speed of the ship divided by the radius of its turn.
Equation 7/24 is now applied around the mass center G of the rotor to give The required bearing reactions become Ans.
We now observe that the forces just computed are those exerted on the rotor shaft by the structure of the ship. Consequently, from the principle of action and reaction, the equal and opposite forces are applied to the ship by the rotor shaft, as shown in the bottom sketch. Therefore, the effect of the gyroscopic couple is to generate the increments R shown, and the bow will tend to fall and the stern to rise (but only slightly).
RA 5890 449 5440 N and RB 3920 449 4370 N R 449 N 1.500(R) 1000(0.200)2(0.0321) 5000(2) 60 [M Ip] 25(0.514) 400 0.0321 rad/s [v ] RA R1 R and RB R2 R 600 mm 900 mm Port (left) p A RA RB B G Starboard (right) Forward Ω M M p y z x B A 1000(9.81) N R1 R1 ΔR ΔR R2 R2 ΔR ΔR G Helpful Hints If the ship is making a left turn, the rotation is counterclockwise as viewed from above, and the preces-sion vector is up by the right-hand rule.
After figuring the correct sense of M on the rotor, the common mis-take is to apply it to the ship in the same sense, forgetting the action-and-reaction principle. Clearly, the results are then reversed. (Be cer-tain not to make this mistake when operating a vertical gyro stabilizer in your yacht to counteract its roll!) Article 7/11 Gyroscopic Motion: Steady Precession 567 SAMPLE PROBLEM 7/9 A proposed space station is closely approximated by four uniform spherical shells, each of mass m and radius r. The mass of the connecting structure and in-ternal equipment may be neglected as a first approximation. If the station is de-signed to rotate about its z-axis at the rate of one revolution every 4 seconds, determine (a) the number n of complete cycles of precession for each revolution about the z-axis if the plane of rotation deviates only slightly from a fixed orien-tation, and (b) find the period of precession if the spin axis z makes an angle of 20 with respect to the axis of fixed orientation about which precession occurs.
Draw the space and body cones for this latter condition.
Solution. (a) The number of precession cycles or wobbles for each revolution of the station about the z-axis would be the ratio of the precessional velocity to the spin velocity p, which, from Eq. 7/30, is The moments of inertia are With very small, cos 1, and the ratio of angular rates becomes Ans.
The minus sign indicates retrograde precession where, in the present case, and p are essentially of opposite sense. Thus, the station will make seven wob-bles for every three revolutions.
(b) For 20 and p 2/4 rad/s, the period of precession or wobble is so that from Eq. 7/30 Ans.
The precession is retrograde, and the body cone is external to the space cone as shown in the illustration where the body-cone angle, from Eq. 7/29, is tan I I0 tan 56/3 32/3 (0.364) 0.637 32.5 2/ ˙, ˙ n ˙ p 56 3 32 3 56 3 7 3 Ixx I0 2(2 3)mr2 2[2 3 mr2 m(2r)2] 32 3 mr2 Izz I 4[2 3 mr2 m(2r)2] 56 3 mr2 ˙ p I (I0 I) cos ˙ O z x r 2r O z = 20° θ = 32.5° β p ψ · Helpful Hint Our theory is based on the assump-tion that Ixx Iyy the moment of inertia about any axis through G perpendicular to the z-axis. Such is the case here, and you should prove it to your own satisfaction.
568 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies PROBLEMS Introductory Problems 7/95 A dynamics instructor demonstrates gyroscopic prin-ciples to his students. He suspends a rapidly spinning wheel with a string attached to one end of its horizon-tal axle. Describe the precession motion of the wheel.
Problem 7/95 7/96 The student has volunteered to assist in a classroom demonstration involving a momentum wheel which is rapidly spinning with angular speed p as shown. The instructor has asked her to hold the axle of the wheel in the horizontal position shown and then attempt to tilt the axis upward in a vertical plane. What motion tendency of the wheel assembly will the student sense?
Problem 7/96 7/97 A car makes a turn to the right on a level road. De-termine whether the normal reaction under the right rear wheel is increased or decreased as a result of the gyroscopic effect of the precessing wheels.
p y z p 7/98 The 50-kg wheel is a solid circular disk which rolls on the horizontal plane in a circle of 600-mm radius.
The wheel shaft is pivoted about the axis O-O and is driven by the vertical shaft at the constant rate rev/min about the Z-axis. Determine the normal force R between the wheel and the horizon-tal surface. Neglect the weight of the horizontal shaft.
Problem 7/98 7/99 The special-purpose fan is mounted as shown. The motor armature, shaft, and blades have a combined mass of 2.2 kg with radius of gyration of 60 mm. The axial position b of the 0.8-kg block A can be adjusted.
With the fan turned off, the unit is balanced about the x-axis when b 180 mm. The motor and fan oper-ate at 1725 rev/min in the direction shown. Deter-mine the value of b which will produce a steady precession of 0.2 rad/s about the positive y-axis.
Problem 7/99 x z y b A O 350 mm 600 mm N O O Z G N 48 Article 7/11 Problems 569 Representative Problems 7/103 A small air compressor for an aircraft cabin consists of the 3.50-kg turbine A which drives the 2.40-kg blower B at a speed of 20 000 rev/min. The shaft of the assembly is mounted transversely to the direc-tion of flight and is viewed from the rear of the air-craft in the figure. The radii of gyration of A and B are 79.0 and 71.0 mm, respectively. Calculate the ra-dial forces exerted on the shaft by the bearings at C and D if the aircraft executes a clockwise roll (rota-tion about the longitudinal flight axis) of 2 rad/s viewed from the rear of the aircraft. Neglect the small moments caused by the weights of the rotors.
Draw a free-body diagram of the shaft as viewed from above and indicate the shape of its deflected centerline.
Problem 7/103 7/104 The two solid cones with the same base and equal altitudes are spinning in space about their common axis at the rate p. For what ratio h/r will precession of their spin axis be impossible?
Problem 7/104 p h h r 150 mm A B C D 7/100 An airplane has just cleared the runway with a takeoff speed v. Each of its freely spinning wheels has a mass m, with a radius of gyration k about its axle. As seen from the front of the airplane, the wheel precesses at the angular rate as the land-ing strut is folded into the wing about its pivot O.
As a result of the gyroscopic action, the supporting member A exerts a torsional moment M on B to prevent the tubular member from rotating in the sleeve at B. Determine M and identify whether it is in the sense of M1 or M2.
Problem 7/100 7/101 An experimental antipollution bus is powered by the kinetic energy stored in a large flywheel which spins at a high speed p in the direction indicated.
As the bus encounters a short upward ramp, the front wheels rise, thus causing the flywheel to pre-cess. What changes occur to the forces between the tires and the road during this sudden change?
Problem 7/101 7/102 The 210-kg rotor of a turbojet aircraft engine has a radius of gyration of 220 mm and rotates counter-clockwise at 18 000 rev/min as viewed from the front. If the aircraft is traveling at 1200 km/h and starts to execute an inside vertical loop of 3800-m radius, compute the gyroscopic moment M trans-mitted to the airframe. What correction to the con-trols does the pilot have to make in order to remain in the vertical plane?
Ω r b A B B M1 M2 O 7/105 The blades and hub of the helicopter rotor weigh 140 lb and have a radius of gyration of 10 ft about the z-axis of rotation. With the rotor turning at 500 rev/min during a short interval following ver-tical liftoff, the helicopter tilts forward at the rate 10 deg/sec in order to acquire forward veloc-ity. Determine the gyroscopic moment M transmit-ted to the body of the helicopter by its rotor and indicate whether the helicopter tends to deflect clockwise or counterclockwise, as viewed by a pas-senger facing forward.
Problem 7/105 7/106 The 4-oz top with radius of gyration about its spin axis of 0.62 in. is spinning at the rate p 3600 rev/min in the sense shown, with its spin axis making an angle 20 with the vertical.
The distance from its tip O to its mass center G is 2.5 in. Determine the precession of the top and explain why gradually decreases as long as the spin rate remains large. An enlarged view of the contact of the tip is shown.
r x z Vertical θ p ˙ Problem 7/106 7/107 The figure shows a gyro mounted with a vertical axis and used to stabilize a hospital ship against rolling. The motor A turns the pinion which pre-cesses the gyro by rotating the large precession gear B and attached rotor assembly about a horizontal transverse axis in the ship. The rotor turns inside the housing at a clockwise speed of 960 rev/min as viewed from the top and has a mass of 80 Mg with radius of gyration of 1.45 m. Calculate the moment exerted on the hull structure by the gyro if the motor turns the precession gear B at the rate of 0.320 rad/s. In which of the two directions, (a) or (b), should the motor turn in order to counteract a roll of the ship to port?
Problem 7/107 A B Vertical (a) (b) Forward Port (left) Starboard (right) z p G r – O Enlarged view of tip contact θ 570 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/11 Problems 571 Problem 7/110 7/111 The primary structure of a proposed space station consists of five spherical shells connected by tubu-lar spokes. The moment of inertia of the structure about its geometric axis A-A is twice as much as that about any axis through O normal to A-A. The station is designed to rotate about its geometric axis at the constant rate of 3 rev/min. If the spin axis A-A precesses about the Z-axis of fixed orienta-tion and makes a very small angle with it, calculate the rate at which the station wobbles. The mass center O has negligible acceleration.
Problem 7/111 A A Z O ˙ 5″ 33″ A B View of wheels and axle Side view of carriage v 4′ 8 ″ 1 – 2 7/108 Each of the identical wheels has a mass of 4 kg and a radius of gyration kz 120 mm and is mounted on a horizontal shaft AB secured to the vertical shaft at O.
In case (a), the horizontal shaft is fixed to a collar at O which is free to rotate about the vertical y-axis. In case (b), the shaft is secured by a yoke hinged about the x-axis to the collar. If the wheel has a large angu-lar velocity p 3600 rev/min about its z-axis in the position shown, determine any precession which oc-curs and the bending moment MA in the shaft at A for each case. Neglect the small mass of the shaft and fitting at O.
Problem 7/108 7/109 If the wheel in case (a) of Prob. 7/108 is forced to precess about the vertical by a mechanical drive at the steady rate 2j rad/s, determine the bend-ing moment in the horizontal shaft at A. In the ab-sence of friction, what torque MO is applied to the collar at O to sustain this motion?
7/110 The figure shows the side view of the wheel carriage (truck) of a railway passenger car where the vertical load is transmitted to the frame in which the jour-nal wheel bearings are located. The lower view shows only one pair of wheels and their axle which rotates with the wheels. Each of the 33-in.-diameter wheels weighs 560 lb, and the axle weighs 300 lb with a diameter of 5 in. Both wheels and axle are made of steel with a specific weight of 489 lb/ft3. If the train is traveling at 80 mi/hr while rounding an 8 curve to the right (radius of curvature 717 ft), calculate the change R in the vertical force sup-ported by each wheel due only to the gyroscopic ac-tion. As a close approximation, treat each wheel as a uniform circular disk and the axle as a uniform solid cylinder. Also assume that both rails are in the same horizontal plane.
A x z p y O 80 mm 400 mm B A x z p y O 80 mm 400 mm B (a) (b) 7/112 The uniform 640-mm rod has a mass of 3 kg and is welded centrally to the uniform 160-mm-radius cir-cular disk which has a mass of 8 kg. The unit is given a spin velocity p 60 rad/s in the direction shown. The axis of the rod is seen to wobble through a total angle of 30. Calculate the angular velocity of precession and determine whether it is or Problem 7/112 7/113 The electric motor has a total weight of 20 lb and is supported by the mounting brackets A and B at-tached to the rotating disk. The armature of the motor has a weight of 5 lb and a radius of gyration of 1.5 in. and turns counterclockwise at a speed of 1725 rev/min as viewed from A to B. The turntable revolves about its vertical axis at the constant rate of 48 rev/min in the direction shown. Determine the vertical components of the forces supported by the mounting brackets at A and B.
Problem 7/113 A B 6′′ 6′′ 8′′ G p 320 mm 160 mm 320 mm 30° ψ1 ψ2 or ψ1 ψ2 · · · · or ˙2. ˙1 ˙ 7/114 The spacecraft shown is symmetrical about its z-axis and has a radius of gyration of 720 mm about this axis. The radii of gyration about the x- and y-axes through the mass center are both equal to 540 mm.
When moving in space, the z-axis is observed to gen-erate a cone with a total vertex angle of 4 as it pre-cesses about the axis of total angular momentum. If the spacecraft has a spin velocity about its z-axis of 1.5 rad/s, compute the period of each full pre-cession. Is the spin vector in the positive or nega-tive z-direction?
Problem 7/114 7/115 The 8-lb rotor with radius of gyration of 3 in. rotates on ball bearings at a speed of 3000 rev/min about its shaft OG. The shaft is free to pivot about the X-axis, as well as to rotate about the Z-axis. Calculate the vector for precession about the Z-axis. Neglect the mass of shaft OG and compute the gyroscopic couple M exerted by the shaft on the rotor at G.
Problem 7/115 9″ 70° i K Z O G Y p X, x Vertical z 2° x y G HG ˙ 572 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/11 Problems 573 7/118 A boy throws a thin circular disk (like a Frisbee) with a spin rate of 300 rev/min. The plane of the disk is seen to wobble through a total angle of 10.
Calculate the period of the wobble and indicate whether the precession is direct or retrograde.
Problem 7/118 7/119 The figure shows a football in three common in-flight configurations. Case (a) is a perfectly thrown spiral pass with a spin rate of 120 rev/min. Case (b) is a wobbly spiral pass again with a spin rate of 120 rev/min about its own axis, but with the axis wobbling through a total angle of 20. Case (c) is an end-over-end place kick with a rotational rate of 120 rev/min. For each case, specify the values of p, , , and as defined in this article. The mo-ment of inertia about the long axis of the ball is 0.3 of that about the transverse axis of symmetry.
Problem 7/119 20° ˙ 10° 7/116 The housing of the electric motor is freely pivoted about the horizontal x-axis, which passes through the mass center G of the rotor. If the motor is turn-ing at the constant rate p, determine the angu-lar acceleration which will result from the application of the moment M about the vertical shaft if 0. The mass of the frame and hous-ing is considered negligible compared with the mass m of the rotor. The radius of gyration of the rotor about the z-axis is kz and that about the x-axis is kx.
Problem 7/116 7/117 The thin ring is projected into the air with a spin velocity of 300 rev/min. If its geometric axis is observed to have a very slight precessional wobble, determine the frequency ƒ of the wobble.
Problem 7/117 300 rev/min b p G z y x M ψ · γ ˙ ˙ ¨ ˙ 7/120 The rectangular bar is spinning in space about its longitudinal axis at the rate p 200 rev/min. If its axis wobbles through a total angle of 20 as shown, calculate the period of the wobble.
Problem 7/120 7/121 The 5-kg disk and hub A have a radius of gyration of 85 mm about the -axis and spin at the rate rev/min. Simultaneously, the assembly rotates about the vertical z-axis at the rate rev/min. Calculate the gyroscopic moment M ex-erted on the shaft at C by the disk and the bending moment in the shaft at O. Neglect the mass of the shaft but otherwise account for all forces acting on it.
Problem 7/121 Ω z z0 p x y A C B O 20° 300 mm 100 mm MO 400 p 1250 z0 20° 4″ 4″ 8″ O p 7/122 The uniform slender bar of mass m and length l is centrally mounted on the shaft A-A, about which it rotates with a constant speed . Simultane-ously, the yoke is forced to rotate about the x-axis with a constant speed . As a function of , deter-mine the magnitude of the torque M required to maintain the constant speed . (Hint: Apply Eq.
7/19 to obtain the x-component of M.) Problem 7/122 7/123 The solid circular disk of mass m and small thick-ness is spinning freely on its shaft at the rate p. If the assembly is released in the vertical position at 0 with 0, determine the horizontal compo-nents of the forces A and B exerted by the respec-tive bearings on the horizontal shaft as the position /2 is passed. Neglect the mass of the two shafts compared with m and neglect all friction.
Solve by using the appropriate moment equations.
Problem 7/123 b A B b l p θ z r ˙ z x 0 ω y A A O M B p l φ φ 0 0 ˙ p 574 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Article 7/11 Problems 575 Problem 7/125 7/126 The solid cylindrical rotor weighs 64.4 lb and is mounted in bearings A and B of the frame which rotates about the vertical Z-axis. If the rotor spins at the constant rate p 50 rad/sec relative to the frame and if the frame itself rotates at the constant rate 30 rad/sec, compute the bending moment M in the shaft at C which the lower portion of the shaft exerts on the upper portion. Also compute the kinetic energy T of the rotor. Neglect the mass of the frame.
Problem 7/126 Z C A B z X Y y r = 3″ 4″ 4″ 60° G x p Ω r h h p 7/124 The earth-scanning satellite is in a circular orbit of period . The angular velocity of the satellite about its y- or pitch-axis is 2/, and the angular rates about the x- and z-axes are zero. Thus, the x-axis of the satellite always points to the center of the earth. The satellite has a reaction-wheel attitude-control system consisting of the three wheels shown, each of which may be variably torqued by its individual motor. The angular rate z of the z-wheel relative to the satellite is 0 at time t 0, and the x- and y-wheels are at rest rela-tive to the satellite at t 0. Determine the axial torques Mx, My, and Mz which must be exerted by the motors on the shafts of their respective wheels in order that the angular velocity of the satellite will remain constant. The moment of inertia of each reaction wheel about its axis is I. The x and z reaction-wheel speeds are harmonic functions of the time with a period equal to that of the orbit.
Plot the variations of the torques and the relative wheel speeds x, y, and z as functions of the time during one orbit period. (Hint: The torque to accel-erate the x-wheel equals the reaction of the gyro-scopic moment on the z-wheel, and vice versa.) Problem 7/124 7/125 The two solid homogeneous right-circular cones, each of mass m, are fastened together at their ver-tices to form a rigid unit and are spinning about their axis of radial symmetry at the rate p 200 rev/min.
(a) Determine the ratio h/r for which the rotation axis will not precess. (b) Sketch the space and body cones for the case where h/r is less than the critical ratio. (c) Sketch the space and body cones when h r and the precessional velocity is 18 rad/s. ˙ y z x y z x O x Ω z Ω y Ω O 576 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies 7/12 Chapter Review In Chapter 7 we have studied the three-dimensional dynamics of rigid bodies. Motion in three dimensions adds considerable complexity to the kinematic and kinetic relationships. Compared with plane mo-tion, there is now the possibility of two additional components of the vectors describing angular quantities such as moment, angular veloc-ity, angular momentum, and angular acceleration. For this reason, the full power of vector analysis becomes apparent in the study of three-dimensional dynamics.
We divided our study of three-dimensional dynamics into kinemat-ics, which is covered in Section A of the chapter, and kinetics, which is treated in Section B.
Kinematics We arranged our coverage of three-dimensional kinematics in order of increasing complexity of the type of motion. These types are: 1. Translation. As in plane motion, covered in Chapter 5 (Plane Kinematics of Rigid Bodies), any two points on a rigid body have the same velocity and acceleration.
2. Fixed-Axis Rotation. In this case the angular-velocity vector does not change orientation, and the expressions for the velocity and ac-celeration of a point are easily obtained as Eqs. 7/1 and 7/2, which are identical in form to the corresponding plane-motion equations in Chapter 5.
3. Parallel-Plane Motion. This case occurs when all points in a rigid body move in planes which are parallel to a fixed plane. Thus, in each plane, the results of Chapter 5 hold.
4. Rotation about a Fixed Point. In this case, both the magnitude and the direction of the angular-velocity vector may vary. Once the angular acceleration is established by careful differentiation of the angular-velocity vector, Eqs. 7/1 and 7/2 may be used to determine the velocity and acceleration of a point.
5. General Motion. The principles of relative motion are useful in analyzing this type of motion. Relative velocity and relative acceler-ation are expressed in terms of translating reference axes by Eqs.
7/4. When rotating reference axes are used, the unit vectors of the reference system have nonzero time derivatives. Equations 7/6 ex-press the velocity and acceleration in terms of quantities referred to rotating axes; these equations are identical in form to the corre-sponding results for plane motion, Eqs. 5/12 and 5/14. Equations 7/7a and 7/7b are the expressions relating the time derivatives of a vector as measured in a fixed system and as measured relative to a rotating system. These expressions are useful in the analysis of gen-eral motion.
Article 7/12 Chapter Review 577 Kinetics We applied momentum and energy principles to analyze three-dimensional kinetics, as follows.
1. Angular Momentum. In three dimensions the vector expression for angular momentum has numerous additional components which are absent in plane motion. The components of angular momentum are expressed by Eqs. 7/12 and depend on both moments and prod-ucts of inertia. There is a unique set of axes, called principal axes, for which the products of inertia are zero and the moments of iner-tia have stationary values. These values are called the principal mo-ments of inertia.
2. Kinetic Energy. The kinetic energy of three-dimensional motion can be expressed either in terms of the motion of and about the mass center (Eq. 7/15) or in terms of the motion about a fixed point (Eq. 7/18).
3. Momentum Equations of Motion. By using the principal axes we may simplify the momentum equations of motion to obtain Euler’s equations, Eqs. 7/21.
4. Energy Equations. The work-energy principle for three-dimen-sional motion is identical to that for plane motion.
Applications In Chapter 7 we studied two applications of special interest, namely, parallel-plane motion and gyroscopic motion.
1. Parallel-Plane Motion. In such motion all points in a rigid body move in planes which are parallel to a fixed plane. The equations of motion are Eqs. 7/23. These equations are useful for analyzing the effects of dynamic imbalance in rotating machinery and in bodies which roll along straight paths.
2. Gyroscopic Motion. This type of motion occurs whenever the axis about which the body is spinning is itself rotating about another axis. Common applications include inertial guidance systems, stabi-lizing devices, spacecraft attitude motion, and any situation in which a rapidly spinning rotor (such as that of an aircraft engine) is being reoriented. In the case where an external torque is present, a basic analysis can be based upon the equation M For the case of torque-free motion of a body spinning about its axis of symmetry, the axis of symmetry is found to execute a coning motion about the fixed angular-momentum vector.
H ˙.
578 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies parallel to the rear axle of the car. The center of mass of the car is a distance h above the road, and the car is rounding an unbanked level turn at a speed v. At what speed p should the rotor turn and in what direction to counteract completely the ten-dency of the car to overturn for either a right or a left turn? The combined mass of car and rotor is m.
7/130 The wheels of the jet plane are spinning at their angu-lar rate corresponding to a takeoff speed of 150 km/h.
The retracting mechanism operates with increasing at the rate of 30 per second. Calculate the angular ac-celeration of the wheels for these conditions.
Problem 7/130 7/131 The motor turns the disk at the constant speed rad/sec. The motor is also swiveling about the horizontal axis B-O (y-axis) at the constant speed rad/sec. Simultaneously, the entire assembly is rotating about the vertical axis C-C at the constant rate rad/sec. For the instant when , determine the angular acceleration of the disk and the acceleration a of point A at the bottom of the disk. Axes x-y-z are attached to the motor housing, and plane O-is horizontal.
Problem 7/131 A x x0 p y r 6″ z B C C O q 6″ θ x0-y 30 q 8 ˙ 2 p 30 x y θ 560 mm REVIEW PROBLEMS 7/127 The cylindrical shell is rotating in space about its geometric axis. If the axis has a slight wobble, for what ratios of l/r will the motion be direct or retro-grade precession?
Problem 7/127 7/128 The solid cube of mass m and side a revolves about an axis M-M through a diagonal with an angular velocity . Write the expression for the angular momentum H of the cube with respect to the axes indicated.
Problem 7/128 7/129 An experimental car is equipped with a gyro stabi-lizer to counteract completely the tendency of the car to tip when rounding a curve (no change in nor-mal force between tires and road). The rotor of the gyro has a mass m0 and a radius of gyration k, and is mounted in fixed bearings on a shaft which is M M z y a a a x ω r p l Article 7/12 Review Problems 579 7/132 The collars at the ends of the telescoping link AB slide along the fixed shafts shown. During an inter-val of motion, vA 5 in./sec and vB 2 in./sec.
Determine the vector expression for the angular velocity n of the centerline of the link for the posi-tion where yA 4 in. and yB 2 in.
Problem 7/132 7/133 The solid cone of mass m, base radius r, and alti-tude h is spinning at a high rate p about its own axis and is released with its vertex O supported by a horizontal surface. Friction is sufficient to pre-vent the vertex from slipping in the x-y plane. De-termine the direction of the precession and the period of one complete rotation about the vertical z-axis.
Problem 7/133 z O y h x p θ r x z y yB yA vA vB A B 6′′ 3′′ 7/134 The rectangular steel plate of mass 12 kg is welded to the shaft with its plane tilted from the plane (x-y) normal to the shaft axis. The shaft and plate are rotating about the fixed z-axis at the rate rev/min. Determine the angular momentum of the plate about the given axes and find its kinetic energy T.
Problem 7/134 7/135 The circular disk of radius r is mounted on its shaft which is pivoted at O so that it may rotate about the vertical z0-axis. If the disk rolls at constant speed without slipping and makes one complete turn around the circle of radius R in time , deter-mine the expression for the absolute angular veloc-ity of the disk. Use axes x-y-z which rotate around the z0-axis. (Hint: The absolute angular ve-locity of the disk equals the angular velocity of the axes plus (vectorially) the angular velocity relative to the axes as seen by holding x-y-z fixed and rotat-ing the circular disk of radius R at the rate of 2/.) Problem 7/135 z O R r y x z0 A N 150 mm 150 mm 120 mm 120 mm O y x z 15° HO N 300 15 580 Chapter 7 Introduction to Three-Dimensional Dynamics of Rigid Bodies Problem 7/140 7/141 Rework Prob. 7/140 if , instead of being constant at 20, is increasing at the steady rate of 120 rev/min.
Find the angular momentum HO of the disk for the instant when 20. Also compute the kinetic energy T of the disk. Is T dependent on ?
7/142 The dynamic imbalance of a certain crankshaft is approximated by the physical model shown, where the shaft carries three small 1.5-lb spheres at-tached by rods of negligible mass. If the shaft ro-tates at the constant speed of 1200 rev/min, calculate the forces RA and RB acting on the bear-ings. Neglect the gravitational forces.
Problem 7/142 8′′ 8′′ 8′′ 8′′ x y z A B 1 1 2 2 3 3 120° End view 120° 120° 6′′ 10″ y z O p C y′ x N 4″ β z′ 7/136 Determine the angular acceleration for the rolling circular disk of Prob. 7/135. Use the results cited in the answer for that problem.
7/137 Determine the velocity v of point A on the disk of Prob. 7/135 for the position shown.
7/138 Determine the acceleration a of point A on the disk of Prob. 7/135 for the position shown.
7/139 A top consists of a ring of mass m 0.52 kg and mean radius r 60 mm mounted on its central pointed shaft with spokes of negligible mass. The top is given a spin velocity of 10 000 rev/min and released on the horizontal surface with the point O remaining in a fixed position. The axis of the top is seen to make an angle of 15 with the vertical as it precesses. Determine the number N of precession cycles per minute. Also identify the direction of the precession and sketch the body and space cones.
Problem 7/139 7/140 The uniform circular disk of 4-in. radius and small thickness weighs 8 lb and is spinning about its y-axis at the rate N 300 rev/min with its plane of rotation tilted at a constant angle 20 from the vertical x-z plane. Simultaneously, the assembly rotates about the fixed z-axis at the rate p 60 rev/min. Calculate the angular momentum HO of the disk alone about the origin O of the x-y-z coordi-nates. Also calculate the kinetic energy T of the disk.
Z O 80 mm 10000 rev/min 15° 60 mm Article 7/12 Review Problems 581 7/143 Each of the two right-angle bent rods weighs 2.80 lb and is parallel to the horizontal x-y plane. The rods are welded to the vertical shaft, which ro-tates about the z-axis with a constant angular speed N 1200 rev/min. Calculate the bending moment M in the shaft at its base O.
Problem 7/143 7/144 Each of the quarter-circular plates has a mass of 2 kg and is secured to the vertical shaft mounted in the fixed bearing at O. Calculate the magni-tude M of the bending moment in the shaft at O for a constant rotational speed N 300 rev/min.
Treat the plates as exact quarter-circular shapes.
x 6″ 6″ z y O 6″ 6″ 6″ 6″ Ν Problem 7/144 7/145 Calculate the bending moment M in the shaft at O for the rotating assembly of Prob. 7/144 as it starts from rest with an initial angular acceleration of 200 rad/s2.
7/146 The half-cylindrical shell of mass m, radius r, and length b revolves about one edge along the z-axis with a constant rate as shown. Determine the angu-lar momentum H of the shell with respect to the x-y-z axes.
Problem 7/146 z y r b x ω 150 mm 150 mm 75 mm 75 mm N O x y z This illustration shows the elements of the left-front suspension on an all-wheel-drive automobile. The spring and shock absorber are coaxial in this McPherson-strut type of suspension.
Courtesy of David R. Kraige 583 8/1 Introduction An important and special class of problems in dynamics concerns the linear and angular motions of bodies which oscillate or otherwise re-spond to applied disturbances in the presence of restoring forces. A few examples of this class of dynamics problems are the response of an engi-neering structure to earthquakes, the vibration of an unbalanced rotat-ing machine, the time response of the plucked string of a musical instrument, the wind-induced vibration of power lines, and the flutter of aircraft wings. In many cases, excessive vibration levels must be re-duced to accommodate material limitations or human factors.
In the analysis of every engineering problem, we must represent the system under scrutiny by a physical model. We may often represent a continuous or distributed-parameter system (one in which the mass and spring elements are continuously spread over space) by a discrete or lumped-parameter model (one in which the mass and spring elements are separate and concentrated). The resulting simplified model is espe-cially accurate when some portions of a continuous system are relatively massive in comparison with other portions. For example, the physical model of a ship propeller shaft is often assumed to be a massless but twistable rod with a disk rigidly attached to each end—one disk repre-senting the turbine and the other representing the propeller. As a sec-ond example, we observe that the mass of springs may often be neglected in comparison with that of attached bodies.
Not every system is reducible to a discrete model. For example, the transverse vibration of a diving board after the departure of the diver is 8/1 Introduction 8/2 Free Vibration of Particles 8/3 Forced Vibration of Particles 8/4 Vibration of Rigid Bodies 8/5 Energy Methods 8/6 Chapter Review CHAPTER OUTLINE 8 Vibration and Time Response a somewhat difficult problem of distributed-parameter vibration. In this chapter, we will begin the study of discrete systems, limiting our discus-sion to those whose configurations may be described with one displace-ment variable. Such systems are said to possess one degree of freedom.
For a more detailed study which includes the treatment of two or more degrees of freedom and continuous systems, you should consult one of the many textbooks devoted solely to the subject of vibrations.
The remainder of Chapter 8 is divided into four sections: Article 8/2 treats the free vibration of particles and Art. 8/3 introduces the forced vibration of particles. Each of these two articles is subdivided into undamped- and damped-motion categories. In Art. 8/4 we discuss the vibration of rigid bodies. Finally, an energy approach to the solu-tion of vibration problems is presented in Art. 8/5.
The topic of vibrations is a direct application of the principles of ki-netics as developed in Chapters 3 and 6. In particular, construction of a complete free-body diagram drawn for an arbitrary positive value of the displacement variable, followed by application of the appropriate gov-erning equations of dynamics, will yield the equation of motion. From this equation of motion, which is a second-order ordinary differential equation, you can obtain all information of interest, such as the motion frequency, period, or the motion itself as a function of time.
8/2 Free Vibration of Particles When a spring-mounted body is disturbed from its equilibrium posi-tion, its ensuing motion in the absence of any imposed external forces is termed free vibration. In every actual case of free vibration, there exists some retarding or damping force which tends to diminish the motion.
Common damping forces are those due to mechanical and fluid friction.
In this article we first consider the ideal case where the damping forces are small enough to be neglected. Then we treat the case where the damping is appreciable and must be accounted for.
Equation of Motion for Undamped Free Vibration We begin by considering the horizontal vibration of the simple frictionless spring-mass system of Fig. 8/1a. Note that the variable x denotes the displacement of the mass from the equilibrium position, which, for this system, is also the position of zero spring deflection.
Figure 8/1b shows a plot of the force Fs necessary to deflect the spring versus the corresponding spring deflection for three types of springs.
Although nonlinear hard and soft springs are useful in some applica-tions, we will restrict our attention to the linear spring. Such a spring exerts a restoring force kx on the mass—that is, when the mass is displaced to the right, the spring force is to the left, and vice versa.
We must be careful to distinguish between the forces of magnitude Fs which must be applied to both ends of the massless spring to cause tension or compression and the force F kx of equal magnitude which the spring exerts on the mass. The constant of proportionality k is called the spring constant, modulus, or stiffness and has the units N/m or lb/ft.
584 Chapter 8 Vibration and Time Response m mg x N x Fs kx k Equilibrium position Hard Linear Fs = kx Soft (a) (b) Figure 8/1 The equation of motion for the body of Fig. 8/1a is obtained by first drawing its free-body diagram. Applying Newton’s second law in the form ΣFx gives (8/1) The oscillation of a mass subjected to a linear restoring force as de-scribed by this equation is called simple harmonic motion and is charac-terized by acceleration which is proportional to the displacement but of opposite sign. Equation 8/1 is normally written as (8/2) where (8/3) is a convenient substitution whose physical significance will be clarified shortly.
Solution for Undamped Free Vibration Because we anticipate an oscillatory motion, we look for a solution which gives x as a periodic function of time. Thus, a logical choice is (8/4) or, alternatively, (8/5) Direct substitution of these expressions into Eq. 8/2 verifies that each expression is a valid solution to the equation of motion. We determine the constants A and B, or C and , from knowledge of the initial dis-placement x0 and initial velocity of the mass. For example, if we work with the solution form of Eq. 8/4 and evaluate x and at time t 0, we obtain Substitution of these values of A and B into Eq. 8/4 yields (8/6) The constants C and of Eq. 8/5 can be determined in terms of given initial conditions in a similar manner. Evaluation of Eq. 8/5 and its first time derivative at t 0 gives x0 C sin and x ˙0 Cn cos x x0 cos nt x ˙0 n sin nt x0 A and x ˙0 Bn x ˙ x ˙0 x C sin (nt ) x A cos nt B sin nt n k/m x ¨ n 2x 0 kx mx ¨ or mx ¨ kx 0 mx ¨ Article 8/2 Free Vibration of Particles 585 Solving for C and yields Substitution of these values into Eq. 8/5 gives (8/7) Equations 8/6 and 8/7 represent two different mathematical expressions for the same time-dependent motion. We observe that C and tan1(A/B).
Graphical Representation of Motion The motion may be represented graphically, Fig. 8/2, where x is seen to be the projection onto a vertical axis of the rotating vector of length C. The vector rotates at the constant angular velocity n which is called the natural circular frequency and has the units radians per second. The number of complete cycles per unit time is the natural frequency ƒn n/2 and is expressed in hertz (1 hertz (Hz) 1 cycle per second). The time required for one complete motion cycle (one rota-tion of the reference vector) is the period of the motion and is given by 1/ƒn 2/n.
k/m, A2 B2 x x0 2 (x ˙0/n)2 sin [nt tan1(x0n/x ˙0)] C x0 2 (x ˙0/n)2 tan1(x0n/x ˙0) 586 Chapter 8 Vibration and Time Response We also see from the figure that x is the sum of the projections onto the vertical axis of two perpendicular vectors whose magnitudes are A and B and whose vector sum C is the amplitude. Vectors A, B, and C ro-tate together with the constant angular velocity n. Thus, as we have al-ready seen, C and tan1(A/B).
Equilibrium Position as Reference As a further note on the free undamped vibration of particles, we see that, if the system of Fig. 8/1a is rotated 90 clockwise to obtain the system of Fig. 8/3 where the motion is vertical rather than horizontal, A2 B2 ψ ωnt ωnt τ C A B x t +x −x −C C x 0 0 x0 2 —– n ω = Figure 8/2 the equation of motion (and therefore all system properties) is un-changed if we continue to define x as the displacement from the equilib-rium position. The equilibrium position now involves a nonzero spring deflection st. From the free-body diagram of Fig. 8/3, Newton’s second law gives At the equilibrium position x 0, the force sum must be zero, so that Thus, we see that the pair of forces kst and mg on the left side of the motion equation cancel, giving which is identical to Eq. 8/1.
The lesson here is that by defining the displacement variable to be zero at equilibrium rather than at the position of zero spring de-flection, we may ignore the equal and opposite forces associated with equilibrium. Equation of Motion for Damped Free Vibration Every mechanical system possesses some inherent degree of fric-tion, which dissipates mechanical energy. Precise mathematical models of the dissipative friction forces are usually complex. The dashpot or vis-cous damper is a device intentionally added to systems for the purpose of limiting or retarding vibration. It consists of a cylinder filled with a viscous fluid and a piston with holes or other passages by which the fluid can flow from one side of the piston to the other. Simple dashpots arranged as shown schematically in Fig. 8/4a exert a force Fd whose magnitude is proportional to the velocity of the mass, as depicted in Fig.
8/4b. The constant of proportionality c is called the viscous damping co-efficient and has units of or lb-sec/ft. The direction of the damp-ing force as applied to the mass is opposite that of the velocity Thus, the force on the mass is Complex dashpots with internal flow-rate-dependent one-way valves can produce different damping coefficients in extension and in compression; nonlinear characteristics are also possible. We will restrict our attention to the simple linear dashpot.
The equation of motion for the body with damping is determined from the free-body diagram as shown in Fig. 8/4a. Newton’s second law gives (8/8) kx cx ˙ mx ¨ or mx ¨ cx ˙ kx 0 cx ˙.
x ˙.
Ns/m mx ¨ kx 0 kst mg 0 k(st x) mg mx ¨ Article 8/2 Free Vibration of Particles 587 For nonlinear systems, all forces, including the static forces associated with equilibrium, should be included in the analysis.
m m k x mg k( st + x) δ δst Equilibrium position Figure 8/3 k x c m Equilibrium position (a) (b) Fd kx cx mg N · x · Figure 8/4 In addition to the substitution n it is convenient, for reasons which will shortly become evident, to introduce the combination of constants The quantity (zeta) is called the viscous damping factor or damping ratio and is a measure of the severity of the damping. You should verify that is nondimensional. Equation 8/8 may now be written as (8/9) Solution for Damped Free Vibration In order to solve the equation of motion, Eq. 8/9, we assume solu-tions of the form Substitution into Eq. 8/9 yields which is called the characteristic equation. Its roots are Linear systems have the property of superposition, which means that the general solution is the sum of the individual solutions each of which corresponds to one root of the characteristic equation. Thus, the general solution is (8/10) Categories of Damped Motion Because 0 , the radicand (2 1) may be positive, negative, or even zero, giving rise to the following three categories of damped motion: I. 1 (overdamped). The roots 1 and 2 are distinct, real, and negative numbers. The motion as given by Eq. 8/10 decays so that x approaches zero for large values of time t. There is no os-cillation and therefore no period associated with the motion.
II. 1 (critically damped). The roots 1 and 2 are equal, real, and negative numbers (1 2 n). The solution to the dif-ferential equation for the special case of equal roots is given by x (A1 A2t)ent A1e(21)nt A2e(21)nt x A1e1t A2e2t 1 n( 2 1) 2 n( 2 1) 2 2n n 2 0 x Aet x ¨ 2nx ˙ n 2x 0 c/(2mn) k/m, 588 Chapter 8 Vibration and Time Response Again, the motion decays with x approaching zero for large time, and the motion is nonperiodic. A critically damped system, when excited with an initial velocity or displacement (or both), will ap-proach equilibrium faster than will an overdamped system. Fig-ure 8/5 depicts actual responses for both an overdamped and a critically damped system to an initial displacement x0 and no ini-tial velocity ( 0).
III. 1 (underdamped). Noting that the radicand (2 1) is negative and recalling that eaeb, we may rewrite Eq.
8/10 as where i It is convenient to let a new variable d repre-sent the combination Thus, Use of the Euler formula eix cos x i sin x allows the previ-ous equation to be written as (8/11) where A3 (A1 A2) and A4 i(A1 A2). We have shown with Eqs. 8/4 and 8/5 that the sum of two equal-frequency harmon-ics, such as those in the braces of Eq. 8/11, can be replaced by a single trigonometric function which involves a phase angle.
Thus, Eq. 8/11 can be written as or (8/12) x Cent sin (dt ) x {C sin (dt )}ent {A3 cos dt A4 sin dt}ent {(A1 A2) cos dt i(A1 A2) sin dt}ent x {A1(cos dt i sin dt) A2(cos dt i sin dt)}ent x {A1eidt A2eidt}ent n1 2.
1.
x {A1ei12nt A2ei12nt}ent e(ab) x ˙0 Article 8/2 Free Vibration of Particles 589 c = 15 N·s/m ( = 2.5), overdamped c = 6 N·s/m ( = 1), critically damped Conditions: m = 1 kg, k = 9 N/m x0 = 30 mm, x0 = 0 x, mm t, s 30 20 10 0 1 0 2 3 4 · ζ ζ Figure 8/5 Equation 8/12 represents an exponentially decreasing harmonic function, as shown in Fig. 8/6 for specific numerical values. The frequency is called the damped natural frequency. The damped period is given by d 2/d It is important to note that the expressions developed for the con-stants C and in terms of initial conditions for the case of no damping are not valid for the case of damping. To find C and if damping is pres-ent, you must begin anew, setting the general displacement expression of Eq. 8/12 and its first time derivative, both evaluated at time t 0, equal to the initial displacement x0 and initial velocity respectively.
Determination of Damping by Experiment We often need to experimentally determine the value of the damp-ing ratio for an underdamped system. The usual reason is that the value of the viscous damping coefficient c is not otherwise well known.
To determine the damping, we may excite the system by initial condi-tions and obtain a plot of the displacement x versus time t, such as that shown schematically in Fig. 8/7. We then measure two successive ampli-tudes x1 and x2 a full cycle apart and compute their ratio The logarithmic decrement is defined as From this equation, we may solve for and obtain For a small damping ratio, x1 x2 and 1, so that /2. If x1 and x2 are so close in value that experimental distinction between them is impractical, the above analysis may be modified by using two ob-served amplitudes which are n cycles apart.
(2)2 2 ln x1 x2 nd n 2 n1 2 2 1 2 x1 x2 Cent1 Cen(t1d) end x ˙0, 2/(n1 2).
d n1 2 590 Chapter 8 Vibration and Time Response Conditions: m = 1 kg, k = 36 N/m x, mm t, s c = 1 N·s /m ( = 0.0833) x0 = 30 mm, x0 = 0 –30 –20 –10 10 20 30 1 2 3 0 0 4 τd Ce – nt ζω · ζ – Ce – nt ζω Figure 8/6 t x t1 t2 2 — d ω τd = x = Ce– nt sin ( dt + ) ζω ω ψ x1 x2 Figure 8/7 SAMPLE PROBLEM 8/1 A body weighing 25 lb is suspended from a spring of constant k 160 lb/ft.
At time t 0, it has a downward velocity of 2 ft/sec as it passes through the posi-tion of static equilibrium. Determine (a) the static spring deflection st (b) the natural frequency of the system in both rad/sec (n) and cycles/sec (ƒn) (c) the system period (d) the displacement x as a function of time, where x is measured from the position of static equilibrium (e) the maximum velocity vmax attained by the mass (ƒ) the maximum acceleration amax attained by the mass.
Solution.
(a) From the spring relationship Fs kx, we see that at equilibrium Ans.
(b) Ans.
Ans.
(c) Ans.
(d) From Eq. 8/6: Ans.
As an exercise, let us determine x from the alternative Eq. 8/7: (e) The velocity is 14.36(0.1393) cos 14.36t 2 cos 14.36t. Because the cosine function cannot be greater than 1 or less than 1, the maximum velocity vmax is 2 ft/sec, which, in this case, is the initial velocity.
Ans.
(ƒ) The acceleration is The maximum acceleration amax is 28.7 ft/sec2.
Ans.
x ¨ 14.36(2) sin 14.36t 28.7 sin 14.36t x ˙ 0.1393 sin 14.36t (0) cos 14.36t 2 14.36 sin 14.36t x x0 cos nt x ˙0 n sin nt 1 ƒn 1 2.28 0.438 sec mg kst st mg k 25 160 0.1562 ft or 1.875 in.
Article 8/2 Free Vibration of Particles 591 k = 160 lb /ft W = 25 lb Equilibrium position mg mg kx ≡ x st δ δ δ Fs = k st k( st + x) Helpful Hints You should always exercise extreme caution in the matter of units. In the subject of vibrations, it is quite easy to commit errors due to mixing of feet and inches, cycles and radians, and other pairs which frequently enter the calculations.
Recall that when we refer the mo-tion to the position of static equilib-rium, the equation of motion, and therefore its solution, for the pres-ent system is identical to that for the horizontally vibrating system.
SAMPLE PROBLEM 8/2 The 8-kg body is moved 0.2 m to the right of the equilibrium position and released from rest at time t 0. Determine its displacement at time t 2 s. The viscous damping coefficient c is 20 and the spring stiffness k is 32 N/m.
Solution.
We must first determine whether the system is underdamped, criti-cally damped, or overdamped. For that purpose, we compute the damping ratio .
Since 1, the system is underdamped. The damped natural frequency is d 1.561 rad/s. The motion is given by Eq. 8/12 and is The velocity is then Evaluating the displacement and velocity at time t 0 gives Solving the two equations for C and yields C 0.256 m and 0.896 rad.
Therefore, the displacement in meters is Evaluation for time t 2 s gives x2 0.01616 m.
Ans.
SAMPLE PROBLEM 8/3 The two fixed counterrotating pulleys are driven at the same angular speed 0. A round bar is placed off center on the pulleys as shown. Determine the nat-ural frequency of the resulting bar motion. The coefficient of kinetic friction be-tween the bar and pulleys is k.
Solution.
The free-body diagram of the bar is constructed for an arbitrary dis-placement x from the central position as shown. The governing equations are Eliminating NA and NB from the first equation yields We recognize the form of this equation as that of Eq. 8/2, so that the natural fre-quency in radians per second is n and the natural frequency in cy-cles per second is Ans.
ƒn 1 2 2k g/a 2k g/a x ¨ 2k g a x 0 aNB a 2 xmg 0 [ΣMA 0] NA NB mg 0 [ΣFy 0] k NA k NB mx ¨ [ΣFx mx ¨] x 0.256e1.25t sin (1.561t 0.896) x0 C sin 0.2 x ˙0 1.25C sin 1.561C cos 0 x ˙ 1.25Ce1.25t sin (1.561t ) 1.561Ce1.25t cos (1.561t ) x Cent sin (dt ) Ce1.25t sin (1.561t ) 21 (0.625)2 n1 2 n k/m 32/8 2 rad/s c 2mn 20 2(8)(2) 0.625 Ns/m, 592 Chapter 8 Vibration and Time Response k x c 8 kg x Equilibrium position mg N cx = 20x · · kx = 32x a 0 ω 0 ω Central position y a –– 2 a –– 2 x A B G NA NB mg k NA μ k NB μ Helpful Hint We note that the exponential factor e1.25t is 0.0821 at t 2 s. Thus, 0.625 represents severe damp-ing, although the motion is still oscillatory.
Helpful Hints Because the bar is slender and does not rotate, the use of a moment equilibrium equation is justified.
We note that the angular speed 0 does not enter the equation of mo-tion. The reason for this is our as-sumption that the kinetic friction force does not depend on the relative velocity at the contacting surface.
PROBLEMS (Unless otherwise indicated, all motion variables are re-ferred to the equilibrium position.) Undamped, Free Vibrations 8/1 When a 3-kg collar is placed upon the pan which is attached to the spring of unknown constant, the addi-tional static deflection of the pan is observed to be 40 mm. Determine the spring constant k in N/m, lb/in., and lb/ft.
Problem 8/1 8/2 Determine the natural frequency of the spring-mass system in both rad/sec and cycles/sec (Hz).
Problem 8/2 8/3 For the system of Prob. determine the displace-ment x of the mass as a function of time if the mass is released from rest at time from a position 2 in.
to the right of the equilibrium position.
8/4 For the system of Prob. determine the displace-ment x of the mass as a function of time if the mass is released at time from a position 2 in. to the left of the equilibrium position with an initial velocity of to the right. Determine the amplitude C of the motion.
7 in./sec t 0 8/2, t 0 8/2, x 64.4 lb k = 24 lb/in.
k 3 kg 40 mm Article 8/2 Problems 593 8/5 For the spring-mass system shown, determine the static defection the system period and the maxi-mum velocity which result if the cylinder is dis-placed 0.1 m downward from its equilibrium position and released.
Problem 8/5 8/6 The cylinder of the system of Prob. is displaced 0.1 m downward from its equilibrium position and is released at time Determine the displacement y and the velocity when What is the maximum acceleration?
8/7 The vertical plunger has a mass of 2.5 kg and is sup-ported by the two springs, which are always in com-pression. Calculate the natural frequency of vibration of the plunger if it is deflected from the equilibrium position and released from rest. Friction in the guide is negligible.
Problem 8/7 Fixed 2.5 kg k1 = 3.6 kN/m k2 = 1.8 kN/m ƒn t 3 s.
v t 0.
8/5 Equilibrium position y m = 4 kg k = 144 N/m vmax , st, Problem 8/10 8/11 A conventional spring scale registers the normal force which it exerts on the feet of the person being weighed. In the orbital environment aboard the space-shuttle orbiter, such normal forces do not exist. Use your knowledge of vibration and explain how an as-tronaut might “weigh” himself or herself.
8/12 During the design of the spring-support system for the weighing platform, it is decided that the frequency of free vertical vibration in the unloaded condition shall not exceed 3 cycles per second. (a) Determine the maximum acceptable spring con-stant k for each of the three identical springs. (b) For this spring constant, what would be the natural fre-quency of vertical vibration of the platform loaded by the truck?
Problem 8/12 8/13 Replace the springs in each of the two cases shown by a single spring of stiffness k (equivalent spring stiffness) which will cause each mass to vibrate with its original frequency.
Problem 8/13 (a) (b) k1 k1 k2 k2 4000 kg k k k 40-Mg ƒn 4000-kg Electromagnet 594 Chapter 8 Vibration and Time Response 8/8 If the 100-kg mass has a downward velocity of as it passes through its equilibrium position, calculate the magnitude of its maximum acceleration.
Each of the two springs has a stiffness Problem 8/8 8/9 Calculate the natural frequency of vertical oscilla-tion of the spring-loaded cylinder when it is set into motion. Both springs are in tension at all times.
Problem 8/9 8/10 An old car being moved by a magnetic crane pickup is dropped from a short distance above the ground.
Neglect any damping effects of its worn-out shock absorbers and calculate the natural frequency in cycles per second of the vertical vibration which occurs after impact with the ground. Each of the four springs on the car has a constant of Because the center of mass is located midway between the axles and the car is level when dropped, there is no rotational motion. State any assumptions.
17.5 kN/m.
1000-kg (Hz) ƒn 10 kg k = 3000 N/m k = 3000 N/m ƒn k k 100 kg k 180 kN/m.
amax 0.5 m/s 8/14 With the assumption of no slipping, determine the mass m of the block which must be placed on the top of the cart in order that the system period be What is the minimum coefficient of static friction for which the block will not slip relative to the cart if the cart is displaced 50 mm from the equi-librium position and released?
Problem 8/14 8/15 An energy-absorbing car bumper with its springs ini-tially undeformed has an equivalent spring constant of If the car approaches a massive wall with a speed of determine (a) the velocity of the car as a function of time during contact with the wall, where is the beginning of the impact, and (b) the maximum deflection of the bumper.
Problem 8/15 8/16 A woman stands in the center of an end-supported board and causes a midspan deflection of 0.9 in. If she flexes her kness slightly in order to cause a vertical vibration, what is the frequency of the motion? Assume elastic response of the board and neglect its relatively small mass.
Problem 8/16 0.9″ ƒn 120-lb 5 mi/hr xmax t 0 v 5 mi/hr, 2500-lb 3000 lb/in.
600 N/m 6 kg m s μ s 0.75 s.
6-kg Article 8/2 Problems 595 8/17 A small particle of mass m is attached to two highly tensioned wires as shown. Determine the system natural frequency for small vertical oscillations if the tension T in both wires is assumed to be con-stant. Is the calculation of the small static deflection of the particle necessary?
Problem 8/17 8/18 The cylindrical buoy floats in salt water (density ) and has a mass of 800 kg with a low cen-ter of mass to keep it stable in the upright position.
Determine the frequency of vertical oscillation of the buoy. Assume that the water level remains undis-turbed adjacent to the buoy.
Problem 8/18 8/19 Shown in the figure is a model of a one-story build-ing. The bar of mass m is supported by two light elas-tic upright columns whose upper and lower ends are fixed against rotation. For each column, if a force P and corresponding moment M were applied as shown in the right-hand part of the figure, the deflection would be given by where L is the effec-tive column length, E is Young’s modulus, and I is the area moment of inertia of the column cross sec-tion with respect to its neutral axis. Determine the natural frequency of horizontal oscillation of the bar when the columns bend as shown in the figure.
Problem 8/19 M m Ground level k k P δ L PL3/12EI, 0.6 m ƒn 1030 kg/m3 m l l n Problem 8/22 8/23 Calculate the natural circular frequency of the system shown in the figure. The mass and friction of the pulleys are negligible.
Problem 8/23 8/24 Derive the differential equation of motion for the system shown in terms of the variable The mass of the linkage is negligible. State the natural fre-quency in rad/s for the case and Assume small oscillations throughout.
Problem 8/24 x1 k1 m1 x2 k2 m2 a b B A O m1 m2 m.
k1 k2 k n x1.
k θ m m n k m b c b c 596 Chapter 8 Vibration and Time Response 8/20 A piece of putty is dropped 2 m onto the initially stationary block, which is supported by four springs, each of which has a constant Determine the displacement x as a function of time during the resulting vibration, where x is measured from the initial position of the block as shown.
Problem 8/20 8/21 Calculate the frequency of vertical oscillation of the block when it is set in motion. Each spring has a stiffness of Neglect the mass of the pulleys.
Problem 8/21 8/22 The weighing platform has a mass m and is con-nected to the spring of stiffness k by the system of levers shown. Derive the differential equation for small vertical oscillations of the platform and find the period Designate as the platform displace-ment from the equilibrium position and neglect the mass of the levers.
y .
k k 50 lb 6 lb/in.
50-lb ƒn 28 kg 3 kg x k k 2 m k 800 N/m.
28-kg 3-kg Damped, Free Vibrations 8/25 Determine the value of the damping ratio for the simple spring-mass-dashpot system shown.
Problem 8/25 8/26 The period of damped linear oscillation for a cer-tain mass is 0.3 s. If the stiffness of the sup-porting linear spring is calculate the damping coefficient 8/27 Viscous damping is added to an initially undamped spring-mass system. For what value of the damping ratio will the damped natural frequency be equal to 90 percent of the natural frequency of the original undamped system?
8/28 The addition of damping to an undamped spring-mass system causes its period to increase by 25 per-cent. Determine the damping ratio 8/29 Determine the value of the viscous damping coeffi-cient c for which the system shown is critically damped.
Problem 8/29 c 80 lb 200 lb/in.
.
d c.
800 N/m, 1-kg d x k = 392 N/m c = 42 N·s/m 2 kg Article 8/2 Problems 597 8/30 The spring-supported cylinder is set into free vertical vibration and is observed to have a period of 0.75 s in part (a) of the figure. The system is then completely immersed in an oil bath in part (b) of the figure, and the cylinder is displaced from its equilib-rium position and released. Viscous damping ensues, and the ratio of two successive positive-displacement amplitudes is 4. Calculate the viscous damping ratio the viscous damping constant c, and the equiva-lent spring constant k.
Problem 8/30 8/31 The figure represents the measured displacement-time relationship for a vibration with small damping where it is impractical to achieve accurate results by measuring the nearly equal amplitudes of two suc-cessive cycles. Modify the expression for the viscous damping factor based on the measured amplitudes and which are N cycles apart.
Problem 8/31 x t x0 t0 t1 t2 tN tN+1 xN xN x0 2.5 kg 2.5 kg Oil (a) (b) , 2.5-kg 8/36 The system shown is released from rest from an initial position Determine the overshoot displacement Assume translational motion in the x-direction.
Problem 8/36 8/37 The mass of a given critically damped system is re-leased at time from the position with a negative initial velocity. Determine the critical value of the initial velocity below which the mass will pass through the equilibrium position.
8/38 The mass of the system shown is released from rest at in. when Determine the displacement x at if (a) and (b) Problem 8/38 8/39 The cannon fires a cannonball with an absolute velocity of at to the horizontal. The combined weight of the cannon and its cart is 1610 lb.
If the recoil mechanism consists of the spring of con-stant and the damper with viscous coefficient determine the maximum recoil deflection of the cannon unit.
Problem 8/39 k c m 20° 800 ft/sec xmax c 600 lb-sec/ft, k 150 lb/in.
20 800 ft/sec 10-lb x c W = 96.6 lb k = 1 lb/in.
c 18 lb-sec/ft.
c 12 lb-sec/ft t 0.5 sec t 0.
x0 6 (x ˙0)c x0 0 t 0 0 x0 x x1 t 0 c = 18 N·s/m m = 3 kg k = 108 N/m x x1.
x0.
598 Chapter 8 Vibration and Time Response 8/32 The mass of Prob. is released from rest at a distance to the right of the equilibrium position.
Determine the displacement x as a function of time t, where is the time of release.
8/33 A damped spring-mass system is released from rest from a positive initial displacement If the suc-ceeding maximum positive displacement is de-termine the damping ratio of the system.
Problem 8/33 8/34 Determine the values of the viscous damping coeffi-cient for which the system has a damping ratio of (a) 0.5 and (b) 1.5.
Problem 8/34 8/35 Further design refinement for the weighing platform of Prob. 8/12 is shown here where two viscous dampers are to be added to limit the ratio of succes-sive positive amplitudes of vertical vibration in the unloaded condition to 4. Determine the necessary viscous damping coefficient c for each of the dampers.
Problem 8/35 4000 kg k c c k k c c 15 lb/in.
40 lb 20° c t x 00 x0 x0/2 x0/2, x0.
t 0 x0 8/25 2-kg 8/40 The owner of a pickup truck tests the action of his rear-wheel shock absorbers by applying a steady force to the rear bumper and measur-ing a static deflection of 3 in. Upon sudden release of the force, the bumper rises and then falls to a maxi-mum of in. below the unloaded equilibrium posi-tion of the bumper on the first rebound. Treat the action as a one-dimensional problem with an equiva-lent mass of half the truck mass. Find the viscous damping factor for the rear end and the viscous damping coefficient for each shock absorber assum-ing its action to be vertical.
Problem 8/40 8/41 Determine the damping ratio of the system depicted in the figure. The mass and friction of the pulleys are negligible, and the cable remains taut at all times.
Problem 8/41 8/42 Derive the differential equation of motion for the system shown in its equilibrium position. Neglect the mass of link AB and assume small oscillations.
c m2 θ m1 k Equil.
position 100 lb 3″ ⎫ ⎬ ⎭ c 1 2 100-lb 3400-lb Article 8/2 Problems 599 Problem 8/42 8/43 Develop the equation of motion in terms of the vari-able x for the system shown. Determine an expres-sion for the damping ratio in terms of the given system properties. Neglect the mass of the crank AB and assume small oscillations about the equilibrium position shown.
Problem 8/43 8/44 Investigate the case of Coulomb damping for the block shown, where the coefficient of kinetic friction is and each spring has a stiffness The block is displaced a distance from the neutral position and released. Determine and solve the differential equation of motion. Plot the resulting vibration and indicate the rate r of decay of the amplitude with time.
Problem 8/44 k/2 k/2 m y x k μ x0 k/2. k k m c O a b B A x x k m1 m2 c O a A B b 8/3 Forced Vibration of Particles Although there are many significant applications of free vibrations, the most important class of vibration problems is that where the motion is continuously excited by a disturbing force. The force may be exter-nally applied or may be generated within the system by such means as unbalanced rotating parts. Forced vibrations may also be excited by the motion of the system foundation.
Harmonic Excitation Various forms of forcing functions F F(t) and foundation displace-ments xB xB(t) are depicted in Fig. 8/8. The harmonic force shown in part a of the figure occurs frequently in engineering practice, and the understanding of the analysis associated with harmonic forces is a nec-essary first step in the study of more complex forms. For this reason, we will focus our attention on harmonic excitation.
We first consider the system of Fig. 8/9a, where the body is subjected to the external harmonic force F F0 sin t, in which F0 is the force am-plitude and is the driving frequency (in radians per second). Be sure to distinguish between n which is a property of the system, and , which is a property of the force applied to the system. We also note that for a force F F0 cos t, one merely substitutes cos t for sin t in the re-sults about to be developed.
k/m, 600 Chapter 8 Vibration and Time Response F(t) or xB(t) (a) Harmonic (b) Periodic Nonharmonic (c) Nonperiodic t t t t F(t) or xB(t) F(t) or xB(t) F(t) or xB(t) Square General Half sine Step Ramp Cycloidal Impulse Random Triangle Saw tooth Figure 8/8 An automobile undergoing vibration testing of its suspension system.
Courtesy of MTS Systems Corporation From the free-body diagram of Fig. 8/9a, we may apply Newton’s second law to obtain In standard form, with the same variable substitutions made in Art. 8/2, the equation of motion becomes (8/13) Base Excitation In many cases, the excitation of the mass is due not to a directly ap-plied force but to the movement of the base or foundation to which the mass is connected by springs or other compliant mountings. Examples of such applications are seismographs, vehicle suspensions, and struc-tures shaken by earthquakes.
Harmonic movement of the base is equivalent to the direct applica-tion of a harmonic force. To show this, consider the system of Fig. 8/9b where the spring is attached to the movable base. The free-body dia-gram shows the mass displaced a distance x from the neutral or equilib-rium position it would have if the base were in its neutral position. The base, in turn, is assumed to have a harmonic movement xB b sin t.
Note that the spring deflection is the difference between the inertial dis-placements of the mass and the base. From the free-body diagram, New-ton’s second law gives or (8/14) We see immediately that Eq. 8/14 is exactly the same as our basic equa-tion of motion, Eq. 8/13, in that F0 is replaced by kb. Consequently, all the results about to be developed apply to either Eq. 8/13 or 8/14.
Undamped Forced Vibration First, we treat the case where damping is negligible (c 0). Our basic equation of motion, Eq. 8/13, becomes (8/15) The complete solution to Eq. 8/15 is the sum of the complementary solution xc, which is the general solution of Eq. 8/15 with the right side set to zero, and the particular solution xp, which is any solution to the complete equation. Thus, x xc xp. We developed the complementary solution in Art. 8/2. A particular solution is investigated by assuming x ¨ n 2x F0 m sin t x ¨ 2nx ˙ n 2x kb sin t m k(x xB) cx ˙ mx ¨ x ¨ 2nx ˙ n 2x F0 sin t m kx cx ˙ F0 sin t mx ¨ Article 8/3 Forced Vibration of Particles 601 x k c m F = F0 sin t ω F0 sin t ω (a) (b) m kx cx · x k B Neutral position xB = b sin t ω c m m k(x – xB) cx · Figure 8/9 that the form of the response to the force should resemble that of the force term. To that end, we assume (8/16) where X is the amplitude (in units of length) of the particular solution.
Substituting this expression into Eq. 8/15 and solving for X yield (8/17) Thus, the particular solution becomes (8/18) The complementary solution, known as the transient solution, is of no special interest here since, with time, it dies out with the small amount of damping which is always unavoidably present. The particular solution xp describes the continuing motion and is called the steady-state solution. Its period is 2/, the same as that of the forcing function.
Of primary interest is the amplitude X of the motion. If we let st stand for the magnitude of the static deflection of the mass under a sta-tic load F0, then st F0/k, and we may form the ratio (8/19) The ratio M is called the amplitude ratio or magnification factor and is a measure of the severity of the vibration. We especially note that M ap-proaches infinity as approaches n. Consequently, if the system pos-sesses no damping and is excited by a harmonic force whose frequency approaches the natural frequency n of the system, then M, and thus X, increase without limit. Physically, this means that the motion ampli-tude would reach the limits of the attached spring, which is a condition to be avoided.
The value n is called the resonant or critical frequency of the sys-tem, and the condition of being close in value to n with the resulting large displacement amplitude X is called resonance. For n, the magnification factor M is positive, and the vibration is in phase with the force F. For n, the magnification factor is negative, and the vibra-tion is 180 out of phase with F. Figure 8/10 shows a plot of the absolute value of M as a function of the driving-frequency ratio /n.
Damped Forced Vibration We now reintroduce damping in our expressions for forced vibra-tion. Our basic differential equation of motion is [8/13] Again, the complete solution is the sum of the complementary solution xc, which is the general solution of Eq. 8/13 with the right side equal to x ¨ 2nx ˙ n 2x F0 sin t m M X st 1 1 (/n)2 xp F0 /k 1 (/n)2 sin t X F0 /k 1 (/n)2 xp X sin t 602 Chapter 8 Vibration and Time Response 0 6 5 4 3 2 1 0 1 |M| n /ω ω 2 3 Figure 8/10 zero, and the particular solution xp, which is any solution to the com-plete equation. We have already developed the complementary solution xc in Art. 8/2. When damping is present, we find that a single sine or co-sine term, such as we were able to use for the undamped case, is not suf-ficiently general for the particular solution. So we try Substitute the latter expression into Eq. 8/13, match coefficients of sin t and cos t, and solve the resulting two equations to obtain (8/20) (8/21) The complete solution is now known, and for underdamped systems it can be written as (8/22) Because the first term on the right side diminishes with time, it is known as the transient solution. The particular solution xp is the steady-state solution and is the part of the solution in which we are primarily interested. All quantities on the right side of Eq. 8/22 are properties of the system and the applied force, except for C and (which are deter-minable from initial conditions) and the running time variable t.
x Cent sin (d t ) X sin (t ) tan1 2/n 1 (/n)2 X F0 /k {[1 (/n)2]2 [2/n]2}1/2 xp X1 cos t X2 sin t or xp X sin (t ) Article 8/3 Forced Vibration of Particles 603 KEY CONCEPTS Magnification Factor and Phase Angle Near resonance the magnitude X of the steady-state solution is a strong function of the damping ratio and the nondimensional fre-quency ratio /n. It is again convenient to form the nondimensional ratio M X/(F0/k), which is called the amplitude ratio or magnification factor (8/23) An accurate plot of the magnification factor M versus the frequency ratio /n for various values of the damping ratio is shown in Fig. 8/11.
This figure reveals the most essential information pertinent to the forced vibration of a single-degree-of-freedom system under harmonic excitation.
It is clear from the graph that, if a motion amplitude is excessive, two possi-ble remedies would be to (a) increase the damping (to obtain a larger value of ) or (b) alter the driving frequency so that is farther from the resonant frequency n. The addition of damping is most effective near resonance.
Figure 8/11 also shows that, except for 0, the magnification-factor curves do not actually peak at /n 1. The peak for any given value of can be calculated by finding the maximum value of M from Eq. 8/23.
M 1 {[1 (/n)2]2 [2/n]2}1/2 The phase angle , given by Eq. 8/21, can vary from 0 to and rep-resents the part of a cycle (and thus the time) by which the response xp lags the forcing function F. Figure 8/12 shows how the phase angle varies with the frequency ratio for various values of the damping ratio .
Note that the value of when /n 1 is 90 for all values of . To fur-ther illustrate the phase difference between the response and the forc-ing function, we show in Fig. 8/13 two examples of the variation of F and xp with t. In the first example, n and is taken to be /4. In the second example, n and is taken to be 3/4.
604 Chapter 8 Vibration and Time Response F, xp F X X F xp xp F xp φ φ 0 2 3 2 3 t ω < = /4 n , ω ω φ t ω F, xp φ 0 t ω F F0 F0 xp φ > = 3 /4 n, ω ω φ t ω Figure 8/13 = 1 = 1 = 0 = 0 0 0 1 2 3 0.1 0.5 0.2 0.1 0.2 0.5 — 2 rad φ n /ω ω ζ ζ ζ ζ Figure 8/12 0.1 0.2 0.5 = 1 = 0 0 6 5 4 3 2 1 0 1 M 2 3 n /ω ω ζ ζ Figure 8/11 Applications Vibration-measuring instruments such as seismometers and ac-celerometers are frequently encountered applications of harmonic exci-tation. The elements of this class of instruments are shown in Fig.
8/14a. We note that the entire system is subjected to the motion xB of the frame. Letting x denote the position of the mass relative to the frame, we may apply Newton’s second law and obtain where (x xB) is the inertial displacement of the mass. If xB b sin t, then our equation of motion with the usual notation is which is the same as Eq. 8/13 if b2 is substituted for F0/m.
Again, we are interested only in the steady-state solution xp. Thus, from Eq. 8/20, we have If X represents the amplitude of the relative response xp, then the nondimensional ratio X/b is where M is the magnification ratio of Eq. 8/23. A plot of X/b as a func-tion of the driving-frequency ratio /n is shown in Fig. 8/14b. The simi-larities and differences between the magnification ratios of Figs. 8/14b and 8/11 should be noted.
X/b (/n)2M xp b(/n)2 {[1 (/n)2]2 [2/n]2}1/2 sin (t ) x ¨ 2nx ˙ n 2x b2 sin t cx ˙ kx m d2 dt2 (x xB) or x ¨ c m x ˙ k m x x ¨B Article 8/3 Forced Vibration of Particles 605 0.1 0.2 0.5 0 6 5 4 3 2 1 0 1 X/b (b) (a) 2 3 m m kx cx · k c x Neutral position Equilibrium position xB = b sin t ω n /ω ω = 0 ζ = 1 ζ Figure 8/14 If the frequency ratio /n is large, then X/b 1 for all values of the damping ratio . Under these conditions, the displacement of the mass relative to the frame is approximately the same as the absolute displace-ment of the frame, and the instrument acts as a displacement meter. To obtain a high value of /n, we need a small value of n which means a soft spring and a large mass. With such a combination, the mass will tend to stay inertially fixed. Displacement meters generally have very light damping.
On the other hand, if the frequency ratio /n is small, then M ap-proaches unity (see Fig. 8/11) and X/b (/n)2 or X b(/n)2. But b2 is the maximum acceleration of the frame. Thus, X is proportional to the maximum acceleration of the frame, and the instrument may be used as an accelerometer. The damping ratio is generally selected so that M ap-proximates unity over the widest possible range of /n. From Fig. 8/11, we see that a damping factor somewhere between 0.5 and 1 would meet this criterion.
Electric Circuit Analogy An important analogy exists between electric circuits and mechani-cal spring-mass systems. Figure 8/15 shows a series circuit consisting of a voltage E which is a function of time, an inductance L, a capacitance C, and a resistance R. If we denote the charge by the symbol q, the equa-tion which governs the charge is (8/24) This equation has the same form as the equation for the mechanical sys-tem. Thus, by a simple interchange of symbols, the behavior of the elec-trical circuit may be used to predict the behavior of the mechanical system, or vice versa. The mechanical and electrical equivalents in the following table are worth noting: Lq ¨ Rq ˙ 1 C q E k/m, 606 Chapter 8 Vibration and Time Response MECHANICAL-ELECTRICAL EQUIVALENTS E C L R Figure 8/15 SAMPLE PROBLEM 8/4 A 50-kg instrument is supported by four springs, each of stiffness 7500 N/m.
If the instrument foundation undergoes harmonic motion given in meters by xB 0.002 cos 50t, determine the amplitude of the steady-state motion of the in-strument. Damping is negligible.
Solution.
For harmonic oscillation of the base, we substitute kb for F0 in our par-ticular-solution results, so that, from Eq. 8/17, the steady-state amplitude becomes The resonant frequency is n 24.5 rad/s, and the im-pressed frequency 50 rad/s is given. Thus, Ans.
Note that the frequency ratio /n is approximately 2, so that the condition of resonance is avoided.
SAMPLE PROBLEM 8/5 The spring attachment point B is given a horizontal motion xB b cos t.
Determine the critical driving frequency c for which the oscillations of the mass m tend to become excessively large. Neglect the friction and mass associated with the pulleys. The two springs have the same stiffness k.
Solution.
The free-body diagram is drawn for arbitrary positive displacements x and xB. The motion variable x is measured downward from the position of sta-tic equilibrium defined as that which exists when xB 0. The additional stretch in the upper spring, beyond that which exists at static equilibrium, is 2x xB.
Therefore, the dynamic spring force in the upper spring, and hence the dynamic tension T in the cable, is k(2x xB). Summing forces in the x-direction gives which becomes The natural frequency of the system is n Thus, Ans.
c n 5k/m 5k/m. x ¨ 5k m x 2kb cos t m 2k(2x xB) kx mx ¨ [ΣFx mx ¨] X 0.002 1 (50/24.5)2 6.32(104) m or 0.632 mm 4(7500)/50 k/m X b 1 (/n)2 Article 8/3 Forced Vibration of Particles 607 xB = b cos t ω k k B T T Equilibrium position Neutral position x m kx (Dynamic forces only) Helpful Hints Note that either sin 50t or cos 50t can be used for the forcing function with this same result.
The minus sign indicates that the motion is 180 out of phase with the applied excitation.
Helpful Hints If a review of the kinematics of con-strained motion is necessary, see Art. 2/9.
We learned from the discussion in Art. 8/2 that the equal and opposite forces associated with the position of static equilibrium may be omitted from the analysis. Our use of the terms dynamic spring force and dy-namic tension stresses that only the force increments in addition to the static values need be considered.
SAMPLE PROBLEM 8/6 The 100-lb piston is supported by a spring of modulus k 200 lb/in. A dash-pot of damping coefficient c 85 lb-sec/ft acts in parallel with the spring. A fluc-tuating pressure p 0.625 sin 30t in lb/in.2 acts on the piston, whose top surface area is 80 in.2 Determine the steady-state displacement as a function of time and the maximum force transmitted to the base.
Solution.
We begin by computing the system natural frequency and damping ratio: The steady-state amplitude, from Eq. 8/20, is The phase angle, from Eq. 8/21, is The steady-state motion is then given by the second term on the right side of Eq. 8/22: Ans.
The force Ftr transmitted to the base is the sum of the spring and damper forces, or The maximum value of Ftr is Ans. 67.9 lb 0.01938[(200)(12)]2 (85)2(30)2 (Ftr)max (kX)2 (cX)2 Xk2 c22 Ftr kxp cx ˙p kX sin (t ) cX cos (t ) xp X sin (t ) 0.01938 sin (30t 1.724) ft 1.724 rad tan1 2(0.492)(30/27.8) 1 (30/27.8)2 tan1 2/n 1 (/n)2 0.01938 ft (0.625)(80)/[(200)(12)] {[1 (30/27.8)2]2 [2(0.492)(30/27.8)]2}1/2 X F0 /k {[1 (/n)2]2 [2/n]2}1/2 c 2mn 85 2 100 32.2(27.8) 0.492 (underdamped) 608 Chapter 8 Vibration and Time Response k W p = p0 sin ωt c Equilibrium position kx x (Dynamic forces only) cx · F = pA sin t ω Helpful Hints You are encouraged to repeat these calculations with the damping coeffi-cient c set to zero so as to observe the influence of the relatively large amount of damping present.
Note that the argument of the in-verse tangent expression for has a positive numerator and a negative denominator for the case at hand, thus placing in the second quad-rant. Recall that the defined range of is 0 .
PROBLEMS (Unless otherwise instructed, assume that the damping is light to moderate so that the amplitude of the forced re-sponse is a maximum at ) Introductory Problems 8/45 Determine the amplitude X of the steady-state motion of the mass if (a) and (b) Problem 8/45 8/46 A viscously damped spring-mass system is excited by a harmonic force of constant amplitude but vary-ing frequency If the amplitude of the steady-state motion is observed to decrease by a factor of 8 as the frequency ratio is varied from 1 to 2, determine the damping ratio of the system.
8/47 The cart is acted upon by the harmonic force shown in the figure. If determine the range of the driving frequency for which the magnitude of the steady-state response is less than 3 in.
Problem 8/47 8/48 If the viscous damping coefficient of the damper in the system of Prob. 8/47 is deter-mine the range of the driving frequency for which the magnitude of the steady-state response is less than 3 in.
8/49 If the driving frequency for the system of Prob. is determine the required value of the damping coefficient if the steady-state amplitude is not to exceed 3 in.
c 6 rad/sec, 8/47 c 2.4 lb-sec/ft, c k = 6 lb/in.
F = 5 cos t lb ω 64.4 lb c 0, 64.4-lb /n .
F0 m = 10 kg k = 100 kN/m F = 1000 cos 120t N c c 0.
c 500 Ns/m 10-kg /n 1.
Article 8/3 Problems 609 8/50 The block of weight is suspended by two springs each of stiffness and is acted upon by the force where t is the time in seconds. Determine the amplitude X of the steady-state motion if the viscous damping coeffi-cient c is (a) 0 and (b) Compare these amplitudes to the static spring deflection Problem 8/50 8/51 An external force sin is applied to the cylinder as shown. What value of the driving fre-quency would cause excessively large oscillations of the system?
Problem 8/51 8/52 A viscously damped spring-mass system is forced harmonically at the undamped natural frequency If the damping ratio is doubled from 0.1 to 0.2, compute the percentage reduction in the steady-state amplitude. Compare with the result of a similar calculation for the condition Verify your results by inspecting Fig. 8/11.
/n 2.
R2 R1 (/n 1).
F m k 2m c t F F0 F k c k W st.
60 lb-sec/ft.
F 75 cos 15t lb k 200 lb/ft W 100 lb 8/57 The variable-speed motorized unit is re-strained in the horizontal direction by two springs, each of which has a stiffness of Each of the two dashpots has a viscous damping coefficient In what ranges of speeds N can the motor be run for which the magnification factor will not exceed 2?
Problem 8/57 8/58 When the person stands in the center of the floor system shown, he causes a static deflection of the floor under his feet. If he walks (or runs quickly!) in the same area, how many steps per second would cause the floor to vibrate with the greatest vertical amplitude?
Problem 8/58 st k k c c N M c 58 Ns/m.
2.1 kN/m.
20-kg 610 Chapter 8 Vibration and Time Response Representative Problems 8/53 A linear spring-mass oscillator has a viscous damp-ing factor and an undamped natural fre-quency By referring to Fig. estimate the range of frequencies ƒ of the periodic applied force for which the amplitude of the oscillator will not exceed twice the static deflection which would be caused by applying a static force equal in magnitude to that of the periodic force. Check your estimate by applying Eq. 8/54 It was noted in the text that the maxima of the curves for the magnification factor M are not located at Determine an expression in terms of the damping ratio for the frequency ratio at which the maxima occur.
8/55 Each ball is attached to the end of the light elastic rod and deflects 4 mm when a force is statically applied to the ball. If the central collar is given a vertical harmonic movement with a fre-quency of 4 Hz and an amplitude of 3 mm, find the amplitude of vertical vibration of each ball.
Problem 8/55 8/56 The motion of the outer frame B is given by b sin For what range of the driving frequency is the amplitude of the motion of the mass m relative to the frame less than 2b?
Problem 8/56 m B k/2 k/2 xB = b sin t ω t.
xB 0.5 kg 0.5 kg y0 2-N 0.5-kg /n 1.
8/23.
8/11, ƒn 6 Hz.
0.2 8/59 The instrument shown has a mass of 43 kg and is spring-mounted to the horizontal base. If the ampli-tude of vertical vibration of the base is 0.10 mm, cal-culate the range of frequencies of the base vibration which must be prohibited if the amplitude of vertical vibration of the instrument is not to ex-ceed 0.15 mm. Each of the four identical springs has a stiffness of Problem 8/59 8/60 Attachment B is given a horizontal motion Derive the equation of motion for the mass m and state the critical frequency for which the oscillations of the mass become excessively large.
Problem 8/60 8/61 Attachment B is given a horizontal motion Derive the equation of motion for the mass m and state the critical frequency for which the oscillations of the mass become excessively large.
What is the damping ratio for the system?
Problem 8/61 k c1 c2 m x B xB = b cos t ω c b cos t.
xB c k1 k2 m x B xB = b cos t ω c b cos t.
xB xB 7.2 kN/m.
ƒn Article 8/3 Problems 611 8/62 Derive an expression for the transmission ratio for the system of the figure. This ratio is defined as the maximum force transmitted to the base divided by the amplitude of the forcing function. Express your answer in terms of and the magnifica-tion factor M.
Problem 8/62 8/63 A device to produce vibrations consists of the two counter-rotating wheels, each carrying an eccentric mass with a center of mass at a distance from its axis of rotation. The wheels are synchronized so that the vertical positions of the un-balanced masses are always identical. The total mass of the device is 10 kg. Determine the two possible values of the equivalent spring constant for the mounting which will permit the amplitude of the pe-riodic force transmitted to the fixed mounting to be 1500 N due to the imbalance of the rotors at a speed of Neglect damping.
Problem 8/63 θ θ ω ω e e m0 m0 1800 rev/min.
k e 12 mm m0 1 kg c k F = F0 sin t ω m Base n, , , F0 T Problem 8/67 8/68 The seismic instrument is mounted on a structure which has a vertical vibration with a frequency of 5 Hz and a double amplitude of 18 mm. The sensing element has a mass and the spring stiff-ness is The motion of the mass rela-tive to the instrument base is recorded on a revolving drum and shows a double amplitude of 24 mm during the steady-state condition. Calculate the viscous damping constant c.
Problem 8/68 Structure 24 mm D c 2 kg k = 1.5 kN/m k 1.5 kN/m.
m 2 kg, C O a m x c k2 k1 b A xB = b0 cos t ω 612 Chapter 8 Vibration and Time Response 8/64 The seismic instrument shown is attached to a struc-ture which has a horizontal harmonic vibration at 3 Hz. The instrument has a mass a spring stiffness and a viscous damping coeffi-cient If the maximum recorded value of x in its steady-state motion is determine the amplitude b of the horizontal movement of the structure.
Problem 8/64 8/65 A device similar to that shown in Prob. is to be used to measure the horizontal acceleration of the structure which is vibrating with a frequency of 5 Hz.
The mass is the spring constant is and the damping factor is If the amplitude of x is 4.0 mm, approximate the maxi-mum acceleration of the structure.
8/66 Derive and solve the equation of motion for the mass which is subjected to the suddenly applied force F that remains constant after application. The dis-placement and velocity of the mass are both zero at time Plot versus for several motion cycles.
Problem 8/66 8/67 Derive and solve the equation of motion for the mass m in terms of the variable x for the system shown.
Neglect the mass of the lever AOC and assume small oscillations.
t x t 0.
amax 0.75.
k 150 N/m, m 0.008 kg, 8/64 x xB k c Structure m xB X 2 mm, c 3 N s/m.
k 20 N/m, m 0.5 kg, Force F k F m Time, t 00 F0 8/69 Determine the amplitude of vertical vibration of the spring-mounted trailer as it travels at a velocity of over the corduroy road whose contour may be expressed by a sine or cosine term. The mass of the trailer is 500 kg and that of the wheels alone may be neglected. During the loading, each 75 kg added to the load caused the trailer to sag 3 mm on its springs. Assume that the wheels are in contact with the road at all times and neglect damping. At what critical speed is the vibration of the trailer greatest?
Problem 8/69 1.2 m 50 mm vc 25 km/h Article 8/3 Problems 613 8/70 Derive the expression for the power loss P averaged over a complete steady-state cycle due to the fric-tional dissipation of energy in a viscously damped linear oscillator. The forcing function is and the displacement-time relation for steady-state motion is where the amplitude X is given by Eq. (Hint: The frictional energy loss during a displacement is where is the vis-cous damping coefficient. Integrate this expression over a complete cycle and divide by the period of the cycle.) c cx ˙ dx, dx 8/20.
xP X sin (t ) F0 sin t, 8/4 Vibration of Rigid Bodies The subject of planar rigid-body vibrations is entirely analogous to that of particle vibrations. In particle vibrations, the variable of interest is one of translation (x), while in rigid-body vibrations, the variable of pri-mary concern may be one of rotation ( ). Thus, the principles of rotational dynamics play a central role in the development of the equation of motion.
We will see that the equation of motion for rotational vibration of rigid bodies has a mathematical form identical to that developed in Arts.
8/2 and 8/3 for translational vibration of particles. As was the case with particles, it is convenient to draw the free-body diagram for an arbitrary positive value of the displacement variable, because a negative displace-ment value easily leads to sign errors in the equation of motion. The practice of measuring the displacement from the position of static equi-librium rather than from the position of zero spring deflection continues to simplify the formulation for linear systems because the equal and op-posite forces and moments associated with the static equilibrium posi-tion cancel from the analysis.
Rather than individually treating the cases of (a) free vibration, un-damped and damped, and (b) forced vibrations, undamped and damped, as was done with particles in Arts. 8/2 and 8/3, we will go directly to the damped, forced problem.
Rotational Vibration of a Bar As an illustrative example, consider the rotational vibration of the uniform slender bar of Fig. 8/16a. Figure 8/16b depicts the free-body dia-gram associated with the horizontal position of static equilibrium.
Equating to zero the moment sum about O yields where P is the magnitude of the static spring force.
Figure 8/16c depicts the free-body diagram associated with an arbi-trary positive angular displacement . Using the equation of rotational motion ΣMO as developed in Chapter 6, we write where IO md2 ml2/12 m(l/6)2 ml2/9 is obtained from the parallel-axis theorem for mass moments of inertia.
For small angular deflections, the approximations sin and cos 1 may be used. With P mg/4, the equation of motion, upon re-arrangement and simplification, becomes (8/25) ¨ c m ˙ 4 k m (F0l/3) cos t ml2/9 I (F0 cos t) l 3 cos 1 9 ml2 ¨ (mg) l 6 cos cl 3 ˙ cos l 3 cos P k 2l 3 sin 2l 3 cos IO ¨ P l 2 l 6 mg l 6 0 P mg 4 614 Chapter 8 Vibration and Time Response O′ y O′ x = 0 Oy O mg mg P (c) (b) (a) Ox F0 cos t ω θ P + k sin 2l — 3 θ c ) = ( cos d — dt l — 3 cl — 3 θ sin θ · θ F0 cos t ω l — 3 l — 3 l — 3 l — 2 l – 6 k c m O Figure 8/16 The right side has been left unsimplified in the form M0(cos t)/IO, where M0 F0l/3 is the magnitude of the moment about point O of the externally applied force. Note that the two equal and opposite mo-ments associated with static equilibrium forces canceled on the left side of the equation of motion. Thus, it is not necessary to include the static-equilibrium forces and moments in the analysis.
Rotational Counterparts of Translational Vibration At this point, we observe that Eq. 8/25 is identical in form to Eq.
8/13 for the translational case, so we may write (8/26) Thus, we may use all of the relations developed in Arts. 8/2 and 8/3 merely by replacing the translational quantities with their rotational counterparts. The following table shows the results of this procedure as applied to the rotating bar of Fig. 8/16: ¨ 2n ˙ n 2 M0 cos t IO Article 8/4 Vibration of Rigid Bodies 615 In the preceding table, the variable k in the expression for represents the equivalent torsional spring constant of the system of Fig. 8/16 and is determined by writing the restoring moment of the spring. For a small angle , this moment about O is Thus, k Note that M0/k is the static angular deflection which would be produced by a constant external moment M0.
We conclude that an exact analogy exists between particle vibra-tion and the small angular vibration of rigid bodies. Furthermore, the utilization of this analogy can save the labor of complete rederivation of the governing relationships for a given problem of general rigid-body vibration.
4 9 kl2.
Mk [k(2l/3) sin ][(2l/3) cos ] (4 9kl2) SAMPLE PROBLEM 8/7 A simplified version of a pendulum used in impact tests is shown in the figure.
Derive the equation of motion and determine the period for small oscillations about the pivot. The mass center G is located a distance 0.9 m from O, and the radius of gyration about O is kO 0.95 m. The friction of the bearing is negligible.
Solution.
We draw the free-body diagram for an arbitrary, positive value of the angular-displacement variable , which is measured counterclockwise for the coor-dinate system chosen. Next we apply the governing equation of motion to obtain or Ans.
Note that the governing equation is independent of the mass. When is small, sin , and our equation of motion may be written as The frequency in cycles per second and the period in seconds are Ans.
For the given properties: Ans.
SAMPLE PROBLEM 8/8 The uniform bar of mass m and length l is pivoted at its center. The spring of constant k at the left end is attached to a stationary surface, but the right-end spring, also of constant k, is attached to a support which undergoes a harmonic motion given by yB b sin t. Determine the driving frequency c which causes resonance.
Solution.
We use the moment equation of motion about the fixed point O to obtain Assuming small deflections and simplifying give us The natural frequency should be recognized from the now-familiar form of the equation to be Thus, c n will result in resonance (as well as violation of our small-angle assumption!).
Ans.
6k/m n 6k/m ¨ 6k m 6kb ml sin t k l 2 sin l 2 cos k l 2 sin yB l 2 cos 1 12 ml2 ¨ ¨ gr kO 2 0 ¨ gr kO 2 sin 0 mgr sin mkO 2 ¨ [ΣMO IO ¨] r 616 Chapter 8 Vibration and Time Response G O Oy Ox θ r – mg θ r – yB = b sin ωt O B m k l — 2 l — 2 k Oy mg Ox θ k ) ( l — 2 sin θ k – yB) ( l — 2 sin θ Helpful Hints With our choice of point O as the moment center, the bearing reac-tions Ox and Oy never enter the equation of motion.
For large angles of oscillation, deter-mining the period for the pendulum requires the evaluation of an elliptic integral.
Helpful Hints As previously, we consider only the changes in the forces due to a move-ment away from the equilibrium position.
The standard form here is where M0 and IO The natural frequency n of a system does not depend on the exter-nal disturbance.
1 12 ml2.
klb 2 M0 sin t IO , n 2 ¨ SAMPLE PROBLEM 8/9 Derive the equation of motion for the homogeneous circular cylinder, which rolls without slipping. If the cylinder mass is 50 kg, the cylinder radius 0.5 m, the spring constant 75 N/m, and the damping coefficient 10 determine (a) the undamped natural frequency (b) the damping ratio (c) the damped natural frequency (d) the period of the damped system.
In addition, determine x as a function of time if the cylinder is released from rest at the position x 0.2 m when t 0.
Solution.
We have a choice of motion variables in that either x or the angular displacement of the cylinder may be used. Since the problem statement in-volves x, we draw the free-body diagram for an arbitrary, positive value of x and write the two motion equations for the cylinder as The condition of rolling with no slip is Substitution of this condition into the moment equation gives F Inserting this expression for the friction force into the force equation for the x-direction yields Comparing the above equation with that for the standard damped oscillator, Eq. 8/9, allows us to state directly (a) Ans.
(b) Ans.
Hence, the damped natural frequency and the damped period are (c) Ans.
(d) Ans.
From Eq. 8/12, the underdamped solution to the equation of motion is At time t 0, x and become The solution to the two equations in C and gives Thus, the motion is given by Ans.
x 0.200e0.0667t sin (0.998t 1.504) m C 0.200 m 1.504 rad x ˙0 0.0667C sin 0.998C cos 0 x0 C sin 0.2 x ˙ 0.998Ce0.0667t cos (0.998t ) x ˙ 0.0667Ce0.0667t sin (0.998t ) The velocity is x Cent sin (dt ) Ce(0.0667)(1)t sin (0.998t ) d 2/d 2/0.998 6.30 s d n1 2 (1)1 (0.0667)2 0.998 rad/s 2n 2 3 c m 1 3 c mn 10 3(50)(1) 0.0667 n 2 2 3 k m cx ˙ kx 1 2 mx ¨ mx ¨ or x ¨ 2 3 c m x ˙ 2 3 k m x 0 1 2 mx ¨.
r ¨.
x ¨ Fr 1 2 mr2 ¨ [ΣMG I ¨] cx ˙ kx F mx ¨ [ΣFx mx ¨] Ns/m, Article 8/4 Vibration of Rigid Bodies 617 x m k c r F O N mg Equilibrium position kx cx · x +θ Helpful Hints The angle is taken positive clock-wise to be kinematically consistent with x.
The friction force F may be as-sumed in either direction. We will find that the actual direction is to the right for x 0 and to the left for x 0; F 0 when x 0.
Problem 8/73 8/74 Determine the natural frequency for small oscilla-tions in the vertical plane about the bearing O for the semicircular disk of radius r.
Problem 8/74 8/75 The thin square plate is suspended from a socket (not shown) which fits the small ball attachment at O. If the plate is made to swing about axis A-A, de-termine the period for small oscillations. Neglect the small offset, mass, and friction of the ball.
Problem 8/75 A A B B O b b b b/2 b/2 O r ƒn b O a 618 Chapter 8 Vibration and Time Response PROBLEMS Introductory Problems 8/71 The light rod and attached sphere of mass m are at rest in the horizontal position shown. Determine the period for small oscillations in the vertical plane about the pivot O.
Problem 8/71 8/72 Derive the differential equation for small oscillations of the spring-loaded pendulum and find the period The equilibrium position is vertical as shown. The mass of the rod is negligible.
Problem 8/72 8/73 A uniform rectangular plate pivots about a horizon-tal axis through one of its corners as shown. Deter-mine the natural frequency of small oscillations.
n m O b k k l .
m O b k k b b 8/76 If the square plate of Prob. is made to oscillate about axis determine the period of small oscil-lations.
8/77 The rectangular frame is formed of a uniform slen-der rod and is suspended from a socket (not shown) which fits the small ball attachment at O. If the rec-tangle is made to swing about axis determine the natural frequency for small oscillations. Neglect the small offset, mass, and friction of the ball.
Problem 8/77 8/78 If the rectangular frame of Prob. is made to oscillate about axis determine the natural fre-quency of small oscillations and compare it with the answer for Prob. 8/79 The flywheel is suspended from its center by a wire from a fixed support, and a period is measured for torsional oscillation of the flywheel about the verti-cal axis. Two small weights, each of mass m, are next attached to the flywheel in opposite positions at a distance r from the center. This additional mass re-sults in a slightly longer period Write an expres-sion for the moment of inertia I of the flywheel in terms of the measured quantities.
Problem 8/79 r r 2.
1 8/77.
B-B, 8/77 A A B B O b b b b 2b A-A, B-B, 8/75 Article 8/4 Problems 619 Representative Problems 8/80 The circular ring of radius r is suspended from a socket (not shown) which fits the small ball attach-ment at O. Determine the ratio R of the period of small oscillations about axis to that about axis Neglect the small offset, mass, and friction of the ball.
Problem 8/80 8/81 A spring-loaded homogeneous plate of mass m pivots freely about a vertical axis through point O. Deter-mine the natural frequency of small oscillations about the equilibrium position shown.
Problem 8/81 O l l/3 2l/3 k k ƒn A A r B B O A-A.
B-B 8/85 When the motor is slowly brought up to speed, a rather large vibratory oscillation of the entire motor about occurs at a speed of which shows that this speed corresponds to the natural fre-quency of free oscillation of the motor. If the motor has a mass of 43 kg and radius of gyration of 100 mm about determine the stiffness k of each of the four identical spring mounts.
Problem 8/85 8/86 The center of mass G of the ship may be assumed to be at the center of the equivalent square sec-tion. The metacentric height h, determined by the intersection M of the ship’s centerline with the line of action of the total buoyancy force acting through the center of buoyancy B, is 3 ft. Determine the pe-riod of one complete roll of the ship if the ampli-tude is small and the resistance of the water is neglected. Neglect also the change in cross section of the ship at the bow and stern, and treat the loaded ship as a uniform solid block of square cross section.
Problem 8/86 M G B W W h 50′ 50′ 50-ft O 200 mm 200 mm O O-O, 360 rev/min, O-O 620 Chapter 8 Vibration and Time Response 8/82 The mass of the uniform slender rod is 3 kg. Deter-mine the position x for the slider such that the system period is 1 s. Assume small oscillations about the horizontal equilibrium position shown.
Problem 8/82 8/83 Determine the expression for the natural frequency of small oscillations of the weighted arm about O.
The stiffness of the spring is k, and its length is ad-justed so that the arm is in equilibrium in the hori-zontal position shown. Neglect the mass of the spring and arm compared with m.
Problem 8/83 8/84 The uniform rod of mass m is freely pivoted about point O. Assume small oscillations and determine an expression for the damping ratio For what value of the damping coefficient c will the system be critically damped?
Problem 8/84 O a b c k ccr .
k m O l b ƒn 0.4 m 3 kg O 1.2 kg 250 N/m x 0.8 m 1.2-kg 8/87 The system of Prob. 8/42 is repeated here. If the link AB now has mass and radius of gyration about point O, determine the equation of motion in terms of the variable x. Assume small oscillations. The damping coefficient for the dashpot is c.
Problem 8/87 8/88 Determine the period of small oscillations of the semicylinder of mass m and radius r as it rolls without slipping on the horizontal surface.
Problem 8/88 O θ G r r – x k m1 m3 m2 c O a A B b kO m3 Article 8/4 Problems 621 8/89 The circular sector of mass m is cut from steel plate of uniform thickness and mounted in a bearing at its center O so that it can swing freely in the vertical plane. If the sector is released from rest with derive its differential equation of motion assuming negligible damping Determine the period for small oscillations about the position Problem 8/89 8/90 Two identical uniform bars are welded together at a right angle and are pivoted about a horizontal axis through point O as shown. Determine the critical driving frequency of the block B which will result in excessively large oscillations of the assembly. The mass of the welded assembly is m.
Problem 8/90 k l l/2 l/2 k B O xB = b sin t ω c θ β β G O r 2.
0, 8/93 The cart B is given the harmonic displacement Determine the steady-state amplitude of the periodic oscillation of the uniform slender bar which is pinned to the cart at P. Assume small angles and neglect friction at the pivot. The torsional spring is undeformed when Problem 8/93 8/94 The circular disk of mass m and moment of inertia I about its central axis is welded to the steel shaft which, in turn, is welded to the fixed block. The disk is given an angular displacement and then re-leased, causing a torsional vibration of the disk with changing between The shaft resists the twist with a moment where J is the polar moment of inertia of the cross section of the shaft about the rotation axis, G is the shear modulus of elasticity of the shaft (resistance to shear stress), is the angle of twist in radians, and L is the length of the twisted shaft. Derive the expression for the natural frequency of the torsional vibration.
Problem 8/94 L θ ƒn M JG /L, 0 and 0. 0 K B P m l θ xB = b sin t ω 0.
xB b sin t.
622 Chapter 8 Vibration and Time Response 8/91 The uniform solid cylinder of mass m and radius r rolls without slipping during its oscillation on the circular surface of radius R. If the motion is confined to small amplitudes determine the period of the oscillations. Also determine the angular velocity of the cylinder as it crosses the vertical centerline.
(Caution: Do not confuse with or with as used in the defining equations. Note also that is not the angular displacement of the cylinder.) Problem 8/91 8/92 The homogeneous solid cylindrical pulley has mass and radius r. If the attachment at B undergoes the indicated harmonic displacement, determine the equation of motion of the system in terms of the variable x. The cord which connects mass to the upper spring does not slip on the pulley.
Problem 8/92 k2 m2 m1 B xB = b cos t ω k1 x O r m2 m1 R O G r θ n ˙ 0, 8/95 The segmented “dummy” of Prob. is repeated here. The hip joint O is assumed to remain fixed to the car, and the torso above the hip is treated as a rigid body of mass m. The center of mass of the torso is at G and the radius of gyration of the torso about is Assume that muscular response acts as an internal torsional spring which exerts a moment on the upper torso, where K is the torsional spring constant and is the angular deflection from the initial vertical position. If the car is brought to a sudden stop with a constant deceleration a, derive the differential equation for the motion of the torso prior to its impact with the dashboard.
Problem 8/95 G O r θ M K kO.
O 6/103 Article 8/4 Problems 623 8/96 The elements of the “swing-axle” type of indepen-dent rear suspension for automobiles are depicted in the figure. The differential D is rigidly attached to the car frame. The half-axles are pivoted at their in-board ends (point O for the half-axle shown) and are rigidly attached to the wheels. Suspension elements not shown constrain the wheel motion to the plane of the figure. The weight of the wheel–tire assembly is and its mass moment of inertia about a diametral axis passing through its mass center G is The weight of the half-axle is negligible.
The spring rate and shock-absorber damping coeffi-cient are and , respec-tively. If a static tire imbalance is present, as represented by the additional concentrated weight as shown, determine the angular velocity which results in the suspension system being dri-ven at its undamped natural frequency. What would be the corresponding vehicle speed Determine the damping ratio Assume small angular deflections and neglect gyroscopic effects and any car frame vi-bration. In order to avoid the complications associ-ated with the varying normal force exerted by the road on the tire, treat the vehicle as being on a lift with the wheels hanging free.
Problem 8/96 l2 = 36″ l1 = 27″ r = 14″ W O k c D w G .
v?
w 0.5 lb c 200 lb-sec/ft k 50 lb/in.
1 lb-ft-sec2.
W 100 lb, 8/5 Energy Methods In Arts. 8/2 through 8/4 we derived and solved the equations of mo-tion for vibrating bodies by isolating the body with a free-body diagram and applying Newton’s second law of motion. With this approach, we were able to account for the actions of all forces acting on the body, in-cluding frictional damping forces. There are many problems where the effect of damping is small and may be neglected, so that the total energy of the system is essentially conserved. For such systems, we find that the principle of conservation of energy may frequently be applied with con-siderable advantage in establishing the equation of motion and, when the motion is simple harmonic, in determining the frequency of vibration.
Determining the Equation of Motion To illustrate this alternative approach, consider first the simple case of the body of mass m attached to the spring of stiffness k and vibrating in the vertical direction without damping, Fig. 8/17. As previously, we find it convenient to measure the motion variable x from the equilib-rium position. With this datum, the total potential energy of the system, elastic plus gravitational, becomes where st mg/k is the initial static displacement. Substituting kst mg and simplifying give Thus, the total energy of the system becomes Because T V is constant for a conservative system, its time deriv-ative is zero. Consequently, Canceling gives us our basic differential equation of motion which is identical to Eq. 8/1 derived in Art. 8/2 for the same system of Fig. 8/3.
Determining the Frequency of Vibration Conservation of energy may also be used to determine the period or frequency of vibration for a linear conservative system, without having to derive and solve the equation of motion. For a system which oscillates with simple harmonic motion about the equilibrium position, from which the mx ¨ kx 0 x ˙ d dt (T V) mx ˙x ¨ kxx ˙ 0 T V 1 2 mx ˙2 1 2 kx2 V 1 2kx2 V Ve Vg 1 2 k(x st)2 1 2 kst 2 mgx 624 Chapter 8 Vibration and Time Response m m k x Equilibrium position st δ Figure 8/17 displacement x is measured, the energy changes from maximum kinetic and zero potential at the equilibrium position x 0 to zero kinetic and maximum potential at the position of maximum displacement x xmax.
Thus, we may write The maximum kinetic energy is and the maximum potential energy is For the harmonic oscillator of Fig. 8/17, we know that the displace-ment may be written as x xmax sin (nt ), so that the maximum ve-locity is nxmax. Thus, we may write where xmax is the maximum displacement, at which the potential energy is a maximum. From this energy balance, we easily obtain This method of directly determining the frequency may be used for any linear undamped vibration.
The main advantage of the energy approach for the free vibration of conservative systems is that it becomes unnecessary to dismember the system and account for all of the forces which act on each member. In Art. 3/7 of Chapter 3 and in Arts. 6/6 and 6/7 of Chapter 6, we learned for a system of interconnected bodies that an active-force diagram of the complete system enabled us to evaluate the work U of the external ac-tive forces and to equate it to the change in the total mechanical energy T V of the system.
Thus, for a conservative mechanical system of interconnected parts with a single degree of freedom where U 0, we may obtain its equa-tion of motion simply by setting the time derivative of its constant total mechanical energy to zero, giving Here V Ve Vg is the sum of the elastic and gravitational potential energies of the system.
Also, for an interconnected mechanical system, as for a single body, the natural frequency of vibration is obtained by equating the expres-sion for its maximum total kinetic energy to the expression for its maxi-mum potential energy, where the potential energy is taken to be zero at the equilibrium position. This approach to the determination of natural frequency is valid only if it can be determined that the system vibrates with simple harmonic motion.
d dt (T V) 0 n k/m 1 2 m(nxmax)2 1 2 k(xmax)2 x ˙max 1 2 k(xmax)2.
1 2 m(x ˙max)2, Tmax Vmax Article 8/5 Energy Methods 625 SAMPLE PROBLEM 8/10 The small sphere of mass m is mounted on the light rod pivoted at O and supported at end A by the vertical spring of stiffness k. End A is displaced a small distance y0 below the horizontal equilibrium position and released. By the energy method, derive the differential equation of motion for small oscillations of the rod and determine the expression for its natural frequency n of vibration.
Damping is negligible.
Solution.
With the displacement y of the end of the bar measured from the equilibrium position, the potential energy in the displaced position for small val-ues of y becomes where st is the static deflection of the spring at equilibrium. But the force in the spring in the equilibrium position, from a zero moment sum about O, is (b/l)mg kst. Substituting this value in the expression for V and simplifying yield The kinetic energy in the displaced position is where we see that the vertical displacement of m is (b/l)y. Thus, with the energy sum constant, its time derivative is zero, and we have which yields Ans.
when is canceled. By analogy with Eq. 8/2, we may write the motion frequency directly as Ans.
Alternatively, we can obtain the frequency by equating the maximum ki-netic energy, which occurs at y 0, to the maximum potential energy, which oc-curs at y y0 ymax, where the deflection is a maximum. Thus, Knowing that we have a harmonic oscillation, which can be expressed as y ymax sin nt, we have ymaxn. Substituting this relation into our energy balance gives us Ans.
as before.
1 2 m b l ymaxn 2 1 2 kymax 2 so that n l b k/m y ˙max Tmax Vmax gives 1 2 m b l y ˙max 2 1 2 kymax 2 n l b k/m y ˙ y ¨ l2 b2 k m y 0 d dt (T V) d dt 1 2 m b l y ˙ 2 1 2 ky2 0 T 1 2 m b l y ˙ 2 V 1 2 ky2 V Ve Vg 1 2 k(y st)2 1 2 kst 2 mg b l y 626 Chapter 8 Vibration and Time Response m O A k b l m O A k y Equilibrium position st δ y b — l Helpful Hints For large values of y, the circular motion of the end of the bar would cause our expression for the deflec-tion of the spring to be in error.
Here again, we note the simplicity of the expression for potential energy when the displacement is measured from the equilibrium position.
SAMPLE PROBLEM 8/11 Determine the natural frequency n of vertical vibration of the 3-kg collar to which are attached the two uniform 1.2-kg links, which may be treated as slender bars. The stiffness of the spring, which is attached to both the collar and the foundation, is k 1.5 kN/m, and the bars are both horizontal in the equilib-rium position. A small roller on end B of each link permits end A to move with the collar. Frictional retardation is negligible.
Solution.
In the equilibrium position, the compression P in the spring equals the weight of the 3-kg collar, plus half the weight of each link or P 3(9.81) 41.2 N. The corresponding static deflection of the spring is st P/k 41.2/1.5(103) 27.5(103) m. With the displacement variable y measured downward from the equilibrium position, which becomes the position of zero po-tential energy, the potential energy for each member in the displaced position is The total potential energy of the system then becomes The maximum kinetic energy occurs at the equilibrium position, where the velocity of the collar has its maximum value. In that position, in which links AB are horizontal, end B is the instantaneous center of zero velocity for each link, and each link rotates with an angular velocity Thus, the kinetic en-ergy of each part is Thus, the kinetic energy of the collar and both links is With the harmonic motion expressed by y ymax sin nt, we have ymaxn, so that the energy balance Tmax Vmax with becomes Ans.
1.9(ymaxn)2 750ymax 2 or n 750/1.9 19.87 Hz y ˙max y ˙ y ˙max T 3 2 y ˙2 2(0.2y ˙2) 1.9y ˙2 1 6 (1.2)y ˙2 0.2y ˙2 (Each link) T 1 2 IB2 1 2 1 3 mll2(y ˙/l)2 1 6 mly ˙2 (Collar) T 1 2 mc y ˙2 3 2 y ˙2 J y ˙/0.3.
y ˙ V 750y2 41.2y 29.4y 2(5.89)y 750y2 J (Each link) Vg ml g y 2 1.2(9.81) y 2 5.89y J (Collar) Vg mc gy 3(9.81)y 29.4y J 750y2 41.2y J 1 2 (1.5)(103)y2 1.5(103)(27.5)(103)y (Spring) Ve 1 2 k(y st)2 1 2 kst 2 1 2 ky2 kst y 2(1 2)(1.2)(9.81) Article 8/5 Energy Methods 627 B A y 1.2 kg 300 mm 1.2 kg k = 1.5 kN/m A B 300 mm 3 kg Helpful Hints Note that the mass center of each link moves down only half as far as the collar.
We note again that measurement of the motion variable y from the equilibrium position results in the total potential energy being simply V Our knowledge of rigid-body kine-matics is essential at this point.
To appreciate the advantage of the work-energy method for this and similar problems of interconnected systems, you are encouraged to ex-plore the steps required for solution by the force and moment equations of motion of the separate parts.
If the oscillations were large, we would find that the angular velocity of each link in its general position would equal which would cause a nonlinear response no longer described by y ymax sin t.
y ˙/0.09 y2, 1 2 ky2.
8/100 A uniform rod of mass m and length l is welded at one end to the rim of a light circular hoop of radius l. The other end lies at the center of the hoop. De-termine the period for small oscillations about the vertical position of the bar if the hoop rolls on the horizontal surface without slipping.
Problem 8/100 8/101 The length of the spring is adjusted so that the equi-librium position of the arm is horizontal as shown.
Neglect the mass of the spring and the arm and cal-culate the natural frequency for small oscillations.
Problem 8/101 Representative Problems 8/102 Calculate the frequency of vertical oscillation of the system shown. The pulley has a radius of gyration about its center O of 200 mm.
Problem 8/102 k = 2 kN/m kO = 200 mm O 300 mm 30 kg 40 kg 40-kg ƒn k m O l b ƒn l 628 Chapter 8 Vibration and Time Response PROBLEMS (Solve the following problems by the energy method of Art. ) Introductory Problems 8/97 The potential energy V of a linear spring-mass sys-tem is given in inch-pounds by where x is the displacement in inches measured from the neutral equilibrium position. The kinetic energy T of the sys-tem in inch-pounds is given by . Determine the differential equation of motion for the system and find the period of its oscillation. Neglect energy loss.
8/98 Derive the equation of motion for the pendulum which consists of the slender uniform rod of mass m and the bob of mass M. Assume small oscillations, and neglect the radius of the bob.
Problem 8/98 8/99 The spoked wheel of radius r, mass m, and cen-troidal radius of gyration rolls without slipping on the incline. Determine the natural frequency of oscillation and explore the limiting cases of and Problem 8/99 θ m r k k r.
k 0 k O M m l θ 8x ˙2 64x2, 8/5.
8/103 By the method of this article, determine the pe-riod of vertical oscillation. Each spring has a stiff-ness of and the mass of the pulleys may be neglected.
Problem 8/103 8/104 The homogeneous circular cylinder of Prob. re-peated here, rolls without slipping on the track of ra-dius R. Determine the period for small oscillations.
Problem 8/104 8/105 The uniform slender rod of length l and mass is secured to the uniform disk of radius and mass If the system is shown in its equilibrium posi-tion, determine the natural frequency and the maximum angular velocity for small oscillations of amplitude about the pivot O.
Problem 8/105 2l/5 l/5 l/5 l/5 k O m1 m2 0 n m1.
l/5 m2 R O G r θ 8/91, k k 50 lb 6 lb/in., Article 8/5 Problems 629 8/106 The ends of the uniform slender bar of mass m and length L move freely in the vertical and horizontal slots under the action of the two precompressed springs each of stiffness k as shown. If the bar is in static equilibrium when determine the nat-ural frequency of small oscillations.
Problem 8/106 8/107 Develop an expression for the natural circular fre-quency of the system of Prob. repeated here. The mass and friction of the pulleys are negligible.
Problem 8/107 k θ m m 8/23, n L k k θ n 0, Problem 8/110 8/111 The thin homogeneous panel of mass m is hinged to swing freely about a fixed axis which makes an angle with the vertical. Determine the period of small oscillations.
Problem 8/111 8/112 The block is supported by the two links with two torsion springs, each of constant m/rad, arranged as shown. The springs are suffi-ciently stiff so that stable equilibrium is established in the position shown. Determine the natural frequency for small oscillations about this equilibrium position.
Problem 8/112 K 5 kg 5 kg 12 kg K 0.8 m ƒn N K 500 5-kg 12-kg c m a b 45° A 300 mm 300 mm 45° Vertical k = 1050 N/m 630 Chapter 8 Vibration and Time Response 8/108 Determine the period of vertical oscillations for the system composed of the frame and two pulleys, each of which has a radius of gyra-tion The flexible wires do not slip on the pulleys.
Problem 8/108 8/109 Derive the natural frequency of the system com-posed of two homogeneous circular cylinders, each of mass M, and the connecting link AB of mass m.
Assume small oscillations.
Problem 8/109 8/110 Each of the two uniform slender bars is hinged freely at A with its small upper-end guide roller free to move in the horizontal guide. The bars are supported in their equilibrium posi-tions by the vertical spring of stiffness If point A is given a very small vertical displacement and then released, calculate the natural frequency of the resulting motion.
1050 N/m.
45 1.5-kg M M m r r0 r0 O B A O r θ θ ƒn k = 4 kN/m kO = 400 mm kO = 400 mm 600 mm 600 mm 80 kg 140 kg 80 kg O O kO 400 mm.
80-kg 140-kg 8/113 The semicylinder of mass m and radius r rolls with-out slipping on the horizontal surface. By the method of this article, determine the period of small oscillations.
Problem 8/113 8/114 The front-end suspension of an automobile is shown. Each of the coil springs has a stiffness of If the weight of the front-end frame and equivalent portion of the body attached to the front end is determine the natural frequency of vertical oscillation of the frame and body in the absence of shock absorbers. (Hint: To relate the spring deflection to the deflection of the frame and body, consider the frame fixed and let the ground and wheels move vertically.) Problem 8/114 8/115 If the spring-loaded frame is given a slight vertical disturbance from its equilibrium position shown, determine its natural frequency of vibration.
The mass of the upper member is 24 kg, and that of the lower members is negligible. Each spring has a stiffness of Problem 8/115 24 kg 150 mm 100 mm 100 mm 150 mm 9 kN/m.
ƒn 18″ 12″ ƒn 1800 lb, 270 lb/in.
O r r _ G θ Article 8/5 Problems 631 8/116 The uniform slender rod of length 2b is supported in the horizontal plane by a bifilar suspension. The rod is set into small angular oscillation about the vertical axis through its center O. Derive the ex-pression for the period of oscillation. (Hint: From the auxiliary sketch note that the rod rises a dis-tance h corresponding to an angular twist Also note that for small angles and that cos may be replaced by the first two terms of its series expansion. A simple harmonic solution of the form may be used for small angles.) Problem 8/116 8/117 The semicircular cylindrical shell of radius r with small but uniform wall thickness is set into small rocking oscillation on the horizontal surface. If no slipping occurs, determine the expression for the period of each complete oscillation.
Problem 8/117 8/118 A hole of radius is drilled through a cylinder of radius R to form a body of mass as shown. If the body rolls on the horizontal surface without slip-ping, determine the period for small oscillations.
Problem 8/118 R/4 O R/4 R/2 m R/4 r A B b b O O B A l b l l θ bθ β lβ h = l(1 – cos ) β 0 sin nt l b .
8/6 Chapter Review In studying the vibrations of particles and rigid bodies in Chapter 8, we have observed that the subject is simply a direct application of the fundamental principles of dynamics as presented in Chapters 3 and 6.
However, in these previous chapters, we determined the dynamic behav-ior of a body only at a particular instant of time or found the changes in motion resulting from only finite intervals of displacement or time.
Chapter 8, on the other hand, has treated the solution of the defining differential equations of motion, so that the linear or angular displace-ment can be fully expressed as a function of time.
Particle Vibration We divided our study of the time response of particles into the two categories of free and forced motion, with the further subdivisions of negligible and significant damping. We saw that the damping ratio is a convenient parameter for determining the nature of unforced but vis-cously damped vibrations.
The prime lesson associated with harmonic forcing is that driving a lightly damped system with a force whose frequency is near the natural frequency can cause motion of excessively large amplitude—a condition called resonance, which usually must be carefully avoided.
Rigid-Body Vibration In our study of rigid-body vibrations, we observed that the equation of small angular motion has a form identical to that for particle vibra-tions. Whereas particle vibrations may be described completely by the equations governing translational motion, rigid-body vibrations usually require the equations of rotational dynamics.
Energy Methods In the final article of Chapter 8, we saw how the energy method can facilitate the determination of the natural frequency n in free vibration problems where damping may be neglected. Here the total mechanical energy of the system is assumed to be constant. Setting its first time de-rivative to zero leads directly to the differential equation of motion for the system. The energy approach permits the analysis of a conservative system of interconnected parts without dismembering the system.
Degrees of Freedom Throughout the chapter, we have restricted our attention to systems having one degree of freedom, where the position of the system can be specified by a single variable. If a system possesses n degrees of freedom, it has n natural frequencies. Thus, if a harmonic force is applied to such a system which is lightly damped, there are n driving frequencies which can cause motion of large amplitude. By a process called modal analysis, a complex system with n degrees of freedom can be reduced to n single-degree-of-freedom systems. For this reason, the thorough understanding of the material of this chapter is vital for the further study of vibrations.
632 Chapter 8 Vibration and Time Response REVIEW PROBLEMS 8/119 Determine the value of the damping coefficient c for which the system is critically damped if kN/m and Problem 8/119 8/120 A I-beam is being hoisted by the cable arrange-ment shown. Determine the period of small oscil-lations about the junction O, which is assumed to remain fixed and about which the cables pivot freely. Treat the beam as a slender rod.
Problem 8/120 8/121 The uniform circular disk is suspended by a socket (not shown) which fits over the small ball attach-ment at O. Determine the period of small motion if the disk swings freely about (a) axis A-A and (b) axis B-B. Neglect the small offset, mass, and fric-tion of the ball.
O 45° 45° 20 m 20-m c c k k k m m 100 kg.
k 70 Article 8/6 Review Problems 633 Problem 8/121 8/122 The block of mass M is suspended by the two uni-form slender rods each of mass m. Determine the natural frequency of small oscillations for the system shown.
Problem 8/122 8/123 The triangular frame is constructed of uniform slender rod and pivots about a horizontal axis through point O. Determine the critical driving fre-quency of the block B which will result in exces-sively large oscillations of the assembly. The total mass of the frame is m.
Problem 8/123 k B d O d d k sB = b sin t ω c l m G M b m b – – 2 b – – 2 a – – 2 a – – 2 n B B A A r O 8/127 Calculate the damping ratio of the system shown if the weight and radius of gyration of the stepped cylinder are and the spring constant is and the damping coeffi-cient of the hydraulic cylinder is The cylinder rolls without slipping on the radius and the spring can support tension as well as compression.
Problem 8/127 8/128 The cylinder A of radius r, mass m, and radius of gyration is driven by a cable-spring system at-tached to the drive cylinder B, which oscillates as indicated. If the cables do not slip on the cylinders, and if both springs are stretched to the degree that they do not go slack during a motion cycle, deter-mine an expression for the amplitude of the steady-state oscillation of cylinder A.
Problem 8/128 8/129 With collar A held in position, a static horizontal force of 3 lb applied to the sphere B gives it a deflection of 0.60 in. against the elastic resistance of the slender rod of negligible mass to which it is attached. If the collar A is given a horizontal har-monic oscillation with a frequency of 2 cycles per second and an amplitude of 0.30 in., calculate the amplitude X of the horizontal vibration of the sphere. Assume negligible damping.
5-lb k = 0 cos wt ω r r0 k A B φ φ max k k r c W r 6 in.
c 2 lb-sec/ft.
k 15 lb/in., k 5.5 in., W 20 lb 634 Chapter 8 Vibration and Time Response 8/124 Determine the period for small oscillations of the assembly composed of two light bars and two parti-cles, each of mass m. Investigate your expression as the angle approaches values of 0 and Problem 8/124 8/125 A slender rod is shaped into the semicircle of radius r as shown. Determine the natural frequency for small oscillations of the rod when it is pivoted on the horizontal knife edge at the middle of its length.
Problem 8/125 8/126 Determine the largest amplitude for which the uniform circular disk will roll without slipping on the horizontal surface.
Problem 8/126 r x m k/2 k/2 s, k μ μ x0 r ƒn α m m l O l 180. Problem 8/129 8/130 The seismic instrument shown is secured to a ship’s deck near the stern where propeller-induced vibration is most pronounced. The ship has a single three-bladed propeller which turns at 180 rev/min and operates partly out of water, thus causing a shock as each blade breaks the surface. The damp-ing ratio of the instrument is and its un-damped natural frequency is 3 Hz. If the measured amplitude of A relative to its frame is 0.75 mm, compute the amplitude of the vertical vibration of the deck.
Problem 8/130 8/131 An experimental engine weighing 480 lb is mounted on a test stand with spring mounts at A and B, each with a stiffness of The radius of gyration of the engine about its mass center G is 4.60 in. With the motor not running, calculate the natural frequency of vertical vibration and of rotation about G. If vertical motion is sup-pressed and a light rotational imbalance occurs, at what speed N should the engine not be run?
(ƒn) (ƒn)y 600 lb/in.
A 0 0.5, A B Article 8/6 Review Problems 635 Problem 8/131 8/132 The uniform bar of mass M and length l has a small roller of mass m with negligible bearing friction at each end. Determine the period of the system for small oscillations on the curved track.
Problem 8/132 m m l R M 10″ A B G y 10″ 8/135 Plot the response x of the body over the time interval second. Determine the maximum and minimum values of x and their re-spective times. The initial conditions are and Problem 8/135 8/136 Shown in the figure are the elements of a displace-ment meter used to study the motion of the base. The motion of the mass relative to the frame is recorded on the rotating drum. If , and determine the range of the spring constant k over which the magnitude of the recorded relative displacement is less than 1.5b. It is assumed that the ratio must re-main greater than unity.
Problem 8/136 Neutral position yB = b sin t ω W k c O l1 l2 l3 /n 10 rad/sec, lb-sec/ft, c 0.1 W 2 lb, l3 2 ft, l2 1.6 ft l1 1.2 ft, yB b sin t 100 lb/in.
18 lb-sec/ft 50 lb F = (160 cos 60t) lb x x ˙0 6 ft/sec.
x0 0 0 t 1 50-lb 636 Chapter 8 Vibration and Time Response Computer-Oriented Problems 8/133 The mass of the system shown is released with the initial conditions and at Plot the response of the system and deter-mine the time(s) (if any) at which the displace-ment Problem 8/133 8/134 The oscillator contains an unbalanced motor whose speed N in revolutions per minute can be varied. The oscillator is restrained in its horizontal motion by a spring of stiffness and by a viscous damper whose piston is resisted by a force of 30 N when moving at a speed of Determine the viscous damping factor and plot the magnification factor M for motor speeds from zero to 300 revolutions per minute. Determine the maximum value of M and the corresponding motor speed.
Problem 8/134 10 kg c k 0.5 m/s.
k 1080 N/m 10-kg 100 N/m x 50 N·s/m 2 kg x 0.05 m.
t 0.
x ˙0 5 m/s x0 0.1 m 8/137 The cylinder is attached to a viscous damper and to the spring of stiffness If the cylinder is released from rest at time from the position where it is displaced a distance from its equilibrium position, plot the displacement y as a function of time for the first second for the two cases where the viscous damping coefficient is (a) and (b) Problem 8/137 8/138 Determine and plot the response as a function of time for the undamped linear oscillator subjected to the force F which varies linearly with time for the first second as shown. The mass is initially at rest with at time Problem 8/138 x F, N t, s k = 90 N/m F 0.75 kg 3/4 6.25 00 t 0.
x 0 3 4 x c 4 kg k = 800 N /m y Equilibrium position c 80 Ns/m.
c 124 Ns/m y 100 mm t 0 k 800 N/m.
4-kg Article 8/6 Review Problems 637 8/139The damped linear oscillator of mass spring constant and viscous damp-ing factor is initially at rest in a neutral position when it is subjected to a sudden impulsive loading F over a very short period of time as shown. If the impulse deter-mine the resulting displacement as a function of time and plot it for the first two seconds following the impulse.
Problem 8/139 c m F x k F t x I F dt 8 Ns, 0.1 k 200 N/m, m 4 kg, 639 See Appendix A of Vol. 1 Statics for a treatment of the theory and calculation of area moments of inertia. Because this quantity plays an important role in the design of structures, especially those dealt with in statics, we present only a brief definition in this Dynamics volume so that the student can appreciate the basic differences between area and mass moments of inertia.
The moments of inertia of a plane area A about the x- and y-axes in its plane and about the z-axis normal to its plane, Fig. A/1, are defined by where dA is the differential element of area and r2 x2 y2. Clearly, the polar moment of inertia Iz equals the sum Ix Iy of the rectangular moments of inertia. For thin flat plates, the area moment of inertia is useful in the calculation of the mass moment of inertia, as explained in Appendix B.
The area moment of inertia is a measure of the distribution of area about the axis in question and, for that axis, is a constant property of the area. The dimensions of area moment of inertia are (distance)4 expressed in m4 or mm4 in SI units and ft4 or in.4 in U.S. customary units. In con-trast, mass moment of inertia is a measure of the distribution of mass about the axis in question, and its dimensions are (mass)(distance)2, which are expressed in in SI units and in lb-ft-sec2 or lb-in.-sec2 in U.S. customary units.
kg m2 Ix y2 dA Iy x2 dA Iz r2 dA y x A dA O x r y Figure A/1 A Area Moments of Inertia 641 B/1 Mass Moments of Inertia about an Axis The equation of rotational motion about an axis normal to the plane of motion for a rigid body in plane motion contains an integral which de-pends on the distribution of mass with respect to the moment axis. This integral occurs whenever a rigid body has an angular acceleration about its axis of rotation. Thus, to study the dynamics of rotation, you should be thoroughly familiar with the calculation of mass moments of inertia for rigid bodies.
Consider a body of mass m, Fig. B/1, rotating about an axis O-O with an angular acceleration . All particles of the body move in parallel planes which are normal to the rotation axis O-O. We may choose any one of the planes as the plane of motion, although the one containing the center of mass is usually the one so designated. An element of mass dm has a component of acceleration tangent to its circular path equal to r, and by Newton’s second law of motion the resultant tangential force on this element equals r dm. The moment of this force about the axis O-O is r2 dm, and the sum of the moments of these forces for all ele-ments is r2 dm.
For a rigid body, is the same for all radial lines in the body and we may take it outside the integral sign. The remaining integral is called the mass moment of inertia I of the body about the axis O-O and is (B/1) This integral represents an important property of a body and is involved in the analysis of any body which has rotational acceleration about a I r2 dm O O m dm r α r dm α Figure B/1 B Mass Moments of Inertia B/1 Mass Moments of Inertia about an Axis B/2 Products of Inertia APPENDIX OUTLINE given axis. Just as the mass m of a body is a measure of the resistance to translational acceleration, the moment of inertia I is a measure of resis-tance to rotational acceleration of the body.
The moment-of-inertia integral may be expressed alternatively as (B/1a) where ri is the radial distance from the inertia axis to the representative particle of mass mi and where the summation is taken over all particles of the body.
If the density is constant throughout the body, the moment of in-ertia becomes where dV is the element of volume. In this case, the integral by itself de-fines a purely geometrical property of the body. When the density is not constant but is expressed as a function of the coordinates of the body, it must be left within the integral sign and its effect accounted for in the integration process.
In general, the coordinates which best fit the boundaries of the body should be used in the integration. It is particularly important that we make a good choice of the element of volume dV. To simplify the inte-gration, an element of lowest possible order should be chosen, and the correct expression for the moment of inertia of the element about the axis involved should be used. For example, in finding the moment of in-ertia of a solid right-circular cone about its central axis, we may choose an element in the form of a circular slice of infinitesimal thickness, Fig.
B/2a. The differential moment of inertia for this element is the expres-sion for the moment of inertia of a circular cylinder of infinitesimal alti-tude about its central axis. (This expression will be obtained in Sample Problem B/1.) Alternatively, we could choose an element in the form of a cylindri-cal shell of infinitesimal thickness as shown in Fig. B/2b. Because all of the mass of the element is at the same distance r from the inertia axis, the differential moment of inertia for this element is merely r2 dm where dm is the differential mass of the elemental shell.
From the definition of mass moment of inertia, its dimensions are (mass)(distance)2 and are expressed in the units in SI units and lb-ft-sec2 in U.S. customary units.
kgm2 I r2 dV I Σ ri 2mi 642 Appendix B Mass Moments of Inertia r (a) (b) Figure B/2 Radius of Gyration The radius of gyration k of a mass m about an axis for which the moment of inertia is I is defined as (B/2) Thus, k is a measure of the distribution of mass of a given body about the axis in question, and its definition is analogous to the definition of the radius of gyration for area moments of inertia. If all the mass m of a body could be concentrated at a distance k from the axis, the moment of inertia would be unchanged.
The moment of inertia of a body about a particular axis is fre-quently indicated by specifying the mass of the body and the radius of gyration of the body about the axis. The moment of inertia is then calcu-lated from Eq. B/2.
Transfer of Axes If the moment of inertia of a body is known about an axis passing through the mass center, it may be determined easily about any parallel axis. To prove this statement, consider the two parallel axes in Fig. B/3, one being an axis through the mass center G and the other a parallel axis through some other point C. The radial distances from the two axes to any element of mass dm are r0 and r, and the separation of the axes is d. Substituting the law of cosines r2 d2 2r0d cos into the def-inition for the moment of inertia about the axis through C gives The first integral is the moment of inertia about the mass-center axis, the second term is md2, and the third integral equals zero, since the u-coordinate of the mass center with respect to the axis through G is zero.
Thus, the parallel-axis theorem is (B/3) Remember that the transfer cannot be made unless one axis passes through the center of mass and unless the axes are parallel.
When the expressions for the radii of gyration are substituted in Eq.
B/3, there results (B/3a) Equation B/3a is the parallel-axis theorem for obtaining the radius of gyration k about an axis which is a distance d from a parallel axis through the mass center, for which the radius of gyration is k.
k2 k2 d2 I I md2 I r0 2 dm d2 dm 2d u dm I r2 dm (r0 2 d2 2r0 d cos ) dm r0 2 Article B/1 Mass Moments of Inertia about an Axis 643 dm m r r0 u u d O C G θ Figure B/3 644 Appendix B Mass Moments of Inertia ry rx z z O y y dm x x rz Figure B/5 For plane-motion problems where rotation occurs about an axis normal to the plane of motion, a single subscript for I is sufficient to designate the inertia axis. Thus, if the plate of Fig. B/4 has plane motion in the x-y plane, the moment of inertia of the plate about the z-axis through O is designated IO. For three-dimensional motion, however, where components of rotation may occur about more than one axis, we use a double subscript to preserve notational symmetry with product-of-inertia terms, which are described in Art. B/2. Thus, the moments of in-ertia about the x-, y-, and z-axes are labeled Ixx, Iyy, and Izz, respectively, and from Fig. B/5 we see that they become (B/4) These integrals are cited in Eqs. 7/10 of Art. 7/7 on angular momentum in three-dimensional rotation.
The defining expressions for mass moments of inertia and area mo-ments of inertia are similar. An exact relationship between the two mo-ment-of-inertia expressions exists in the case of flat plates. Consider the flat plate of uniform thickness in Fig. B/4. If the constant thickness is t and the density is , the mass moment of inertia Izz of the plate about the z-axis normal to the plate is (B/5) Thus, the mass moment of inertia about the z-axis equals the mass per unit area t times the polar moment of inertia Iz of the plate area about the z-axis. If t is small compared with the dimensions of the plate in its Izz r2 dm t r2 dA tIz Izz rz 2 dm (x2 y2) dm Iyy ry 2 dm (z2 x2) dm Ixx rx 2 dm (y2 z2) dm x x t O dm z y y r Figure B/4 plane, the mass moments of inertia Ixx and Iyy of the plate about the x- and y-axes are closely approximated by (B/6) Thus, the mass moments of inertia equal the mass per unit area t times the corresponding area moments of inertia. The double subscripts for mass moments of inertia distinguish these quantities from area mo-ments of inertia.
Inasmuch as Iz Ix Iy for area moments of inertia, we have (B/7) which holds only for a thin flat plate. This restriction is observed from Eqs. B/6, which do not hold true unless the thickness t or the z-coordinate of the element is negligible compared with the distance of the element from the corresponding x- or y-axis. Equation B/7 is very useful when dealing with a differential mass element taken as a flat slice of differential thickness, say, dz. In this case, Eq. B/7 holds exactly and becomes (B/7a) for axes x and y in the plane of the plate.
Composite Bodies As in the case of area moments of inertia, the mass moment of iner-tia of a composite body is the sum of the moments of inertia of the indi-vidual parts about the same axis. It is often convenient to treat a composite body as defined by positive volumes and negative volumes.
The moment of inertia of a negative element, such as the material re-moved to form a hole, must be considered a negative quantity.
A summary of some of the more useful formulas for mass moments of inertia of various masses of common shape is given in Table D/4, Appendix D.
The problems which follow the sample problems are divided into the categories Integration Exercises and Composite and Parallel-Axis Exercises. The parallel-axis theorem will also be useful in some of the problems in the first category. dIzz dIxx dIyy Izz Ixx Iyy Iyy x2 dm t x2 dA tIy Ixx y2 dm t y2 dA tIx Article B/1 Mass Moments of Inertia about an Axis 645 SAMPLE PROBLEM B/1 Determine the moment of inertia and radius of gyration of a homogeneous right-circular cylinder of mass m and radius r about its central axis O-O.
Solution.
An element of mass in cylindrical coordinates is dm dV tr0 dr0 d, where is the density of the cylinder. The moment of inertia about the axis of the cylinder is Ans.
The radius of gyration is Ans.
SAMPLE PROBLEM B/2 Determine the moment of inertia and radius of gyration of a homogeneous solid sphere of mass m and radius r about a diameter.
Solution.
A circular slice of radius y and thickness dx is chosen as the volume element. From the results of Sample Problem B/1, the moment of inertia about the x-axis of the elemental cylinder is where is the constant density of the sphere. The total moment of inertia about the x-axis is Ans.
The radius of gyration about the x-axis is Ans.
Ixx 2 r r (r2 x2)2 dx 8 15 r5 2 5 mr2 dIxx 1 2 (dm)y2 1 2 (y2 dx)y2 2 (r2 x2)2 dx I r0 2 dm t 2 0 r 0 r0 3 dr0 d t r4 2 1 2 mr2 646 Appendix B Mass Moments of Inertia t O O dr0 r0 r dθ y r x x dx y Helpful Hints If we had started with a cylindrical shell of radius r0 and axial length t as our mass element dm, then dI dm directly. You should evaluate the integral.
The result I applies only to a solid homogeneous circular cylinder and cannot be used for any other wheel of circular periphery.
1 2 mr2 r0 2 Helpful Hint Here is an example where we utilize a previous result to express the mo-ment of inertia of the chosen ele-ment, which in this case is a right-circular cylinder of differential axial length dx. It would be foolish to start with a third-order element, such as dx dy dz, when we can eas-ily solve the problem with a first-order element.
SAMPLE PROBLEM B/3 Determine the moments of inertia of the homogeneous rectangular paral-lelepiped of mass m about the centroidal x0- and z-axes and about the x-axis through one end.
Solution.
A transverse slice of thickness dz is selected as the element of vol-ume. The moment of inertia of this slice of infinitesimal thickness equals the mo-ment of inertia of the area of the section times the mass per unit area dz. Thus, the moment of inertia of the transverse slice about the y-axis is and that about the x-axis is As long as the element is a plate of differential thickness, the principle given by Eq. B/7a may be applied to give These expressions may now be integrated to obtain the desired results.
The moment of inertia about the z-axis is Ans.
where m is the mass of the block. By interchange of symbols, the moment of in-ertia about the x0-axis is Ans.
The moment of inertia about the x-axis may be found by the parallel-axis theo-rem, Eq. B/3. Thus, Ans.
This last result may be obtained by expressing the moment of inertia of the ele-mental slice about the x-axis and integrating the expression over the length of the bar. Again, by the parallel-axis theorem Integrating gives the result obtained previously: The expression for Ixx may be simplified for a long prismatic bar or slender rod whose transverse dimensions are small compared with the length. In this case, a2 may be neglected compared with 4l2, and the moment of inertia of such a slender bar about an axis through one end normal to the bar becomes I By the same approximation, the moment of inertia about a centroidal axis nor-mal to the bar is I 1 12 ml2.
1 3 ml2.
Ixx ab l 0 a2 12 z2 dz abl 3 l2 a2 4 1 12 m(a2 4l2) dIxx dIxx z2 dm ( dz)( 1 12 a3b) z2ab dz ab a2 12 z2 dz Ixx Ix0 x0 m l 2 2 1 12 m(a2 4l2) Ix0 x0 1 12 m(a2 l2) Izz dIzz ab 12 (a2 b2) l 0 dz 1 12 m(a2 b2) dIzz dIxx dIyy ( dz) ab 12 (a2 b2) dIxx ( dz)( 1 12 a3b) dIyy ( dz)( 1 12 ab3) Article B/1 Mass Moments of Inertia about an Axis 647 l/2 z dz x0 y0 x′ x y y′ z a b G l/2 Helpful Hint Refer to Eqs. B/6 and recall the ex-pression for the area moment of in-ertia of a rectangle about an axis through its center parallel to its base.
648 Appendix B Mass Moments of Inertia SAMPLE PROBLEM B/4 The upper edge of the thin homogeneous plate of mass m is parabolic with a vertical slope at the origin O. Determine its mass moments of inertia about the and z-axes.
Solution.
We begin by clearly establishing the function associated with the upper boundary. From evaluated at we find that so that We choose a transverse slice of thickness for the integrations leading to and The mass of this slice is and the total mass of the plate is The moment of inertia of the slice about the x-axis is For the entire plate, we have In terms of the mass m: Ans.
The moment of inertia of the element about the y-axis is For the entire plate, Ans.
For thin plates which lie in the plane, Ans.
Izz m h2 5 3b2 7 Izz Ixx Iyy 1 5mh2 3 7mb2 x-y Iyy dIyy b 0 t h bx 3 dx 2 7thb3 m 2 3thb 3 7mb2 dIyy dm y2 (ty dx)y2 ty3 dx Ixx 2 15th3b m 2 3thb 1 5mh2 Ixx dIxx b 0 1 3t h b x 3 dx 2 15th3b dIxx 1 3 dm y2 1 3(ty dx)y2 1 3ty3 dx m dm ty dx b 0 t h bx dx 2 3 thb dm ty dx Iyy.
Ixx dx y h bx.
k h/b (x, y) (b, h), y kx x-, y-, Helpful Hints If we have saying that “y gets large faster than x” helps establish that the parabola opens upward.
Here, we have which says that “x gets large faster than y”, helping establish that the parabola opens rightward.
For a full b by h rectangular plate of thickness t, the mass would be (density times volume). So the factor of for the parabolic plate makes sense.
Recall that for a slender rod of mass m and length l, the moment of iner-tia about an axis perpendicular to the rod and passing through one end is Note that is independent of the width b.
Note that is independent of the height h.
Iyy Ixx 1 3ml2.
2 3 thb y2 k2x, y kx2, O b dx x x h Parabolic z y y t Article B/1 Mass Moments of Inertia about an Axis 649 SAMPLE PROBLEM B/5 The radius of the homogeneous solid of revolution is proportional to the square of its x-coordinate. If the mass of the body is m, determine its mass mo-ments of inertia about the x- and y-axes.
Solution.
We begin by writing the boundary in the plane as The constant k is determined by evaluating this equation at the point which gives so that As is usually convenient for bodies with axial symmetry, we choose a disk-shaped slice as our differential element, as shown in the given figure. The mass of this element is where represents the density of the body. The moment of inertia of the ele-ment about the x-axis is The mass of the entire body is and the moment of inertia of the entire body is All that remains is to express more conventionally in terms of its mass. We do so by writing Ans.
By the parallel-axis theorem, the moment of inertia of the disk-shaped element about the y-axis is For the entire body, we have Finally, we multiply by the same unit expression as above to obtain a result in terms of the body mass m.
Ans.
Iyy r2h r2 36 h2 7 m 1 5 r2h 5m r2 36 h2 7 r2h r2 36 h2 7 Iyy dIyy h 0 r2 h4 1 4 r2 h4 x8 x6 dx r2 h4 1 4 r2 h4 x9 9 x7 7 h 0 r h2 x2 2 1 4 r2 h4 x4 x2 dx r2 h4 1 4 r2 h4 x8 x6 dx dm 1 4 r h2 x2 2 x2 y2 dx 1 4 r2 h4 x4 x2 dIyy dIyy x2 dm 1 4 dm y2 x2 dm Ixx 1 18r4h m 1 5r2h 5 18mr2 Ixx Ixx dIxx h 0 1 2y4 dx h 0 1 2 r h2 x2 4 dx 1 18r4h m dm h 0 y2 dx h 0 r h2 x2 2 dx r2 h4 x5 5 h 0 1 5r2h dIxx 1 2 dm y2 1 2(y2 dx) y2 1 2y4 dx dm y2 dx y r h2 x2.
k r/h2, (h, r): r kh2, (x, y) y kx2.
x-y Helpful Hints The volume of a disk is the area of its face times its thickness. Then density times volume gives mass.
From Sample Problem the mass moment of inertia of a uniform cylinder (or disk) about its longitudi-nal axis is Remember to regard an integral op-eration as an infinite summation.
The parenthetical expression here is unity, because its numerator and denominator are equal.
We note that is independent of h.
So the body could be compressed to or elongated to a large value of h with no resulting change in This is true because no particle of the body would be changing its dis-tance from the x-axis.
Ixx.
h 0 Ixx 1 2 mr2.
B/1, y y′ z x dx r h O x y Problem B/3 B/4 Determine the mass moment of inertia of the uniform thin triangular plate of mass m about the x-axis. Also determine the radius of gyration about the x-axis. By analogy state . Then determine Problem B/4 B/5 Calculate the moment of inertia of the tapered steel rod of circular cross section about an axis normal to the rod through O. Note that the rod diameter is small compared with its length.
Problem B/5 100 mm 10 mm 5 mm O x z y b h m Izz and kz.
Iyy and ky y m x L — 2 L — 2 β 650 Appendix B Mass Moments of Inertia PROBLEMS Integration Exercises B/1 Use the mass element where is the mass per unit length, and determine the mass moments of inertia and of the homogeneous slender rod of mass and length l.
Problem B/1 B/2 In order to better appreciate the greater ease of inte-gration with lower-order elements, determine the mass moment of inertia of the homogeneous thin plate by using the square element (a) and then by using the rectangular element (b). The mass of the plate is m. Then by inspection state and finally, determine Problem B/2 B/3 Determine the mass moments of inertia about the x-, axes of the slender rod of length L and mass m which makes an angle with the axis as shown.
x- y-, and z-y (a) (b) x z b — 2 a — 2 a — 2 b — 2 t dx dx dy Izz.
Iyy, Ixx y x dx dm l/2 x l/2 y′ m Iyy Iyy dm dx, B/6 Determine the mass moment of inertia of the uniform thin equilateral triangular plate of mass m about the x-axis. Also determine the corresponding radius of gyration.
Problem B/6 B/7 Determine the mass moment of inertia about the y-axis for the equilateral triangular plate of the previ-ous problem. Also determine its radius of gyration about the y-axis.
B/8 Determine the mass moments of inertia of the thin parabolic plate of mass m about the and axes.
Problem B/8 B/9 Determine the mass moment of inertia of the uniform thin parabolic plate of mass m about the axis. State the corresponding radius of gyration.
x-y x z b h y = kx2 z-x-, y-, x z y b b b – – 2 b – – 2 Article B/1 Problems 651 Problem B/9 B/10 Determine the mass of inertia about the axis for the parabolic plate of the previous problem. State the radius of gyration about the axis.
B/11 Calculate the moment of the homogeneous right-circular cone of mass m, base radius r, and altitude h about the cone axis x and about the axis through its vertex.
Problem B/11 y x r h y-y-y-x z y m b – – 2 b – – 2 Parabolic h Problem B/15 B/16 Determine the radius of gyration about the z-axis of the paraboloid of revolution shown. The mass of the homogenous body is m.
Problem B/16 B/17 Determine the moment of inertia about the axis for the paraboloid of revolution of Prob. B/16.
B/18 Determine the mass moment of inertia about the axis of the solid spherical segment of mass m.
Problem B/18 x y R — 2 R — 2 x-y-r z x y h π x –– a a b m x y z y = b sin 652 Appendix B Mass Moments of Inertia B/12 Determine the mass moment of inertia of the uni-form thin elliptical plate (mass m) about the axis.
Then, by analogy, state the expression for Finally, determine Problem B/12 B/13 Determine the mass moment of inertia of the homogeneous solid of revolution of mass m about the axis.
Problem B/13 B/14 Determine the mass moment of inertia of the homo-geneous solid of revolution of the previous problem about the and axes.
B/15 Determine the mass moment of inertia about the axis for the uniform thin plate of mass m shown.
x-z-y-y z x r h m y = kx1.5 x-a b m x y z Elliptical Izz.
Iyy.
x-B/19 Determine the moment of inertia about the axis of the homogeneous solid semiellipsoid of revolution having mass m.
Problem B/19 B/20 A homogeneous solid of mass m is formed by revolv-ing the right triangle about the axis. Determine the radius of gyration of the solid about the axis.
Problem B/20 B/21 Determine by integration the moment of inertia of the half-cylindrical shell of mass m about the axis The thickness of the shell is small compared with r.
Problem B/21 B/22 Determine the moment of inertia about the generat-ing axis of a complete ring (torus) of mass m having a circular section with the dimensions shown in the sectional view.
r a a l/2 l/2 a-a.
r a a a z z-z-45 y z x b a x-Article B/1 Problems 653 Problem B/22 B/23 The plane area shown in the top portion of the figure is rotated about the axis to form the body of revolution of mass m shown in the lower portion of the figure. Determine the mass moment of inertia of the body about the axis.
Problem B/23 2b b b m 2b y x L – – 2 L – – 2 L – – 2 L – – 2 Parabolic y x x-x-180 a R Problem B/28 B/29 Determine the moment of inertia of the one-quarter-cylindrical shell of mass m about the axis. The thickness of the shell is small compared with r.
Problem B/29 B/30 A shell of mass m is obtained by revolving the quarter-circular section about the axis. If the thickness of the shell is small compared with a and if a/3, determine the radius of gyration of the shell about the axis.
Problem B/30 r a z z-r z-r x x b – 2 b – 2 x-x x z y r 654 Appendix B Mass Moments of Inertia B/24 Determine for the homogen eous body of revolu-tion of the previous problem.
B/25 The thickness of the homogeneous triangular plate of mass m varies linearly with the distance from the ver-tex toward the base. The thickness a at the base is small compared with the other dimensions. Determine the moment of inertia of the plate about the axis along the centerline of the base.
Problem B/25 B/26 Determine the moment of inertia of the triangular plate described in Prob. B/25 about the axis.
B/27 Determine the moment of inertia, about the generat-ing axis, of the hollow circular tube of mass m ob-tained by revolving the thin ring shown in the sectional view completely around the generating axis.
Problem B/27 B/28 Determine the moments of inertia of the half-spherical shell with respect to the and axes. The mass of the shell is m, and its thickness is negligible com-pared with the radius r.
z-x-a R z-a z x y h b — 2 b — 2 y-Iyy Composite and Parallel-Axis Exercises B/31 The two small spheres of mass m each are connected by the light rigid rod which lies in the plane. De-termine the mass moments of inertia of the assembly about the and axes.
Problem B/31 B/32 State without calculation the moment of inertia about the axis of the thin conical shell of mass m and ra-dius r from the results of Sample Problem applied to a circular disk. Observe the radial distribution of mass by viewing the cone along the z-axis.
Problem B/32 B/33 The moment of inertia of a solid homogeneous cylin-der of radius r about an axis parallel to the central axis of the cylinder may be obtained approximately by multiplying the mass of the cylinder by the square of the distance d between the two axes. What percentage error e results if (a) and (b) ?
d 2r d 10r h z O r B/1 z-m m y z L L x L L z-x-, y-, x-z Article B/1 Problems 655 Problem B/33 B/34 Every “slender” rod has a finite radius r. Refer to Table and derive an expression for the percentage error e which results if one neglects the radius r of a homogeneous solid cylindrical rod of length l when calculating its moment of inertia . Evaluate your expression for the ratios Problem B/34 B/35 Calculate the mass moment of inertia about the axis O-O for the uniform 10-in. block of steel with cross-section dimensions of 6 and 8 in.
Problem B/35 O O 3″ 3″ 8″ 6″ 10″ r z l — 2 l — 2 r/l 0.01, 0.1, and 0.5.
Izz D/4 d r B/39 Determine the moment of inertia of the half-ring of mass m about its diametral axis a-a and about axis b-b through the midpoint of the arc normal to the plane of the ring. The radius of the circular cross section is small compared with r.
Problem B/39 B/40 The semicircular disk weighs 5 lb, and its small thickness may be neglected compared with its 10-in.
radius. Compute the moments of inertia of the disk about the and axes.
Problem B/40 B/41 A 6-in. steel cube is cut along its diagonal plane. Cal-culate the moment of inertia of the resulting prism about the edge .
Problem B/41 x 6″ 6″ 6″ x x-x 10″ x y z y′ z-x-, y-, y-, b r b a a m 656 Appendix B Mass Moments of Inertia B/36 Determine for the cylinder with a centered circu-lar hole. The mass of the body is m.
Problem B/36 B/37 Determine the mass moment of inertia about the axis for the right-circular cylinder with a central longitudinal hole.
Problem B/37 B/38 Determine the moment of inertia of the mallet about the axis. The density of the wooden handle is and that of the soft-metal head is The longitudinal axis of the cylindrical head is normal to the axis. State any assumptions.
Problem B/38 300 mm 40 mm 30 mm 50 mm 50 mm x x-9000 kg/m3.
800 kg/m3 x-m 2r r L – – 2 L – – 2 z z-r1 r2 x Ixx B/42 Determine the length L of each of the slender rods of mass which must be centrally attached to the faces of the thin homogeneous disk of mass m in order to make the mass moments of inertia of the unit about the and axes equal.
Problem B/42 B/43 A badminton racket is constructed of uniform slen-der rods bent into the shape shown. Neglect the strings and the built-up wooden grip and estimate the mass moment of inertia about the axis through O, which is the location of the player’s hand. The mass per unit length of the rod material is Problem B/43 L — 4 L — 4 L — 8 y O L .
y-L m — 2 m — 2 L x m r z z-x-m/2 Article B/1 Problems 657 B/44 As a sorting-machine part, the steel half-cylinder is subject to rapid angular acceleration and decelera-tion about the y-axis, and its moment of inertia about this axis is required in the design of the machine.
Calculate Use Tables and as needed.
Problem B/44 B/45 Calculate the moment of inertia of the steel control wheel, shown in section, about its central axis.
There are eight spokes, each of which has a constant cross-sectional area of 200 . What percent n of the total moment of inertia is contributed by the outer rim?
Problem B/45 Dimensions in millimeters 50 100 300 400 120 200 mm2 75 mm2 y 90 mm 40 mm 80 mm x z D/4 D/1 Iyy.
B/48 The clock pendulum consists of the slender rod of length l and mass m and the bob of mass 7m. Neglect the effects of the radius of the bob and determine in terms of the bob position x. Calculate the ratio R of evaluated for to evaluated for .
Problem B/48 B/49 Determine the moment of inertia about the axis of the portion of the homogeneous sphere shown. The mass of the sphere portion is m.
Problem B/49 B/50 A square plate with a quarter-circular sector re-moved has a net mass m. Determine its moment of inertia about axis normal to the plane of the plate.
Problem B/50 A A O a x y z a a A-A x y R — 2 R — 2 x-x O l 7m m x l IO x 3 4l IO IO 658 Appendix B Mass Moments of Inertia B/46 The uniform rod of length 4b and mass m is bent into the shape shown. The diameter of the rod is small compared with its length. Determine the mo-ments of inertia of the rod about the three coordi-nate axes.
Problem B/46 B/47 Calculate the moment of inertia of the solid steel semicylinder about the axis and about the paral-lel axis. (See Table for the density of steel.) Problem B/47 x0 60 mm 100 mm 60 mm x0 x x D/1 x0-x0 x-x z b x O y b b b B/51 Determine the moments of inertia about the tangent axis for the full ring of mass and the half-ring of mass Problem B/51 B/52 The slender metal rods are welded together in the configuration shown. Each 6-in. segment weighs 0.30 Compute the moment of inertia of the as-sembly about the axis.
Problem B/52 B/53 The welded assembly shown is made from a steel rod which weighs 0.667 per foot of length. Calculate the moment of inertia of the assembly about the axis.
Problem B/53 x x 8″ 8″ 8″ 8″ 8″ x-x lb 6″ 6″ 6″ 6″ z 6″ x y y-lb.
m2.
m1 x-x Article B/1 Problems 659 B/54 The machine element is made of steel and is designed to rotate about axis O-O. Calculate its radius of gy-ration about this axis.
Problem B/54 B/55 Determine for the cone frustum, which has base radii and and mass m.
Problem B/55 B/56 By direct integration, determine the moment of iner-tia about the axis of the thin semicircular disk of mass m and radius R inclined at an angle from the plane.
Problem B/56 R z X x Z y X-y Z-x h r1 r2 y – G r2 r1 Ixx O O 30 80 Dimensions in millimeters 80 40 kO r x x r x m1 x m2 B/2 Products of Inertia For problems in the rotation of three-dimensional rigid bodies, the expression for angular momentum contains, in addition to the moment-of-inertia terms, product-of-inertia terms defined as (B/8) These expressions were cited in Eqs. 7/10 in the expansion of the ex-pression for angular momentum, Eq. 7/9.
The calculation of products of inertia involves the same basic proce-dure which we have followed in calculating moments of inertia and in evaluating other volume integrals as far as the choice of element and the limits of integration are concerned. The only special precaution we need to observe is to be doubly watchful of the algebraic signs in the ex-pressions. Whereas moments of inertia are always positive, products of inertia may be either positive or negative. The units of products of iner-tia are the same as those of moments of inertia.
We have seen that the calculation of moments of inertia is often simplified by using the parallel-axis theorem. A similar theorem exists for transferring products of inertia, and we prove it easily as follows. In Fig. B/6 is shown the x-y view of a rigid body with parallel axes x0-y0 passing through the mass center G and located from the x-y axes by the distances dx and dy. The product of inertia about the x-y axes by defini-tion is The last two integrals vanish since the first moments of mass about the mass center are necessarily zero.
Similar relations exist for the remaining two product-of-inertia terms. Dropping the zero subscripts and using the bar to designate the mass-center quantity, we obtain (B/9) Iyz Iyz mdydz Ixz Ixz mdxdz Ixy Ixy mdxdy Ix0 y0 mdx dy x0 y0 dm dx dy dm dx y0 dm dy x0 dm Ixy xy dm (x0 dx)(y0 dy) dm Iyz Izy yz dm Ixz Izx xz dm Ixy Iyx xy dm 660 Appendix B Mass Moments of Inertia dx dm y G O x x0 x0 y0 y0 dy Figure B/6 These transfer-of-axis relations are valid only for transfer to or from parallel axes through the mass center.
With the aid of the product-of-inertia terms, we can calculate the moment of inertia of a rigid body about any prescribed axis through the coordinate origin. For the rigid body of Fig. B/7, suppose we must deter-mine the moment of inertia about axis O-M. The direction cosines of O-M are l, m, n, and a unit vector along O-M may be written li mj nk. The moment of inertia about O-M is where r r sin h. The cross product is and, after we collect terms, the dot-product expansion gives Thus, with the substitution of the expressions of Eqs. B/4 and B/8, we have (B/10) This expression gives the moment of inertia about any axis O-M in terms of the direction cosines of the axis and the moments and products of inertia about the coordinate axes.
Principal Axes of Inertia As noted in Art. 7/7, the array whose elements appear in the expansion of the angular-momentum ex-pression, Eq. 7/11, for a rigid body with attached axes, is called the iner-tia matrix or inertia tensor. If we examine the moment- and product-of-inertia terms for all possible orientations of the axes with re-spect to the body for a given origin, we will find in the general case an orientation of the x-y-z axes for which the product-of-inertia terms van-ish and the array takes the diagonalized form IM Ixxl2 Iyym2 Izzn2 2Ixylm 2Ixzln 2Iyzmn 2 xylm 2 xzln 2 yzmn (r ) (r ) h2 (y2 z2)l2 (x2 z2)m2 (x2 y2)n2 (r ) (yn zm)i (zl xn)j (xm yl)k IM h2 dm (r ) (r ) dm Article B/2 Products of Inertia 661 r y O x dm h M z θ Figure B/7 Such axes x-y-z are called the principal axes of inertia, and Ixx, Iyy, and Izz are called the principal moments of inertia and represent the maximum, minimum, and intermediate values of the moments of inertia for the particular origin chosen.
It may be shown that for any given orientation of axes x-y-z the so-lution of the determinant equation (B/11) for I yields three roots I1, I2, and I3 of the resulting cubic equation which are the three principal moments of inertia. Also, the direction cosines l, m, and n of a principal inertia axis are given by (B/12) These equations along with l2 m2 n2 1 will enable a solution for the direction cosines to be made for each of the three I’s.
To assist with the visualization of these conclusions, consider the rectangular block, Fig. B/8, having an arbitrary orientation with respect to the x-y-z axes. For simplicity, the mass center G is located at the ori-gin of the coordinates. If the moments and products of inertia for the block about the x-y-z axes are known, then solution of Eq. B/11 would give the three roots, I1, I2, and I3, which are the principal moments of in-ertia. Solution of Eq. B/12 using each of the three I’s, in turn, along with l2 m2 n2 1, would give the direction cosines l, m, and n for each of the respective principal axes, which are always mutually perpendicular.
From the proportions of the block as drawn, we see that I1 is the maxi-mum moment of inertia, I2 is the intermediate value, and I3 is the mini-mum value. Izxl Izym (Izz I)n 0 Iyxl (Iyy I)m Iyzn 0 (Ixx I)l Ixym Ixzn 0 662 Appendix B Mass Moments of Inertia See, for example, the first author’s Dynamics, SI Version, 1975, John Wiley & Sons, Art. 41.
3 1 2 3 1 2 G z y x Figure B/8 SAMPLE PROBLEM B/6 The bent plate has a uniform thickness t which is negligible compared with its other dimensions. The density of the plate material is . Determine the prod-ucts of inertia of the plate with respect to the axes as chosen.
Solution.
Each of the two parts is analyzed separately.
Rectangular part.
In the separate view of this part, we introduce parallel axes x0-y0 through the mass center G and use the transfer-of-axis theorem. By symmetry, we see that 0 so that Because the z-coordinate of all elements of the plate is zero, it follows that Ixz Iyz 0.
Triangular part.
In the separate view of this part, we locate the mass center G and construct x0-, y0-, and z0-axes through G. Since the x0-coordinate of all ele-ments is zero, it follows that 0 and 0. The transfer-of-axis theorems then give us We obtain Iyz by direct integration, noting that the distance a of the plane of the triangle from the y-z plane in no way affects the y- and z-coordinates. With the mass element dm t dy dz, we have Adding the expressions for the two parts gives Ans.
Ans.
Ans. Iyz 0 1 8 tb2c2 1 8 tb2c2 Ixz 0 1 6 tabc2 1 6 tabc2 Ixy 1 4 ta2b2 1 3 tab2c 1 12 tab2(3a 4c) tc2 2b2 b 0 y3 dy 1 8 tb2c2 Iyz t b 0 cy/b 0 yz dz dy t b 0 y z2 2 cy/b 0 dy Iyz yz dm Ixz 0 t b 2 c(a) c 3 1 6 tabc2 [Ixz Ixz mdxdz] Ixy 0 t b 2 c(a) 2b 3 1 3 tab2c [Ixy Ixy mdxdy] Ix0 z0 Ixz Ix0 y0 Ixy [Ixy Ixy mdxdy] Ixy 0 taba 2 b 2 1 4 ta2b2 Ix0 y0 Ixy Article B/2 Products of Inertia 663 c a b y x z 2c/3 c/3 b/3 2b/3 z a G y0 x0 y x G y0 z0 dm x0 y x Helpful Hints We must be careful to preserve the same sense of the coordinates. Thus, plus x0 and y0 must agree with plus x and y.
We choose to integrate with respect to z first, where the upper limit is the variable height z cy/b. If we were to integrate first with respect to y, the limits of the first integral would be from the variable y bz/c to b.
SAMPLE PROBLEM B/7 The angle bracket is made from aluminum plate with a mass of 13.45 kg per square meter. Calculate the principal moments of inertia about the origin O and the direction cosines of the principal axes of inertia. The thickness of the plate is small compared with the other dimensions.
Solution.
The masses of the three parts are Part 1 Part 2 Ixz Ixz mdxdz 0 0.0518(0.16)(0.05) 4.14(104) kgm2 Ixy 0 Iyz 0 13.41(104) kg m2 Izz 1 4 mr2 mdx 2 0.0518 (0.035)2 4 (0.16)2 14.86(104) kg m2 0.0518 (0.035)2 2 (0.16)2 (0.05)2 Iyy 1 2 mr2 m(dx 2 dz 2) 1.453(104) kg m2 Ixx 1 4 mr2 mdz 2 0.0518 (0.035)2 4 (0.050)2 0 m a 2 b 2 0.282(0.105)(0.05) 14.83(104) kgm2 Ixz Ixz mdx dz Ixy 0 Iyz 0 Izz 1 3 ma2 1 3 (0.282)(0.21)2 41.5(104) kgm2 Iyy 1 3 m(a2 b2) 1 3 (0.282)[(0.21)2 (0.1)2] 50.9(104) kgm2 Ixx 1 3 mb2 1 3 (0.282)(0.1)2 9.42(104) kgm2 m3 13.45(0.12)(0.11) 0.1775 kg m2 13.45(0.035)2 0.0518 kg m1 13.45(0.21)(0.1) 0.282 kg 664 Appendix B Mass Moments of Inertia z O x y 160 110 120 50 50 70 50 Dimensions in millimeters z x O y z O x y 2 1 3 r = 35 b = 100 a = 210 c = 110 dz = 50 dx = 160 d = 120 Helpful Hints Note that the mass of the hole is treated as a negative number.
You can easily derive this formula.
Also check Table D/4.
SAMPLE PROBLEM B/7 (Continued) Part 3 Totals Substitution into Eq. B/11, expansion of the determinant, and simplification yield Solution of this cubic equation yields the following roots, which are the principal moments of inertia Ans.
The direction cosines of each principal axis are obtained by substituting each root, in turn, into Eq. B/12 and using l2 m2 n2 1. The results are Ans.
The bottom figure shows a pictorial view of the bracket and the orientation of its principal axes of inertia. n1 0.839 n2 0.312 n3 0.445 m1 0.410 m2 0.1742 m3 0.895 l1 0.357 l2 0.934 l3 0.01830 I3 43.4(104) kg m2 I2 11.82(104) kgm2 I1 48.3(104) kg m2 I3 103.5(104)I2 3180(108)I 24 800(1012) 0 Izz 43.8(104) kg m2 Ixz 10.69(104) kgm2 Iyy 43.2(104) kgm2 Iyz 0 Ixx 16.48(104) kgm2 Ixy 5.86(104) kgm2 Iyz 0 Ixz 0 0 m c 2 d 2 0.1775(0.055)(0.06) 5.86(104) kgm2 Ixy Ixy mdx dy 15.68 (104) kgm2 Izz 1 3 m(c2 d2) 1 3 (0.1775)[(0.11)2 (0.12)2] Iyy 1 3 mc2 1 3 (0.1775)(0.11)2 7.16(104) kgm2 Ixx 1 3 md2 1 3 (0.1775)(0.12)2 8.52(104) kgm2 Article B/2 Products of Inertia 665 z x O y z O x y 2 1 3 r = 35 b = 100 a = 210 c = 110 dz = 50 dx = 160 d = 120 z 3 1 y x O 2 A computer program for the solution of a cubic equation may be used, or an algebraic solution using the for-mula cited in item 4 of Art. C/4, Ap-pendix C, may be employed.
B/59 Determine the products of inertia of the uniform slender rod of mass m about the coordinate axes shown.
Problem B/59 B/60 Determine the products of inertia about the coordi-nate axes for the thin plate of mass m which has the shape of a circular sector of radius a and angle as shown.
Problem B/60 B/61 Determine the products of inertia about the coordi-nate axes for the thin square plate with two circular holes. The mass of the plate material per unit area is .
Problem B/61 b – 4 b – 4 b – 4 b – 4 b – 8 b – 4 y x b – 8 b – 4 b – 4 b – 4 m a y z x β z y x a m h b 666 Appendix B Mass Moments of Inertia PROBLEMS Introductory Problems B/57 Determine the products of inertia about the coordi-nate axes for the unit which consists of three small spheres, each of mass m, connected by the light but rigid slender rods.
Problem B/57 B/58 Determine the products of inertia about the coordi-nate axes for the unit which consists of four small particles, each of mass m, connected by the light but rigid slender rods.
Problem B/58 x z y m m m m l l l l l l l l l l l l l m l y x m m z B/62 The slender rod of mass m is formed into a quarter-circular arc of radius r. Determine the products of inertia of the rod with respect to the given axes.
Problem B/62 B/63 The uniform rectangular block weighs 50 lb. Calcu-late its products of inertia about the coordinate axes shown.
Problem B/63 B/64 Determine the product of inertia Ixy for the slender rod of mass m.
Problem B/64 y x l/2 l/2 θ x y z 8″ 6″ 4″ x z r y 45° Article B/2 Problems 667 B/65 The semicircular disk of mass m and radius R, in-clined at an angle from the X-y plane, of Prob. B/56 is repeated here. By the methods of this article, de-termine the moment of inertia about the Z-axis.
Problem B/65 B/66 Determine the products of inertia for the rod of Prob. B/46, repeated here.
Problem B/66 z b x O y b b b R z X x Z y B/70 Determine the moment of inertia of the solid cube of mass m about the diagonal axis A-A through oppo-site corners.
Problem B/70 B/71 The steel plate with two right-angle bends and a central hole has a thickness of 15 mm. Calculate its moment of inertia about the diagonal axis through the corners A and B.
Problem B/71 B/72 Prove that the moment of inertia of the rigid assem-bly of three identical balls, each of mass m and ra-dius r, has the same value for all axes through O.
Neglect the mass of the connecting rods.
Problem B/72 m z y x b b b m O m 150 500 Dimensions in millimeters 500 A B 300 300 A A a a a 668 Appendix B Mass Moments of Inertia Representative Problems B/67 The S-shaped piece is formed from a rod of diameter d and bent into the two semicircular shapes. Deter-mine the products of inertia for the rod, for which d is small compared with r.
Problem B/67 B/68 Determine the three products of inertia with respect to the given axes for the uniform rectangular plate of mass m.
Problem B/68 B/69 For the slender rod of mass m bent into the configu-ration shown, determine its products of inertia Ixy, Ixz, and Iyz.
Problem B/69 b b b b y x z 45° 45° y x z b h θ r d y x z r Computer-Oriented Problems B/73 Each sphere of mass m has a diameter which is small compared with the dimension b. Neglect the mass of the connecting struts and determine the principal moments of inertia of the assembly with respect to the coordinates shown. Determine also the direction cosines of the axis of maximum mo-ment of inertia.
Problem B/73 B/74 Determine the moment of inertia I about axis O-M for the uniform slender rod bent into the shape shown. Plot I versus from 0 to 90 and de-termine the minimum value of I and the angle which its axis makes with the x-direction. (Note: Because the analysis does not involve the z-coordi-nate, the expressions developed for area moments of inertia, Eqs. A/9, A/10, and A/11 in Appendix A of Vol. 1 Statics, may be utilized for this problem in place of the three-dimensional relations of Appen-dix B.) The rod has a mass per unit length.
Problem B/74 x y O r r M θ b b b O m m y x z m b b Article B/2 Problems 669 B/75 The assembly of three small spheres connected by light rigid bars of Prob. B/57 is repeated here. De-termine the principal moments of inertia and the direction cosines associated with the axis of maxi-mum moment of inertia.
Problem B/75 B/76 The bent rod of Probs. B/46 and B/66 is repeated here. Its mass is m, and its diameter is small com-pared with its length. Determine the principal mo-ments of inertia of the rod about the origin O. Also find the direction cosines for the axis of minimum moment of inertia.
Problem B/76 z b x O y b b b l l l l l m l y x m m z B/78 The slender rod has a mass per unit length and is formed into the shape shown. Determine the prin-cipal moments of inertia about axes through O and calculate the direction cosines of the axis of mini-mum moment of inertia.
Problem B/78 z x y O b b 670 Appendix B Mass Moments of Inertia B/77 The thin plate has a mass per unit area and is formed into the shape shown. Determine the prin-cipal moments of inertia of the plate about axes through O.
Problem B/77 2b b z y x b b O 671 C/1 Introduction Appendix C contains an abbreviated summary and reminder of se-lected topics in basic mathematics which find frequent use in mechanics.
The relationships are cited without proof. The student of mechanics will have frequent occasion to use many of these relations, and he or she will be handicapped if they are not well in hand. Other topics not listed will also be needed from time to time.
As the reader reviews and applies mathematics, he or she should bear in mind that mechanics is an applied science descriptive of real bodies and actual motions. Therefore, the geometric and physical inter-pretation of the applicable mathematics should be kept clearly in mind during the development of theory and the formulation and solution of problems.
C/2 Plane Geometry 1. When two intersect-ing lines are, respec-tively, perpendicular to two other lines, the angles formed by the two pairs are equal.
2. Similar triangles 3. Any triangle Area 1 2 bh x b h y h 4. Circle 5. Every triangle inscribed within a semicircle is a right triangle.
6. Angles of a triangle 4 1 2 1 2 3 180 Sector area 1 2r2 Arc length s r Area r2 Circumference 2r 1 = θ 2 θ 1 θ 2 θ x y h b h b r s θ 1 θ 2 θ 1 + = /2 θ 2 θ π 1 θ 3 θ 4 θ 2 θ C Selected Topics of Mathematics C/3 Solid Geometry 672 Appendix C Selected Topics of Mathematics 1. Sphere 2. Spherical wedge Volume 2 3r3 Surface area 4r2 Volume 4 3r3 3. Right-circular cone 4. Any pyramid or cone where B area of base Volume 1 3Bh L r2 h2 Lateral area rL Volume 1 3r2h r r θ r h L B h C/4 Algebra 1. Quadratic equation 2. Logarithms Natural logarithms 3. Determinants 2nd order 3rd order log10x 0.4343 ln x log 1 0 log an n log a log (1/n) log n log (a/b) log a log b log (ab) log a log b ex y, x log e y ln y b e 2.718 282 bx y, x log b y x b b2 4ac 2a , b2 4ac for real roots ax2 bx c 0 4. Cubic equation For general cubic equation Substitute x x0 a/3 and get Ax0 B.
Then proceed as above to find values of x0 from which x x0 a/3.
x0 3 x3 ax2 bx c 0 x1 2q1/3, x2 x3 q1/3 roots equal) Case III: q2 p3 0 (three roots real, two x1 (q q2 p3)1/3 (q q2 p3)1/3 roots imaginary) Case II: q2 p3 positive (one root real, two x3 2p cos (u/3 240) x2 2p cos (u/3 120) x1 2p cos (u/3) cos u q/(p p), 0 u 180 distinct) Case I: q2 p3 negative (three roots real and Let p A/3, q B/2.
x3 Ax B C/5 Analytic Geometry 1. Straight line 2. Circle x x y y r r b a x2 + y2 = r2 (x – a)2 + (y – b)2 = r2 m x x y y 1 a a b y = a + mx x – a y – b + = 1 Article C/6 Trigonometry 673 3. Parabola 4. Ellipse 5. Hyperbola x y a b x2 — a2 y2 — b2 + = 1 b b a a y x x y x2 — a2 y = b y2 — b2 x = a a a a y x x b x2 — a2 y2 — b2 – = 1 y xy = a2 C/6 Trigonometry 1. Definitions 2. Signs in the four quadrants θ (+) (+) I θ (+) (–) II θ (–) (–) III θ (+) (–) IV tan a/b cot b/a cos b/c sec c/b sin a/c csc c/a a c b θ 3. Miscellaneous relations cos (a b) cos a cos b sin a sin b sin (a b) sin a cos b cos a sin b cos 2 cos2 sin2 sin 2 2 sin cos 1 cot2 csc2 1 tan2 sec2 sin2 cos2 1 674 Appendix C Selected Topics of Mathematics 4. Law of sines 5. Law of cosines c2 a2 b2 2ab cos D c2 a2 b2 2ab cos C a b sin A sin B A C D B c b a C/7 Vector Operations 1. Notation.
Vector quantities are printed in boldface type, and scalar quantities appear in lightface italic type. Thus, the vector quantity V has a scalar magnitude V. In longhand work vector quantities should always be consistently indicated by a symbol such as to distinguish them from scalar quantities.
2. Addition 3. Subtraction 4. Unit vectors i, j, k where 5. Direction cosines l, m, n are the cosines of the angles between V and the x-, y-, z-axes. Thus, so that and l2 m2 n2 1 V V(li mj nk) l Vx/V m Vy/V n Vz/V V V Vx 2 Vy 2 Vz 2 V Vxi Vy j Vzk P Q P (Q) Associative law P (Q R) (P Q) R Commutative law P Q Q P Parallelogram addition P Q R Triangle addition P Q R V or V l P P P P Q –Q P – Q R R Q Q V k i j z x y kVz iVx jVy 6. Dot or scalar product This product may be viewed as the magnitude of P multiplied by the component Q cos of Q in the direction of P, or as the magni-tude of Q multiplied by the component P cos of P in the direction of Q.
From the definition of the dot product It follows from the definition of the dot product that two vec-tors P and Q are perpendicular when their dot product vanishes, .
The angle between two vectors P1 and P2 may be found from their dot product expression , which gives where l, m, n stand for the respective direction cosines of the vec-tors. It is also observed that two vectors are perpendicular to each other when their direction cosines obey the relation l1l2 m1m2 n1n2 0.
7. Cross or vector product.
The cross product P Q of the two vectors P and Q is defined as a vector with a magnitude and a direction specified by the right-hand rule as shown. Reversing the vector order and using the right-hand rule give Q P P Q.
Distributive law P (Q R) P Q P R From the definition of the cross product, using a right-handed coordinate system, we get i j k j k i k i j j i k k j i i k j i i j j k k 0 Q PQ sin P Distributive law P (Q R) P Q P R cos P1 P2 P1P2 P1xP2x P1yP2y P1zP2z P1P2 l1l2 m1m2 n1n2 P1 P2 P1P2 cos P Q 0 P P Px 2 Py 2 Pz 2 PxQx PyQy PzQz P Q (Pxi Pyj Pzk) (Qxi Qyj Qzk) ij ji i k k i jk kj 0 ii jj k k 1 Commutative law PQ QP P Q PQ cos Article C/7 Vector Operations 675 P P P θ θ θ θ Q Q cos θ P cos Q Q θ θ Q Q P P P × Q Q × P = –P × Q With the aid of these identities and the distributive law, the vector product may be written The cross product may also be expressed by the determinant 8. Additional relations Triple scalar product . The dot and cross may be interchanged as long as the order of the vectors is main-tained. Parentheses are unnecessary since is meaning-less because a vector P cannot be crossed into a scalar . Thus, the expression may be written The triple scalar product has the determinant expansion Triple vector product (P Q) R R (P Q) R (Q P). Here we note that the parentheses must be used since an expression P Q R would be ambiguous because it would not identify the vector to be crossed. It may be shown that the triple vector product is equivalent to or The first term in the first expression, for example, is the dot prod-uct , a scalar, multiplied by the vector Q.
9. Derivatives of vectors obey the same rules as they do for scalars. d(P Q) dt P Q ˙ P ˙ Q d(PQ) dt PQ ˙ P ˙ Q d(Pu) dt Pu ˙ P ˙u dP dt P ˙ P ˙xi P ˙yj P ˙zk R P P (Q R) P RQ PQR (P Q) R R PQ R QP P QR PQ R QR P (QR) (P Q) R R (P Q) (PyQz PzQy)i (PzQx PxQz)j (PxQy PyQx)k P Q (Pxi Pyj Pzk) (Qxi Qyj Qzk) 676 Appendix C Selected Topics of Mathematics 10. Integration of vectors.
If V is a function of x, y, and z and an el-ement of volume is d dx dy dz, the integral of V over the volume may be written as the vector sum of the three integrals of its com-ponents. Thus, V d i Vx d j Vy d k Vz d Article C/9 Derivatives 677 C/8 Series (Expression in brackets following series indicates range of convergence.) [Fourier expansion for l x l] where an 1 l l l ƒ(x) cos nx l dx, bn 1 l l l ƒ(x) sin nx l dx ƒ(x) a0 2 n1 an cos nx l n1 bn sin nx l cosh x ex ex 2 1 x2 2! x4 4! x6 6! [x2 ] sinh x ex ex 2 x x3 3! x5 5! x7 7! [x2 ] cos x 1 x2 2! x4 4! x6 6! [x2 ] sin x x x3 3! x5 5! x7 7! [x2 ] (1 x)n 1 nx n(n 1) 2! x2 n(n 1)(n 2) 3! x3 [x2 1] C/9 Derivatives , d sinh x dx cosh x, d cosh x dx sinh x, d tanh x dx sech2 x d sin x dx cos x, d cos x dx sin x, d tan x dx sec2 x lim xl0 cos x cos dx 1 lim xl0 sin x sin dx tan dx dx d u v dx v du dx u dv dx v2 dxn dx nxn1, d(uv) dx u dv dx v du dx C/10 Integrals 678 Appendix C Selected Topics of Mathematics Article C/10 Integrals 679 680 Appendix C Selected Topics of Mathematics C/11 Newton’s Method for Solving Intractable Equations Article C/11 Newton’s Method for Solving Intractable Equations 681 Frequently, the application of the fundamental principles of me-chanics leads to an algebraic or transcendental equation which is not solvable (or easily solvable) in closed form. In such cases, an iterative technique, such as Newton’s method, can be a powerful tool for obtain-ing a good estimate to the root or roots of the equation.
Let us place the equation to be solved in the form ƒ(x) 0. Part a of the accompanying figure depicts an arbitrary function ƒ(x) for values of x in the vicinity of the desired root xr. Note that xr is merely the value of x at which the function crosses the x-axis. Suppose that we have avail-able (perhaps via a hand-drawn plot) a rough estimate x1 of this root.
Provided that x1 does not closely correspond to a maximum or minimum value of the function ƒ(x), we may obtain a better estimate of the root xr by extending the tangent to ƒ(x) at x1 so that it intersects the x-axis at x2. From the geometry of the figure, we may write where ƒ(x1) denotes the derivative of ƒ(x) with respect to x evaluated at x x1. Solving the above equation for x2 results in The term ƒ(x1)/ƒ(x1) is the correction to the initial root estimate x1.
Once x2 is calculated, we may repeat the process to obtain x3, and so forth.
Thus, we generalize the above equation to xk1 xk ƒ(xk) ƒ(xk) x2 x1 ƒ(x1) ƒ(x1) tan ƒ(x1) ƒ(x1) x1 x2 x3 ƒ(x) ƒ(x) x ƒ(x) x x2 x1 x1 x2 x1 x2 xr2 xr1 xr xr Tangent to ƒ(x) at x = x1 ƒ(x) x (c) (b) (a) where This equation is repeatedly applied until ƒ(xk1) is sufficiently close to zero and xk1 xk. The student should verify that the equation is valid for all possible sign combinations of xk, ƒ(xk), and ƒ(xk).
Several cautionary notes are in order: 1. Clearly, ƒ(xk) must not be zero or close to zero. This would mean, as restricted above, that xk exactly or approximately corresponds to a minimum or maximum of ƒ(x). If the slope ƒ(xk) is zero, then the tangent to the curve never intersects the x-axis. If the slope ƒ(xk) is small, then the correction to xk may be so large that xk1 is a worse root estimate than xk. For this reason, experienced engineers usu-ally limit the size of the correction term; that is, if the absolute value of ƒ(xk)/ƒ(xk) is larger than a preselected maximum value, that maximum value is used.
2. If there are several roots of the equation ƒ(x) 0, we must be in the vicinity of the desired root xr in order that the algorithm actually converges to that root. Part b of the figure depicts the condition when the initial estimate x1 will result in convergence to rather than .
3. Oscillation from one side of the root to the other can occur if, for example, the function is antisymmetric about a root which is an inflection point. The use of one-half of the correction will usually prevent this behavior, which is depicted in part c of the accompa-nying figure.
Example: Beginning with an initial estimate of x1 5, estimate the sin-gle root of the equation ex 10 cos x 100 0.
The table below summarizes the application of Newton’s method to the given equation. The iterative process was terminated when the ab-solute value of the correction ƒ(xk)/ƒ(xk) became less than 106.
k xk ƒ(xk) ƒ(xk) 1 5.000 000 45.576 537 138.823 916 0.328 305 2 4.671 695 7.285 610 96.887 065 0.075 197 3 4.596 498 0.292 886 89.203 650 0.003 283 4 4.593 215 0.000 527 88.882 536 0.000 006 5 4.593 209 2(108) 88.881 956 2.25(1010) xk1 xk ƒ(xk) ƒ(xk) xr1 xr2 ƒ(xk) the function derivative evaluated at x xk ƒ(xk) the function ƒ(x) evaluated at x xk xk the kth estimate of the desired root xr xk1 the (k 1)th estimate of the desired root xr 682 Appendix C Selected Topics of Mathematics C/12 Selected Techniques for Numerical Integration 1. Area determination.
Consider the problem of determining the shaded area under the curve y ƒ(x) from x a to x b, as depicted in part a of the figure, and suppose that analytical integration is not feasi-ble. The function may be known in tabular form from experimental measurements or it may be known in analytical form. The function is taken to be continuous within the interval a x b. We may divide the area into n vertical strips, each of width x (b a)/n, and then add the areas of all strips to obtain A y dx. A representative strip of area Ai is shown with darker shading in the figure. Three useful numerical approximations are cited. In each case the greater the number of strips, the more accurate becomes the approximation geometrically. As a gen-eral rule, one can begin with a relatively small number of strips and increase the number until the resulting changes in the area approxima-tion no longer improve the accuracy obtained.
Article C/12 Selected Techniques for Numerical Integration 683 I. Rectangular [Figure (b)] The areas of the strips are taken to be rectangles, as shown by the representative strip whose height ym is cho-sen visually so that the small cross-hatched areas are as nearly equal as possible. Thus, we form the sum Σym of the effective heights and multi-ply by x. For a function known in analytical form, a value for ym equal to that of the function at the midpoint xi x/2 may be calculated and used in the summation.
II. Trapezoidal [Figure (c)] The areas of the strips are taken to be trapezoids, as shown by the representative strip. The area Ai is the average yi + 1 x0 x1 x2 x3 xi x xi + 1 y = ƒ(x) = a xn ym = b y y0 y1 yi y2 y3 yn yi yi + 1 yn – 1 Ai (a) (b) Rectangular Ai Ai = ym x A = y dx ym x x x height (yi yi 1)/2 times x. Adding the areas gives the area approxima-tion as tabulated. For the example with the curvature shown, clearly the approximation will be on the low side. For the reverse curvature, the ap-proximation will be on the high side.
684 Appendix C Selected Topics of Mathematics III. Parabolic [Figure (d)] The area between the chord and the curve (neglected in the trapezoidal solution) may be accounted for by approximating the function by a parabola passing through the points defined by three successive values of y. This area may be calculated from the geometry of the parabola and added to the trapezoidal area of the pair of strips to give the area A of the pair as cited. Adding all of the A’s produces the tabulation shown, which is known as Simpson’s rule.
To use Simpson’s rule, the number n of strips must be even.
Example: Determine the area under the curve from x 0 to x 2. (An integrable function is chosen here so that the three approximations can be compared with the exact value, which is 1) 3.393 447).
NUMBER OF AREA APPROXIMATIONS SUBINTERVALS RECTANGULAR TRAPEZOIDAL PARABOLIC 4 3.361 704 3.456 731 3.392 214 10 3.388 399 3.403 536 3.393 420 50 3.393 245 3.393 850 3.393 447 100 3.393 396 3.393 547 3.393 447 1000 3.393 446 3.393 448 3.393 447 2500 3.393 447 3.393 447 3.393 447 1 3(55 1 3(1 x2)3/22 0 A 2 0 x 1 x2 dx y x 1 x2 yi + 1 yi (c) Parabolic Trapezoidal yi + yi + 1 –——––– 2 Ai = x y0 — + y1 + y2 + … + yn – 1 + 2 yn — 2 A = y dx ( ) x x A x yi (d) yi + 1 yi + 2 1 –– 3 A = x ( yi + 4yi + 1 + yi + 2) A = y dx ≅1 –– 3 (y0 + 4y1 + 2y2 + 4y3 + 2y4 + … + 2yn – 2 + 4yn – 1 + yn) x x Δ Δ Δ Δ Δ Δ Δ Δ Δ Ai Note that the worst approximation error is less than 2 percent, even with only four strips.
2. Integration of first-order ordinary differential equations.
The application of the fundamental principles of mechanics frequently results in differential relationships. Let us consider the first-order form dy/dt ƒ(t), where the function ƒ(t) may not be readily integrable or may be known only in tabular form. We may numerically integrate by means of a simple slope-projection technique, known as Euler integra-tion, which is illustrated in the figure.
Article C/12 Selected Techniques for Numerical Integration 685 Beginning at t1, at which the value y1 is known, we project the slope over a horizontal subinterval or step (t2 t1) and see that y2 y1 ƒ(t1)(t2 t1). At t2, the process may be repeated beginning at y2, and so forth until the desired value of t is reached. Hence, the general expres-sion is If y versus t were linear, i.e., if ƒ(t) were constant, the method would be exact, and there would be no need for a numerical approach in that case. Changes in the slope over the subinterval introduce error. For the case shown in the figure, the estimate y2 is clearly less than the true value of the function y(t) at t2. More accurate integration techniques (such as Runge-Kutta methods) take into account changes in the slope over the subinterval and thus provide better results.
As with the area-determination techniques, experience is helpful in the selection of a subinterval or step size when dealing with analytical functions. As a rough rule, one begins with a relatively large step size and then steadily decreases the step size until the corresponding changes in the integrated result are much smaller than the desired ac-curacy. A step size which is too small, however, can result in increased error due to a very large number of computer operations. This type of error is generally known as “round-off error,” while the error which re-sults from a large step size is known as algorithm error.
yk1 yk ƒ(tk)(tk1 tk) t1 y1 y2 y3 y4 t t2 t3 t4 y(t) y(t) Accumulated algorithmic error etc.
Slope = ƒ(t3) Slope = ƒ(t2) Slope = ƒ(t1) dy –– dt Slope = ƒ(t) Example: For the differential equation dy/dt 5t with the initial condi-tion y 2 when t 0, determine the value of y for t 4.
Application of the Euler integration technique yields the following results: NUMBER OF SUBINTERVALS STEP SIZE y at t 4 PERCENT ERROR 10 0.4 38 9.5 100 0.04 41.6 0.95 500 0.008 41.92 0.19 1000 0.004 41.96 0.10 This simple example may be integrated analytically. The result is y 42 (exactly).
686 Appendix C Selected Topics of Mathematics 687 TABLE D/1 PHYSICAL PROPERTIES Density (kg/m3) and specific weight (lb/ft3) kg/m3 lb/ft3 kg/m3 lb/ft3 Air 1.2062 0.07530 Lead 11 370 710 Aluminum 2 690 168 Mercury 13 570 847 Concrete (av.) 2 400 150 Oil (av.) 900 56 Copper 8 910 556 Steel 7 830 489 Earth (wet, av.) 1 760 110 Titanium 3 080 192 (dry, av.) 1 280 80 Water (fresh) 1 000 62.4 Glass 2 590 162 (salt) 1 030 64 Gold 19 300 1205 Wood (soft pine) 480 30 Ice 900 56 (hard oak) 800 50 Iron (cast) 7 210 450 At 20C (68F) and atmospheric pressure Coefficients of friction (The coefficients in the following table represent typical values under normal working conditions. Actual coefficients for a given situation will depend on the exact nature of the contacting surfaces. A variation of 25 to 100 percent or more from these values could be expected in an actual application, depending on prevail-ing conditions of cleanliness, surface finish, pressure, lubrication, and velocity.) TYPICAL VALUES OF COEFFICIENT OF FRICTION CONTACTING SURFACE STATIC, s KINETIC, k Steel on steel (dry) 0.6 0.4 Steel on steel (greasy) 0.1 0.05 Teflon on steel 0.04 0.04 Steel on babbitt (dry) 0.4 0.3 Steel on babbitt (greasy) 0.1 0.07 Brass on steel (dry) 0.5 0.4 Brake lining on cast iron 0.4 0.3 Rubber tires on smooth pavement (dry) 0.9 0.8 Wire rope on iron pulley (dry) 0.2 0.15 Hemp rope on metal 0.3 0.2 Metal on ice 0.02 D Useful Tables TABLE D/2 SOLAR SYSTEM CONSTANTS Universal gravitational constant G 6.673(1011) m3/ 3.439(108) ft4/(lbf-s4) Mass of Earth me 5.976(1024) kg 4.095(1023) lbf-s2/ft Period of Earth’s rotation (1 sidereal day) 23 h 56 min 4 s 23.9344 h Angular velocity of Earth 0.7292(104) rad/s Mean angular velocity of Earth–Sun line 0.1991(106) rad/s Mean velocity of Earth’s center about Sun 107 200 km/h 66,610 mi/h (kgs2) 688 Appendix D Useful Tables 1 Mean distance to Earth (center-to-center) 2 Diameter of sphere of equal volume, based on a spheroidal Earth with a polar diameter of 12 714 km (7900 mi) and an equatorial diameter of 12 756 km (7926 mi) 3 For nonrotating spherical Earth, equivalent to absolute value at sea level and latitude 37.5 4 Note that Jupiter is not a solid body.
BODY MEAN DISTANCE TO SUN km (mi) ECCENTRICITY OF ORBIT e PERIOD OF ORBIT solar days MEAN DIAMETER km (mi) MASS RELATIVE TO EARTH SURFACE GRAVITATIONAL ACCELERATION m/s2 (ft/s2) ESCAPE VELOCITY km/s (mi/s) Sun Moon Mercury Venus Earth Mars Jupiter4 — 384 3981 (238 854)1 57.3 106 (35.6 106) 108 106 (67.2 106) 149.6 106 (92.96 106) 227.9 106 (141.6 106) 778 106 (483 106) — 0.055 0.206 0.0068 0.0167 0.093 0.0489 — 27.32 87.97 224.70 365.26 686.98 4333 1 392 000 (865 000) 3 476 (2 160) 5 000 (3 100) 12 400 (7 700) 12 7422 (7 918)2 6 788 (4 218) 139 822 (86 884) 333 000 0.0123 0.054 0.815 1.000 0.107 317.8 274 (898) 1.62 (5.32) 3.47 (11.4) 8.44 (27.7) 9.8213 (32.22)3 3.73 (12.3) 24.79 (81.3) 616 (383) 2.37 (1.47) 4.17 (2.59) 10.24 (6.36) 11.18 (6.95) 5.03 (3.13) 59.5 (36.8) TABLE D/3 PROPERTIES OF PLANE FIGURES Arc Segment Circular Area Semicircular Area Quarter-Circular Area Quarter and Semicircular Arcs — — — – r – y C C C r r r α α – r = r sin –––––– α – y = 2r –– – y = 4r –– 3 – y = – x = 4r –– 3 Ix = Iy = r4 ––– 4 α Area of Circular Sector – x r α α C r r – y y y C x x r4 ––– 2 Iz = r4 ––– 4 Iz = Ix = Iy = r4 ––– 8 – y – x x x C y y C – – 8 – Ix = ( ) 8 –– 9 r4 r4 ––– 8 Iz = Ix = Iy = r4 ––– 16 — 16 – Ix = Iy =( ) 4 –– 9 r4 α – x = r sin –––––– α 2 – 3 1 – 2 1 – 2 1 – 2 r4 –– 4 Ix = ( – sin 2 ) α r4 –– 4 Iy = ( + sin 2 ) α Iz = α α α r4 FIGURE CENTROID AREA MOMENTS OF INERTIA – – Useful Tables 689 690 Appendix D Useful Tables TABLE D/3 PROPERTIES OF PLANE FIGURES Continued Rectangular Area Triangular Area Area of Elliptical Quadrant Subparabolic Area – x = – x = 4a –– 3 – y = 4b –– 3 – x = 3a –– 8 – y = 3b –– 5 – x = 3a –– 4 – y = 3b –– 10 y x bh3 ––– 3 a + b –––– 3 – y = h –– 3 bh3 ––– 12 Ix = x x x1 x y x0 y0 y – Ix = FIGURE CENTROID AREA MOMENTS OF INERTIA h h C C b b a — bh –– 12 – Iz = (b2 + h2) ab3 ––––, 16 –– 16 – ( ) 4 –– 9 –– 16 – ( ) 4 –– 9 ab3 Ix = – Ix = a3b a3b ––––, 16 Iy = – Iy = Iz = ab ––– 16 (a2 + b2) ab3 ––– 21 Ix = a3b ––– 5 Iy = a3 –– + 5 ( ) b2 –– 21 Iz = ab 2ab3 –––– 7 Ix = 2a3b –––– 15 Iy = a2 –– + 15 ( ) b2 –– 7 Iz = 2ab bh3 ––– 12 bh3 ––– 36 Ix = bh3 ––– 4 Ix1 = – Ix = – y – x C b a b a – y – x – y – x C y = kx2 = x2 b — a2 ab –– 3 Area A = Parabolic Area x y b a – y – x C y = kx2 = x2 b — a2 2ab ––– 3 Area A = Useful Tables 691 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS (m mass of body shown) MASS MASS MOMENTS BODY CENTER OF INERTIA Circular Cylindrical — Shell Half Cylindrical Shell Circular Cylinder — Semicylinder Rectangular Parallelepiped — Iy2 y2 1 3 m(b2 l2) Iy1 y1 1 12 mb2 1 3 ml2 Izz 1 12 m(a2 b2) Iyy 1 12 m(b2 l2) Ixx 1 12 m(a2 l2) Izz 1 2 16 92 mr2 Izz 1 2 mr2 1 4 mr2 1 3 ml2 x 4r 3 Ix1x1 Iy1y1 1 4 mr2 1 12 ml2 Ixx Iyy Izz 1 2 mr2 Ix1x1 1 4 mr2 1 3 ml2 Ixx 1 4 mr2 1 12 ml2 Izz 1 4 2 mr2 Izz mr2 1 2 mr2 1 3 ml2 x 2r Ix1x1 Iy1 y1 1 2 mr2 1 12 ml2 Ixx Iyy Izz mr2 Ix1x1 1 2 mr2 1 3 ml2 Ixx 1 2 mr2 1 12 ml2 – – x1 z x y G l – 2 l – 2 r x1 z x y G l – 2 l – 2 r y1 x1 z x G l – 2 l – 2 r x1 z x y G l – 2 l – 2 r y1 z x y y2 y1 G l – 2 l – 2 b a 692 Appendix D Useful Tables TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m mass of body shown) MASS MASS MOMENTS BODY CENTER OF INERTIA Spherical — Shell Hemispherical Shell Sphere — Hemisphere Uniform Slender Rod — Iy1 y1 1 3 ml2 Iyy 1 12 ml2 Iyy Izz 83 320mr2 x 3r 8 Ixx Iyy Izz 2 5mr2 Izz 2 5mr2 Iyy Izz 5 12mr2 x r 2 Ixx Iyy Izz 2 3mr2 Izz 2 3mr2 – – – – r G z z x G r y r G z z x G r y y G l – 2 l – 2 y1 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m mass of body shown) Useful Tables 693 MASS MASS MOMENTS BODY CENTER OF INERTIA Quarter-Circular Rod Elliptical Cylinder — Conical Shell Half Conical Shell Right-Circular Cone Iyy 3 20 mr2 3 80 mh2 Izz 3 10 mr2 Iy1y1 3 20 mr2 1 10 mh2 z 3h 4 Iyy 3 20 mr2 3 5 mh2 Izz 1 2 16 92 mr2 Izz 1 2 mr2 z 2h 3 1 4 mr2 1 6 mh2 Ix1x1 Iy1y1 x 4r 3 1 4 mr2 1 2 mh2 Ixx Iyy Iyy 1 4mr2 1 18mh2 Izz 1 2mr2 Iy1y1 1 4mr2 1 6mh2 z 2h 3 Iyy 1 4mr2 1 2mh2 Iy1y1 1 4mb2 1 3ml2 Izz 1 4m(a2 b2) Iyy 1 4mb2 1 12ml2 Ixx 1 4ma2 1 12ml2 Izz mr2 2r Ixx Iyy 1 2mr2 x y – – – y _ y _ x z x r G z x G l – 2 l – 2 y1 y b a z y1 y G h r y1 y h z x r x1 G z y1 y G h r TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m mass of body shown) 694 Appendix D Useful Tables MASS MASS MOMENTS BODY CENTER OF INERTIA Half Cone Semiellipsoid Elliptic Paraboloid Rectangular Tetrahedron Half Torus Izz mR2 3 4ma2 Ixx Iyy 1 2mR2 5 8ma2 x a2 4R2 2R Izz 3 80m(a2 b2) Iyy 3 80m(a2 c2) z c 4 Ixx 3 80m(b2 c2) y b 4 Izz 1 10m(a2 b2) x a 4 Iyy 1 10m(a2 c2) Ixx 1 10m(b2 c2) Iyy 1 6m(a2 1 3c2) Ixx 1 6m(b2 1 3c2) Izz 1 6m(a2 b2) z 2c 3 Iyy 1 6ma2 1 2mc2 Ixx 1 6mb2 1 2mc2 Iyy 1 5m(a2 19 64c2) Ixx 1 5m(b2 19 64c2) Izz 1 5m(a2 b2) z 3c 8 Iyy 1 5m(a2 c2) Ixx 1 5m(b2 c2) Izz 3 10 1 2 mr2 Izz 3 10 mr2 3 20 mr2 1 10 mh2 z 3h 4 Ix1x1 Iy1y1 x r 3 20 mr2 3 5 mh2 Ixx Iyy – – – – – – – – y1 y h z x r x1 G y z G x a b c x2 — a2 y2 — b2 + z2 — c2 + = 1 x z G y b a c x2 — a2 y2 — b2 + z – c = x y b a c z G y z x G R R a 695 Absolute measurements, 5, 119 Absolute motion, 22, 88, 338, 349, 372 Absolute system of units, 7, 120 Acceleration: absolute, 89, 119, 372, 388 angular, 328, 330, 517 average, 23, 41 constant, 25, 328 Coriolis, 119, 388 cylindrical components of, 79 function of displacement, 26 function of time, 25 function of velocity, 26 graphical determination of, 24 due to gravity, 9, 10, 120 instantaneous, 23, 42 normal components of, 55 polar components of, 67 rectangular components of, 43, 79 relative to rotating axes, 387, 528 relative to translating axes, 89, 244, 372, 527 spherical components of, 80 tangential component of, 55 vector representation of, 42 from work-energy principle, 477 Acceleration-displacement diagram, 24 Acceleration-time diagram, 24 Accelerometer, 606 Action and reaction, principle of, 6 Active-force diagram, 160, 412, 461 Addition of vectors, 5, 674 Amplitude ratio, 602, 603 Amplitude of vibration, 586 Angular acceleration, 328, 330, 517 Angular displacement, 327 Angular impulse, 206, 487 Angular momentum: applied to fluid streams, 290 conservation of, 208, 230, 276, 489 of a particle, 205, 246 relative, 246, 274 of a rigid body, 413, 487, 539 of a system, 272 units of, 205 vector representation of, 205 Angular motion: of a line, 327 vector representation of, 329, 514 Angular velocity, 327, 328, 514, 517 absolute, 542, 372 of the earth, 119, 688 vector representation of, 329, 516, 517 Apogee velocity, 234 Area moments of inertia, 639 Associative law, 674 Astronomical frame of reference, 4, 118, 244 Axes: rotating, 385, 528 translating, 88, 244, 348, 527 Balancing in rotation, 552 Base units, 7, 120 Bodies, interconnected, 270, 418, 477, 488 Body, rigid, 5, 267, 326, 411, 514 Body centrode, 363 Body cone, 517, 565 Cajori, F., 4 Center: of curvature, 55 of mass, motion of, 269 of percussion, 431 Central-force motion, 230 Index Centrifugal force, 245 Centrode, body and space, 363 Centroids, table of, 689 Circular frequency, natural, 586 Circular motion, 56, 68 Coefficient: of friction, 687 of restitution, 218, 490 of viscous damping, 587 Commutative law, 674, 675 Complementary solution, 601 Computer-oriented problems, 14, 112, 262, 407, 509, 636, 669 Cone, body and space, 517, 565 Conservation: of energy, 178, 275, 625 of momentum, 193, 208, 218, 230, 276, 489, 564 Conservative force, 178 Conservative system, 275, 462, 625 Constant of gravitation, 8, 688 Constrained motion, 22, 98, 123, 417, 444 Constraint, equations of, 98, 478 Coordinates: cartesian, 22 choice of, 22, 43, 88, 106, 122, 123, 642 cylindrical, 79 normal and tangential, 54 polar, 66 rectangular, 43, 79 rotating, 385, 528 spherical, 80 transformation of, 81, 107 translating, 88, 244, 348, 527 Coriolis, G., 4, 388 Coriolis acceleration, 119, 388 Couple: gyroscopic, 560 resultant, 477, 551 work of, 459 Critical frequency, 602 Cross or vector product, 205, 329, 385, 675 Curvature: center of, 55 radius of, 54, 680 Curvilinear motion: in cylindrical coordinates, 79 in normal and tangential coordinates, 54, 138 of a particle, 40, 79, 138 in polar coordinates, 66, 138 in rectangular coordinates, 43, 79, 138 in spherical coordinates, 80 Curvilinear translation, 326, 421, 514 D’Alembert, J., 4, 245 D’Alembert’s principle, 244 Damped forced vibration, 602 Damped free vibration, 587 Damping: coefficient, 587 critical, 588 ratio, 588 viscous or fluid, 587 Dashpot, 587 Degrees of freedom, 98, 123, 584 Densities, table of, 687 Derivative: table of, 677 transformation of, 387, 529 of a vector, 41, 676 Descartes, R., 22 Diagram: acceleration-displacement, 24 acceleration-time, 24 active-force, 160, 412, 461 displacement-time, 24 force-displacement, 155, 157 force-time, 193 free-body, 14, 123, 413, 461, 584 impulse-momentum, 192, 487 kinetic, 413, 414, 415, 416, 421, 431, 443 velocity-displacement, 24 velocity-time, 24 Dimensions, homogeneity of, 11 Direction cosines, 674 Discrete or lumped-parameter model, 583 Displacement: angular, 327 in curvilinear motion, 40 graphical determination of, 24 linear, 22 virtual, 478 Displacement meter, 606 Displacement-time diagram, 24 Distance, 40 Distributed-parameter system, 583 Distributive law, 675 Dot or scalar product, 154, 155, 159, 675 Dynamical energy, conservation of, 178, 275 Dynamic balance in rotation, 552 Dynamic equilibrium, 245 Dynamics, 3 Earth, angular velocity of, 119, 688 Earth satellites, equations of motion for, 230 Efficiency, 161 Einstein, A., 4, 120 Elastic impact, 219 Elastic potential energy, 176, 461 Electric circuit analogy, 606 Energy: conservation of, 178, 275, 625 kinetic, 159, 269, 460, 542 potential, 175, 461, 624 in satellite motion, 233 in vibration, 624 Equations of constraint, 98, 478 Equations of motion: for fixed-axis rotation, 431 for particles, 125, 138 for plane motion, 414, 443 in polar coordinates, 138 for rectilinear and curvilinear translation, 421 for a rigid body, 413, 550 696 Index for rotation about a point, 515 for a system of particles, 269 Equilibrium, dynamic, 245 Euler, L., 4, 551 Euler’s equations, 551 Fluid damping, 587 Fluid streams, momentum equations for, 289, 290 Foot, 7 Force: centrifugal, 245 concept of, 5 conservative, 178 external, 269 inertia, 245 internal, 268 gravitational, 8, 11, 121 resultant, 6, 124, 159, 191, 269, 413, 477, 551 units of, 7 work of, 154, 459 Force-displacement diagram, 155, 157 Forced vibration, 600, 614 damped, 602 equation for, 601 frequency ratio for, 602 magnification factor for, 602, 603 resonant frequency of, 602 steady-state, 602, 603 undamped, 601 Force field, conservative, 178 Force-time diagram, 193 Forcing functions, 600 Formulation of problems, 12 Frame of reference, 6, 88, 244, 246, 385, 527, 528 Free-body diagram, 14, 123, 413, 461, 584 Freedom, degrees of, 98, 123, 584 Free vibration: damped, 587 energy solution for, 624 equations for, 584, 587 undamped, 584 vector representation of, 586 Frequency: critical, 602 damped, 590 natural and circular, 586 Frequency ratio, 602 Friction: coefficients of, 687 work of, 462 Galileo, 3 Gradient, 179 Graphical representation, 14, 24, 155, 328, 350, 373, 586 Gravitation: constant of, 8, 688 law of, 8 Gravitational force, 8, 11, 121 Gravitational potential energy, 175, 461 Gravitational system of units, 7, 120 Gravity: acceleration due to, 9, 10, 120 International Formula for, 10 Gyration, radius of, 643 Gyroscope, 558 Gyroscopic couple, 560 Gyroscopic motion, equation of, 559 Harmonic motion, simple, 29, 585 Hertz (unit), 586 Hodograph, 42 Horsepower, 161 Huygens, C., 3 Imbalance, rotational, 552 Impact, 217, 490 classical theory of, 219 direct central, 217 elastic, 219 energy loss in, 219 inelastic or plastic, 219 oblique, 219 Impulse: angular, 206, 487 linear, 192, 486 Impulse-momentum diagram, 192, 487 Impulse-momentum equation, 192, 206, 246 Inertia, 5, 118 area moments of, see Moments of inertia of area mass moments of, see Moments of inertia of mass principal axes of, 541, 661 products of, 540, 660 Inertia force, 245 Inertial system, 4, 88, 89, 118, 119 Inertia tensor or matrix, 541, 661 Instantaneous axis of rotation, 362, 516 Instantaneous center of zero velocity, 362 Integrals, table of selected, 678 Integration, numerical techniques for, 683, 685 of vectors, 677 Interconnected bodies, 270, 418, 477, 488 International Gravity Formula, 10 International System of units, 6 Joule (unit), 155 Kepler, J., 230 Kepler’s laws of motion, 230, 232 Kilogram, 7, 120 Kinematics, 3, 21, 325, 514 of angular motion, 328, 514 of curvilinear motion, 40 of rectilinear motion, 22 of relative motion, 22, 88, 244, 348, 372, 385, 527 of rigid bodies, 325, 514 Kinetic diagram, 413, 414, 415, 416, 421, 431, 443 Kinetic energy: of a particle, 159 of plane motion, 460 of rotation, 460, 542 of space motion, 543 Index 697 Kinetic energy: (continued) of a system of particles, 270, 542 of translation, 460 units of, 159 Kinetic friction, coefficient of, 687 Kinetics, 3, 21, 117, 267, 411, 539 of particles, 117 of rigid bodies, in plane motion, 411, 552 in rotation, 431, 539 in space motion, 539 Kinetic system of units, 120 Lagrange, J. L., 4 Lagrange’s equations, 419 Laplace, P., 4 Law: associative, 674 commutative, 674, 675 of conservation of dynamical energy, 178, 275, 625 distributive, 675 of gravitation, 8 Laws of motion: Kepler’s, 230, 232 Newton’s, 6, 117, 244, 268, 486 Light, speed of, 4, 120 Line, angular motion of, 327 Linear displacement, 22 Linear impulse, 192, 486 Linear momentum: applied to fluid streams, 289 conservation of, 193, 218, 276, 489 moment of, 205 of a particle, 191 relative, 246 of a rigid body, 486 of a system, 271 Logarithmic decrement, 590 Lumped-parameter or discrete model, 583 Magnification factor, 602, 603 Mass, 5, 117 steady flow of, 288 unit of, 7, 120 variable, 303 Mass center, motion of, 269 Mass flow, equations of motion for, 289, 290 Mass moments of inertia, see Moments of inertia of mass Mathematical model, 12 Mathematics, selected topics in, 671 Matrix, inertia, 541, 661 Measurements: absolute, 5, 119 relative, 88, 244, 246, 348, 372, 385, 527 Mechanics, 3 Meter, 7 Metric units, 6, 120 Moment center, choice of, 416, 421, 443 Moment equation of motion, 206, 272, 273, 274, 275, 414, 416, 417, 550 Moment of linear momentum, 205 Moments of inertia of area, 639, 644, 689 Moments of inertia of mass, 414, 539, 641 choice of element of integration for, 642 for composite bodies, 645 about any prescribed axis, 661 principal axes for, 541, 661 radius of gyration for, 643 table of, 691 transfer of axes for, 643 Momentum: angular, 205, 246, 272, 290, 487, 539 conservation of, 193, 208, 218, 230, 276, 489, 564 equations for mass flow, 289, 290 linear, 191, 271, 486 moment of, 205 rate of change of, 6, 191, 206, 271, 272, 273, 274, 275, 486, 487, 488, 550 vector representation of, 191, 205, 486, 487, 542 Motion: absolute, 22, 88, 338, 349, 372 angular, 327, 329, 514, 515, 517, 518 central-force, 230 circular, 56, 68 constrained, 22, 98, 123, 417, 444 curvilinear, 40, 79, 138 in cylindrical coordinates, 79 general space, 527 graphical representation of, 14, 24, 155, 328, 350, 373, 586 gyroscopic, 558 of mass center, 269 Newton’s laws of, 6, 117, 244, 268, 486 in normal and tangential coordinates, 54 parallel-plane, 515, 552 plane, 22, 40, 326, 411, 443 planetary and satellite, 230 in polar coordinates, 66 in rectangular coordinates, 43, 79 rectilinear, 22, 124 relative, 22, 88, 244, 348, 372, 385, 527 rotational, 327, 431, 514, 515 simple harmonic, 29, 585 in spherical coordinates, 80 of a system of particles, 267 unconstrained, 22, 123, 444 Natural frequency, 586 Newton, Isaac, 3, 4 Newton (unit), 7 Newtonian frame of reference, 246, 268, 272 Newtonian mechanics, 120, 268 Newton’s laws, 6, 117, 244, 268, 486 Newton’s method, 681 Notation for vectors, 5, 41, 674 Numerical integration, 683, 685 Nutation, 561 Oblique central impact, 219 Orbit, elliptical, 232 Osculating plane, 22 Parallel-axis theorems, for mass moments of inertia, 643 Parallel-plane motion, 515, 552 698 Index Particles, 5, 21 curvilinear motion of, 40, 79, 138 equations of motion of, 125, 138 kinematics of, 21 kinetics of, 117 motion of system of, 267 Particle vibration, 584 Particular solution, 601, 603 Path variables, 22, 55 Percussion, center of, 431 Perigee velocity, 234 Period: of orbital motion, 232, 235 of vibration, 586, 590 Phase angle, 603 Plane motion, 22, 40, 326, 411, 443 curvilinear, 40 equations of motion for, 414, 443 general, 326, 443 kinematics of, 40, 88, 326 kinetic energy of, 460 kinetics of, 411, 552 Planetary motion: Kepler’s laws of, 230, 232 period of, 232 Poinsot, L., 4 Polar moment of inertia, 644 Position vector, 41 Potential energy, 175, 461, 624 Potential function, 178 Pound force, 7, 120, 122 Pound mass, 8, 121, 122 Power, 160, 462 Precession: defined, 517, 558 direct and retrograde, 565 steady, 558, 560, 563 velocity of, 558 with zero moment, 564 Primary inertial system, 4, 88, 118 Principal axes of inertia, 541, 661 Principia, 4 Principle: of action and reaction, 6 of conservation of momentum, 193, 208, 218, 230, 276, 489, 564 D’Alembert’s, 245 of motion of mass center, 269 Products of inertia, 540, 660 Products of vectors, 154, 205, 675 Projectile motion, 44 Propulsion, rocket, 305 Radius: of curvature, 54, 680 of gyration, 643 Rectilinear motion of a particle, 22, 124 Rectilinear translation, 326, 420, 514 Reference frame, 6, 88, 244, 246, 385, 527, 528 Relative acceleration, rotating axes, 387, 528 translating axes, 89, 244, 372, 527 Relative angular momentum, 246, 274 Relative linear momentum, 246 Relative motion, 22, 88, 244, 348, 372, 385, 527 Relative velocity: rotating axes, 386, 528 translating axes, 89, 348, 527 Relativity, theory of, 120 Resonance, 602 Restitution, coefficient of, 218, 490 Resultant: couple, 477, 551 force, 6, 124, 159, 191, 269, 413, 477, 551 Right-hand rule, 205, 675 Rigid bodies: kinematics of, 325, 514 kinetics of, 411, 552 Rigid body, 5, 267, 326, 411, 514 Rigid-body motion, general moment equations for, 413, 416, 417 Rigid-body vibration, 614 Rocket propulsion, 305 Rotating axes, 385, 528 Rotation: equations of motion for, 417, 431 finite, 515 fixed-axis, 326, 329, 431, 514 fixed-point, 515 infinitesimal, 516 instantaneous axis of, 362, 516 kinematics of, 327, 329, 517 kinetic energy of, 460, 542 of a line, 327 of a rigid body, 326, 431, 514, 515 Rotational imbalance, 552 Satellite, motion of, 230 Scalar, 5 Scalar or dot product, 154, 155, 159, 675 Second, 7 Series, selected expansions, 677 Simple harmonic motion, 29, 585 SI units, 6, 7 Slug, 7, 120, 121 Solar system constants, 688 Solution, method of, 12 Space, 4 Space centrode, 363 Space cone, 517, 565 Space motion, general, 527 Speed, 41 Spin axis, 558 Spin velocity, 558 Spring: constant or stiffness of, 156, 584, 625 potential energy of, 176 work done by, 156 Standard conditions, 10, 120 Static friction, coefficient of, 687 Steady mass flow, force and moment equations for, 289, 290 Steady-state vibration, 602, 603 Subtraction of vectors, 674 Index 699 System: conservative, 275, 462, 625 of interconnected bodies, 270, 418, 477, 488 of particles: angular momentum of, 272 equation of motion for, 269 kinetic energy of, 270, 542 linear momentum of, 271 of units, 6, 8, 120 Table: of area moments of inertia, 689 of centroids, 689 of coefficients of friction, 687 of densities, 687 of derivatives, 677 of integrals, 678 of mass centers, 691 of mass moments of inertia, 691 of mathematical relations, 671 of solar-system constants, 688 of units, 7 Tensor, inertia, 541, 661 Theory of relativity, 120 Thrust, rocket, 305 Time, 5, 7, 120 Time derivative, transformation of, 387, 529 Transfer of axes: for moments of inertia, 643 for products of inertia, 660 Transformation of derivative, 387, 529 Transient solution, 602, 603 Translating axes, 88, 244, 348, 527 Translation, rectilinear and curvilinear, 326, 420, 514 Triple scalar product, 543, 676 Triple vector product, 676 Two-body problem: perturbed, 235 restricted, 235 Unconstrained motion, 22, 123, 444 Units, 6, 8, 120 kinetic system of, 120 Unit vectors, 43, 54, 66 derivative of, 54, 66, 385, 528 U.S. customary units, 6, 8, 120 Variable mass, force equation of, 304 Vectors, 5, 674 addition of, 5, 674 cross or vector product of, 205, 329, 385, 675 derivative of, 41, 676 dot or scalar product of, 154, 155, 159, 675 integration of, 677 notation for, 5, 41, 674 subtraction of, 674 triple scalar product of, 676 triple vector product of, 676 unit, 43, 54, 66, 79, 674 Velocity: absolute, 88 angular, 327, 328, 514, 517 average, 23, 40 cylindrical components of, 80 defined, 23, 41 graphical determination of, 24 instantaneous, 23, 41 instantaneous axis or center of, 362 in planetary motion, 234 polar components of, 66 rectangular components of, 43, 79 relative to rotating axes, 386, 527 relative to translating axes, 89, 348 spherical components of, 80 tangential component of, 54 vector representation of, 41 Velocity-displacement diagram, 24 Velocity-time diagram, 24 Vibration: amplitude of, 586 damped, 587 energy in, 624 forced, 600, 614 free, 584 frequency of, 586, 589 over- and underdamped, 588, 589 period of, 586, 590 reduction of, 603 simple harmonic, 585 steady-state, 602, 603 transient, 602, 603 work-energy solution for, 624 Virtual displacement, 478 Virtual work, 154, 478 Viscous damping coefficient, 587 Watt, 161 Weight, 7, 8, 11, 121 Work, 154, 459 of a constant force, 156 of a couple, 459 an exact differential, 178 examples of, 155 of a force, 154, 459 of friction, 462 graphical representation of, 155 of a spring force, 156 units of, 155 virtual, 154, 478 of weight, 157 Work-energy equation, 160, 177, 246, 270, 461, 552 700 Index 701 Problem Answers Chapter 1 1/1 1/2 1/3 1/4 1/5 (a) (b) 1/6 1/7 1/8 1/9 1/10 1/11 022 or 432 348 km 1/12 1/13 On earth: On moon: 1/14 1/15 [MLT1] [MLT1] RB 2.21 RA 2.19, Rem 0.001677 Rem 286 000 Res 1656 d 346 1.770 Wh 186.0 lb gh 29.9 ft/sec2, Wrel 882.2 N Wabs 883.9 N, h 0.414R F (5.73i 3.31j)109 N 3.14(1011)i N 1.255(1010)i N 27, 1.392i 18j, 19.39i 6j, 178.7k, 21.5 m 0.01294 slugs 0.1888 kg, W 1.853 N W 14 720 N 3310 lb, m 102.8 slugs W 801 N 180-lb person: m 5.59 slugs 81.6 kg Chapter 2 2/1 2/2 7.89 sec 2/3 2/4 2/5 2/6 2/7 2/8 2/9 2/10 2/11 2/12 2/13 2/14 2/15 2/16 2/17 (a) (b) 2/18 2/19 2/20 2/21 2/22 2/23 2/24 2/25 2/26 (a) (b) t 0.0555 sec t 0.0370 sec, s 2250 m s 330 m s 713 m v 99.8 mi/hr a 1.168 ft/sec2, t 3.26 s s 3.26 m, v 0.8 m/s s 64 m a 0.5 m/s2, s 1.819 km t 65.5 s, v 25.6 m/s v 21.9 m/s, v 1.25 m/s vav 0.75 m/s, tacc 0.8 sec a 31.2 ft/sec2, v 25 ft/sec, tAC 2.39 sec vB 56.4 ft/sec down t 4.24 sec, h 49.4 ft, t 40.8 s h 2040 m, v2 139.0 ft/sec s 213 ft s 5 3t 15t2 2 3 t3 m v 3 – 30t 2t2 m/s D 1.419 m s 1.248 m, a 3.61g s 24 m a constant D 45 mm, s 27 mm, a 150 mm/s2 a 15 m/s2 v 42 m/s, s 72 m, t 2.11 sec, v 75 m/s (When a problem asks for both a general and a specific result, only the specific result might be listed below.) 702 Problem Answers 2/27 2/28 2/29 2/30 2/31 2/32 2/33 2/34 2/35 2/36 2/37 2/38 2/39 2/40 Particle 1: Particle 2: Particle 3: 2/41 2/42 2/43 2/44 2/45 2/46 2/47 2/48 2/49 (a) (b) 2/50 (a) (b) 2/51 2/52 2/53 (a) (b) 2/54 2/55 2/56 2/57 2/58 2/59 2/60 2/61 2/62 2/63 2/64 2/65 2/66 2/67 2/68 2/69 2/70 2/71 2/72 2.57 ft above B 2/73 right of B 2/74 2/75 2/76 2/77 2/78 2/79 2/80 2/81 2/82 2/83 2/84 2/85 2/86 2/87 2/88 2/89 2/90 2/91 2/92 2/93 2/94 2/95 2/96 2/97 2/98 2/100 2/101 2/102 2/103 2/104 2/105 2/106 an 36.7 ft/sec2, at 20 ft/sec2 tA 8.97 s, tB 8.89 s, 2.50 m tA 8.97 s, tB 9.92 s, 23.9 m 266 m a 0.269 m/s2 v 5.30 ft/sec, an 25.0 ft/sec2 v 71.3 km/h at 8.39 ft/sec2 vA 11.75 m/s, vB 13.46 m/s x 1242 ft, y 62.7 ft 90 2 , 45, 60, 67.5 h 583 ft, tƒ 12.59 sec, d 746 ft vx l 0, vy l g k (v0 sin g k )(1 ekt) g k t y 1 k vy v0 sin g k ekt g k x v0 cos k (1 ekt) vx (v0 cos )ekt, ƒ2 1ƒ1, ƒ2 1 2 Ri 0.667v0 2 g , ti 1.155v0 g s 1.046 km, t 17.75 s h 1.227 m 0.445, h 500 ft 7.76 ft 31.0 34.3 or 53.1 54.7 umax 1.135 m/s, umin 0.744 m/s 20.6 v0 22.4 ft/sec s 455 m, t 13.35 s R 46.4 m, 23.3 1 26.1, 2 80.6 u 14.41 m/s R 2970 m 48.7 or 53.6 v 70.0 ft/sec, s 11.85 ft 5.57 21.7 14.91 50.6, 2.50 ft v0 77.1 ft/sec, 31.1 v0 16.33 ft/sec, 66.8 u 343 m/s Rmax v0 2 g v0 3.67 m/s, d 1.340 m (y 2)3 144x2, v 30 ft/sec t 24.7 sec, h 1.786 mi v 13.45 in./sec, a 26.8 in./sec2 v 24.2 ft/sec, a 25.3 ft/sec2 x 26.6 a 4.47 mm/s2, v 8.94 mm/s, x 63.4 v 6.20i 3.36j m/s, 27.9 aav 5 m/s2, 53.1 vav 20.6 m/s, 76.0 t 105 sec, amax 11.73 ft/sec2 v g c(ect 1) k c b(ebt ect) vƒ 78.5 ft/sec h 120.8 ft, x 0.831 ft D 0.693/k, t 1 kv0 s 1268 m s 1206 m, v v0 kx x v0 k (1ekt), v v0ekt, v [v0 1n c(n1)t]1/(1n) v 4990 ft/sec v 6490 ft/sec, v 12,290 ft/sec v 13,040 ft/sec, s 416 m t 10 s, D 1 2C2 ln 1 C2 C1 v0 2 vmax 18 m/s c 3v0 2 6gym 2ym 3 D 65.3 ft, s 6.67 ft, s5 3.67 ft t 50.8 s K 1.073(103) ft1, t 25.4 sec v 2 K(L D/2) LD s v0 k sin (kt), v v 2 0 ks2 s v0t 1 6kt3, v v0 1 2kt2 s v0 k (1 ekt), v v0ekt c v0 2 2gym ym 2 D 3710 ft s 5810 ft v 178.9 ft/sec t 2.74 sec a 8.72 ft/sec2, t 16.67 s a 6.67 m/s2, t l vmax 35.9 ft/sec v 38.9 km/h v 1.587 in./sec t 0.917 s s 972 ft h 1.934 km vm 120 m/s, Problem Answers 703 2/107 2/108 2/109 2/110 2/111 (a) (b) (c) 2/112 2/113 2/114 2/115 2/116 (a) (b) 2/117 (a) (b) 2/118 2/119 2/120 2/121 , 2/122 (a) (b) (c) 2/123 2/124 2/125 2/126 2/127 2/128 2/129 2/130 2/131 2/133 2/134 2/135 2/136 2/137 2/138 2/139 2/140 2/141 2/142 , 2/143 2/144 2/145 2/146 2/147 v 161.7 ft/sec, ¨ 0.0808 rad/sec 0, ˙ v0 sin d , ¨ 1 d 2v0 2 d cos sin g r d, r ˙ v0 cos , r ¨ v0 2 sin2 d r ¨ 4.62 m/s2, v 960 km/h r ˙ 1.512 m/s, ˙ 0.0495 rad/s v 1200 ft/sec, a 67.0 ft/sec2 ¨ 23.4 rad/sec2 r ¨ 11.69 ft/sec2 r ˙ 1.5 ft/sec, ˙ 4.50 rad/sec 19.44 a 0.272 m/s2 at v 0.377 m/s at 260 a 2K 2R2 r0 2 l ˙ 32.8 mm/s r ¨ 11.52 m/s2, ¨ 0.0813 rad/s2 r ˙ 42.5 m/s, ˙ 0.1403 rad/s v 545 mm/s, a 632 mm/s2 r ¨ 2.07 m/s2, ¨ 1.653 rad/s2 r ˙ 9.31 m/s, ˙ 0.568 rad/s r ˙ 47.7 ft/sec, ˙ 41.0 deg/sec ar 12.80 m/s2, a 8.80 m/s2 xC 22.5 m, yC 22.9 m 1.25 m L 46.1 m 437 mm, an 8.74 mm/s2, at 36.3 mm/s2 18 480 km a 9.39 ft/sec2 tA 10.52 s, tB 10.86 s t 1.2 s: a 19.62 m/s2, x 180 t 0.8 s: a 73.1 m/s2, x 128.1 a 97.3 m/s2, x 168.4 a 38.9 m/s2, x 59.7 a 2g right, x 0 v ˙ 12.65 m/s2 1907 km an 1.838 ft/sec2 aP1 338 m/s2, aP2 1.5 m/s2 41.7 in.
149.7 ft, at 8.75 ft/sec2 142.2 ft, at 6.58 ft/sec2 133.4 ft, v ˙ 0 243 ft, v ˙ 18.47 ft/sec2 v 72 km/h a2 85.4 m/s2 P2: an 80 m/s2, P1: v 2 m/s, a1 50 m/s2 N 3.36 rev/min a 16en 16.10et ft/sec2 a 8.82 ft/sec2 a 17.97 ft/sec2 a 7 ft/sec2, v 27.8(103) km/h a 0.0260 m/s2 v 356 m/s, vA 25.8 m/s, vB 39.6 m/s B 163.0 m 2/148 2/149 2/150 2/151 2/152 2/153 2/154 2/155 2/156 , 2/157 2/158 2/159 2/160 2/161 2/162 2/163 2/164 2/165 2/166 2/167 2/168 2/169 2/170 2/171 2/172 2/173 ¨ 0.0238 rad/s2 R ¨ 20.1 m/s2, ¨ 0 ˙ 0.0731 rad/s R ˙ 92.0 km/h, ˙ 0.1988 rad/s az 0.386 ft/sec2 ar 19.82 ft/sec2, a 2.91 ft/sec2 aP ( l ¨ (l0 l)2)2 4l ˙ 22 h ¨2 vP l ˙2 (l0 l)22 h ˙2 amax r24 16n44z0 2 v u cos sin v u sin , vR u cos cos a 27.5 m/s2 az 32.2 ft/sec2 ax ay 0, vy 85.5 ft/sec, vz 211 ft/sec vx 235 ft/sec, z 2220 ft x 4700 ft, y 1710 ft, 74.6, v ˙ 1.571 m/s2, 8.59 m ¨ 0.660 rad/sec2 ˙ 0.334 rad/sec 31.9, r ¨ 11.35 ft/sec2 r 51.0 ft, r ˙ 91.4 ft/sec, ¨ 1.398(107) rad/sec2 ˙ 3.48(104) rad/sec r ˙ 8910 ft/sec, r ¨ 1.790 ft/sec2 26.6, ˙ 0.06 rad/s, ¨ 0.0518 rad/s2 r 224 m, r ˙ 6.71 m/s, r ¨ 4.59 m/s2 ¨ 1510 rad/s2 r ¨ 315 m/s2, r ˙ 3.58 m/s, ˙ 17.86 rad/s ¨ 9.01(105) rad/s2 ˙ 0.00312 rad/s 43.2, r 21 900 m, r ˙ 73.0 m/s, r ¨ 2.07 m/s2 ˙ att 2b att2 4b , r 2b sin att2 4b, r ˙ att cos att2 4b ¨ 0.0390 rad/s2 vr 96.2 m/s, v 55.6 m/s, ar 10.29 m/s2 0.289j m/s, a 0.328i 0.1086j m/s2 v 0.064i v 0.296 m/s, a 0.345 m/s2, x 225 a A 7.54 m/s2 vA 1.190 m/s, x 125.2 ˙ 6.46 rad/s r 0.256 m, r ˙ 4.72 m/s, 38.7 ¨ 38.5 rad/s2 r ˙ 1.732 m/s, r ¨ 3.33 m/s2 v 69.9 mi/hr v b sin , a b22 h cot3 v 529 m/s, 48.9, a 9.76 m/s2 a 10.75K a 24c2 4bc cos b2 a 8.62 ft/sec2, ¨ 0.01832 rad/sec2 704 Problem Answers 2/174 2/175 2/176 2/177 2/178 2/179 2/180 2/181 2/182 where 2/183 2/184 2/185 west of south 2/186 2/187 2/188 2/189 2/190 2/191 2/192 2/193 2/194 2/195 2/196 2/197 2/198 2/199 2/200 2/201 2/202 2/203 2/204 2/205 (a) (b) 2/206 2/207 2/208 vA 1.8 m/s up, aA 3 m/s2 down vA 0.4 m/s down vA/B 71.5i 47.4j ft/sec v ˙r 0.884 m/s2, r 5660 m vA/B 50i 50j m/s, aA/B 1.25j m/s2 r ¨ 0.637 m/s2, ¨ 1.660(104) rad/s2 33.3, vA/B 73.1i 73.1j ft/sec aB/A 0.733i 29.2j ft/sec2 aB/A 0.787 m/s2, 93.5 vr 924 km/h, vn 354 km/h r ¨ r ˙2, ¨ 2r ˙ ˙/r 55.6 aB 1.389 m/s2 vB 206 km/h, aB 0.457 m/s2 vB 523i 16.67j ft/sec 17.82 east of north) (89.3 mi/hr, vB/W 27.3i 85j mi/hr 18.87 28.7 below normal vB 6.43 m/s 23.8 vC 1.383 knots, 231 0.628j m/s2 aA/B 3.63i vA/B 3.00i 1.999j m/s 7.18, vA 79.4 mi/hr 6.48 north of west) (12.37 mi/hr, vW/R 12.29i 1.396j mi/hr west of south) (22.3 mi/hr, 33.4 vW/R 12.29i 18.60j mi/hr vA/B 1442 km/h, 33.7 aA/B 10.86 ft/sec2 vA/B 15i 22.5j m/s, aA/B 4.5j m/s2 tan1(b/h) ar b ˙2(tan2 sin2 1)e tan sin v h/1 (h/2R)2 vR 0, v R1 (h/2R)2 a R 5.10 m/s2, a 7.64 m/s2, a 0.3 m/s2 vP 2.85 m/s, aP 5.80 m/s2 vA 1.347 m/s, aA 8.41 m/s2 v 2.96 m/s, a 0.672 m/s2 v hu cos 2 4b2 sin2 2 h2 2 vR bu sin 4b2sin2 2 h2 , v u sin x 3 2 r cos vt r , y r sin vt r , z 1 2r cos vt r v c2 K 2l2 sin2 , a K sin K 2l2 4c2 2/209 2/210 2/211 2/212 2/213 2/214 2/215 2/216 2/217 2/218 2/219 2/220 2/221 2/222 2/223 2/224 2/225 2/226 2/227 2/228 2/229 2/230 2/231 2/232 2/233 2/234 2/235 2/236 2/237 2/238 2/239 2/240 2/241 2/242 2/243 2/244 (a) (b) 2/245 vB 46.8 mm/s up a bKK2 420 2 a bK4 40 2 cos2 at 4 m/s2, 129.9 m ¨ 0.0352 rad/s2, an 6.93 m/s2 ˙ 0.325 rad/s r ˙ 15 m/s, r ¨ 4.44 m/s2, 1.499 m, ax 5 m/s2 aA/B 4.58 m/s2, 20.6 west of north (1255, 1193, 0) ft 12.09 an 103 ft/sec2, 63 5 ft ar 53 ft/sec2, ay 53 ft/sec2, at 0 t 53.5 sec vP 2.72 m/s 9.53 km 34.3 ft ˙ 13.86 rad/sec, ¨ 215 rad/sec2 r r0 1 1 ˙ 2 2 t 208 s, h 418 km t1 2.27 sec, t2 8.48 sec (311, 20) ft v 15 ft/sec, vB 61.8 ft/sec v 7.27 m/s vB vA 2x2 h2 x2 h2 aB 11.93 mm/s2 up vB 62.9 mm/s up vA 2vB cos 2 sin 2 vB s 2x x 2s vA ax L2v 2 A (L2 y2)3/2 v 83.8 mm/s vA 2.76 m/s vB/A 1 ft/sec, aB/A 2 ft/sec2, vC 4 ft/sec vA 2x2 h2 x vB vB 3yvA 2y2 b2 vA 4vB 2vC 0, two vA 4vB 0, one v 1.5 m/s up t 200 s h 400 mm 7aA aB 0 aA 2 ft/sec2 up Problem Answers 705 2/246 2/247 2/248 2/249 (a) (b) (c) 2/250 90 at t 0.526 sec at t 0.324 sec ˙max 3.79 rad/s max 110.4 at t 0.802 sec a 30.3 m/s2 a 0 a 30.3 m/s2 v 113.5 ft/sec k 0.00323 ft1, vt 99.8 ft/sec, t 1.473 s, x 0.1178 m aB 7.86 mm/s2 up 3/1 3/2 (a) (b) Crate does not stop 3/3 (a) no motion, (b) down incline 3/4 3/5 3/6 3/7 3/8 up 3/9 up incline 3/10 3/11 (a) , (b) 3/12 (a) (b) 3/13 3/14 3/15 3/16 3/17 3/18 up 3/19 3/20 3/21 (a) (b) right 3/22 Not possible 3/23 3/24 (a) (b) 3/25 3/26 3/27 3/28 3/29 down incline up, 3/30 3/31 3/32 right down, 3/33 down, T 8.21 N aB 2.37 m/s2 T 46.6 N aB 9.32 m/s2 aA 1.364 m/s2 v 2P kgL x 201 m T 105.4 N aB 0.725 m/s2 aA 1.450 m/s2 T 171.3 N k 5 lb/in.
n 66.0% tan1 a g sin g cos aA aB 0.667 m/s2 aA 1.095 m/s2, aB 0.981 m/s2 k 0.555 a 1.390 m/s2 a 0, k 0.429 TA 75.0 N, TB 55.4 N FA 4080 lb F 0.0206 lb a 5.66 m/s2 T1 39,200 lb, T100 392 lb T 1042 N 12 s a 0.513 m/s2 a 0.257 m/s2, a 16.10 ft/sec2 a 6.44 ft/sec2 a g(sin 1 sin 2) a 4.96 m/s2 a 3.58 ft/sec2 su 807 m, sd 751 m n sin a g F 2890 N R 846 N, L 110.4 N; R, L l 0 a 3.45 ft/sec2 t 5.59 s, x 19.58 m t 1.784 s, x 6.24 m 3/34 3/35 (a) (b) left 3/36 Case (a), 3/37 3/38 3/39 (a) (b) 3/40 (a) (b) 3/41 3/42 3/43 No motion 3/44 3/45 3/46 (a) (b) 3/47 3/48 (a) (b) 3/49 3/50 (a) (b) 3/51 3/52 3/53 3/54 3/55 3/56 3/57 (a) (b) 3/58 3/59 3/60 3/61 3/62 3/63 3/64 3/65 3/66 3/67 3/68 F 165.9 N T 1.76 N, F 3.52 N D 45.0 kN, L 274 kN NA 3380 N, NB 1617 N F 920 lb s v0 /kg k cos2 vA 140.7 ft/sec, vB 163.8 ft/sec at 22.0 ft/sec2 an 0.818g, F 2460 lb ˙ 3.37 deg/s NA 241 N vB 54.2 m/s, 1.064 rad/s N 8.63 rev/min s 0.540 N 0.0241 lb P 4 lb (side A) 45.3 R 0.271 lb R 0.25 lb, R 1.173 N, at 7.21 m/s2 v ˙ 16.10 ft/sec2 N 1.374 lb, NA 10.89 N, NB 8.30 N T 16.14 N T 8.52 N, T 138.0 N, a 0.766 m/s2 v0 11.19 km/s P 9.66 aB 0.322 aA 4.83 ft/sec2 P 54 lb: aA aB 0.1789P 3.22 27 P 54 lb: 0 P 27 lb: 5.88, 47.2 v 2100 m/s t 0.0768 s, y 0.01529 m vs 0.327 m/s h 127.4 m h 55.5 m, tan1 1 s 2 a 1.406 m/s2 v 7.43 m/s a 0.714 m/s2 a 0, ax 32.2(14 – 30x), v 14.47 ft/sec Chapter 3 2/251 2/252 2/253 2/254 vmax 10 in./sec at t 0.330 sec, x 2.45 in.
amax 10.76 m/s2 at 0 and t 0 amin 9.03 m/s2 at 44.3 and t 0.237 s 42.2, R 101.3 m sB 150 m t 10 s and 2.52 m/s2 at sB 0, (aA/B)min sB 557 m, (aA/B)max 6.12 m/s2 at t 0 and s B 1264 m, (vA/B)min 10 m/s at t 23.6 s and (vA/B)max 70 m/s at t 47.1 s and 706 Problem Answers 3/69 3/70 3/71 3/72 3/73 3/74 Dynamic: Static: 3/75 3/76 3/77 3/78 3/79 3/80 3/81 3/82 3/83 3/84 3/85 3/86 3/87 3/88 3/89 (a) (b) (c) 3/90 3/91 3/92 3/93 3/94 3/95 3/96 3/97 (a) (b) 3/98 3/99 3/100 3/101 3/102 3/103 3/104 3/105 (a) and (b) 3/106 3/107 (a) (b) 3/108 (a) (b) 3/109 (a) (b) 3/110 (a) (b) 3/111 3/112 3/113 3/114 3/115 3/116 3/117 3/118 3/119 At Halfway: 3/120 (a) (b) (c) (d) 3/121 3/122 3/123 (a) (b) (c) 3/124 3/125 (a) (b) , (c) 3/126 3/127 3/128 3/129 3/130 3/131 3/132 3/133 3/134 3/135 3/136 (a) (b) 3/137 (a) (b) (c) 3/138 3/139 3/140 3/141 (a) (b) 3/142 (a) (b) (c) down 3/143 3/144 3/145 W 88.1 lb N m5g kR m (3 – 22) vC 4gR kR2 m (3 – 22) vB 2gR kR2 m (3 – 22) y 0.224 m NE 35.3 N NC 10.19 N NC 77.7 N, 54.2 mm vB 9.40 m/s, v 3.43 m/s, x 48.5 mm v 3.65 ft/sec v 5.46 ft/sec xss 93.2 mm xmax 186.4 mm v 0.496 m/s, v 70.9 mi/hr Pup 35.2 hp, Pdown 3.17 hp P30 5 hp, P60 16 hp 29.4 mm v 1.734 m/s, v 1.889 m/s v 7.80 ft/sec k 8.79 kN/m Pin 36.8 kW 0.1445 m vC 3.59 m/s vA 3.44 m/s v 17.48 ft/sec v 5.30 m/s s 4R 1 k3 NC 7mg NB 4mg, k 5mg(h d) d5 P 0 P 287 W, P 0, v0 6460 m/s v 13.37 ft/sec P 2670 hp P 6530 hp, P 3270 hp F 61,200 lb, F 6960 N P 136.7 kW B: P 193.4 kW, e 0.892 Q 1620 J R 4.05 kN, R 4.05 kN, P 52.2 N v 7.08 ft/sec Q 903 kJ e 0.764 v 1.881 m/s P 0.393 hp, P 293 W v 6.55 ft/sec v 5.93 ft/sec, x 98.9 mm v 2.56 m/s, s 1.226 m s 1.853 m, s 0.349 m s 0.663 m, P 0.400 hp v 8.10 m/s Q 1.835 J v 2gh R 3340 lb k 974 lb/in.
v 17.18 ft/sec Uƒ 672 ft-lb vB 3.05 m/s U1–2 2.35 J U1–2 60 J, N mg (1 4k2x2)3/2 0 s r 2k ln v0 2 v0 4 r 2g2 rg N 81.6 N, R 38.7 N v 5.52 m/s max 2, T mg(3 sin 3 cos 2) T 2.53 N, R 1.028 N (lower side) r ¨ 1.153 ft/sec2, ¨ 2.72(108) rad/sec2 r ˙ 9620 ft/sec, ˙ 1.133(104) rad/sec F 1562 lb F 2260 lb F 1562 lb, at 10 m/s2, NA NB 4.83 N, T 5.23 N at 10 m/s2, NA NB 2 N, T 2.83 N P 8.62 lb v r00 cosh 0t vr r00 2 (e0t e0t), r r0 2 (e0t e0t) P 2.21 N, N 14.22 N k 1 R2 g , W 99.655 N 4.77 rad/s N 1 4 sg r 2 1 T 2.52 lb, N 0.326 lb (side B) F 4.39 N x 118.8 mm, N 25.3 N R 2.14 1.913 cos N, vB 2.06 m/s v 149.4 ft/sec, vmin 0, vmax 345 ft/sec v L 2 g b, N 2mg Fr 5.89 N, F 10.19 N Fr 4.79 N, F 14.00 N 3000 km, v ˙ 6.00 m/s2 P 27.0 N, Ps 19.62 N 1.414 t 5 s T1 0.0707 0.0354t2 N 0.0707 0.0354t2 0 t 1.414 s T2 0.0707 0.0354t2 N 0.0707 0.0354t2 5.01 rad/s v gr tan Problem Answers 707 3/146 3/147 3/148 3/149 (a) (b) 3/150 3/151 3/152 3/153 3/154 3/155 (a) (b) 3/156 3/157 3/158 3/159 3/160 3/161 3/162 3/163 3/164 3/165 3/166 3/167 3/168 (a) (b) 3/169 3/170 (a) (b) 3/171 3/172 3/173 3/174 3/175 3/176 3/177 3/178 3/179 down 3/180 3/181 (a) (b) 3/182 left 3/183 3/184 3/185 left 3/186 3/187 3/188 (a) (b) right 3/189 3/190 3/191 right 3/192 3/193 3/194 3/195 3/196 3/197 3/198 3/199 3/200 (a) (b) (c) 3/201 3/202 3/203 3/204 3/205 3/206 3/207 3/208 knots 3/209 3/210 3/211 3/212 (a) (b) (c) 3/213 3/214 3/215 3/216 (a) (b) (c) 3/217 3/218 3/219 3/220 3/221 (a) (b) 3/222 3/223 3/224 3/225 3/226 3/227 3/228 3/229 3/230 3/231 (a) (b) 3/232 HO 1 2mgut2k HO 2mv0 3 sin2 cos g HO 0, A: (H ˙O)z 0, B: (H ˙O)z 0.120 kg m2/s2 vB 47 850 m/s, vP 58 980 m/s 0/4, n 3/4 0.1721 rad/s CW d dr 2 r H ˙B 1.113k lb-ft vP 17,723 mi/hr t 15.08 s ˙ 2m1 m1 4m2 v1 L HO 2mrgr, H ˙O 0 HO mr2gr, H ˙O mgr 5v 3L HO mv(ci ak), H ˙ O F(bi aj) (HO)2 34i 0.1333j 19.6k kg m2/s H 389 Nms, M 260 N m T 24 J HO 23.2k kg m2/s G 8.49i 8.49j kg m/s HO 69.3 kgm2/s v2 40.0 mm/s vB mA mB 2gl 1 mA/mB v 9.10 m/s v 7.23 m/s v 3.10 m/s, R 43.0 N, 8.68 Rx 559 lb, Ry 218 lb s r k mA mA mC 2 v 5.20 R 472 lb, a 4660g, d 0.900 in.
v1 0, v3 0.468 m/s, v5 5.30 m/s, v7 0 v 6.61 m/s v 15.62 ft/sec, 50.2 v 17.82 mi/hr, 54.7 v 0.1935 km/h v 0.663 km/h R 12,150 lb aB 195.6 ft/sec2 aA 97.8 ft/sec2, v 13.33 mi/hr v2 1.423 m/s down incline, t 8.25 s v 3.42 m/s t 4 min 33 sec v F0 mb 1 ebt, s F0 mb t 1 b (ebt 1) R 14.96 kN v1 0, v3 2.42 ft/sec, v5 6.44 ft/sec, v7 0 vƒ 0.00264 m/s, Fav 59.5 N v1 0.417 m/s, v3 8.96 m/s x ˙1 2.05 m/s left, x ˙2 0.878 m/s v 190.9 m/s, E 17.18(103) J T 2780 N v4 2.69 m/s v1 0, v 5.86 ft/sec t 12.18 min x ˙1 2.90 m/s right, x ˙2 0.483 m/s v 2.10 m/s R 423 lb vC 1.231 m/s E 2230 ft-lb v 2.44 mi/hr, t 7.73 s vB 1.652 m/s k 0.302 R 568 N E 13 480 J, n 99.9% v2 188.5i 74j 47k m/s F 3.03 kN 1.7 N s v gr 2 4 ˙ 1.045 rad/s v 0.522 m/s k 111.9 N/m, v 1.143 m/s v 1.005 m/s m 0.528 kg, 43.8 (vB)max 0.962 m/s k 155.1 N/m v2 35.1 km/h vm/k v 20.4 mi/hr xmax 105.9 mm, vmax 1.493 m/s, x 27.2 mm vB 26 300 km/h v 3.43 ft/sec v 4.93 m/s P 2.86 lb v 0.331 m/s v 1.641 ft/sec v 3.06 ft/sec, k 86.8 N/m, v 1.371 m/s vB 8.54 ft/sec vA 0.616 m/s, vB 0.924 m/s ˙ 4.22 rad/s k 393 N/m, v 1.370 m/s, ˙ 2.28 rad/s x 0.510 in.
v 3.84 ft/sec, Uƒ 2.36 J, Fav 3.38 N v0 6460 m/s v 1.248 ft/sec 708 Problem Answers 3/233 3/234 3/235 3/236 3/237 3/238 3/239 3/240 3/241 right 3/242 left 3/243 3/244 3/245 3/246 3/247 3/248 3/249 3/251 (a) (b) 3/252 3/253 3/254 3/255 3/256 3/257 (a) (b) 3/258 3/259 at at 3/260 3/261 at at 3/262 3/263 3/264 3/265 3/266 3/267 3/268 3/269 3/270 3/271 (a) (b) 3/272 h 88.0 km vrel 18,306 mi/hr vrel 16,227 mi/hr v 1.987 mi/sec See Prob. 1/14 and its answer v 7569 m/s v 18.51 mi/sec 11.37, 78.6 h2 0.385 m 82.3, 22.3 e h h h , vx g 2 d h h h (vB)x 6.99 m/s, (vB)y 3.84 m/s (vA)x 1.672 m/s, (vA)y 1.649 m/s B 50.2, n 34.6% vB 6.51 m/s A 180 vA 6.83 m/s 2.92(104) B 270, n 44.4% vB 3.22 ft/sec A 63.5 vA 6.73 ft/sec vB 21.7 km/h h 16.25 in.
h 14.98 in., L2 eL1 R 1.613 m vn 1 e 2 n1 v1 e 0.434 e 0.333 x 0.286d x d 3, m 90 kg, v 2.66 m/s, E 2470 J h 10.94 in., h2 7.43 in.
e h2 h 1/4 vA 0.633v, vB 0.733v 1 e 2 v m k v u 4 (1 e)2 m1 m2 e F 107.0 N v1 4.52 m/s left, v2 2.68 m/s v0 4.20 m/s e 0.829, n 31.2% 3.00 rad/s, U 5.34 J 52.9 TB 0.745 lb 2.77 rad/s CCW, 52.1 vr 88,870 ft/sec, v 125,700 ft/sec C: H A 0.714mg3 CCW, H D 1.126mg3 CCW B: HA 0, HD mg3 CCW 3/273 3/274 3/275 3/276 3/277 (a) (b) (c) , (d) 3/278 , 3/279 3/280 3/281 3/282 3/283 3/284 3/285 3/286 3/287 3/288 3/289 3/290 3/291 3/292 3/293 3/294 3/295 3/296 3/297 3/298 3/299 3/300 3/301 3/302 3/303 3/304 3/305 3/307 3/308 3/309 (a) and (b) 3/310 3/311 3/312 3/313 (a) and (b) 3/314 3/315 grel 9.825 – 0.03382 cos 2 m/s2 vA [v0 2 2gl sin 2v0 cos 2gl sin ]1/2 h2 e2h1 Prel 0.1206 hp T0 m(g a0)(3 – 2 cos 0) T 3ma0 sin , T/2 90 N T 112 J a0 16.99 m/s2, R 0 (vrel)max a0m/k xC/T 2.83 m, vrel 2.46 m/s F 194.0 kN F 376 lb P 66.9 kN (HB)rel 1.5k kgm2/s HO 4.5k kgm2/s T 13.5 J, Trel 1.5 J G 9i kg m/s, Grel 3i kgm/s k 0.382 arel k 1 m1 1 m2 rmin 6.49(106) m rmax 6.66(106) m, vp 7890 m/s va 7690 m/s, 5301 s, e 0.01284 a 6572 km (parallel to the x-axis) vA 2370 m/s, vB 1447 m/s h 922,000 mi t 162.5 s v 2940 m/s p 0.0514 rad/s vA R g R H 1 R R H ƒ 658.69 h, nƒ 654.68 h 153.3 3.39 v 148 ft/sec v 302 ft/sec tan e sin 1 e cos 1 h 36 min 25 s, 6 min 4 s vB 10,551 ft/sec, b 17,833 mi vP 1683 m/s, vA 1609 m/s, v 18.35 m/s t 32.9 s t 71.6 sec t 64.6 days hmax 899 mi 90 v 3217 m/s, v 7767 m/s v 10 668 m/s v 10 398 m/s v 7912 m/s v 7544 m/s, ha 32 600 km v 534 m/s vP 3745 mi/hr e 0.01196, 1 h 30 min 46 s Problem Answers 709 3/317 3/318 3/319 3/320 3/321 3/322 3/323 3/324 3/325 3/326 3/327 3/328 3/329 3/330 3/331 3/332 (a) (b) 3/333 left 3/334 mg k d2 2 2(1 k), d 22 vrel 7 3 gl at 14.89 m/s2 at 10.75 m/s2, T 424 N v 6.55 ft/sec, x 0.316 ft, n 0.667 F 2650 lb u 5 2gR, xmin 2R t 2.02 s TB mu2 r 2g sin , TC mu2 r 5g sin rmax 5.29(109) km mgR(5 2 k) k vA 7451 m/s, e 0.0295 R 46.7 N k 3.18 kN/m tan1 a g P 1.936 kW vB 2.87 m/s, vC 1.533 m/s Uƒ 7.54 J 4/1 4/2 4/3 4/4 4/5 4/6 4/7 4/8 4/9 4/10 Mass-center accelerations are identical 4/11 4/12 4/13 4/14 4/15 4/16 4/17 4/18 4/19 4/20 4/21 4/22 4/23 4/24 (both spheres) 4/25 (a) (b) 4/26 4/27 (a) , (b) 4/28 4/29 4/30 4/31 4/32 4/33 Q 0.571gr2 v 3.92 ft/sec vx 2gl, ˙ 2 2g l v m0 m0 2m v0, ˙ v0 b m0 m0 2m v 72.7 km/h v 0.877 m/s ¨ 2Fb mL2 a F 2m Q 2.52 J, Ix 12.87 Ns v 3gr/2 Pmin 9mg , v 4.71 m/s x 0.316 ft, no s (m1 m2)x1 m2l m0 m1 m2 v 0.355 mi/hr, n 95.0% v 0.205 m/s vA 1.015 m/s, vB 1.556 m/s t 2.72 s ˙ 80.7 rad/s t 4mr2 M a 13.42 ft/sec2 HO 3.3k kgm2/s HO 2m(r2 vy)k v 1.137gr, R 2.29mg F 2.92 N a 15.19 m/s2 aC F 2m g sin a 4 m/s2 T 58.3 lb ay 5.19 m/s2 H ˙G 2Fd 7 (2i 3j) HG mvd 7 (72i 24j 28k) H ˙O Fdj r ¨ Fk 7m, T 13mv2, HO mvd(12i 6j 2k) r ˙ v 7(4i 2j 6k) r d 7(i 4j 6k), HG 12 7 mvdk, H ˙ G Fd 7 k r ¨ Fi 7m, T 6mv2, HO 4mvdk, H ˙ O Fdk r ˙ v 7(4i 4j) r d 7(10i 6j), Chapter 4 3/335 3/336 3/337 3/338 3/339 3/340 3/341 3/342 3/343 3/344 3/345 3/346 3/347 3/348 3/349 3/350 3/351 3/352 3/353 3/354 Nmax 2.75 lb at 66.2 vmax 5.69 ft/sec at 50.8 t 0.408 s vmax 1.283 m/s at 17.40 e 0.610, y 1.396 ft t 1.069 s, 30.6 21.7 min 0.622 at 121.9 min 40.8 ˙max 1.072 rad/s at 65.6, t 3.40 s, 663 vB max 3.25 m/s at sB sB0 0.0635 m vA max 3.86 m/s at sB sB0 0.0767 m ˙ 8.090.5 sin cos 1 rad/s, max 53.1 38.7 65.8 r g 20 2 (cosh cos ) P30 3 hp, P60 16 hp, t 205 sec, s 5900 ft v 182.9 mi/hr v1 1.741 m/s, v2 0.201 m/s t 8.50 sec s 2.28 m Fav 428 lb 2.55 in.
710 Problem Answers 4/34 4/35 4/36 4/37 4/38 4/39 4/40 4/41 4/42 4/43 4/44 4/45 4/46 4/47 4/48 4/49 4/50 4/51 4/52 4/53 4/54 4/55 4/56 4/57 4/58 4/59 4/60 4/61 4/62 4/63 4/64 4/65 4/66 4/67 4/68 4/69 4/70 4/71 4/72 (a) (b) 4/73 4/74 4/75 4/76 4/77 4/78 4/79 4/80 4/81 4/82 4/83 (a) (b) 4/84 4/85 4/86 4/87 4/88 (a) (b) (c) 4/89 4/90 4/91 4/92 4/93 4/94 4/95 4/96 4/97 4/98 4/99 4/100 4/101 4/102 4/103 4/104 4/105 4/106 4/107 M 1837 lb-ft a 2.67 m/s2, ¨ 15.40 rad/s2, a 5.33 m/s2 R 3 2(a g)2t2 F 3gx T 21.1 kN, F 12.55 kN P 1 4v2 F 159.8 lb v 13.90 ft/sec at t 231 sec a 5.64 m/s2 at t 60 s, amax 69.3 m/s2 an 4.67 m/s2, at 19.34 m/s2 a 64.4 ft/sec2, ¨ 325 rad/sec2 v u ln m0 m0 mt gt F 812 lb 12.37 rad/sec a 53.7 ft/sec2 v 2gx 3 , a g 3, Q gL2 6 T1 1 2 g x(L x/2) L x x(L x/2) L x , a g1 x(L x/2) (L x)2 , R 1 2 g(L x) R 1 2g(L 3x), T1 gx, Q 1 4gL2 v v0 1 2L/m, x m 1 2v0t m 1 Q ghL h 2 v2 2gh[1 ln (L/h)] v1 2gh ln (L/h) a P m0x, v v0 2 2p ln m0 m0 x, T v2 x 6.18 m P v2 g(h y), R g(L h y) v 13.83 ft/sec a 1.104 m/s2 P 6.75 N, a 1.603 m/s2 P 20.4 N F s ˙2 (L s)s ¨ P 963 lb a 0.498 ft/sec2 R gx v2 P 209 N F (xx ¨ x ˙2) m m0e a g u t P 76.5 N P 1242 lb P 1113 lb, a 4.70 m/s2, m 448 kg/s an 8.31 m/s2, at 21.2 m/s2 m 10.86 Mg u 131.0 m/s d 165.3 mm, D 2.55 m C 100.3 lb 38.2 M Q Qr 4A r2 b2, Qr 4A(r2 b2) Rx 311 lb right, Ry 539 lb down 2.31, ay 1.448 m/s2 v 1 r mg , P mg 2r mg a 14.68 ft/sec2 m 184.3 kg P 0.671 kW P 5.56 kN, R 8.49 kN T 12,610 lb, V 162.3 lb, M 1939 lb-ft p 34.2 lb/in.2, M 461 lb-in.
R mg d2 4 u(u v cos ) F 2.66 lb p 0.1556 lb/in.2 v 56.4 m/s, M 29.8 kNm R 5980 lb F 4.44 lb T 6530 lb R d2 4 B 1 B Av2 (pB pA) p 840 kPa n 0.638 h 18.57 ft Q2 Q 2 (1 cos ) F Av2 sin , Q1 Q 2 (1 cos ) Fx Fy 442 N T 32.6 kN F 202,000 lb T 2.85 kN R 1885 N F 750 N F 4.56 lb 17.22 vc 2gl(1 cos ) (m2/m1)(1 m2/m1) vb/c (1 m1/m2)2gl(1 cos ) Problem Answers 711 Chapter 5 5/1 5/2 5/3 5/4 5/5 5/6 5/7 5/8 5/9 5/10 5/11 5/12 5/13 5/14 5/15 5/16 5/17 (a) (b) (c) 5/18 5/19 5/20 5/21 5/22 5/23 5/24 5/25 5/26 5/27 5/28 5/29 5/30 5/31 5/32 5/33 5/34 5/35 5/36 5/37 5/38 5/39 5/40 5/41 5/42 5/43 5/44 5/45 5/46 5/47 5/48 5/49 5/50 5/52 5/53 5/54 5/55 5/56 2 1.923 rad/s vC vB 2 8 sec2 2 0.825 rad/sec CW ˙ 6.28 cos 0.278 1.939 cos rad/sec a tv2 2r2 CB 6.30 rad/s 1.056 rad/s CW, 0.500 rad/s2 CCW 17.95 rad/sec CW 4 3 rad/sec CCW, 1 rad/sec2 CCW v 2s ˙b2 L2 2bL cos L tan v 1 2 vA tan v x(x/r)2 1 ax e2 sin rh0 x2 h2 x r x2 r2 vO 1.2 m/s, 1.333 rad/s CCW v 7 2u cot 2 OA hv h2 s2 aB 789 mm/s2 down v vO2(1 sin ), a vO 2/r toward O 1.2 rad/sec, vO 3.4 ft/sec 3v 2L1 3xA 2 4L2 v r sin , a r sin r2 cos v 6.28 m/s t 66.7 sec 2ax 4b2 x2 OA vd s2 d2 NB 415 rev/min N 513 rev/min aC 149.6 m/s2 24.6k rad/sec 0 250 rev, 0 187.5 rev 0.596 rad a 0.1965i 0.246j m/s2 v 0.0464i 0.1403j m/s a 3.02i 1.683j m/s2 v 0.223i 0.789j m/s 0.605j m/s2 a 0.757i v 0.374i 0.1905j m/s aC 2(11i 5j) in./sec2 2k rad/sec, 3 2k rad/sec2 r 3 in.
aC 22.5 m/s2 aB 37.5 m/s2 300 rad/s2, t 0.1784 sec 3.95 rad/s2 ˙ 30.4 rad/s, ¨ 346 rad/s2 9 rad 244 rad 0 10.99 rad, t 1.667 s 4.57j m/s2 aA 16.34i vA 1.777i 2.70j m/s N 300 rev b 180.6 mm v 5 m/s, a 50 m/s2 0.411 rad/sec, av 0.344 rad/sec max ¨ max 00 2 at 0 ˙ max 00 at 0 N 33.3 rev 9.16j m/s2 aA 6.42i vA 1.332i 2.19j m/s (h2 b)j aA (b2 h)i vA (hi bj) 0.76j m/s2 aA 0.32i vA 0.32i 0.08j m/s2 4/108 (a) 4/109 4/110 4/111 4/112 R 2g[H (2h2/H)] a (h/H)g, v hg/H v gx, R gL 3 2 x C 4340 N up, D 3840 N down R gx 4L 3x 2(L x) a gx L, (b) T gx1 x L, (c) v gL 712 Problem Answers 5/57 5/58 5/59 5/60 (a) (b) (c) 5/61 5/62 5/63 5/64 5/65 5/66 5/67 5/68 5/69 5/70 5/71 5/72 5/73 5/74 5/75 5/76 5/77 5/78 5/79 5/80 5/81 5/82 5/83 5/84 5/85 5/86 5/87 5/88 5/89 5/90 5/91 5/92 5/93 5/94 5/95 5/96 (a) (b) 5/97 5/98 (a) (b) (c) 5/99 5/100 5/101 5/102 5/103 5/104 5/105 5/106 5/107 5/108 5/109 5/110 5/111 5/112 5/113 5/114 5/115 5/116 5/117 5/118 5/119 5/120 (a) (b) 5/121 5/122 (a) (b) (c) 5/123 (a) (b) (c) 5/124 5/125 5/126 5/127 5/128 5/129 5/130 5/131 5/132 5/133 5/134 5/135 5/136 5/137 5/138 OA 396 rad/s2 CCW 73.6j ft/sec2 aD 265i aA 24i 270j ft/sec2 2l2/r CW AB 3.640 2 CCW, aB 6.82r0 2 up 0.733j ft/sec2, aL j ft/sec2 aD 2i aC 0.267i 3j ft/sec2 (aB)t 2.46 m/s2 left AB 4k rad/s2, aA 1.6i m/s2 AB 2 v 3.67 mi/hr aA 26.6 m/s2 aB 0.0279i m/s2 0.286 rad/s2 CCW, aA 0.653 m/s2 down vO 0.6i m/s, aO 1.8i m/s2 sin1 r R, vO R r aO 4 R2 r2 aA 5 m/s2 aA 6.2 m/s2 aA 4.39 m/s2 aA 0.2 m/s2, d 1.5 ft aC 0.625 ft/sec2 up, 0.0833 rad/sec2 CCW aA 9.58 m/s2, aB 9.09 m/s2 B 600 rev/min B 360 rev/min, 10.73 rad/s CW 1.10 rad/sec CW vA 5.95 m/s vA 0.278 m/s AD 12.5 rad/s CCW, BD 7.5 rad/s CCW vD 4.50 m/s, 7.47 rad/s CCW AB 19.38 rad/sec CW AB 1.414 rad/sec CCW , BD 3.77 rad/sec CW vC vB 2sec2 2 8 CA 0.429k rad/s v 10.71 mi/hr, vs 6.98 ft/sec vD 2.31 m/s, 13.33 rad/s CW 15 rad/s CW, vP 1.897 m/s vE 0.1386 m/s, B 0.289 rad/s CW 60 AB 0.966 CCW, vB 1.414r AB CW, vB 2.58r down BC 2.77 rad/sec CCW vB 0.884 m/s, 3.20 rad/s CCW BD 1.2 rad/s CCW, AD 1.333 rad/s CCW vA 0.707 m/s, vP 1.581 m/s vA 9.04 in./sec, vC 6.99 in./sec vD vP 0 vB 2v, vC v (all right) vA v, u v r CCW l v r CW, vA 0.408 m/s down vA 15j in./sec, vB 75j in./sec vA 20j in./sec, vB 40j in./sec vG 277 mm/s 0.15 m below P, vP 0.3 m/s, vA 0.806 m/s OB 8.59 rad/s CCW 0.5 m below G, vA 1.949 m/s, vB 2.66 m/s 0.5 m above G, vA vB 2.33 m/s vB 3.97 m/s CA 0.429k rad/s AB 1.725 rad/s CCW, BC 4 rad/s CCW 2 1.923 rad/s CCW vC 6.24 ft/sec vD 9 m/s AB 19.38 rad/sec CW 8.59 rad/s CCW 11.55 rad/s CW, vG 1.155 m/s 1.394 rad/s CCW, vA 0.408 m/s down vA 9.04 in./sec, vC 6.99 in./sec BC 3 rad/sec CW AC 0.295 rad/sec CCW 60 AB 0.9660 CCW, vB 2r0 AB 0 CW, vB 2.58r0 down vP 0.900 m/s BC 2.77 rad/sec CCW vB 4.38 m/s, 3.23 rad/s CCW vA/B 1.2(i j) m/s, vP 1.2i 0.8j m/s OA 3.33k rad/s vO 8.49 m/s right, 26.1 rad/s CW vA/B 23.7i 31.0j in./sec vO 0.6 m/s, vB 0.849 m/s 0.375 rad/s CCW AB 0.96 rad/s CCW vA 58.9 mm/s vA 0.7i 0.4j m/s, vP 0.3i m/s 6.65 rad/sec CW vC/D 0.579 m/s vC 1672i 107 257j km/h vB 105 585j km/h, vD 108 929j km/h vA 1672i 107 257j km/h N 45.8 rev/min CW N 45.8 rev/min CCW N 91.7 rev/min CCW vB 1.386i 1.2j m/s 0.1408 rad/sec2 CCW r2 l2 1 1 r2 l2 sin2 3/2 AB r0 2 l sin AB r0 l cos 1 r2 l2 sin2 Problem Answers 713 5/139 5/140 5/141 5/142 5/143 5/144 5/145 5/146 5/147 5/148 5/149 5/150 5/151 5/152 5/153 5/154 5/155 5/156 5/157 5/158 5/159 5/160 5/161 5/162 5/163 (a) (b) 5/164 5/165 5/166 (a) (b) 5/167 5/168 5/169 5/170 5/171 5/172 5/173 5/176 5/177 5/178 5/179 5/180 (a) (b) 5/181 5/182 5/183 5/184 5/185 5/186 5/187 5/188 5/189 5/190 5/191 5/192 5/193 5/194 5/195 5/196 (a) (b) 5/197 5/198 5/199 5/200 5/201 5/202 5/203 5/204 5/205 at 5/206 5/207 5/208 5/209 5/210 5/211 5/212 5/213 72.3 (vA)max 69.6 ft/sec at 72.3 vA r sin 1 cos (l/r)2 sin 2 ˙/ ˙ 2 cos 1 5 – 4 cos 2 2 cos ( ) 2 cos cos ( ) BC max 11.83 rad/s at 216 AB max 10.15 rad/s at 203 BC max 112.2 rad/s2 at 182.1 AB max 88.6 rad/s2 at 234 7.47 rad/s at 215 BC max AB max 6.54 rad/s at 202 46.1 251 (vrel)min 2 m/s at 109.5, (vrel)max 2 m/s 6.25 rad/s2 CW vrel 50.3i 87.1j km/h DB 41.2 rad/s2 CW DB 3.24 rad/s CW DE 2.45 rad/s2 CCW v cos r , AC v sin 2 D r cos vB 288 mm/s OA 0.966vd s2 d2 0.518sd aA 17.44 ft/sec2 aA 8.08 ft/sec2, BC 2 rad/s CW aB 5.25 in./sec2 left AB 0.354 rad/sec CW, vO 7.88 in./sec 1089j ft/sec vB 176i vA 1249i 189.1k ft/sec aC 83.1 ft/sec2 up AB 1.203 rad/s CCW 60 0 1 0t tan , 0 1 0t tan 2 tan 0.75 rad/s2 CCW vP 4.27 ft/sec 0 k K t 1 Kk tan1 vrel 26 220i km/h, arel 8.02j m/s2 2 16.53 rad/s2 CCW 5 rad/s CW, arel 8660i mm/s2 a 156.5 ft/sec2 vrel 7380i km/h vrel 5250i 5190j km/h 119.0 rad/s2 CW BC 1.046 rad/s CW, BC 19.11 vrel 7.71 m/s and arel 15.66 m/s2 BC 170.0 rad/s2 CW BC 1.429 rad/s CCW 19.11 vrel 3.93 m/s and arel 15.22 m/s2 27.0j ft/sec2 arel 16.06i vrel 22.3i 65.7j ft/sec 4 rad/s CW, 64.0 rad/s2 CCW arel r22 4u2 0.00918j m/s2 arel 0.854i vrel 1.136i 0.537j m/s 0.0642j m/s2 arel 0.864i vrel 2.71i 0.259j m/s 0.1350 in.
120 BC 0.634 rad/s CCW, v rel 0.483 m/s arel 4.69k m/s2 aA 10.42er 5.70e ft/sec2 aA 10.42i 5.70j ft/sec2 aA/B 1.76i 0.70j ft/sec2 aCor 2ui vrel dj, no aCor 0.0203 m/s2 aCor 0, vA 3.4i m/s, aA 2i 0.667j m/s2 aA 10.06 ft/sec2 38.2j ft/sec2 aA 48.7i vA 4.38i 7.58j ft/sec 12.67j ft/sec2 aA 34.5i vA 3.33i 4.5j ft/sec 139.4 aCor 0.4j m/s2, aA 0.35i 0.3j m/s2 vA 0.1i 0.25j m/s, 68.2 aE 0.285 m/s2 right aD 0.568 m/s2 down BD 46.9 rad/s2 CW aA 4.89 m/s2 right, AB 0.467 rad/s2 CCW AB 1.688 rad/s2 CCW 345 AB 36.6 rad/s2 CCW, aB 1.984 m/s2 OA 0, aD 480i 360j m/s2 aA/B 0.711j ft/sec2 0.0986 rad/s2 CW AB 16.02 rad/s2 CW, BC 13.31 rad/s2 CCW aP 3.62 m/s2 CB 5.76 rad/sec2 CW OB v 2 A rl BC 2.08 rad/sec2 CCW AB 22, BC 0 aB 13.89i 3.33j m/s2 aA 8.33i 10j m/s2, aP 8.33i m/s2 (aB)t 23.9 m/s2, 36.2 rad/s2 CW 105 8.00 rad/s2 CW, (aB)t 8.90 m/s2 714 Problem Answers Chapter 6 6/1 6/2 6/3 6/4 6/5 6/6 6/7 (a) and (b) 6/8 6/9 6/10 6/11 6/12 6/13 6/14 (a) (b) 6/15 (a) (b) 6/16 6/17 6/18 6/19 6/20 6/21 6/22 6/23 6/24 6/25 6/26 6/27 (a) (b) 6/28 (a) (b) Slips first if Tips first if 6/29 (a) (b) 6/30 6/31 6/32 6/33 6/34 6/35 6/36 6/37 6/38 (a) (b) 6/39 6/40 6/41 6/42 6/43 6/44 6/45 6/46 6/47 6/48 (a) (b) 6/49 6/50 (a) (b) 6/51 6/52 6/53 6/54 6/55 6/56 6/57 6/58 6/59 6/60 6/61 6/62 FA 108.3 N, FB 141.6 N A 22.1 N, B 11.03 N b 40.7 mm, R 167.8 N g 2r CCW, A 0.593mg O 1 4mgcos2 100 sin2 t 78.6 s 6 7 g l 12 7 k m (5 3) CW x b/6, 3 2 g b CW x l 23 , g3 l CW M mr2, R 22mr 2 4 Mƒ 0.1045 lb-ft, Mmot 0.836 lb-ft Ot 8.66 lb at all times 11.16 rad/s2 CCW 8.46 rad/s2 CCW R 18 lb 6.28 rad/s2 CCW 7.85 rad/s2 CCW On 2mg sin , Ot mg 2 cos M d 1 2r4 4lt 1 3l2 rl r2 0.389 g b CW 15 47 g r CW AA: 32g 5b ; BB: 32g 7b /2: 8g 3b CW; : 8g 3b CW Oy 1 32 92mg up 8g 3b CW, Ox 32mg 92 left A 56.3 N R 3.57 lb 2g 3r CW, O mg/3 g 2r CW, O mg/2 9.12 rad/sec2 CW IO 1453 lb-ft-sec2 1.193 rad/s2 CCW, FA 769 N FA FB 24.5 N R 49.0 N A 2.02 kN B 188.3 N 0.964 (nose up) D 2178 N D 1714 N, v2 gr b 2h tan 1 b 2h tan b 2h and tan b 2h v2 gr tan 1 tan b 2h and tan tan1 v2 gr 24.8, a 5 4g 51.3, N 257 kN D 234 N, 5.87 rad/s2 CW 0.598 F 78.3 N, M 28.7 N m M 196.0 Nm CCW t 3.41 s B 410 lb, Bst 417 lb P 118.7 lb, 49.2 TA 12.99 lb, TB 39.0 lb, 8.05 rad/sec2 a 161 ft/sec2 Ay 1389 N down A 1192 N NA 2550 lb (79.6%), NB 652 lb (20.4%) NA 1920 lb (60%), NB 1280 lb (40%) NA 1908 lb (59.6%), N B 1292 lb (40.4%) NA 1280 lb (40%), NB 1920 lb (60%) T 27.3 N, Ax 18.34 N right, Ay 15.57 N up a 4.14 m/s2 F A 1110 N, Ox 45 N right, O y 667 N down a 0.706 m/s2 right a 1.306 m/s2 right P 3(M m)g P mg c b d kh/2 a 0.268g P k(M m)g cos B 1.5 lb a 3g a g3 Problem Answers 715 6/63 6/64 6/65 6/66 (a) (b) (c) 6/67 (a) (b) 6/68 6/69 (a) (b) 6/70 (a) (b) 6/71 6/72 6/73 6/74 6/75 6/76 6/77 6/78 6/79 6/80 6/81 6/82 6/83 6/84 6/85 6/86 6/87 6/88 6/89 6/90 6/91 6/92 6/93 6/94 6/95 6/96 6/97 6/98 6/99 6/100 6/101 6/102 6/103 6/104 6/105 6/106 (a) (b) 6/107 6/108 , 6/109 6/110 6/111 6/112 6/113 6/114 6/115 6/116 6/117 6/118 6/119 6/120 6/121 6/122 6/123 6/124 6/125 6/126 6/127 6/128 N 346 lb E 0.435 ft-lb vA 2.45 m/s 4.59 rad/s v 6gb sin 2 M 28.4 lb-in. CW 3.31 rad/s l0 90.0 mm k 92.6 N/m, 2.42 rad/s CW N 3240 rev/min 13.19 rad/s h 54.5 mm A: vA 2gx sin , B: vB gx sin O 91 27 mg up 33.2 v 3.01 m/s 0.839 g b CW v 2.97 ft/sec 48g 7L 24g 7L CW, vG 3 14 gL aB 56.9 ft/sec2 down incline aA 65.0 ft/sec2 down 255 RB 0.359 lb 18.18 rad/sec2 CCW, RA 1.128 lb right 3g 2l CW ( s)min 0.589 0.00581 g r CCW m 3.50M, A 347 lb N mg 1 3 sin2 , 6g sin L(1 3 sin2 ) CW v 11.73 m/s max 53.1 On mg cos a sin 1 2r2 kO 2 2g r2 kO 2 Ot m(g sin a cos )1 r 2 kO 2 ¨ r kO 2 (a cos g sin ) s 18.66 ft aA 14 109 g down incline 2.97 rad/sec CCW 84 65 a L CCW aA 5.93 m/s2 left 5a 7r CW, 10 7r g(1 cos ) a sin aA 1.143g down incline MB 3.55 Nm CCW aO 3.73 m/s2 down incline B 36.4 N BC 4.03 N (tension) 12bg 7b2 3h2 CW, TA 3mg(b2 h2) 7b2 3h2 up F 17.62 N 0.295 rad/s2 CCW, a 1.027 m/s2 right F 19.38 N 2.12 rad/s2 CW, a 0.425 m/s2 right 22.8 rad/s2 CW, s 0.275 N mg r22 R r T 23 13 mg T 20.8 N F 3 8mg, N 13 16mg B: B g 2r sin , s 1 2 tan A: A g r sin , s 0 1 kO 2ar aO 2.41 m/s2 up incline, 93.8 rad/s2 CW a 13.31 ft/sec2, F 0.693 lb a 13.80 ft/sec2, F 1.714 lb ( s)min 0.775 aO 7.02 m/s2, 9.08 rad/s2 CCW 0.310 rad/s2, ax 6.87 m/s2, ay 2.74 m/s2 8.61 rad/s2 CCW, aG 5j m/s2 48.8 rad/s2 CW, aG 5j m/s2 aA P 10m (37i 9j) aB P 2m (3i j) 54.6 s 0.229, 53.1 s 0.1880, R 101.3 N MB 5.51 Nm CCW MA 109.8 N m CCW /3 A 10 6 ml F ml/6, 3.84 rad/sec2 CW, t 34.9 sec T 987 N, A 1.007 kN B 25.5 rad/s2 CCW 716 Problem Answers 6/129 (a) (b) 6/130 6/131 6/132 6/133 6/134 6/135 6/136 (a) (b) 6/137 (a) (b) 6/138 6/139 6/140 6/141 6/142 6/143 6/144 6/145 6/146 6/147 6/148 6/149 6/150 6/151 6/152 6/153 6/154 6/155 6/156 6/157 6/158 6/159 6/160 6/161 6/162 6/163 6/164 6/165 (a) (b) 6/166 (a) CW (b) 6/167 6/168 6/169 6/170 6/171 6/172 6/173 (a) CW (b) 6/174 6/175 6/176 6/177 6/178 6/179 6/180 6/181 6/182 6/183 6/184 6/185 6/186 6/187 (a) (b) and (c) 6/188 5v0 k/r 7 k 2 tan t 2v0 g(7 k cos 2 sin ), v 5v0 k 7 k 2 tan 3 1.757 rad/s 2 6.57 rad/s, 2 2[(It Id)1 Id] 2It Id N2 2.59 rev/min v1 4.88 m/s RA 27.2 lb, RB 18.77 lb, both up T 0.750 0.01719t N k 0.204, v 3 m/s (2)max 1.718 v1 L CW, x 0.291L N2 2.04 rev/s (vG)y 27 59 v1 down 2 72 59 v1 b CW, (vG)x 18 59 v1 right 2 3mv1 (M m)L CW, Oxt M 2(M m) mv1 right t m1 2m1 m2 r kg t m1 2m1 m2 r kg 12vm L m 4M 7m CCW vx M M m vM, vy m M m vm 2 1.220 rad/sec CW 2 1.166 rad/sec 1.202 rad/sec CW t I 0k 23.4 rad/sec CW H 2.66(1040) kg m2/s v 0.379 m/s up, 56.0 rad/s CW 1.093 rad/s 72 rad/s CCW 119.0 rad/s HO 0.373 kg m2/s CW HO 0.587 kg m2/s CW a 0.341 m/s2 down incline 27.3 rad/s2 CW M 0, MB 11.44 kNm CW 10.54 rad/sec2 CCW 135.3 rad/s2 CCW N 133.0 rev/min a 3 8 P m 3g 2 a F 2m g P(2 cos2 1) mb(8 cos2 1) CW 64.3 [M mg(b cos a sin )]/IA CCW a M 2mb1 (h/2b)2 g a 2P 5m tan 2 g 33.7 rad/sec2 CW 7.08 3g cos 2b M mb2cos2 1 3 CW cos ; otherwise 0 if P m 2 m0 g P m 2 m0g cos b(m m0) CCW v 5.95 m/s vmax 1.325 m/s vA 3gl v 8.02 ft/sec 3g 2b (22 1) N 3720 rev/min 3.11 rad/sec v 2.29 m/s P (mg sin M0/r)v 3 9 2 15 g r 2 3 9 8 57 g r 1.484 rad/s CW k 93.3 N/m, 9.51 rad/s k 24.0 lb/in.
2.36 rad/s 2.11, motor shaft turns CW v 12gb c 2b 3c 4b cos 2 2.23 rad/s CW 4.36 rad/s CW m 1.196 kg, Problem Answers 717 6/189 6/190 6/191 (a) 6/192 6/193 6/194 6/195 6/196 6/197 6/198 6/199 6/200 6/201 6/202 6/203 6/204 6/205 6/206 6/207 6/208 6/209 6/210 (a) (b) 6/211 6/212 6/213 6/214 (a) CW (b) 6/215 6/216 6/217 6/218 6/219 6/220 6/221 6/222 6/223 6/224 6/225 6/226 6/227 6/228 , 6/229 6/230 t 2.85 sec, vA 75.7 ft/sec 1.586 rad/s at 90 min 0.910 rad/s at 74.0 max 45.9 max 0.680 rad/s at 22.4 max 39.9, 4.50 rad/s CW Oy mg 4 (3 cos 1)2 Ox 3mg 4 sin (3 cos 2) max 23.0, ˙ max 0.389 rad/s 12.17 (vA)max 7.57 ft/sec at 48.2 v1 15.581 cos (45 ) ft/sec max 2.35 rad/s CW at 18.88 2 2.50 rad/s CW x l/3 Mz 18.41 lb-in.
Mx 46.0 lb-in., My 16.11 lb-in.
r ˙ r0 IO IO mr2 b L 1 n 19.26 6.25 rad/s CW 4.94 rad/s t 1206 s B 2130 lb M 2mrr ˙ FO 76.6 lb FO 6.68 lb, 4.15 rad/s CCW s 2 5 tan 3b2gh 2(b2 c2), n 62.5% t l 3g 0 d cos cos HO vr m 2 m0 2 l2 12r2 CW a 2.22 m/s2 right N 504 rev/min, E 98.1 J 0.604 rad/s2 CCW hmax 1 2 b mg P (c b) hmin 1 2 b mg P (c b) 1.135 rad/s v r k2 r2 rh2gh(k2 r2) v v 3 (1 2 cos ), n 0.0202 0.308 rad/s CCW v 9v2 4 sin2 3gL cos 109.6 rev/min, E 1.298 J N 63.8 rev/min I 3.45 kg m2 0.974 rad/sec s Mt (I Iw), w/s I Iw Mt (I Iw) N 4.78 rev/sec 20(sin k cos ) (2 sin 7 k cos ) v 2r0(sin k cos ) (2 sin 7 k cos ) t 2r0 g(2 sin 7 k cos ) Chapter 7 7/1 Finite rotations cannot be added as proper vectors 7/2 Infinitesimal rotations add as proper vectors 7/3 7/4 7/5 7/7 7/8 7/9 7/10 7/11 , 7/12 7/13 (a) (b) 7/14 7/15 7/16 7/17 1.386i rad/s2 0.693j 2.40k rad/s 2.5 rad/s, 3j rad/s2 1.5i 0.8j 2.60k rad/s2 0.4i 2.69k rad/s, 0.8j rad/s2 17.32 rad/s 26.5 rad/s, 50 23 i k rad/sec2 a 52(25j 18k) in./sec2 3k) in./sec 122 j rad/sec2, v 5(4i 6j 1.2i rad/sec2, aP 35.8j 80k in./sec2 pj 0k, p0 i (h sin2 d cos sin )k] a 2[(h sin cos d cos2 )i lj v [l cos i (d cos h sin )j l sin k] N2 440 rev/min vB 2.62 m/s vA 0.8i 1.5j 2k m/s aB 1285 m/s2 aP 11,080 in./sec2 vz 6 ft/sec, R 2.81 in.
718 Problem Answers 7/18 7/19 7/20 7/21 7/22 (a) (b) 7/23 7/24 (a) (b) 7/25 7/26 7/27 7/28 7/29 7/30 7/31 7/32 7/33 7/34 7/35 7/36 7/37 7/38 7/39 7/40 (a) (b) 7/41 7/42 7/43 7/44 7/45 7/46 7/47 7/48 7/49 7/50 , 7/51 7/52 7/53 7/54 7/55 7/56 7/57 7/58 7/59 7/60 7/61 7/62 7/63 7/64 7/65 7/66 7/67 7/68 7/69 T mr2 8 1 2 1 4b2 r22 2 2p2 HO 1 4mr2 1i 1 4b2 r22 j 2pk HG 2mƒ(k2 sin i k2 cos k) mk2pk HO 2m 1 3 l2 sin2 b2k T 3 10mr2 1 4 h2 r22 1 2 p2 HO 3 10mr2 1 2 6h2 r2 i pk 4.96 H 1 4mr2 ( sin cos )i (sin 2 2 cos2 )k H (0.4j 0.6k) Nms, T 59.2 J T 805 ft-lb HO 2.38j 25.6k lb-ft-sec HO 0.01i 0.0045j 0.0576k lb-ft-sec where xi y j zk HG Ip(i j k) 2(I mb2) HO b2 3 r2 4 h2 mi 1 2mr2pj H mr2(2c b) 3 j r 2 k H 1 3ml2 sin (j sin k cos ) HG 1 4 b3(i j 2k) T 4 3b32 HO b31 2 i 3 2 j 8 3k HA 2.70j 744k kgm2/s HG 1.613j 744k kgm2/s HO mb2 i 2j 1 6 l2 b2 2k G mb2, HO 3mb2 n 10 49 40 3 i 2j 6k rad/s vB 2 3 k m/s 42.8 rad/s2 Rp2 r cos i pj cos k sin R r 2(43i 9j 33k) rad/sec2 (3i 3j 5k) rad/sec a 2(b2 2r1)i r(1 2 p2)j 2rp2k v rpi (r1 b2)k aA 0.313i 2.43j 0.1083k ft/sec2 ˙ 1 8i rad/sec2 aA 22(2.4i 4j 5k) m/s2 402i rad/s2 b ˙2 sin ]j b ˙2 cos k aA 2b ˙ cos i [2(R b sin ) vA (R b sin )i b ˙ cos j b ˙ sin k a 2(6.32i 0.1k) m/s2 v (0.1i 0.8j 0.08k) m/s a 2(6.32i 0.1k) m/s2 v (0.1i 0.8j 0.08k) m/s 3i 4j rad/sec2 p2i p1j 12k 3.49(103)j rad/s2 (3.88i 3.49j)103 rad/s2 vA 6j m/s vA 0.636i 4.87j 1.273k m/s vA 6.8 ft/sec, aA 20.8 ft/sec2 p sin i ˙(p cos )j p ˙ sin k n 1 49(3i 20j 9k) rad/sec 1.2(3i k) rad/s2 aA 34.8j 6.4k m/s2 vA 3i 1.6j 1.2k m/s 40j 6k rad/s2 10.77 rad/s, 40j rad/s2 pqj p 28.2 rad/s, vB/A 4.10i m/s 1 r2 1 h2 i,
2 h2 r h h rj 31.0j rad/sec2 v 14.35j in./sec, a 338i 194.8k in./sec2 6.32 rad/s2 11.44 rad/s2 0.785i 2.60j 2.5k rad/s 8i rad/s2, aA 10.67 m/s2 6j 8k rad/s2, aA 21.2 m/s2 a 11.70i m/s2 v 0.691j 1.448k m/s k, 83 3 j 2 3 j k, 23 3 i v 3.48 m/s, a 1.104 m/s2 v 3.95 m/s, a 72.2 m/s2 a 2 2 R R r r R i k v 2R i j r R k 2 2 R r i Problem Answers 719 7/70 7/71 7/72 7/73 7/74 7/75 7/76 7/77 7/78 7/79 7/80 7/81 7/82 7/83 7/84 7/85 7/86 7/87 7/88 7/89 7/90 7/91 7/92 7/93 7/94 7/95 CCW as viewed from above 7/96 Tendency to rotate to student’s right 7/97 Decreased 7/98 7/99 7/100 7/101 Right-side normal forces are increased 7/102 left rudder 7/103 7/104 7/105 CCW 7/106 7/107 7/108 (a) No precession, (b) 7/109 7/110 7/111 (retrograde precession) 7/112 (direct precession) 7/113 7/114 negative 7/115 7/116 7/117 7/118 retrograde precession 7/119 (a) (b) (c) 7/120 7/121 7/122 7/123 7/124 7/125 (a) 7/126 7/127 Direct precession, Retrograde precession, 7/128 7/129 7/130 77.9i rad/s2 p mvh m0k2, opposite to car wheels H ma2 63 i j k l r 6 l r 6 M 97.9i lb-ft, T 73.8 ft-lb h r/2 Mz I0 sin t My 0 Mx I0 cos t where ˙ 2 2gl r2 4l2 Bz m ˙ 2 r2 2b p l ˙ Az m ˙ 2 r2 2b p l ˙ M 1 12ml2p0 sin 2 M 196.3i Nm, MO 319 Nm 0.443 sec p 0, 90, ˙ 4 rad/s ˙ 5.47 rad/s p 4 rad/s, 10, 3.03 p 4 rad/s, 0, ˙ 0 0.0996 s, ƒ 10 Hz ¨ M/m kx 2 cos2 kz 2 sin2 1.230K rad/sec, M 67.7i lb-in.
z-direction 1.831 s, RA 7.80 lb, RB 12.20 lb ˙ ˙1 124.2 rad/s ˙ 6 rev/min R 98.1 lb MA 30.9 Nm, MO 0 0.723 rad/s, MA 3.14 N m MA 12.56 Nm M 5410 kNm, (b) 6.67k rad/sec h/r 3/2 C D 948 N M 1681 N m, M M1 mk2 v r b 216 mm R 712 N A mg 3 7a 2b 2a b A mg/6 M 2mr (g 2r2) 2 3g l Mz 1 12mb2(1 4 sin2 ) My 1 6mb2 sin 2 Mx 1 6mb22 sin2 2 NB mg 2 m2v2 2r NA mg 2 m2v2 2r Otherwise 0 cos1 4g 5R2 if 2 4g 5R M 1 8mr22 sin 2j sin1 3g 22 b2 c2 b3 c3 FA 91.7j N, FB 91.7j N, M 10.8 N m FA 1608i N, FB 1608i N M 79.0i Nm M IO cos2 I sin2 M 4MO 3 M 2 mr22 A 576 N, B 247 N M 2b32 Bx 3Mb 2lc sin , By 3Mb 2lc cos B mbl2 2c (i sin j cos ) Ax Bx 0, Ay mR2 3 , By mR2 3 M 1 2 mbl2 a2 ack m c2 3 b2 3 cos2 HO m 6 b2 sin 2i 2b2k 2 5 r2 1 3 c2 cos2 HO m 1 6 c2 sin 2 j T 148.1 ft-lb HO 0.1626(i 8j) lb-ft-sec 720 Problem Answers Chapter 8 8/1 8/2 8/3 8/4 8/5 8/6 8/7 8/8 8/9 8/10 8/12 (a) (b) 8/13 (a) (b) 8/14 8/15 (a) 8/16 8/17 8/18 8/19 8/20 8/21 8/22 8/23 8/24 8/25 8/26 8/27 8/28 8/29 8/30 8/31 8/32 8/33 8/34 (a) (b) 8/35 8/36 8/37 8/38 (a) (b) 8/39 8/40 8/41 8/42 8/43 x ¨ b2 a2 c mx ˙ k mx 0, cb2 2a2km m1 a2 b2 m2x ¨ a2 b2 cx ˙ kx 0 c 4k(m1 4m2) 0.274, c 39.9 lb-sec/ft xmax 0.286 ft x 4.72 in.
x 4.42 in., (x ˙0)c nx0 x1 0.1630x0 c 16.24(103) Ns/m c 22.4 lb-sec/ft c 7.48 lb-sec/ft, 0.1097 x x0 (cos 9.26t 1.134 sin 9.26t)e10.5t N (2N)2 2 N , where N ln x0 xN 0.215, k 175.5 N/m, c 9.02 Ns/m c 154.4 lb-sec/ft 0.6 0.436 c 38.0 N s/m 0.75 n k m m1 b2 a2 m2x ¨1 k1 b2 a2 k2x1 0 n k 5m y ¨ k m c b 4 y 0, 2 b c 2 m k ƒn 3.07 Hz 59.7(103) sin 10.16t m x 9.20(103)(1 cos 10.16t) n 2 6EI mL3 ƒn 0.301 Hz n 2T ml ƒn 3.30 Hz v 88.0 cos 21.5t in./sec, (b) xmax 4.09 in.
m 2.55 kg, s 0.358 1 k 1 k1 1 k2 k k1 k2, ƒn 0.905 Hz k 474 kN/m, ƒn 1.332 Hz ƒn 3.90 Hz amax 30 m/s2 ƒn 7.40 Hz amax 3.6 m/s2 y 0.0660 m, v 0.451 m/s, both down st 0.273 m, 3 s, vmax 0.6 m/s x 2.08 sin (12t 1.287) in., C 2.08 in.
x 2 cos 12t in.
n 12 rad/sec, ƒ n 6 Hz k 736 N/m, 4.20 lb/in., 50.4 lb/ft 7/131 7/132 7/133 7/134 , 7/135 7/136 7/137 7/138 7/139 7/140 7/141 7/142 7/143 7/144 7/145 7/146 H mr b 2i b j 2rk M 2.70 Nm M 13.33 Nm M 271 lb-ft RA RB 159.3 lb T 11.85 ft-lb, no HO 0.0867i 0.421j 1.281k lb-ft-sec T 11.30 ft-lb HO 0.421j 1.281k lb-ft-sec N 1.988 cycles/min 2r3 R2k 3r R2 r aA 2 2R2 r2 2r2 R2 3j vA 4 R r2 R i 2 2 R2 r2 r i 2 R r r Rj R2 r2 R k T 69.9 J HO 0.707j 4.45k kgm2/s k, 4r2p/(5gh) n 9 49(2i k) rad/sec 83i 1203j 52k rad/sec2 a 2090i 369j 4810k in./sec2 Problem Answers 721 8/44 8/45 (a) (b) 8/46 8/47 8/48 8/49 8/50 (a) (b) 8/51 8/52 8/53 8/54 8/55 8/56 8/57 8/58 8/59 8/60 8/61 8/62 8/63 8/64 8/65 8/66 8/67 8/68 8/69 8/70 8/71 8/72 8/73 8/74 8/75 8/76 8/77 8/78 8/79 8/80 8/81 8/82 8/83 8/84 8/85 8/86 8/87 8/88 8/89 8/90 8/91 8/92 8/93 8/94 8/95 8/96 8/97 8/98 8/99 8/100 2 2l 3g n k m1 k2 r2 ¨ 3g(m 2M) 2l(m 3M) 0 x ¨ 8x 0, 2.22 sec 1.707 n 10.24 rad/sec, v 11.95 ft/sec mkO 2 ¨ K mr(g sin a cos ) 0 ƒn 1 2 JG IL 3 2 b l2 n 2 2, where n 3 ml2 3g 2l (m2 1 2 m1)x ¨ (k1 k2)x k1b cos t 2 3(R r) 2g , 0 r 2g(R r)/3 c 6 5 2k m g l 3r g sin 7.78 r g m1 a2 b2 m2 kO 2 b2 m3x ¨ a2 b2 cx ˙ kx 0 13.05 sec k 3820 N/m 1 2 a2 b2 c 3 km , ccr 2b2 a2 km 3 ƒn b 2l k m x 0.558 m ƒn 1 4 13k m R 2/3 I 2mr2 ( 2/ 1)2 1 n g 2b n 3 2 g 2b 2 5b 6g 2 2b 3g ƒn 1 2g 3r n 3g 2 4 a2 b2 ¨ g l 2kb2 ml2 0, 2 g l 2kb2 ml2 6 m 5k P cX 22/2 X 14.75 mm, vc 15.23 km/h c 44.6 N s/m mx ¨ cx ˙ k1 k2b2 a2 x k2 b a b0 cos t x F0 k (1 cos nt) amax 75.0 m/s2 b 1.886 mm k 227 kN/m or 823 kN/m T M1 2 n 2 c k m, c1 c2 2km mx ¨ (c1 c2)x ˙ kx c2b sin t c k1 k2 m mx ¨ cx ˙ (k1 k2)x k2b cos t 2.38 ƒn 5.32 Hz ƒ 1 2 g st N 108.1 rev/min or N 153.5 rev/min n 2 3 and n 2 y0 8.15 mm n 1 – 22 ƒ 4.68 Hz and ƒ 6.66 Hz R1 50%, R2 2.52% c k 6m st 0.25 ft X 0.0791 ft, X 0.251 ft, c 3.33 lb-sec/ft 5.32 rad/sec and 6.50 rad/sec 5.10 rad/sec and 6.78 rad/sec 0.1936 X 2.27(102) m X 1.344(102) m, r 2 kg m k 722 Problem Answers 8/101 8/102 8/103 8/104 8/105 , 8/106 8/107 8/108 8/109 8/110 8/111 8/112 8/113 8/114 8/115 8/116 8/117 8/118 8/119 8/120 8/121 (a) (b) 8/122 8/123 8/124 8/125 8/126 8/127 8/128 8/129 8/130 8/131 8/132 8/133 8/134 , 8/135 8/136 8/137 (a) (b) 8/138 8/139 x 0.284e0.707t sin 7.04t x 0.0926(t 0.0913 sin 10.95t) m y 0.1414e10t sin (10t 0.785) m y 0.1722e9.16t 0.0722e21.8t m 0 k 1.895 lb/ft xmin 0.0792 ft at t 0.1923 sec xmax 0.1955 ft at t 0.0462 sec 0.289 Mmax 1.809 at N 90.6 rev/min t 0.0544 s or 0.442 s 2 2mR2 MR2 1 6 Ml2 g4R2 l2 m M 2 N 645 rev/min (ƒn)y 4.95 Hz, (ƒn) 10.75 Hz 0 0.712 mm X 0.507 in.
max 0 r0/r 1 n 2, where n r k 2k m 0.0697 x0 3 smg k ƒn 1 2 g 2r 2 l g cos (/2) c 2g 3d 4k m n g l M m M 2m 3 n 2g 3r n 2 g 5r, 7.33 s c 4580 Ns/m 41.4 R g 2 ( 2)r g 2 l 3g ƒn 2.62 Hz ƒn 1.142 Hz 7.78r/g ƒn 1.496 Hz 2 2b 3g sin ƒn 3.65 Hz ƒn 1 2 mgr0 3Mr2 m(r r0)2 1/2 0.957 s n k 5m n 6k m 3g 2l 30 6k 3m1 26m2 n 3 6k 3m1 26m2 6(R r) g 0.326 sec ƒn 1.519 Hz ƒn 1 2 b l k m Appendix B B/1 B/2 B/3 B/4 B/5 B/6 B/7 B/8 B/9 B/10 B/11 B/12 Ixx 1 4mb2, Iyy 1 4ma2, Izz 1 4m(a2 b2) Ixx 3 10mr2, Iyy 3 5m r2 4 h2 Iyy 1 20mb2, ky 0.224b Ixx 3 7mh2, kx 0.655h Izz m 3b2 5 h2 7 Ixx 1 7mh2, Iyy 3 5mb2 Ixx 1 8mb2, kx 0.354b Ixx 1 8mb2, kx 0.354b IO 158.9 kg mm2 Izz 1 6m(h2 b2), kz 0.408h2 b2 Iyy 1 6mb2, ky b/6 Ixx 1 6mh2, kx h/6 Izz 1 12mL2 Ixx 1 12mL2 sin2 , Iyy 1 12mL2 cos2 Ixx 1 12ma2, Iyy 1 12mb2, Izz 1 12m(a2 b2) Iyy 1 12ml2, Iyy 1 3ml2 Problem Answers 723 B/13 B/14 B/15 B/16 B/17 B/18 B/19 B/20 B/21 B/22 B/23 B/24 B/25 B/26 B/27 B/28 B/29 B/30 B/31 B/32 B/33 (a) (b) B/34 B/35 B/36 B/37 B/38 B/39 B/40 B/41 B/42 B/43 B/44 B/45 B/46 B/47 B/48 B/49 B/50 B/51 B/52 B/53 B/54 B/55 B/56 B/57 B/58 B/59 B/60 B/61 B/62 B/63 B/64 B/65 B/66 B/67 B/68 B/69 B/70 B/71 B/73 B/74 B/75 B/76 B/77 B/78 l 0, m 0.550, n 0.835 I1 1.448b3, I2 0.360b3, I3 1.142b3 I1 3.78b4, I2 0.612b4, I3 3.61b4 l1 0.1903, m1 0.963, n1 0.1903 I1 0.750mb2, I2 0.799mb2, I3 0.1173mb2 l1 0.816, m1 0.408, n1 0.408 I1 9ml2, I2 7.37ml2, I3 1.628ml2 Imin 0.1870r3 at 38.6 l1 0.521, m1 0.756, n1 0.397 I1 7.53mb2, I2 6.63mb2, I3 1.844mb2 IAB 2.58 kg m2 IAA 1 6ma2 Ixy mb2 42 , Ixz 1 12mb2, Iyz mb2 42 Iyz 1 4mbh sin Ixy 1 6mb2 sin 2, Ixz 1 4mbh cos Ixy 2mr2/, Ixz Iyz 0 Ixy 1 8mb2, Ixz 0, Iyz 1 8mb2 IZZ 1 4mR2 (1 cos2 ) Ixy 1 24ml2 sin 2 Ixz 1.035 lb-in.-sec2 Iyz 0.776 lb-in.-sec2 Ixy 1.553 lb-in.-sec2 Ixy 1 4mr2, Ixz Iyz 1 2 mr2 Ixy b4 512 , Ixz Iyz 0 Ixz Iyz 0, Ixy ma2 sin2 4 Ixy mab, Iyz 1 2mbh, Ixz 1 2mah Ixy 2ml2, Ixz 4ml2, Iyz 0 Ixy 0, Ixz Iyz 2ml2 IZZ 1 4mR2 (1 cos2 ) Ixx 3 10m r2 5 r1 5 r2 3 r1 3 kO 97.5 mm Ixx 0.410 lb-in.-sec2 Iyy 0.01398 lb-ft-sec2 Ixx 3 2m1r2, Ixx m2r2 3 2 4 IA 0.1701ma2 Ixx 1 4mR2 IO m(7x2 1 3l2), R 0.582 Ix0x0 0.1010 kg m2 Ixx Izz 3 4mb2, Iyy 1 6mb2 I 1.031 kg m2, n 97.8% Iyy 0.0250 kg m2 Iyy 43 192 83 128L3 L r3/2 Ixx 1.898 lb-in.-sec2 Iyy 0.0433 lb-ft-sec2, Izz 0.0539 lb-ft-sec2 Ixx Iyy 0.0270 lb-ft-sec2 Iaa 1 2mr2, Ibb 2mr21 2 Ixx 0.1220 kg m2 Izz m 12 15r2 L2 Ixx 1 2mr2 2 r1 2 IOO 1.767 lb-ft-sec2 e 100 1 1/3 (r/l)2 (in percent) e 11.11% e 0.498%, Izz 1 2mr2 Ixx 2mL2, Iyy 4mL2, Izz 2mL2 kz 0.890a Ixx 1 2mr2 b2 6 Ixx Izz 2 3mr2 I 1 2m2R2 3a2 Izz 1 10m b2 2 h2 Iyy 1 10mh2 Iyy m 83 147 b2 23 196 L2 Ixx 166 147mb2 I mR2 3 4 a2 Iaa m 2 r2 l2 6 k a 2 39 5 Ixx 1 5m(a2 b2) Ixx 53 200mR2 Iyy 1 2mh2 r2 3 kz r/3 Ixx 2 9mb2 Iyy Izz m r2 7 2h2 3 Ixx 2 7mr2 Conversion Factors U.S. Customary Units to SI Units To convert from To Multiply by (Acceleration) foot/second2 (ft/sec2) meter/second2 (m/s2) 3.048 101 inch/second2 (in./sec2) meter/second2 (m/s2) 2.54 102 (Area) foot2 (ft2) meter2 (m2) 9.2903 102 inch2 (in.2) meter2 (m2) 6.4516 104 (Density) pound mass/inch3 (lbm/in.3) kilogram/meter3 (kg/m3) 2.7680 104 pound mass/foot3 (lbm/ft3) kilogram/meter3 (kg/m3) 1.6018 10 (Force) kip (1000 lb) newton (N) 4.4482 103 pound force (lb) newton (N) 4.4482 (Length) foot (ft) meter (m) 3.048 101 inch (in.) meter (m) 2.54 102 mile (mi), (U.S. statute) meter (m) 1.6093 103 mile (mi), (international nautical) meter (m) 1.852 103 (Mass) pound mass (lbm) kilogram (kg) 4.5359 101 slug (lb-sec2/ft) kilogram (kg) 1.4594 10 ton (2000 lbm) kilogram (kg) 9.0718 102 (Moment of force) pound-foot (lb-ft) newton-meter (N m) 1.3558 pound-inch (lb-in.) newton-meter (N m) 0.1129 8 (Moment of inertia, area) inch4 meter4 (m4) 41.623 108 (Moment of inertia, mass) pound-foot-second2 (lb-ft-sec2) kilogram-meter2 (kg m2) 1.3558 (Momentum, linear) pound-second (lb-sec) kilogram-meter/second (kg m/s) 4.4482 (Momentum, angular) pound-foot-second (lb-ft-sec) newton-meter-second (kg m2/s) 1.3558 (Power) foot-pound/minute (ft-lb/min) watt (W) 2.2597 102 horsepower (550 ft-lb/sec) watt (W) 7.4570 102 (Pressure, stress) atmosphere (std)(14.7 lb/in.2) newton/meter2 (N/m2 or Pa) 1.0133 105 pound/foot2 (lb/ft2) newton/meter2 (N/m2 or Pa) 4.7880 10 pound/inch2 (lb/in.2 or psi) newton/meter2 (N/m2 or Pa) 6.8948 103 (Spring constant) pound/inch (lb/in.) newton/meter (N/m) 1.7513 102 (Velocity) foot/second (ft/sec) meter/second (m/s) 3.048 101 knot (nautical mi/hr) meter/second (m/s) 5.1444 101 mile/hour (mi/hr) meter/second (m/s) 4.4704 101 mile/hour (mi/hr) kilometer/hour (km/h) 1.6093 (Volume) foot3 (ft3) meter3 (m3) 2.8317 102 inch3 (in.3) meter3 (m3) 1.6387 105 (Work, Energy) British thermal unit (BTU) joule (J) 1.0551 103 foot-pound force (ft-lb) joule (J) 1.3558 kilowatt-hour (kw-h) joule (J) 3.60 106 Exact value SI Units Used in Mechanics Quantity Unit SI Symbol (Base Units) Length meter m Mass kilogram kg Time second s (Derived Units) Acceleration, linear meter/second2 m/s2 Acceleration, angular radian/second2 rad/s2 Area meter2 m2 Density kilogram/meter3 kg/m3 Force newton N ( kg m/s2) Frequency hertz Hz ( 1/s) Impulse, linear newton-second N s Impulse, angular newton-meter-second N m s Moment of force newton-meter N m Moment of inertia, area meter4 m4 Moment of inertia, mass kilogram-meter2 kg m2 Momentum, linear kilogram-meter/second kg m/s ( N s) Momentum, angular kilogram-meter2/second kg m2/s ( N m s) Power watt W ( J/s N m/s) Pressure, stress pascal Pa ( N/m2) Product of inertia, area meter4 m4 Product of inertia, mass kilogram-meter2 kg m2 Spring constant newton/meter N/m Velocity, linear meter/second m/s Velocity, angular radian/second rad/s Volume meter3 m3 Work, energy joule J ( N m) (Supplementary and Other Acceptable Units) Distance (navigation) nautical mile ( 1.852 km) Mass ton (metric) t ( 1000 kg) Plane angle degrees (decimal) Plane angle radian — Speed knot (1.852 km/h) Time day d Time hour h Time minute min Also spelled metre.
SI Unit Prefixes Multiplication Factor Prefix Symbol 1 000 000 000 000 1012 tera T 1 000 000 000 109 giga G 1 000 000 106 mega M 1 000 103 kilo k 100 102 hecto h 10 10 deka da 0.1 101 deci d 0.01 102 centi c 0.001 103 milli m 0.000 001 106 micro 0.000 000 001 109 nano n 0.000 000 000 001 1012 pico p Selected Rules for Writing Metric Quantities 1. (a) Use prefixes to keep numerical values generally between 0.1 and 1000.
(b) Use of the prefixes hecto, deka, deci, and centi should generally be avoided except for certain areas or volumes where the numbers would be awkward otherwise.
(c) Use prefixes only in the numerator of unit combinations. The one exception is the base unit kilogram. (Example: write kN/m not N/mm; J/kg not mJ/g) (d) Avoid double prefixes. (Example: write GN not kMN) 2. Unit designations (a) Use a dot for multiplication of units. (Example: write N m not Nm) (b) Avoid ambiguous double solidus. (Example: write N/m2 not N/m/m) (c) Exponents refer to entire unit. (Example: mm2 means (mm)2) 3. Number grouping Use a space rather than a comma to separate numbers in groups of three, counting from the decimal point in both directions. (Example: 4 607 321.048 72) Space may be omitted for numbers of four digits. (Example: 4296 or 0.0476) Conversion Charts Between SI and U.S. Customary Units Conversion Charts Between SI and U.S. Customary Units (cont)
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[brag-ă-doh-chi-oh]
A cowardly but boastful man who appears as a stock character in many comedies; or the empty boasting typical of such a braggart. This sort of character was known in Greek comedy as the alazon. When he is a soldier, he is often referred to as the miles gloriosus (‘vainglorious soldier’) after the title of a comedy by the Roman dramatist Plautus. The most famous example in English drama is Shakespeare's Falstaff.
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BRIEF RESEARCH REPORT
published: 05 May 2020 doi: 10.3389/fenrg.2020.00074
Frontiers in Energy Research | www.frontiersin.org 1May 2020 | Volume 8 | Article 74
Edited by:
Jun Wang, University of Wisconsin-Madison, United States
Reviewed by:
Mingjun Wang, Xi’an Jiaotong University, China Claudio Tenreiro, University of Talca, Chile
Correspondence:
Xi-dao Mao maoxidao@126.com
Specialty section:
This article was submitted to Nuclear Energy, a section of the journal Frontiers in Energy Research
Received: 25 February 2020
Accepted: 14 April 2020
Published: 05 May 2020
Citation:
Mao X and Xia H (2020) Natural Convection Heat Transfer of the Horizontal Rod-Bundle in a Semi-closed Rectangular Cavity. Front. Energy Res. 8:74. doi: 10.3389/fenrg.2020.00074
Natural Convection Heat Transfer of the Horizontal Rod-Bundle in a Semi-closed Rectangular Cavity
Xi-dao Mao 1,2 and Hong Xia 1
1College of Nuclear Science and Technology, Harbin Engineering University, Harbin, China, 2China Nuclear Power Engineering Co., Ltd., Beijing, China
During the refueling of PWR, the spent fuel assemblies are transferred from the reactor building to the fuel building through an underwater tube. The heat transfer characteristics of the spent fuel assembly in the top-corner area of the carrier with poor heat transfer conditions is important for the safety design. Experiments were carried out and single phase natural convection and pool boiling heat transfer coefficients of the three fuel rods in the top-corner area of the carrier under different heat flux was measured and obtained. The new correlations were presented. The results can provide a reference for evaluating the thermal safety state and the maximum surface temperature of the fuel assembly during the transportation process in future engineering applications.
Keywords: fuel transfer tube, fuel assembly, natural convection, pool boiling heat transfer, horizontal bundle
INTRODUCTION
During a refueling of the light water reactor, the spent fuel assemblies are transferred from the reactor building to the fuel building through an underwater tube, as shown in Figure 1A . When the fuel assembly is transferred horizontally through the transfer tube, it is carried and protected by the perforated carrier and the transport trolley ( Guo and Wang, 2013 ), as shown in Figure 1B .It is important for the design of a fuel transfer device to consider the hypothetical accidents such as the mechanical failure and loss of power supply. In this situation, the spent fuel assembly may be trapped in the transfer tube. Therefore, to avoid the overheating of the assembly, it is necessary to study the heat transfer of the fuel assembly in the transfer tube to find out that whether the decay heat of the fuel assembly can be removed only rely on natural convection cooling. The driving force of natural circulation in the transfer tube comes from the density difference of hot and cold fluids. The cold water from the pool flows from the lower part of the tube to the fuel assembly. The water flows into the fuel assembly through the circular holes on both sides of the carrier. The cooling water is heated in the fuel assembly and then flows out of the assembly through the top holes of the carrier due to the buoyancy. The water flows back to the pool through the upper part of the tube. In the previous analysis, the flow field in the transfer tube can be decomposed into two directions: longitudinal flow along the transfer tube and crosswise flow in the cross section of the transfer tube, respectively ( Xi-dao et al., 2018 ). The flow of cooling water in the cross section of the fuel assembly in the transfer tube is shown in Figure 1B . The natural convection in the cross section is more complicated compared with the longitudinal flow along the transfer tube. It is a key factor affecting the cladding temperature of the fuel assembly. The fuel assembly usually employs an arrangement of 17 × 17. The core of the cooling problem of the fuel assembly in the transfer channel is the natural convection heat transfer of the horizontal Mao and Xia Natural Convection Heat Transfer
FIGURE 1 | Schematic of fuel transfer system. (A) Schematic of fuel transfer system. (B) Schematic of carrier and transport trolley in the transfer tube. The symbol “1” refers to fuel assembly; the symbol “2” refers to carrier; the symbol “3” refers to transport trolley.
rod bundle. However, the fuel assembly is located in the carrier in the transfer tube, which is a semi-closed rectangular cavity. This would result to a weakened heat exchange. Therefore, the natural convection heat transfer of the fuel assembly in the carrier is different from the natural convection in an infinite space or in enclosure. When the temperature of cooling water is low, the flow in the tube bundles is single-phase natural convection. The disturbance and the pre-heating effects of the lower tube will enhance and weaken the heat transfer ability of the upper tube bundles, respectively. The natural convective heat transfer characteristics of the tube bundles are very different from a single rod. There are some researches focused on the heat transfer of the tube bundle and the single tube under different conditions, respectively. Studies of the steady state free convection heat transfer from horizontal isothermal cylinders with low Rayleigh numbers were presented, such as Sadeghipour and Asheghi (1994) and
Tokanai et al. (1997) . The correlations were presented to calculate the Nusselt numbers in terms of Rayleigh numbers in these studies, which are in the condition of infinite space. The pitch-to-diameter ratio has a significant influence on the heat transfer behavior. There are some results of the natural convection of the tube bundles in enclosure too. Keyhani and Dalton (1996) studied the natural convection heat transfer in enclosed horizontal 3 × 3, 5
× 5, and 7 × 7 rods with a pitch-to-diameter ratio of 1.35. Each array was positioned in an isothermal square enclosure with a width-to-diameter ratio of 20.6. The experimental results show that the heat transfer behavior is related to the relative size of the closed boundary and the position of the tube bundles. A two-dimensional solution for natural convection in an enclosure with 2 × 2 tube bundles was obtained by Park et al. (2015) . The numerical results show the similar conclusions with Keyhani that locations of the cylinders affect the heat transfer in the enclosure.
Ashjaee et al. (2008) studied the steady two-dimensional free convection heat transfer from a horizontal, isothermal cylinder located underneath an adiabatic ceiling using a Mach–Zehnder interferometer. When the Rayleigh number is 14,500 to 40,000,
Frontiers in Energy Research | www.frontiersin.org 2May 2020 | Volume 8 | Article 74 Mao and Xia Natural Convection Heat Transfer
and the ratio of the height of the rod top and the plate to the diameter (L/D) of the rod ranges from 0.5 to 1.5, the local Nusselt number at the rod top is the minimum value of the entire rod surface. Sebastian ( Sebastian and Shine, 2015 ) developed a two-dimensional model for natural convection heat transfer of a single rod with a horizontal confinement at the top. The calculation results are in good agreement with the experimental results. The results show that the local Nusselt number at the top of the rod is the minimum value of the entire rod surface within a certain L/D range. However, the boundary conditions studied by Ashjaee and Sebastian are oversimplified compared to the boundary conditions of the fuel assembly in the carrier. The difference in heat transfer characteristics between single rods and rod bundles is also very big, and it is difficult to directly use their results to evaluate safety of fuel assembly. On the other side, when the temperature reaches the saturation temperature, the single-phase convection in the fuel assembly will turn into the pool boiling. There are many results present to study the different factors affecting boiling, such as wall materials, surface characteristics, and physical properties of working fluids. The Rohsenow pool boiling model is widely used for heat transfer calculation of the flat plate and the single tube. However, the model error of Rohsenow is as much as ±50% in some conditions due to the complexity of the boiling (Pioro et al., 2004a,b ). Therefore, different specialized experiments are developed to get better estimates for different conditions. The pool boiling of the horizontal bundle is more complicated than the condition of a single tube. The heat transfer of the upper parts of the bundles can be enhanced or weaken under different conditions. Kang studied the 2 × 1 tube bundle to reveal the buddle effect ( Kang, 2015, 2016 ). Zhang studied the multiple tube bundles to reveal the nucleate pool boiling ( Zhang et al., 2018 ). These studies have similar results that the heat transfer of the upper part is enhanced when the bubbles generated from the lower parts of the bundles scour the upper parts and thin the liquid film as well as the bubbles gathered and fallen. Under the condition of high heat flux, experimental results show that the upper tube will be coated by the bubbles generated in the lower part of the tube bundle, which can weaken the heat transfer of the upper tube ( Swain and Das, 2018 ). The pool boiling of the fuel bundle in the transfer tube is different with the bundle in the reboiler or the steam generator. The carrier influences the inflow and outflow of the fluid. Some of the fuel elements on the top-corner of the fuel assembly are in a semi-closed state, and bubbles are likely to accumulate at this position and affect heat transfer. The heat transfer of the fuel assembly in the transfer tube under natural convection is different with the regular horizontal tube bundle, including the single-phase condition and the two-phase condition. The natural convection heat transfer of fuel assembly in the transfer tube is a key factor to effect the temperature of the spent fuel and the cladding integrity. However, the heat transfer of the fuel assembly in the carrier of the fuel transfer tube is not specially studied yet. This paper present an experiment to study the natural convection heat transfer on the cross-section which is affected by the carrier. New correlations are also presented both for single-phase and the two-phase conditions.
EXPERIMENTAL SETUP
The experiment of full-scale ratio is not realistic due to the expensiveness and long-period. Therefore, a simplified test facility is developed based on the similar rules and the theoretical analysis ( Xi-dao et al., 2018 ). A set of electrical heating rods is used to simulate the fuel assembly. The heat flux of the heating rod is uniform. Considering the heating uniformity of the real fuel element, 1.65 or 2.62 multiply the power density in the test with the mean heat flux of the fuel element conservatively. The test facility is shown in Figure 2A . The symbol A in this figure refers to an electric heating rod with a length of 750mm, which is used to simulate the fuel assembly near the central symmetry plane. The symbol B refers to the carrier. The symbol C refers to the transfer trolley. The symbol D refers to the cooler, which is used to simulate the heat exchange of at the inlet of the transfer tube. A mesh resistance element was added between the heating rod and the cooler to simulate the pressure loss caused by the transshipment tube and the transfer device out of the experimental section. The main geometric parameters of the test facility is listed in Table 1 .According to the previous research results ( Guo and Wang, 2013 ), the maximum temperature of the cladding of the fuel assembly is located on the top of the fuel rod at the top corner of the carrier. In order to reduce the interference to the flow field of the top-corner area of the carrier, thermocouples are only fixed on the top of 1#, 3#, and 5# heating rod at intervals, as shown in
Figure 2B .
Experimental Parameters for Single-Phase Natural Convection
The temperature of inlet of the fuel assembly is controlled as 60, 70, and 80 ◦C, respectively. The heat flux of the heating rod is controlled as 3900, 4300, 6200, and 6800 W/m 2. T-Type thermocouples and pressure gauges with high precision were employed. The instruments were calibrated. The overall error of the temperature measurement is about ±1.5 ◦C, and the over error of the pressure measurement is about ±0.14%. The data acquisition system was Solartron IMP 35951C. In the test, the power of the heating rod is set to a fixed value, and then the temperatures of the both sides of the fuel assembly simulated are adjusted by controlling the cooler. The data are acquired when the system and all the measurement are stable.
Parameters for Pool Boiling Test
During the pool boiling experiment, the pressure is controlled at 0.1MPa. The inlet subcooled temperatures of the assembly are controlled under 0.5 ◦C. The range of heat flux on the surface of the heating rod is 2,400–20,000 W/m 2. The heating rod is made of polished 316L stainless steel with a surface roughness Ra ≈0.8 μm. Thermocouples were fixed on top of 1#, 3#, and 5# heating rods as before.
Frontiers in Energy Research | www.frontiersin.org 3May 2020 | Volume 8 | Article 74 Mao and Xia Natural Convection Heat Transfer
FIGURE 2 | 3D model and tube wall temperature measurement of the test facility. (A) 3D model of the Test facility. (B) Schematic diagram of wall temperature measurement in upper corner of the fuel carrier.
In order to exclude the influence of the unstable vaporized core on the wall surface of the heating pipe on the test results, the surface aging of the heating pipe was carried out before the formal test. The tests were performed from low heat flux to high heat flux. The power of the heating rod and the temperature of each thermocouple were recorded when the condition is stable.
EXPERIMENTAL RESULTS AND DISCUSSION Single Phase Experimental Results
In the data processing, the local Nusselt number Nu top on the top corner of the fuel rod in the natural convection is calculated from the following:
Frontiers in Energy Research | www.frontiersin.org 4May 2020 | Volume 8 | Article 74 Mao and Xia Natural Convection Heat Transfer
TABLE 1 | The main geometric parameters of the test facility.
Item Unit Test facility
Heating rod OD mm 9.5 Pitch mm 12.6 Heating rod length mm 750 Length of the fuel transfer tube m∼2Diameter of the fuel transfer tube mm 508 The diameter of the hole in the carrier mm 160
Nu top = htop D
λf
(1) The temperature used to define the physical property parameters is defined as tm = (tw + tin )/2, where the tw is the top wall temperature of the rod in the top-corner of the carrier, and the tin
is the temperature of cooling water at the inlet of the carrier. λf
is the thermal conductivity of water evaluated at the temperature
tm. The characteristic length D employs the hydraulic diameter of the fuel rod. A series of experiments were conducted on the test facility to get the local heat transfer data on the top of the rod at the top-corner of the carrier. Figure 3A shows a comparison of the heat transfer coefficient for the different rod location. For the same Rayleigh Number (Ra number), the Nu top at the top of the 1 # rod is the smallest, and the Nu number at the top of the 5 # rod is the largest. From the experiment data, it is found that the heat transfer coefficient is relative with heating power and distance from the top-corner of the carrier. When the heat flux of fuel rod and water temperature at the inlet of the carrier are the same, the closer the rod to the top-corner of the carrier, the smaller the natural convective heat transfer coefficient.
Pooling Boiling Experimental Results
In the test, the system pressure is atmospheric pressure. The characteristic temperature is the saturation temperature of the water at the test pressure. The characteristic length of the pool boiling is defined as: l∗ = σ/ [g(ρl − ρg)]. where σ is the surface tension, with the unit of N · m−1, ρg is the density in gaseous phase, ρl is the density in liquid phase. The unit of the density is kg · m−3.
Figure 3B shows the pool boiling heat transfer coefficients at the top of 1 #, 3 #, and 5 # rods. The solid black line in the figure is the single-rod pool boiling heat transfer coefficient calculated by the Rohsenow correlation ( Rohsenow, 1952 ), which is used to compare with the fuel assembly test results. Although the bubble movement is hindered by the carrier, it can be seen that the boiling heat transfer coefficient at the top of the fuel assembly 1 #, 3 #, and 5 # heating rods is still higher than that of a single rod. With the increase of heat flux density, the rod bundle effect of boiling heat transfer coefficient of 1 #, 3 #, and 5 # rods gradually weakened. Under the condition of heat flux about 2 × 10 4W/m2,the boiling heat transfer coefficient of 1 # rod closest to the top corner of the carrier is almost equal to that of a single rod.
FIGURE 3 | Heat transfer characteristics of fuel assemblies in transfer tube.
(A) The Nusselt number of single-phase natural convection. (B) The pool boiling heat transfer coefficient.
NEW CORRELATIONS FOR LOCAL HEAT TRANSFER COEFFICIENTS
The published correlations were developed using the data for plates, wires, and outside surfaces of tubes ( Pioro et al., 2004b; Shah, 2017 ). Since the correlations do not contain the effects of the semi-closed rectangular cavity and the lower tube heat flux, the calculated heat transfer coefficients does not agree with the present experimental data. Therefore, it is not reasonable to use the correlation equations available on the literature to predict the present experimental data.
Single-Phase Heat Transfer Correlations
The Nusselt Number of the top corner of the fuel element in the natural convection can be expressed with a function of Rayleigh Number Ra . Through the regression analysis of the experimental data with the help of a computer program (which uses the least square method as a regression technique) three correlations were determined as follows: Nu 1, top = 0.0176(Ra) 0.319 2.54 × 10 6 ≤ Ra ≤ 6.03 × 10 6 (2) Nu 3,top = 0.0091(Ra) 0.372 2.38 × 10 6 ≤ Ra ≤ 5.09 × 10 6 (3) Nu 5,top = 0.0099(Ra) 0.382 2.02 × 10 6 ≤ Ra ≤ 4.28 × 10 6 (4)
Frontiers in Energy Research | www.frontiersin.org 5May 2020 | Volume 8 | Article 74 Mao and Xia Natural Convection Heat Transfer
FIGURE 4 | Comparison of calculated data with experimental data. (A)
Comparison of calculated heat transfer coefficients with single phase experimental data. (B) Comparison of calculated heat transfer coefficients with pool boiling experimental data.
A comparison between the measured Nusselt and the calculated values by Equation (2) to (4) is shown in Figure 4A . The newly developed correlation predicts the present experimental data within ±10%.
Pooling Boiling Heat Transfer Correlations
A Pioro -type ( Pioro et al., 2004b ) correlation is introduced here to predict local pool boiling heat transfer coefficients.
htop = C k
l∗
{
q
hfg ρ0.5 g [σ g(ρl − ρg)] 0.25
}n
Pr m (5) where htop is the boiling heat transfer coefficient of the top corner of the fuel elements with unit of W · m−2 · K−1; q is the heat flux of the heating rod surface with the unit of W · m−2; k is liquid heat transfer coefficient with the unit of W · m−1 · K−1; hfg is the latent heat of vaporization with the unit of J · kg −1;Pr is the Prandtl Number; l∗ is the characteristic length. C, n and m are to be solved.
htop , q, Pr and the physical property parameters can be calculated based on the experiment data, and then the coefficient
C, n and m can be solved by multiple regression. The coefficient
C, n and mare solved, and the boiling heat transfer coefficient can be expressed by the equation of (6)–(8) for the position of 1#, 3#, and 5#, respectively:
h1,top = 984.5 k
l∗
{
q
hfg ρ0.5 g [σ g(ρl − ρg)] 0.25
}0.593
Pr −1.1 (6)
h3,top = 909.9 k
l∗
{
q
hfg ρ0.5 g [σ g(ρl − ρg)] 0.25
}0.556
Pr −1.1 (7)
h5,top = 697.8 k
l∗
{
q
hfg ρ0.5 g [σ g(ρl − ρg)] 0.25
}0.493
Pr −1.1 (8)
Figure 4B plots the error range of the fitted equations of (6) to (8). As shown in the figure, the errors of the predicted quantities from these equations are <10%, which means that the precision of the fitting process of these equations was good.
CONCLUSIONS
An experimental heat transfer investigation for the horizontal rod-bundle in a semi-closed rectangular cavity submerged in the cooling water tube was conducted to find the characteristics of the fuel transfer device under natural circulation conditions. The major conclusions of the present study are as follows: In the single-phase natural convection test and the pool-boiling test, the heat transfer coefficient of the 1# heating rod closest to the top corner of the carrier was the smallest. Within the parameters of the pool-boiling test, the heating rod in the upper row of the heating rod is affected by the steam generated from the lower heating tube, and the boiling heat transfer coefficient is higher than that of the single tube pool boiling heat transfer coefficient. As the heating power increases, the boiling heat transfer coefficient of the 1 # heating rod is closer to that of a single rod. Through the regression analysis of the experimental data, new empirical correlations suitable for single-phase and pool boiling heat transfer were developed, which can provide a reference for evaluating the thermal safety status and maximum surface temperature of fuel assembly during transfer.
DATA AVAILABILITY STATEMENT
All datasets generated for this study are included in the article/supplementary material.
AUTHOR CONTRIBUTIONS
HX is graduate advisor of XM.
FUNDING
This work was supported by China Nuclear Power Engineering Co Ltd.
Frontiers in Energy Research | www.frontiersin.org 6May 2020 | Volume 8 | Article 74 Mao and Xia Natural Convection Heat Transfer
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Ashjaee, M., Eshtiaghi, A.H, Yaghoubi, M., and Yousefi, T. (2008). Experimental investigation on free convection from a horizontal cylinder beneath an adiabatic ceiling. Exp. Therm. Fluid Sci . 32, 614–623. doi: 10.1016/j.expthermflusci.2007.07.004 Guo, Q., and Wang, H. (2013). “Numerical simulation on 3D flow in the fuel transfer canal and local flow field analysis,” in ASME Proceedings of the 21th International Conference on Nuclear Engineering (Chengdu: ASME). Kang, M. G. (2015). Effects of elevation angle on pool boiling heat transfer of tandem tubes. Int. J. Heat Mass Transf . 85, 918–923. doi: 10.1016/j.ijheatmasstransfer.2015.02.041 Kang, M. G. (2016). Pool boiling heat transfer from an inclined tube bundle. Int. J. Heat Mass Transf. 101, 445–451. doi: 10.1016/j.ijheatmasstransfer.2016.05.099 Keyhani, M., and Dalton, T. (1996). Natural convection heat transfer in horizontal rod-bundle enclosures. J. Heat Transfer 118, 598–605. doi: 10.1115/1.2822674 Park, Y. G., Ha, M. Y., and Park, J. (2015). Natural convection in a square enclosure with four circular cylinders positioned at different rectangular locations. Int. J. Heat Mass Transf . 81, 490–511. doi: 10.1016/j.ijheatmasstransfer.2014.10.065 Pioro, I. L., Rohsenow, W., and Doerffer, S. S. (2004a). Nucleate pool-boiling heat transfer, I: Review of parametric effects of boiling surface. Int. J. Heat Mass Transf . 47, 5033–5044. doi: 10.1016/j.ijheatmasstransfer.2004.06.019 Pioro, I. l., Rohsenow, W., and Doerffer, S. S. (2004b). Nucleate pool-boiling heat transfer, II: Assessment of prediction methods. Int. J. Heat Mass Transf . 47, 5045–5057. doi: 10.1016/j.ijheatmasstransfer.2004.06.020 Rohsenow, W. M. (1952). A method of correlating heat transfer data for surface boiling of liquids. Transac. ASME . 74, 969–976. Sadeghipour, M. S., and Asheghi, M. (1994). Free convection heat transfer from arrays of vertically separated horizontal cylinders at low Rayleigh numbers. Int. J. Heat Mass Transf. 37, 103–109. doi: 10.1016/0017-9310(94)90165-1 Sebastian, G., and Shine, S. R. (2015). Natural convection from horizontal heated cylinder with and without horizontal confinement. Int. J. Heat Mass Transf. 82, 325–334. doi: 10.1016/j.ijheatmasstransfer.2014.11.063 Shah, M. M. (2017). A correlation for heat transfer during boiling on bundles of horizontal plain and enhanced tubes. Int. J. Refrigerat . 78, 47–59. doi: 10.1016/j.ijrefrig.2017.03.010 Swain, A., and Das, M. K. (2018). Performance of porous coated 5 ×3 staggered horizontal tube bundle under flow boiling. Appl. Therm . Eng . 128, 444–452. doi: 10.1016/j.applthermaleng.2017.09.038 Tokanai, H., Kuriyama, M., Harada, E., Konno, H. (1997). Natural convection heat transfer from vertical and inclined arrays of horizontal cylinders to air. J. Chem. Eng. Japan 30, 728–734. doi: 10.1252/jcej.30.728 Xi-dao, M., Yang, L., Hai-jun, J., and Qiang, G. (2018). “Design and distortion analysis of thermal-hydraulics test facility for the fuel transfer tube,” in
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(London: ASME). Zhang, K., Hou, Y. D., Tian, W. X., Zhang, Y. P., Su, G. H., and Qiu, S. Z. (2018). Experimental investigation on steam-water two-phase flow boiling heat transfer in a staggered horizontal rod bundle under cross-flow condition.
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Conflict of Interest: XM was employed by the company China Nuclear Power Engineering Co., Ltd. The authors declare that this study received funding from China Nuclear Power Engineering Co., Ltd. The funder had the following involvement with the study: Experimental study of thermo-hydraulic characteristics in the transfer channel of double pressure containment (No. KY514).
Copyright © 2020 Mao and Xia. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
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Blade area ratio is defined as the ratio of the total blade area to the total rotor area of a propeller, which is a critical parameter in the design of propellers, particularly in the Newton-Rader series where it ranged from about 0.5 to 1.0.
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Chapter
Propulsion
2013, Introduction to Naval Architecture (Fifth Edition)Eric C. Tupper BSc, CEng, RCNC, FRINA, WhSch
Blade Area
Blade area is defined as a ratio of the total area of the propeller disc.
The usual form is
Developed blade area ratio=developed blade area disc area
In some earlier work, the developed blade area was increased to allow for a nominal area within the boss. The allowance varied with different authorities and care is necessary in using such data. Sometimes the projected blade area is used, leading to a projected blade area ratio.
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Book 2013, Introduction to Naval Architecture (Fifth Edition)Eric C. Tupper BSc, CEng, RCNC, FRINA, WhSch
Chapter
Glossary of terms and definitions
2008, The Maritime Engineering Reference Book
Propeller geometry
A D
Developed blade area
A E
Expanded area
A O
Disc area
A P
Projected blade area
BAR
Blade area ratio
b
Span of aerofoil or hydrofoil
c
Chord length
d
Boss or hub diameter
D
Diameter of propeller
f M
Camber
P
Propeller pitch
R
Propeller radius
t
Thickness of aerofoil
Z
Number of blades of propeller
α
Angle of attack
φ
Pitch angle of screw propeller
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Book 2008, The Maritime Engineering Reference Book
Chapter
Propeller Design
2019, Marine Propellers and Propulsion (Fourth Edition)J.S. Carlton FREng
22.6.2 Determination of Mean Pitch Ratio
Assuming that the propeller B p value together with its constituent quantities and the behind hull diameter D b are known, then the mean pitch ratio can be evaluated of the standard series equivalent propeller.
The behind hull value of delta (δ b) is calculated as
(22.3)δ b=ND b V a
from which this value together with the power coefficientB p is entered onto the B p-δ chart, as shown in Fig. 22.11 B. From this chart, the equivalent pitch ratio (P/D) can be read off directly. As in the case of propeller diameter, this process should be repeated for a range of blade area ratios to interpolate for the required blade area ratio. It will be found, however, that P/D is relatively insensitive to blade area ratio under normal circumstances.
In the case of the Wageningen B series, all the propellers have constant pitch except for the four-blade series, where there is a reduction of pitch toward the root as indicated in Chapter 6. In this latter case, the P/D value derived from the chart needs to be reduced by 1.5% to arrive at the mean pitch.
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Book 2019, Marine Propellers and Propulsion (Fourth Edition)J.S. Carlton FREng
Chapter
Powering
2008, The Maritime Engineering Reference Book
5.6.5.2 Determination of mean pitch ratio
Assuming that the propeller B p value together with its constituent quantities and the behind hull diameter D b axe known, then to evaluate the mean pitch ratio of the standard series equivalent propeller is an easy matter. First the behind hull δ value is calculated as
(5.100)δ b=N D b V a
from which this value together with the power coefficientB p is entered onto the B p−δ chart, as shown in Figure 5.97(b). From this chart the equivalent pitch ratio (P/D) can be read off directly. As in the case of propeller diameter this process should be repeated for a range of blade area ratios in order to interpolate for the required blade area ratio. It will be found, however, that P/D is relatively insensitive to blade area ratio under normal circumstances.
In the case of the Wageningen B series all of the propellers have constant pitch with the exception of the four-blade series, where there is a reduction of pitch towards the root. In this latter case, the P/D value derived from the chart needs to be reduced by 1.5 per cent in order to arrive at the mean pitch.
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Book 2008, The Maritime Engineering Reference Book
Chapter
Propeller Design
2012, Marine Propellers and Propulsion (Third Edition)J.S. Carlton FREng
Design Considerations 445
22.7.1
Direction of Rotation 445
22.7.2
Blade Number 448
22.7.3
Diameter, Pitch–Diameter Ratio and Rotational Speed448
22.7.4
Blade Area Ratio 448
22.7.5
Section Form 449
22.7.6
Cavitation 449
22.7.7
Skew 449
22.7.8
Hub Form 449
22.7.9
Shaft Inclination 450
22.7.10
Duct Form450
22.7.11
The Balance Between Propulsion Efficiency and Cavitation Effects 450
22.7.12
Propeller Tip Considerations 451
22.7.13
Propellers Operating in Partial Hull Tunnels 451
22.7.14
Composite Propeller Blades452
22.7.15
The Propeller Basic Design Process 452
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Book 2012, Marine Propellers and Propulsion (Third Edition)J.S. Carlton FREng
Chapter
Propeller Blade Vibration
2019, Marine Propellers and Propulsion (Fourth Edition)J.S. Carlton FREng
21.3 The Effect of Immersion in Water
The principal effect of immersing the propeller in water is to cause a reduction in the frequency at which a particular mode of vibration occurs. The reduction factor is not a constant value for all modes of vibration and appears to be larger for the lower modes than for the higher modes. To investigate this effect in global terms, a frequency reduction ratio Λ can be defined as:
(21.1)Λ=frequency of mode in water frequency of mode in air
Burrill (1949) investigated this relationship for the propeller whose vibratory characteristics in air are shown by Fig. 21.3. The results of his investigation are shown in Table 21.2 for both the flexural and torsional modes of vibration. From the table, it can be seen that for this particular propeller, the value of Λ increases with the modal number for each of the flexural and torsional modes. For the higher blade area ratio propeller results of Hughes, shown in Fig. 21.4, Table 21.3 shows the corresponding trends. Again, from this table, the general trend of increasing values of Λ with increasing complexity of the modal form is clearly seen. Hughes also investigated the effect of pitch on response frequency by comparing the characteristics of pitched and flat-plate blades. He found that for most modes, except for the first torsional mode, the pitched blade had a frequency of around 10% higher than the flat-plate blade and in the case of the first torsional mode the increase in frequency was of the order of 60%. The influence of water immersion on these variations was negligible. In this study, it was also shown that for a series of other blade forms, having broadly similar dimensions except for blade area and outline, a reasonable correlation existed between the value of Λ and the frequency of vibration in air. It is likely, however, that such a correlation would not be generally applicable.
Table 21.2. The Effect on Modal Frequency of Immersion in Water for a Four-Bladed Propeller with a BAR of 0.524 and P/D=0.65
| Flexural Vibration Modes | Frequency (Hz) | Λ |
---
| In Air | In Water |
| Fundamental | 160 | 100 | 0.625 |
| One-node mode | 230 | 161 | 0.700 |
| Two-node mode | 460 | 375 | 0.815 |
| Three-node mode | 710 | 625 | 0.880 |
| Four-node mode | 1020 | 1000 | 0.980 |
| Torsional vibration modes |
| One-node mode | 400 | 265 | 0.662 |
| Two-node mode | 670 | 490 | 0.731 |
| Three-node mode | 840 | – | – |
Compiled from Burrill, L.C., 1949. Underwater propeller vibration tests. Trans. NECIES 65.
Table 21.3. The Effect of Modal Frequency of Immersion for a Four-Bladed Propeller with a BAR of 0.85 and a P/D=1.0
| Mode Shape | Frequency (Hz) | Λ |
---
| In Air | In Water |
| i | 310 | 200 | 0.645 |
| ii | 395 | 280 | 0.709 |
| iii | 550 | 395 | 0.718 |
| iv | 805 | 605 | 0.751 |
| v | 808 | 650 | 0.804 |
| vi | 1055 | 810 | 0.768 |
| vii | 1180 | 910 | 0.771 |
| viii | 1345 | 1055 | 0.784 |
| ix | 1690 | 1330 | 0.786 |
| x | 1805 | 1435 | 0.795 |
Compiled from Burrill, L.C., 1949. Underwater propeller vibration tests. Trans. NECIES 65.
The influence of immersing a blade in water is chiefly to introduce an added mass term due to the water, which is set in motion by the blade. If a blade is considered as a single degree of freedom system at each of the critical frequencies, then the following relation holds from simple mathematical analysis for undamped motion,
(21.2)f=1 2 π k m
By assuming that the stiffness remains unchanged, then by combining Eqs. (21.1), (21.2) we have
Λ=equivalent mass of the blade equivalent mass of the blade+added mass due to water
The effect of the modal frequency on the value of Λ can be explained by considering the decrease in virtual inertia due to the increased crossflow induced by the motion of adjacent blade areas, which are vibrating out of phase with each other: the greater the number of modal lines, the greater is this effect.
With respect to the effect of immersing the propeller blades on the mode shapes, Fig. 21.6 shows that this is generally small in the examples taken from Burrill’s and Hughes’ work. While the basic mode shape is preserved, it is seen that there is sometimes a shift in position of the modal line on the blade.
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Fig. 21.6. Mode shapes in air and water for the two different propeller forms.
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Book 2019, Marine Propellers and Propulsion (Fourth Edition)J.S. Carlton FREng
Chapter
Propeller Design
2012, Marine Propellers and Propulsion (Third Edition)J.S. Carlton FREng
22.6.4 To Find the rpm of a Propeller to give the Required P D or P E
In this example, which is valuable in power absorption studies, a propeller would be defined in terms of its diameter, pitch ratio and blade area ratio and the problem is to define the rpm to give a particular delivered power P D or, by implication, P E. In addition it is necessary to specify the speed of advance V a either as a known value or as an initial value to converge in an iterative loop.
The procedure is to form a series of rpm values, N j (where j=1, . . ., k), from which a corresponding set of δ j can be produced. Then, by using the B p–δ chart in association with the known P/D values, a set of B p j values can be produced, as seen in Figure 22.11(c). From these values the delivered powers P D j can be calculated, corresponding to the initial set of Nj, and the required rpm can be deduced by interpolation to correspond to the particular value of P D required. The value of P D is, however, associated with the blade area ratio of the chart, and consequently this procedure needs to be repeated for a range of A E /A O values to allow the unique value of P D to be determined for the actual A E /A O of the propeller.
By implication this can be extended to the production of the effective power to correlate with the initial value of V a chosen. To accomplish this the open water efficiency needs to be read off at the same time as the range of B p j values to form a set of η o j values. Then the efficiency η o can be calculated to correspond with the required value of P D in order to calculate the effective power P E as,
P E=η o η H η r P D
Figure 22.12 demonstrates the algorithm for this calculation, which is typical of many similar procedures that can be based on standard series analysis to solve particular problems.
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FIGURE 22.12. Calculation algorithm for power absorption calculations by hand calculation.
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Book 2012, Marine Propellers and Propulsion (Third Edition)J.S. Carlton FREng
Chapter
Propeller performance characteristics
2007, Marine Propellers and Propulsion (Second Edition)JS Carlton
6.5.6 Newton–Rader series
The Newton–Rader series embraces a relatively limited set of twelve, three-bladed propellers intended for high-speed craft. The series (Reference 14) was designed to cover pitch ratios in the range 1.0 to 2.0 and blade area ratios from about 0.5 to 1.0.
The parent model of the series, based on a design for a particular vessel, had a diameter of 254 mm (10 in.). The principal features of the parent design were a constant face pitch ratio of 1.2 and a blade area ratio of 0.750, together with a non-linear blade thickness distribution having a blade thickness fraction of 0.06. The blade section form was based on the NACA a = 1.0 mean line with a quasi-elliptic thickness form superimposed. The series was designed in such a way that the propellers in the series should have the same camber ratio distribution as the parent propeller. Since previous data and experience was limited with this type of propeller it was fully expected that the section form would need to be modified during the tests. This expectation proved correct and the section form was modified twice on the parent screw to avoid the onset of face cavitation; the modification essentially involved the cutting back and 'lifting' of the leading edge. These modifications were carried over onto the other propellers of the series, which resulted in the series having the characteristics shown in Table 6.12 and the blade form shown in Figure 6.14.
Table 6.12. Extent of the Newton–Rader series
| BAR | Empty Cell | Face pitch ratio | Empty Cell |
--- |
| 0.48 | 1.05 | 1.26 | 1.67 | 2.08 |
| 0.71 | 1.05 | 1.25 | 1.66 | 2.06 |
| 0.95 | 1.04 | 1.24 | 1.65 | 2.04 |
Note: Box indicates resultant parent form.
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Figure 6.14. Newton–Rader series blade form
(Reproduced with permission from Reference 14)
Each of the propellers of the series was tested in a cavitation tunnel at nine different cavitation numbers; 0.25, 0.30, 0.40, 0.50, 0.60, 0.75, 1.00, 2.50 and 5.5. For the tests the Reynolds number ranged from about 7.1 χ 10 5 for the narrow-bladed propeller through to 4.5 χ 10 6 for the wide-bladed design at 0.7 R. Theresults of the series are presented largely in tabular form by the authors.
This series is of considerable importance for the design of propellers, usually for relatively small craft, where significant cavitation is likely to be encountered.
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Chapter
Propulsion
2013, Introduction to Naval Architecture (Fifth Edition)Eric C. Tupper BSc, CEng, RCNC, FRINA, WhSch
Propeller Features
The diameter of a propeller (Figure 8.4) is that of a circle passing tangentially through the tips of the blades. At their inner ends the blades are attached to a boss, the diameter of which is kept as small as possible consistent with strength. Blades and boss are often one casting for fixed pitch propellers. The boss diameter is usually expressed as a fraction of the propeller diameter.
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Figure 8.4. (A) View along shaft axis and (B) Side elevation.
The blade outline can be defined by its projection on to a plane normal to the shaft. This is the projected outline. The developed outline is the outline obtained if the circumferential chord of the blade, i.e. the circumferential distance across the blade at a given radius, is set out against radius. The shape is often symmetrical about a radial line called the median. In some propellers the median is curved back relative to the rotation of the blade. Such a propeller is said to have skew back. Skew is expressed in terms of the circumferential displacement of the blade tip. Skew back can be advantageous where the propeller is operating in a flow with marked circumferential variation. In some propellers the face in profile is not normal to the axis and the propeller is said to be raked. It may be raked forward or backward, but generally the latter to improve the clearance between the blade tip and the hull. Rake is usually expressed as a percentage of the propeller diameter.
Blade Sections
A section (Figure 8.5) is a cut through the blade at a given radius, i.e. it is the intersection between the blade and a circular cylinder. The section can be laid out flat. Early propellers had a flat face and a back in the form of a circular arc. Such a section was completely defined by the blade width and maximum thickness.
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Figure 8.5. (A) Flat face, circular back; (B) aerofoil; and (C) cambered face.
Modern propellers use aerofoil sections. The median or camber line is the line through the mid-thickness of the blade. The camber is the maximum distance between the camber line and the chord which is the line joining the forward and trailing edges. The camber and the maximum thickness are usually expressed as percentages of the chord length. The maximum thickness is usually forward of the mid-chord point. In a flat face circular back section the camber ratio is half the thickness ratio. For a symmetrical section the camber line ratio would be zero. For an aerofoil section the section must be defined by the ordinates of the face and back as measured from the chord line.
The maximum thickness of blade sections decreases towards the tips of the blade. The thickness is dictated by strength requirements and does not necessarily vary in a simple way with radius. In simple, small, propellers thickness may reduce linearly with radius. This distribution gives a value of thickness that would apply at the propeller axis were it not for the boss. The ratio of this thickness, t o, to the propeller diameter is termed the blade thickness fraction.
Pitch Ratio
The ratio of the pitch to diameter is called the pitch ratio. When pitch varies with radius, that variation must be defined. For simplicity a nominal pitch is quoted being that at a certain radius. A radius of 70% of the maximum is often used for this purpose.
Blade Area
Blade area is defined as a ratio of the total area of the propeller disc.
The usual form is
Developed blade area ratio=developed blade area disc area
In some earlier work, the developed blade area was increased to allow for a nominal area within the boss. The allowance varied with different authorities and care is necessary in using such data. Sometimes the projected blade area is used, leading to a projected blade area ratio.
Handing of Propellers
If, when viewed from aft, a propeller turns clockwise to produce ahead thrust it is said to be right handed. If it turns anti-clockwise for ahead thrust it is said to be left handed. In twin screw ships the starboard propeller is usually right handed and the port propeller left handed. In that case the propellers are said to be outward turning. Should the reverse apply they are said to be inward turning. With normal ship forms inward turning propellers sometimes introduce manoeuvring problems which can be solved by fitting outward turning screws. Tunnel stern designs can benefit from inward turning screws.
Forces on a Blade Section
From dimensional analysis it can be shown that the force experienced by an aerofoil can be expressed in terms of its area, A, chord, c, and its velocity, V, as
F ρ A V 2=f(v V c)=f(R n)
Another factor affecting the force is the attitude of the aerofoil to the velocity of flow past it. This is the angle of incidence or angle of attack. Denoting this angle by α, the expression for the force becomes
F ρ A V 2=f(R n,α)
This resultant forceF (Figure 8.6) can be resolved into two components. That normal to the direction of flow is termed the lift, L, and the other in the direction of the flow is termed the drag, D. These two forces are expressed non-dimensionally as
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Figure 8.6. Forces on blade section.
C L=L 1 2 ρ A V 2 and C D=D 1 2 ρ A V 2
Each of these coefficients will be a function of the angle of incidence and Reynolds number. For a given Reynolds number they depend on the angle of incidence only and a typical plot of lift and drag coefficients against angle of incidence is presented in Figure 8.7.
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Figure 8.7. Lift and drag curves.
Initially the curve for the lift coefficient is practically a straight line starting from a small negative angle of incidence called the no lift angle. As the angle of incidence increases further the curve reduces in slope and then the coefficient begins to decrease. A steep drop occurs when the angle of incidence reaches the stall angle and the flow around the aerofoil breaks down. The drag coefficient has a minimum value near the zero angle of incidence, rises slowly at first and then more steeply as the angle of incidence increases.
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Book 2013, Introduction to Naval Architecture (Fifth Edition)Eric C. Tupper BSc, CEng, RCNC, FRINA, WhSch
Chapter
Propeller performance characteristics
2007, Marine Propellers and Propulsion (Second Edition)JS Carlton
6.5.3 Gawn series
This series of propellers whose results were presented by Gawn (Reference 9) comprised a set of 37 three-bladed propellers covering a range of pitch ratios from 0.4 to 2.0 and blade area ratios from 0.2 to 1.1.
The propellers of this series each had a diameter of 503 mm (20 in.), and by this means many of the scale effects associated with smaller diameter propeller series have been avoided. Each of the propellers has a uniform face pitch; segmental blade sections; constant blade thickness ratio, namely 0.060, and a boss diameter of 0.2D. The developed blade outline was of elliptical form with the inner and outer vertices at 0.1 R and the blade tip, respectively. Figure 6.12 shows the outline of the propellers in this series. The entire series were tested in the No. 2 towing rank at A.E.W. Haslar within a range of slip from zero to 100 per cent: to achieve this the propeller rotational speed was in the range 250 to 500 rpm. No cavitation characteristics are given for the series.
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Figure 6.12. Blade outline of the Gawn series
(Reproduced with permission from Reference 9)
The propeller series represents a valuable data set, despite the somewhat dated propeller geometry, for undertaking preliminary design studies for warships and other high-performance craft due to the wide range of P/D and A E/A O values covered. Blount and Hubble (Reference 10) in considering methods for the sizing of small craft propellers developed a set of regression coefficients of the form of equation (6.17) to represent the Gawnseries. The coefficients for this series are given in Table 6.9 and it is suggested that the range of applicability of the regression study should be for pitch ratio values from 0.8 to 1.4, although the study was based on the wider range of 0.6 to 1.6. Inevitably, however, some regression formulations of model test data tend to deteriorate towards the outer limits of the data set, and this is the cause of the above restriction.
Table 6.9. Blount and Hubble coefficients for Gawn propeller series – equation (6.17)
| Thrust(K T) |
| n | c n | s(J) | t (P/D) | u (EAR) | v (Z) |
| 1 | −0.0558636300 | 0 | 0 | 0 | 0 |
| 2 | −0.2173010900 | 1 | 0 | 0 | 0 |
| 3 | 0.2605314000 | 0 | 1 | 0 | 0 |
| 4 | 0.1581140000 | 0 | 2 | 0 | 0 |
| 5 | −0.1475810000 | 2 | 0 | 1 | 0 |
| 6 | −0.4814970000 | 1 | 1 | 1 | 0 |
| 7 | 0.3781227800 | 0 | 2 | 1 | 0 |
| 8 | 0.0144043000 | 0 | 0 | 0 | 1 |
| 9 | −0.0530054000 | 2 | 0 | 0 | 1 |
| 10 | 0.0143481000 | 0 | 1 | 0 | 1 |
| 11 | 0.0606826000 | 1 | 1 | 0 | 1 |
| 12 | −0.0125894000 | 0 | 0 | 1 | 1 |
| 13 | 0.0109689000 | 1 | 0 | 1 | 1 |
| 14 | −0.1336980000 | 0 | 3 | 0 | 0 |
| 15 | 0.0024115700 | 0 | 6 | 0 | 0 |
| 16 | −0.0005300200 | 2 | 6 | 0 | 0 |
| 17 | 0.1684960000 | 3 | 0 | 1 | 0 |
| 18 | 0.0263454200 | 0 | 0 | 2 | 0 |
| 19 | 0.0436013600 | 2 | 0 | 2 | 0 |
| 20 | −0.0311849300 | 3 | 0 | 2 | 0 |
| 21 | 0.0124921500 | 1 | 6 | 2 | 0 |
| 22 | −0.0064827200 | 2 | 6 | 2 | 0 |
| 23 | −0.0084172800 | 0 | 3 | 0 | 1 |
| 24 | 0.0168424000 | 1 | 3 | 0 | 1 |
| 25 | −0.0010229600 | 3 | 3 | 0 | 1 |
| 26 | −0.0317791000 | 0 | 3 | 1 | 1 |
| 27 | 0.0186040000 | 1 | 0 | 2 | 1 |
| 28 | −0.0041079800 | 0 | 2 | 2 | 1 |
| 29 | −0.0006068480 | 0 | 0 | 0 | 2 |
| 30 | −0.0049819000 | 1 | 0 | 0 | 2 |
| 31 | 0.0025963000 | 2 | 0 | 0 | 2 |
| 32 | −0.0005605280 | 3 | 0 | 0 | 2 |
| 33 | −0.0016365200 | 1 | 2 | 0 | 2 |
| 34 | −0.0003287870 | 1 | 6 | 0 | 2 |
| 35 | 0.0001165020 | 2 | 6 | 0 | 2 |
| 36 | 0.0006909040 | 0 | 0 | 1 | 2 |
| 37 | 0.0042174900 | 0 | 3 | 1 | 2 |
| 38 | 0.0000565229 | 3 | 6 | 1 | 2 |
| 39 | −0.0014656400 | 0 | 3 | 2 | 2 |
| Torque (KQ) |
| n | C n | s(J) | t (P/D) | u (EAR) | v (Z) |
| 1 | 0.0051589800 | 0 | 0 | 0 | 0 |
| 2 | 0.0160666800 | 2 | 0 | 0 | 0 |
| 3 | −0.0441153000 | 1 | 1 | 0 | 0 |
| 4 | 0.0068222300 | 0 | 2 | 0 | 0 |
| 5 | −0.0408811000 | 0 | 1 | 1 | 0 |
| 6 | −0.0773296700 | 1 | 1 | 1 | 0 |
| 7 | −0.0885381000 | 2 | 1 | 1 | 0 |
| 8 | 0.1693750200 | 0 | 2 | 1 | 0 |
| 9 | −0.0037087100 | 1 | 0 | 0 | 1 |
| 10 | 0.0051369600 | 0 | 1 | 0 | 1 |
| 11 | 0.0209449000 | 1 | 1 | 0 | 1 |
| 12 | 0.0047431900 | 2 | 1 | 0 | 1 |
| 13 | −0.0072340800 | 2 | 0 | 1 | 1 |
| 14 | 0.0043838800 | 1 | 1 | 1 | 1 |
| 15 | −0.0269403000 | 0 | 2 | 1 | 1 |
| 16 | 0.0558082000 | 3 | 0 | 1 | 0 |
| 17 | 0.0161886000 | 0 | 3 | 1 | 0 |
| 18 | 0.0031808600 | 1 | 3 | 1 | 0 |
| 19 | 0.0129043500 | 0 | 0 | 2 | 0 |
| 20 | 0.0244508400 | 1 | 0 | 2 | 0 |
| 21 | 0.0070064300 | 3 | 0 | 2 | 0 |
| 22 | −0.0271904600 | 0 | 1 | 2 | 0 |
| 23 | −0.0166458600 | 3 | 1 | 2 | 0 |
| 24 | 0.0300449000 | 2 | 2 | 2 | 0 |
| 25 | −0.0336974900 | 0 | 3 | 2 | 0 |
| 26 | −0.0035002400 | 0 | 6 | 2 | 0 |
| 27 | −0.0106854000 | 3 | 0 | 0 | 1 |
| 28 | 0.0011090300 | 3 | 3 | 0 | 1 |
| 29 | −0.0003139120 | 0 | 6 | 0 | 1 |
| 30 | 0.0035895000 | 3 | 0 | 1 | 1 |
| 31 | −0.0014212100 | 0 | 6 | 1 | 1 |
| 32 | −0.0038363700 | 1 | 0 | 2 | 1 |
| 33 | 0.0126803000 | 0 | 2 | 2 | 1 |
| 34 | −0.0031827800 | 2 | 3 | 2 | 1 |
| 35 | 0.0033426800 | 0 | 6 | 2 | 1 |
| 36 | −0.0018349100 | 1 | 1 | 0 | 2 |
| 37 | 0.0001124510 | 3 | 2 | 0 | 2 |
| 38 | −0.0000297228 | 3 | 6 | 0 | 2 |
| 39 | 0.0002695510 | 1 | 0 | 1 | 2 |
| 40 | 0.0008326500 | 2 | 0 | 1 | 2 |
| 41 | 0.0015533400 | 0 | 2 | 1 | 2 |
| 42 | 0.0003026830 | 0 | 6 | 1 | 2 |
| 43 | −0.0001843000 | 0 | 0 | 2 | 2 |
| 44 | −0.0004253990 | 0 | 3 | 2 | 2 |
| 45 | 0.0000869243 | 3 | 3 | 2 | 2 |
| 46 | −0.0004659000 | 0 | 6 | 2 | 2 |
| 47 | 0.0000554194 | 1 | 6 | 2 | 2 |
(taken from Reference 10)
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