EXPERIMENT NO. 16
SURFACE TENSION.

THEORY AND METHOD.

The contractile force of a liquid surface per cm. length of surface is called the *surface tension* of the liquid.

If a metal is touched to the surface of water and then pulled away the water in close contact with the metal sticks firmly to it. The pulling away of the metal necessitates the separation of the water which sticks to the metal from the main portion of the liquid. This separation of water from water is opposed by the contractile force at the water-air surfaces. This contractile force is easily measured by the apparatus shown in Fig. 25.

A light metal ring is suspended from the end of a long light spiral spring, to which is attached a pointer P. By means of this pointer the extension of the spring caused by the surface tension can be measured on the meter stick AB.

A vessel containing the liquid whose surface tension is to be measured is raised until the liquid surface SS touches the metal ring all around the circumference. If the vessel is then slowly lowered the ring is pulled down until the upward force due to the extension of the spring is sufficient to overcome the surface tension force of the two liquid surfaces which extend downward, one from the inside and the other from the outside of the metal ring. As soon as these surfaces are broken the ring flies back almost to its original position. If B is the lowest position of the pointer and A its height just after the spring is at rest, A-B is the extension of the spring caused by the force of surface tension. The surface tension pull in grams can then be found by calibrating the extension of the spring with weights.

FIGURE 25

The total surface tension force thus measured is that of two surfaces, the length of each being the same as that of the circumference of the ring. To get the *surface tension* per cm. of length the total pull must be divided by twice the circumference of the ring.

Theoretically there are two corrections which should be applied to make this measurement exact. The first is due to the fact that there is some water lifted between the two surfaces attaching to the ring, and its weight makes the extension of the spring larger than it ought to be. The second is due to the fact that the pull of the surface tension on the ring is not exactly vertically downward and its effect on the spring is only its vertical component, making the extension less than it ought to be. Measurements by more accurate methods however show that these two sources of error nearly balance each other.

EXPERIMENT NO. 1

THE COMPONENT OF A FORCE.

DEFINITION.

The (rectangular) *component* of a force in any given direction is the force in that direction which would have the same effect as the actual force has, or

The *component* of a force in any given direction is the *effective-value* of the force in that direction.

MAGNITUDE OF COMPONENT.

Suppose AB (Figure 1) represents a certain force both in magnitude and direction. The component or effective value of AB in any direction AC is represented by the line AD, the projection of AB in the direction of AC. The point D is obtained by drawing a line from B perpendicular to the direction AC.

But

$$
AD = AB \cos BAD
$$

hence, the *magnitude* of the *component* of a force in any direction is equal to the *magnitude* of the force times the cosine of the angle between the direction of the force and the direction of the component.

OBJECT AND METHOD OF EXPERIMENT.

This experiment is to illustrate the meaning of the above definition, and to verify
the above rule for calculating the magnitude of a component. If the car of figure 2
were on the plane without any upward force applied at F, it would run down the
plane just as if a push were applied to it parallel to the plane. The only outside force
acting however is the weight of the car, which is known to be a *vertically* downward
force, and may be represented by the vertical line AB. Evidently this vertical force
has an *effect* in the direction AD. This effective force in the direction AD, or the
component of AB in the direction of the plane, can be balanced by an *equal* but oppo-
sitely directed force F. The size of F, required to balance the force AD is easily
found by experiment. It should be found within the limit of experimental errors, to
be equal to the component AD as calculated by the above rule.

FIGURE 1

FIGURE 2

DIRECTIONS.

Weigh the car and the scale pan separately to the nearest gram on an open
balance.

Incline the plane at 20 degrees and measure the lengths HL, HK, and KL, to the near-
est millimeter. Calculate what balancing force, F, will be required for this inclination
both for the empty car and when it is loaded with a 200 gram weight.

Calculate also what F ought to be for an incline of 30 degrees, the car being empty.
You are then to check these three calculated results experimentally, as in the following
directions.

Find what weight in the scale pan is just sufficient to keep the car moving up the
plane without change of speed after it has been started slowly. Find weight which
will just allow car to move down the plane without change of speed after it has been

Cp

started slowly downward. Add the weight of the scale pan in each case and take
the mean. The "up" weight is equal to F plus friction. The "down" weight is equal
to F minus friction. The mean of these two is the force which would just balance
the car if there were no friction.

Repeat with 200 grams added to the car.

Repeat with empty car with plane at a different angle.

COMPUTATIONS AND DRAWINGS.

From the measured values of HK and HL find the values of the cosines of angle KHL and then use the formula given on preceding page to get each of the components AD. Calculate per cent difference between these and measured values of F.

The values of F can also be found graphically as follows. Scale on graph paper (four small divisions per cm.) a triangle similar to each of the two experimental triangles HKL. On the line HK (extended if necessary) scale a vertical line to represent the weight of the car and its load. From the lower end of this vector draw line at right angles to the hypotenuse of the triangle, using a rectangular sheet of paper to get the perpendicular direction. Then measure from the top of vertical vector along the hypotenuse to its point of intersection with the perpendicular. This length represents the component of the weight of the car in the direction of the inclined plane and should of course be equal to the measured value of F.

On the graph be sure to mark both number of division of paper of each line and
the number of centimeters of length or grams force it corresponds to in the experiment.

PROBLEMS.

1. How much force does it take parallel to the incline to push a 4000 lb. car up a hill which has a vertical rise of 15 feet for every 100 feet along incline?

2. With what force does the above car push perpendicularly to the roadway?

EXPERIMENT NO. 2

SEVERAL FORCES IN EQUILIBRIUM

RESULTANT OF A NUMBER OF FORCES.

By definition, the resultant of a number of forces is the single force which is
equivalent to, and therefore can replace the several forces.

FIGURE 3

FIGURE 4

The resultant of two forces at right angles to each other, such as OX and OY of
Figure 3, is found to be equal, both in magnitude and direction, to OR, the diagonal
of the rectangle having OX and OY as two adjacent sides.

The resultant of several forces in general, such as OA, OB and OC, is found as
follows: Draw an X-axis and a Y-axis at right angles to each other. Obtain the
x-components of each force, Ox₁, Ox₂ and Ox₃ and their y-components, Oy₁, Oy₂ and
Oy₃. Find the algebraic sum of the x-components, which will be in Fig. 3, the force
OX. Find the algebraic sum of the y-components which in Fig. 3 will be OY. The
three forces OA, OB and OC have now been replaced by the two forces OX and
OY. These two forces, OX and OY, can however be replaced by their resultant OR.
The force OR is therefore the equivalent of the three forces OA, OB and OC and is
by definition the resultant of these three forces.

PROPOSITIONS TO BE TESTED EXPERIMENTALLY.

(1) Whenever a number of forces are in equilibrium their resultant is equal to zero and therefore the sum of the x-components is equal to zero and the sum of the y-components is equal to zero.

(2) When a number of forces are in equilibrium the resultant of all but one of the forces is equal in magnitude but opposite in direction to the remaining force.

DIRECTIONS.

Take three strings tied together at a point O, of Figure 4, with their free ends each attached to a spring balance. Clamp the balances to the table as in Fig. 4 adjusting the lengths of the strings or positions of clamps so that each balance will read more than 500 grams. Each balance should register a different amount of force.

Mark the intersection of two heavy lines near the center of a graph paper with a fine pencil point and lay this paper with the marked point directly under O where the cords meet.

Near the edges of the graph sheet write the value of each force, marking them with the letters A, B and C. Make a pencil point mark directly under each string to fix its direction.

Remove the paper and draw fine lines from each direction mark to O. On these force-direction lines scale lengths from O to represent the forces, indicating the end of each force vector by an arrow.

From the end of each force vector draw its x and y components, and along each component line write its computed value in grams with the proper plus or minus sign. Record both forces and components on the data sheet and add x and y components separately to test proposition (1).

For test of proposition (2) proceed as follows. Plot the sum of the x and also of the y components of the A and B forces and then plot the resultant of these two forces. Measure its length and calculate its value in grams and compare with the measured value of the third force C.

Repeat the experiment using four forces instead of three. Test the two propositions in a similar manner.

EXPERIMENT NO. 3
CRANE STRESSES.

OBJECT.

To illustrate the *Component* and the *Graphic* methods of computing the forces of tension and compression in a simple crane.

THEORY.

When the weight W is held out from the wall AB by means of a rigid rod AO and a string BO, the rod AO becomes a little shorter, and the string BO a little longer, than before the weight was suspended.

The pushing of the molecules in AO closer together gives rise to repulsive forces between them which tend to restore AO to its original length. The stress in the rod due to its compression is called a force of compression, and the amount of this force in this experiment is designated by the letter C. This force C pushes at O horizontally to the right.

The stress set up in the string BO (or in both string and spring balance St) due to the lengthening of the arm BO, is called a force of tension, and the amount of this tension force is designated in this experiment by the letter T. This force T pulls at O in the direction of OT upward.

The ring O is held in Equilibrium by the three forces W, T and C; or in other words, W, T, and C constitute a set of balanced forces.

The values of the forces T and C can be computed for any known value of W, if the lengths of the sides of the triangle AOB are known.

The tension force T can be observed directly on the spring balance St placed in the arm BO. The force C could also be observed directly by putting a compression spring balance in the arm AO, but balances of this kind are subject to considerable friction. More accurate values for C can be observed by finding with the spring balance Sc, the horizontal pull to the right which will just replace the force C.

DIRECTIONS.

The spring balance St should have a maximum reading of 2000 gms. Before the weight W is attached, the balance should be observed with no weight on it, to find its zero correction and to find which side of its pointer it is best to use for reading.

For the first set of observations make W about 1000 grams.

Before reading the scale St make the rod AO as nearly horizontal as you can by testing with a square or rectangular block.

FIGURE 5

The force T should be read to at least one-half of the smallest scale divisions, and corrected for your observed zero error, if there is any.

Measure the length of the rod OA from the center of the ring at O to the wall at A; the length OB from the center of ring at O to the point at which the string (or the line OB extended) meets the wall; and the length AB.

The above values for the length of the sides of the triangle and the value of T should be obtained before the scale Sc is attached at O. Then find the horizontal pull on the balance Sc which is just sufficient to free the nail at A from the wall. As you pull on Sc see that you do not change the reading of the balance St. Also hold your finger under the rod AO near the end A, so that you can tell when the rod at A becomes free.

Repeat with a larger weight W. On account of the stretch of the spring St, this will require new readings for BO and AB.

Repeat with a different length of the string BO.

## COMPUTATION OF T and C by COMPONENTS.

From the measured lengths of the sides of the triangle AOB calculate the numerical values of the cosines of the angles x and y. Set down expressions for the values of the vertical and horizontal components of the three forces T, C, and W, with T and C as the only unknown forces.

Express in the form of an equation the fact that the sum of vertical components (each with its proper plus or minus sign) is equal to zero.

As a second equation set the sum of the horizontal components equal to zero.

Solve these two equations for T and C and compare with the observed values.

Using calculated values of T and C, find sum of components of T, C and W, in direction BO. What should this sum be? Why?

## GRAPHIC METHOD FOR FINDING T AND C.

Scale a triangle similar to AOB on cross-section paper. This is done easily by first laying off a length to represent the side BA to as large a scale as conveniently possible, and then the length AO to the same scale. Then connect the points B and O by the line BO.

Measure from B along the direction BA a length BW representing to some convenient scale the value of the force W. From the point W draw a horizontal line, marking the point at which it intersects the line BO, or the line BO extended, with the letter C. The length WC represents the value of the force C, and the length BC the force T, on the same scale that you have chosen for representing the force W by the length BW.

gives an equation containing the *two* unknown quantities, C and F. If, however, an axis is then thought of as passing thru the point B, the same law of moments will give a second equation containing the same two unknown quantities C and F. From these two equations both C and F can be computed.

### DIRECTIONS.

A 5-lb. and a 2-lb. weight are suitable for $W_1$ and $W_2$. For the beam A-B use a meter stick on edge.

First adjust the pointer on one of the balances, by means of the screw on the top, so that it reads zero with nothing on it. Then weigh the weights $W_1$ and $W_2$ which you are going to use, recording their weights in ounces.

Place the triangular blocks and the beam on the balances as shown in the figure. Place the balances at such a distance that the edges of the blocks at C and F will be directly under the 10-cm. and 90-cm. mark of the meter stick. Before suspending the weight $W_1$ and $W_2$, adjust the screws on both balances until the scale pointers read exactly zero. This is done in order to get rid of the effect of the weight of the beam and the blocks.

Hang the weights $W_1$ and $W_2$ from any two points D and E, and record the distances DC, EC, and FC, to the nearest millimeter; record also the balance readings at C and at F, in ounces.

Write down the equations given by the law of moments: (1) for an axis thru C; (2) for an axis thru F; (3) for an axis thru A; (4) for an axis thru B.

Compute C and F from the equations (1) and (2); also from the equations (3) and (4).

Repeat the observations and calculations of F and C, with the positions of D and E considerably changed.

### OBJECT—PART 2.

To show how the law of moments may be applied to calculate the weight of a beam if one has a known weight and an edge upon which the beam may be balanced.

### DIRECTIONS.

With a triangular block on one of the balances adjust the pointer to zero. Locate the center of gravity, the point G of figure 7, by balancing the meter stick on the edge of this triangular block.

Having found the location of the center of gravity, balance the meter stick as shown in figure 7, using about 100 grams for the weight W. Record the distances CD, GD and AC. The weight of the meter stick, designated by X in the figure, acts as a single downward vertical force at G.

Write the equation for the equality of moments with respect to an axis thru D, and from this equation calculate the weight of the meter stick.

The weight of the meter stick is also to be calculated from the equality of moments equation regarding A as the axis. Three forces act on the stick which have moments around A. They are the weight W, the weight X of the stick, and the upward force exerted by the balance on the stick at D. This latter force is equal to $(W + X)$.

Repeat the balancing and calculation of the weight of the stick for a second position of the weight W.

Find the weight of the meter stick in grams on an open balance and compute per cent error of average calculated weight.

EXPERIMENT NO. 5
COEFFICIENT OF FRICTION.

THEORY.

For surfaces of given material and certain degree of smoothness the force of sliding friction is proportional to the force normal to the sliding surfaces, that is to the force which presses the sliding surface together.

Designating the frictional force by F and the normal force by N the above statement is expressed by the equation.

$$F = CN.$$

C is the proportionality factor and is ordinarily called the Coefficient of Friction. If this coefficient C is known, the force of friction is obtained by multiplying the normal force by the coefficient.

If the area of the contact surface is reduced while the normal force remains the same, the friction force per unit area is increased but the amount of area is decreased in the same proportion, so that, "the force of friction does not depend upon the size of the area of the surfaces in contact."

For example, the block A of Figure 8 will have about the same friction when sliding on its narrow as on its broader side.

That the amount of frictional force is independent of the size of contact area may seem to be in contradiction to the well known practice of reducing the frictional effect on axle bearings by reducing the size of the axle. In such cases however it is not the force of friction but the moment of the force of friction which is reduced.

The purpose of this experiment is to determine the coefficient of friction for wood on wood by two different methods and to show that the friction remains approximately the same when the area of contact is changed.

FIGURE 8

DIRECTIONS.

First Method.

Weigh the block A to the nearest gram.

(1) Lay the block A, broad side down, on the horizontal board BB, and find what weights must be put on pan W to make A move without change of speed after it has been slowly started. On account of the variation of smoothness on different portions of the board BB the motion of A should be observed only over a small distance and always over the same portion.

(2) Repeat with a 100 gram weight on the block A.

(3) Repeat with a 200 gram weight on the block A.

(4) Repeat with the block A, on its narrow side without load.

In this method the normal force is the weight of the block A plus any load it may have and the friction force is the weight in the pan plus the weight of the pan.

Compute C, the coefficient of friction, for each case, from the equation given above. This coefficient should be approximately the same for all four cases.

FIGURE 9

Second Method.

(1) With the block A placed on the same part of the board BB as before tilt the latter, as in Figure 9, until the block seems to move without change of speed after being started slowly down. The block A should be allowed to move only a short distance on account of the irregularities of the surface of BB.

When the proper inclination of the plane has been found measure the lengths of the sides of the triangle, BH, BK, and KH.

In Figure 9 the vector AW represents the weight of the sliding block (and its load); AD the component of AW in the direction of the plane; AF the force of friction; and AN the component of AW normal to the plane. When the block slides with constant speed down the plane the force of friction AF must equal the component force AD, and the coefficient of friction equals AD divided by AN.

From the known weight of the block A and the measured lengths of the sides of the triangle BHK compute the component forces AD and AN and the coefficient of friction.

Repeat the above observations and computations with 200 grams placed on the sliding block.

PROBLEM.

Calculate what force in the direction AF will be required to pull this block up the plane at constant speed when the angle at B is made 30 degrees. Use in this calculation the mean coefficient of friction that you have obtained by first method. Check this computation by setting the plane BB 30 degrees from the horizontal and observing what force is required. Explain calculation on inside page.

Calculated force.

Observed force.

Per cent Difference.

EXPERIMENT NO. 14

ARCHIMEDES PRINCIPLE.

THE DENSITY OF A SOLID AND OF A LIQUID.

THEORY AND METHOD.

The purpose of this experiment is to find the density of a solid body, and of some liquid other than water. The density is the mass of one unit volume.

The density of a regular solid body can be obtained by finding its volume from measurement of its dimensions and its mass by weighing. If a body is small or of irregular shape it is difficult to determine its volume from a measurement of its dimensions, but it can be readily obtained by the use of the Principle of Archimedes. This principle is "A body immersed in a fluid is buoyed up by a force equal to the weight of a quantity of the fluid of the same volume as that of the immersed body."

The method of finding the volume by this principle is as follows. The body whose volume is to be found, B of figure 21, is first weighed in air, and then when immersed in water. Its pull on the balance arm in the latter case is less than its weight by the buoyant force of the water. The difference between the balance weight in air and when the body is immersed in water is therefore the buoyant or upward force of the water on the body B. But according to the above principle this is the weight of a volume of water equal to the volume of the body B. The density, or weight of one unit of volume, of water, at the temperature of the experiment, having been found from tables, the volume of water displaced by the body B, is calculated by dividing its weight by the weight of one unit of volume. The volume of the body B is of course equal to the volume of the water it displaces.

This Principle can be used to calculate the density of a liquid as follows. The solid B having been weighed in air and in water and its volume calculated as explained above, it is weighed when immersed in the liquid whose density is to be found. The difference between the weight of B in air and in this liquid, is the weight of that volume of the liquid equal to the volume of the solid. Knowing the weight of this volume of the liquid, the weight of a unit volume of it, which is its density, is found by division.

In the above descriptions it is assumed that the buoyant force of the air on the solid body may be neglected. The weight of displaced air, its buoyant force, is in most cases so small that it may be neglected, unless great accuracy is desired, or the solid body has a density which is not large in comparison with the density of air.

FIGURE 21

DIRECTIONS.

The weighings should be made with an enclosed balance. An accuracy of one milligram is sufficient. This may be attained by taking the mean of two successive swings of the balance pointer, one on either side of the middle.

The balance pans are first to be dusted with a brush, and the zero-point for empty pans found.

The solid whose density is to be found, unless otherwise directed by the instructor, is a glass bottle stopper. The stopper is to be hung from the hook of the balance arm, by a light thread, at such a height, that when immersed in a glass of water as shown in the figure, it will swing up or down about one cm. without touching either the surface of the water or the bottom of the glass. This height should be tested with the glass in position but without water in it.

The weight of the stopper in air is then to be found, by finding the weight to the nearest milligram, which will bring the pointer of the balance as near as possible to the zero-point which you have found with the pans empty. Be careful to see that the wires of the balance pan, as the balance swings, do not touch the sides of the glass jar.

Then weigh the stopper when immersed in water to the nearest milligram.

Take the temperature of the water and find its density at this temperature from some book of Physical or Chemical Constants.

Weigh the stopper when immersed in alcohol.

Calculate density of the stopper and also of the alcohol.

PROBLEMS.

1. A certain body weighs 8.576 grams in air, 4.965 grams in water at 20° Cent., and 4.215 grams in a second liquid. Find the density of the solid body and of the second liquid.

2. Taking density of air as 0.0012 grams/cm³, calculate the buoyant effect of the air on the glass stopper you have weighed. Taking account of this buoyant effect recalculate the density of the glass stopper.

RECORD.

Weight of stopper in air
" " " water
Weight of equal volume of water
Temperature of water
Density of water
Volume of displaced water
Volume of the glass stopper
Density of the glass stopper

Weight of stopper in alcohol
Weight of alcohol displaced
Volume of alcohol displaced
Density of alcohol

EXPERIMENT NO. 18

BOYLE'S LAW

THEORY AND METHOD.

If the volume of a definite quantity of gas is diminished or increased while the temperature remains constant, the pressure changes inversely as the volume. This is known as Boyle's Law and is expressed by the equation

$$PV = K$$

in which P is the pressure, V the corresponding volume and K a constant.

This law holds true for all gases very approximately except when they approach the state of saturation.

The *object* of this experiment is to verify this law for a quantity of air enclosed in a glass tube.

The quantity of air whose pressure and volume is to be changed is contained in the upper part of the tube AC, and the stopcock at C is to be kept closed during the whole of the experiment. The lower part of the tube as well as the rubber tube and part of the tube B is to be filled with mercury. The glass tubes AC and B are arranged so that they can be clamped at various heights on a wooden stand. The tube B is open at its top so that the pressure at the top of the mercury in it is always the same as the atmospheric pressure which is to be read on the barometer. If the tube AC is kept in one position and the tube B raised the gas in AC will be diminished in volume and its pressure increased, while when B is lowered the gas in AC will be increased in volume and its pressure decreased.

FIGURE 27

At any setting of the tubes the pressure of the gas in AC will be practically the same as the pressure at the top of the mercury column A. When the mercury is higher

in B than A, the pressure of the gas in AC is equal to the atmospheric pressure plus pressure due to mercury column of height AB. When the mercury level B is lower than A the pressure of gas in AC is equal to atmospheric pressure minus that due to column of mercury of height AB. In this way the pressure of the gas in AC can be obtained for each setting by reading the atmospheric pressure and then adding to it or subtracting from it the difference in level of mercury at A and B.

If the tube AC is of uniform bore with cross-section equal to S the volume of the enclosed gas column of height H is equal to HS. Substituting this value for V in the above equation gives

$$PHS = K$$

Since S, the cross-section, is constant the law is proven true if

$$PH = K^1$$

where K¹ is another constant equal to K/S.

If the relation between P and H is shown by a curve, using the values H as abscissas and those of P as ordinates, the form of the equation (PH = K) shows that the curve should be one branch of an hyperbola symmetrically situated with respect to the two rectangular axes.

## DIRECTIONS.

Read the laboratory barometer (do not remove it from its position on the wall) at the beginning of the experiment and also at the end. If there is any difference in the readings at the two times take the mean of the two readings as the pressure of the atmosphere during the experiment.

Clamp the tube AC so that the top of the air column at C is about the 80 mark of the scale. Clamp the tube B as high as you can without allowing mercury level in B to run down into the rubber tubing.

Using a paper straight edge running across tubes and scale, read accurately the height of top of air column at C and the mercury levels at A and B.

Leaving the tube C clamped in the same position get readings of the mercury levels for eight different positions of the tube B, lowering the latter tube about 10 cms. between each set of readings.

Compute the pressure of gas in AC for each setting as explained above then multiply each pressure by its corresponding length of gas column H. These products should be approximately the same for all the different settings.

## PROBLEMS.

1. Owing to the weight of air in column AC, the air pressure is slightly larger at A than C. Assuming the density of air to be 0.0012 grams per cm³, calculate this difference in pressure 10 cms. of mercury, when the column of air AC is 15 cms. long. Is this enough for you to read in this experiment?

2. Calculate the lbs. per square inch pressure of the air in AC, when at its least and greatest volumes. Use density of mercury at the temperature of the experiment.

3. Using lengths of the air column AC as abscissas, and pressures of gas in AC as ordinate, plot curve showing the variation of the pressure with the volume of the gas.

4. Using mean value of PH from your observation calculate values of H, for P = 120, 100, and 40 and values of P for H = 40, 60, 80, 100 and 120. Plot these values using scale of one small division equal to 2 cms. for both H and P.

EXPERIMENT NO. 6
LINEAR ACCELERATION.

FIGURE 10

FIGURE 11

THEORY AND METHOD.

If two equal masses, M and M¹, are hung by a paper tape over the wheel W, the gravity forces on the two equal masses balance each other and they will be held in equilibrium. But if a small additional mass, m₁, be added to M both the masses M and M¹ will move with a *constant* acceleration due to the unbalanced weight of the mass m₁.

This acceleration can be calculated from the general equation,

$$F = kma \quad (Equation 1)$$

In this case F, the unbalanced force causing the acceleration, is the weight of m₁ in
grams or pounds; k is 1/980 or 1/32.16; m is the total mass accelerated which is

$$M^1 + M + m_1 + \text{Equivalent mass of wheel.}$$

By "equivalent mass of wheel" is meant the mass which would have a linear inertia
(resistance to change of speed) equal to that of the rotating wheel. It can be calculated
from the mass and shape of the wheel and *will be found marked upon the wheel*.

Equation 1 is to be tested experimentally by means of the apparatus shown in Fig-
ure 10. V is vibrating rod which makes a vibration back and forth in exactly 1/5th of
a second. An inked brush, B, is attached to this vibrator, and traces a curve on the tape
similar to that of Figure 11, when the tape is moved with constant increase of speed.
From this curve on the tape the actual acceleration (i. e, increase in speed per second)
can be calculated, and compared with the accelerations calculated from the equation
by substituting the known values of F, k, and m.

["DIRECTIONS.", "An additional tape (not shown in the Figure) is to be hung from the bottom of the weight M and M¹. Before using the inked brush see to it that, as the weights are released, the tape will run straight and without touching the sides of the frame or the swinging weight holder, since a little bit of friction will spoil the measurement.", "Before adding the rider m₁ a small weight is to be added to M¹ to overcome friction. The size of this weight is to be found by trial. It should be just enough to make M¹ go down without decrease or increase of speed, after being slowly started.", "Get the first curve with a rider of at least 10 grams—and a second with 0.02 lbs. or more.", "Mark each curve with its proper values of m₁, M, M¹, and Equiv. Mass.", "Starting with the second or third vibration, mark every fourth center of the curve (the points a, c, e, etc. in Figure) with a pencil dot. Measure the succeeding lengths a-c, c-e, etc., and record them in the column marked \"distances.\"", "COMPUTATIONS.", "To calculate the acceleration from experiment.", "In the first column are to be placed the distances traversed in successive intervals of 2/5 sec. as measured on the tape. In the second column are to be placed the average velocity in cms. per sec. during each time interval. Then calculate the change in velocity for each successive 2/5 second and the increase in velocity in cms. per sec. per sec.", "Finding the Value of \"g\"—the acceleration of gravity.", "The above measurements may also be used for calculating the value of \"g\" the acceleration given a freely falling body by gravity.", "The constant k of the above equation when m is measured in grams or pounds, F in grams-force or pounds-force and a in cms. or feet per second per second is equal to 1/g, hence the equation may be written.", "$$F = ma/g \\quad (Equation 2)$$", "Substituting experiment values for F and m and the value obtained for a from tape measurements calculate value of g from Equation 2. Compute per cent error as compared with known value—980 or 32.16.", "<table><thead><tr><th colspan=\"4\">First Tape m₁ = ______ grams</th><th colspan=\"4\">Second tape m₁ = ______ pounds</th></tr><tr><th>Distances</th><th>Average Velocity</th><th>Change in Velocity</th><th>Acceleration</th><th>S</th><th>Average Velocity</th><th>Change in Velocity</th><th>Acceleration</th></tr></thead><tbody><tr><td></td><td></td><td>X</td><td>X</td><td></td><td></td><td>X</td><td>X</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr

EXPERIMENT NO. 7

FORCE OF A HAMMER BLOW.

VELOCITY, ACCELERATION AND DISTANCE.

THEORY.

While the force of a hammer blow varies from the beginning to the end of the blow, the average force during the blow can be calculated if we know the mass of the hammer, and the acceleration (loss of speed per second) with which it is stopped, by use of the equation

$F = k m a$

The acceleration, *a*, of stopping, can be calculated from a knowledge of the speed of the hammer at the beginning of the blow and the distance it goes during the blow.

FIGURE 12

The speed of a hammer at the end of its free fall can be calculated from the distance it falls and the known value of the acceleration of gravity—980 cms. per second per second. If the hammer M falls the distance H before hitting a nail, and the time taken for this fall is t seconds, the average speed during the fall is H/t per sec. Calling the speed as it strikes the nail V, and remembering that since the initial speed is zero, the final speed will be twice the average speed, we get

$V = 2H/t$

But since the increase of speed per second of fall is 980 cms. per second,

$V = 980 t$

Multiplying the two above equations gives

$V^2 = 2H \ 980 \quad \text{Equa. 1}$

Suppose the hammer drives the nail a distance of d cms. in t¹ seconds. If the hammer, during the blow, loses the speed V cms. per second, its negative acceleration designated by *a*, is

$a = V/t^1$ from which $t^1 = V/a$

The average speed during the blow is V/2, and the time of the blow V/a, hence the distance traveled is given by equation

$d = (V/2) (V/a) = V^2/2a$, hence $V^2 = 2ad$ \quad Equa. 2

Since the V in both equations is the same, we have

$2ad = 2 H \ 980 \text{ or } a = 980 \ \text{H/d}$

In case of a freely falling hammer the acceleration with which it is stopped can be calculated by this equation, if H and d are measured.

DIRECTIONS.

Weigh the iron block used as a hammer, to the nearest gram.

Start nail into wood about ½ inch, normal to the surface of the block, and measure
its projecting length to the nearest ½ millimeter.

Place the block with the nail directly under center of hammer. Measure the dis-
tance from the bottom of the hammer to the top of the nail. Allow hammer to fall
freely from this height. Take the nail and block out of frame and again measure
projecting length.

The difference between the two projecting lengths is the distance the nail goes in.

Raise the hammer to the same distance above top of nail as before and repeat four or five times.

Calculate H/d for each blow and multiply by 980 to get the acceleration in cms.
per sec. per sec. Substitute the values of k, m, and a in the equation F = kma to get
the force in grams of each blow.

Also calculate the force of each blow without finding the acceleration, on the as-
sumption that the work done by the gravity force as the hammer falls equals the work
done in driving the nail into the wood. Remember that in each case work equals force
times distance.

Weight of hammer ........................grams ........................lbs.

Height of fall ............................cms. ............................ft.

PROBLEMS.

Write out in full the solution of the following problems without the use of any
formula. Use 32.2 ft. per sec. per sec. as the acceleration of a freely falling body and the
fact that *average* speed by time always gives distance.

1. Find the speed of a body at the end of 4 seconds of fall.

2. Find the distance a body falls in three seconds.

3. Find the distance a body falls *during* the first second of its fall.

4. Find the distance a body falls *during* the fourth second of its fall.

5. Find the time for a body to fall 100 feet.

6. Find speed a body will have at the end of a 100 foot fall.

EXPERIMENT NO. 8
CENTRIPETAL AND CENTRIFUGAL FORCE

THEORY AND METHOD

To keep a body moving with the constant speed V in a circle of radius r requires a constant acceleration towards the center equal to V²/r. If the body makes N revolutions per second around the circle its speed V is given by the equation,

$$V = 2\pi r N$$

and its acceleration toward the center is

$$V^2/r = (2\pi N)^2 r$$

If the revolving body has a mass of M we obtain the force towards the center required to keep it moving in the circle from the general equation

$$F = k M a$$

by substituting the above value for the acceleration towards the center for a. That is

$$F = k M (2\pi N)^2 r$$

where F is the force pulling the body towards the center, as well as the force with which the mass M pulls away from the center.

This equation is to be tested in this experiment by means of the apparatus shown in Figure 13. The mass M is to be revolved at a constant speed which is measured by means of the revolution counter attached at W. M¹ is another weight fixed to the rod R at the same distance from the center as the mass M in order to prevent vibration of the apparatus. The mass M slides freely on the rod R and is kept from sliding too far by means of a double cord passing over the pulley wheels at C and connected with the spring balance B. V is a swivel joint which allows the lower cord to revolve without rotation of the pointer P and the balance B. The amount of the centrifugal force exerted by M is read directly on the spring balance. The position of the pointer serves to indicate when the speed of rotation is constant.

FIGURE 13

DIRECTIONS.

Unless otherwise informed by the instructor, the rotating mass M is to be considered 1 lb.

See that the balance height is so adjusted that the body M does not have to go quite to the end of the rod R to give 8 lbs. on the balance.

Measure to the nearest millimeter, the distance from the center of gravity of M to the axis of rotation for a balance reading of 6 lbs. This is the distance "r" to be used in the formula for calculating centrifugal force, but it must be reduced to feet before substituting, if the force F is to be in lbs.

Set the fixed mass $M^1$ at the same distance on the other side of the axis.

If two persons are working on the experiment one should turn the wheel and observe the balance reading while the other observes the time by means of the second hand of a watch or clock.

Turning steadily in order to make the centrifugal force just 6 lbs., find the number of revolutions made in one minute, by means of the revolution counter. Since this number is hard to observe with accuracy, at least three separate trials for the same force, should be recorded, and the average used in the computation. Dividing this average number of revolutions per minute by 60 gives the number of revolutions per second, which is the N of the formula.

Repeat the experiment, turning fast enough to give a centrifugal force of eight pounds.

Compute each force by the equation given, using $1/32.2$ as the value of "k" in the formula.

## PROBLEMS.

1. How many revolutions per second would the rotating weight of this experiment at a radius of 6 inches have to make to give a 80 lb. pull on the string?

2. Suppose an 80-ton locomotive goes around a railroad curve of 600 feet radius at a speed of 60 feet per second. If the track is horizontal what is the horizontal push of the flanges of the wheels on the outer rail? For a track width of 4 ft. 8 in. how much should the outer rail be elevated, to make the total thrust of the engine normal to the track?

## DERIVATION.

Prove by the vector diagram method that whenever a body moves with a constant speed V around a circle of radius r, it has an acceleration towards the center equal to $V^2/r$.

Substitute a 1000 gram weight for the 1500 grams and obtain a second trace.

## CALCULATIONS.

The calculation of the linear acceleration of a point on the circumference of the wheel from the trace on the paper is made in the same way as in Experiment No. 6, except that in this case the vibrator makes a complete vibration in exactly 1/6th of a second.

Measure the successive wave lengths on the tapes and enter them as distances. They are the linear distances traversed by a point on the circumference while the vibrator makes one vibration. Dividing each distance in cms. by the time of one vibration in seconds gives the average velocity in cms. per second during each vibration. Subtract each of these average velocities from the following one in order to get the change in velocity in cms. per second during one vibration period.

Divide the linear acceleration in cms. per sec. by R, the radius of the wheel, to obtain the angular acceleration in radians per sec. per sec. This is the A of the equation.

To obtain the moment of the force it must be remembered that the force F producing the acceleration is the tension of the string and that this tension is not as great as it would be if the weight were not moving with acceleration. The actual tension can be calculated from the weight and the acceleration with which it is moving, as follows.

The unbalanced force giving the weight its acceleration is the force of gravity on it, m grams, minus the tension of the string, F. Therefore, calling the acceleration of the weight a¹

$$m - F = k ma = ma^2/980 \\ F = m - (ma^2/980) = m (1 - a^2/980)$$

To obtain a¹ multiply the angular acceleration in radians, as found above, by r, the radius of the axle on which the string is wound.

## PROOFS.

(a) Explain why multiplying the difference of measured lengths by 36 gives acceleration of tape.

(b) Give derivation of Equation 1.

EXPERIMENT NO. 9
SIMPLE HARMONIC MOTION.

DEFINITIONS.

A Simple Harmonic Motion is a back and forth motion in a straight line whose acceleration is variable but always proportional to, and oppositely directed to, the displacement from the midpoint of the vibration.

Since the acceleration of a body having any mass is always proportional to the unbalanced force acting on it, a body will vibrate in simple harmonic motion provided the unbalanced force acting on it is always proportional to, and oppositely directed to, its displacement from its midpoint of vibration.

THEORY AND METHOD.

In all simple harmonic motions the frequency, or the number of vibrations per second, is independent of the amplitude, or width of the vibration.

A general formula can be derived for the frequency which can be applied to all simple harmonic motions. The derivation of this equation, using figure 14, is as follows.

In this figure, if the point P¹ be conceived of as moving around the circle of radius "r", with a constant speed "V", the point P which is the horizontal projection of the moving point P¹ on the diameter AB, will move up and down the line AB with a motion which is simple harmonic, according to the first of the above definitions.

The number of vibrations down and up again of P in a second is the same as the number of revolutions of the point P¹.

Since the point P moves up and down just as fast as the point P¹ moves up or down, the acceleration of P is at any instant equal to the vertical component of the acceleration of P¹. Although the acceleration of P¹ is always constant in amount it is always directed towards the center, and its vertical component is therefore always changing in amount.

Designating the acceleration of P¹ by the letter a¹, at the instant P¹ is in the position shown in the figure, its acceleration will be represented by the diagonal of the smaller of the two right angle triangles of the figure. At the same instant its vertical component will be equal to "a", the vertical leg of the same triangle. This side "a" evidently represents the acceleration at this instant of the point P which is moving with the simple harmonic motion. At the same instant "x", the vertical side of the larger of the two right angle triangles, represents the distance of the point P from its midpoint, or the displacement of P. From the similarity of the two triangles,

$$ \frac{\bar{a}}{a^1} = \frac{\bar{x}}{r} \text{ or } a = a^1 \frac{x}{r} $$

Since P¹ is moving with the constant speed V in the circle its acceleration is always equal to V²/r, or 4π²N²r. Substituting this value for a¹, we have

$$ a = 4\pi^2 N^2 x \text{ or } N^2 = \frac{a}{x 4\pi^2} $$

FIGURE 14

FIGURE 15

This equation shows that the frequency of any simple harmonic motion can be calculated if the ratio of its acceleration to its corresponding displacement is known.

For *bodies* vibrating with simple harmonic motions it is generally simpler to find the force acting on the body than its acceleration. The above equation is transformed into one giving the frequency in terms of the unbalanced force which causes the motion, and the mass of the vibrating body. If force is measured in grams-force, mass in grams-mass, and acceleration in cms. per sec. per sec.

$$F = \frac{ma}{980}, \text{ or } a = \frac{F}{m} 980$$

Substituting this value of a in the above equation for frequency gives

$$N^2 = \frac{F}{x} \frac{980}{4\pi^2 m}$$

as the general form of the frequency equation for a simple harmonic motion of a body whose mass is m.

In this experiment the above equation is to be tested by observing and calculating the frequency of a known mass vibrating up and down as it hangs from a spring

whose $\frac{F}{x}$ can be easily measured.

## DIRECTIONS.

First find the value of F/x as follows.

The apparatus to be used is similar to that shown in figure 15, but it does not have the horizontal pointer running off sidewise from the top of the weight pan to the scale. Instead, the height of the spring with reference to the scale is observed by sighting along the top of the round part of the rod of the weight hanger. When this top is in line of sight with its image in a mirror scale, the scale height is determined without parallax.

Observe this height on the scale first with 200 grams on the weight hanger, and then with 300, 400, and 500 grams, respectively, on the hanger. The differences between successive heights gives three observations for the extension of the spring for 100 grams, from which F/x is calculated. This F/x is evidently the force required to extend the spring one centimeter. The mean of the three values obtained is to be used in the above equation for calculating N.

Having found this force-constant for the spring, observe the vibration frequency for each of the cases when 300, 400 and 500 grams are on the weight hanger. This should be done by pulling the weight down just a little and then let it vibrate. Count the number of times it goes down and back in 100 seconds. You should count this number for each weight at least three times and record the average.

Substitute in each case for "m" in the equation, the mass on the hanger plus the mass of the hanger itself, and calculate the frequency you should have for each mass.

EXPERIMENT NO. 12
PENDULUM.

THEORY AND METHOD.

A rigid body, as AD, Figure 19, supported on a horizontal axis so that it will oscillate back and forth when pulled to one side from its equilibrium position and then let go, is called a *physical pendulum*. For small oscillation the time for one oscillation of such a pendulum is independent of the arc through which it swings.

Theory shows that the motion is *rotary harmonic* and its period given by the equation

$$T = 2\pi\sqrt{kI/MX}$$

where T is the time in seconds for one oscillation back and forth, k the constant 1/980, I the moment of inertia, M the mass of the pendulum in grams and X the distance from the axis of suspension to the center of gravity, G, of the pendulum.

If the position of the axis is changed, both I and X change and in general there is a different oscillation period for each axis position; but in case the pendulum is a uniform bar for each axis position near its end, as A, there is a second axis position, as B, about 1/3 way down on the bar for which the period is the same. There are two corresponding axis positions, below the center of the bar, such as C and D, for which with the bar inverted the oscillation period will also be the same as for the axis A.

The position C is called the center of oscillation or the center of percussion corresponding to the axis of suspension A.

A *simple* pendulum is defined as one which has all its weight located at one point. This can never be actually attained in practice but is approximated by a small ball suspended by a light thread, as in Figure 20.

If the distance from the point of the suspension to the center of gravity of the small ball is called L the moment of inertia of such a pendulum will be approximately equal to ML². Substituting this value for I in the above equation, and for X the length L, the above equation becomes

$$T = 2\pi\sqrt{kL}$$

or as it is more often written, substituting 1/g for k,

$$T = 2\pi\sqrt{L/g}$$

FIGURE 19

FIGURE 20

The distance A-C, from the axis of suspension A in the physical pendulum to the axis position C on the other side of the center of gravity for which the oscillation period

of the inverted pendulum is the same as with the axis at A, is equal to the length of the ideal simple pendulum which vibrates with the same oscillation time as the bar AD when suspended from either A or C.

OBJECT.

The purpose of this experiment is (1) to verify the above equation for the period of a physical pendulum, (2) to test the statement that the length AC is equal to that of the simple pendulum that has the same period, and (3) to calculate the value of "g" from observations of the period and length of the simple pendulum.

DIRECTIONS.

Find the center of gravity of bar to be used as pendulum by balancing on a thin edge.

Observe and record the time of 50 complete vibrations, when suspended from a nail through the topmost hole, A. Measure the distance X from the top of this hole to the center of gravity.

Repeat for an axis about two inches from end A, for axis no more than an inch from middle and for axis at C, two-thirds way down from A.

Observe the time of 50 complete vibrations of the simple pendulum whose length L is exactly 100 cms., and also when its length is exactly 25 cms., and also for length equal to AC of the physical pendulum.

CALCULATIONS.

The I, the moment of inertia for a bar of length L for an axis thru its center of gravity is $ML^2/12$. For an axis at distance X from the center of gravity $I=ML^2/12+MX^2$. Substituting this value for I in the equation for vibration period of a pendulum gives

$$T = 2\pi\sqrt{k}\frac{(L^2/12 + X^2)}{X} = 2\pi\sqrt{k}\sqrt{\frac{L^2}{12X} + X}$$

In this $2\pi\sqrt{k}$ and $\frac{L^2}{12}$ being the same for all axes, should be worked out first.

Calculate the value of "g" from the observation of time for each of the three lengths of simple pendulums.

DERIVATION.

Starting with the equation for the frequency of a Simple Harmonic motion given in experiment No. 9, *derive* the equation for the period of a physical pendulum.

Experiment No. 17

Density of Gases.

A flask of known volume is evacuated for 30 seconds and weighed as accurately as possible. The weighing is to be done by the method of vibrations used in Experiment 13. The second determination of the Rest Point of the balance should check with the first to a division or sc. If it does not it indicates leakage in the pet cocks and the flask should be re-evacuated. Record the weights on the balance. This is roughly the weights of the flask and is to be corrected by use of the sensitivity of the balance. One of the pet cocks is now opened without removing the flask from the balance. The cock should be left open so that it will fill with air at atmospheric pressure and room temperature. The flask, filled with air, is now weighed as before and the rest point determination checked by a second set of oscillation readings. The sensibility of the balance is now found by adding one milligram and determining the new rest point. The change in rest point for one milligram is now known and the first two weighing may be made exact by determining the additional weight necessary to bring the rest points back to the original zero of the balance. The flask is now filled with illuminating gas by connecting one opening to the pump and one to the gas tube. First evacuate. Second close the pet cock next to the pump and open the pet cock next to the gas. Close this cock and open the other to re-evacuate and repeat this cycle three times ending with the flask full of pure illuminating gas. Remove the tubing and open one pet cock for an instant only to assure the gas being at atmospheric pressure. Weigh as in the case of air.

Find the density of air and gas at observed temperature and pressure. Reduce these values to densities at 0°C and 76 cm. of mercury by the relation

$$d_o = dx \frac{76}{P} \times \frac{273 + t}{273}$$

**Problem:** Measure dimensions of the laboratory and, using the density of air as you find it, calculate the weight of air in the laboratory in grams and in pounds.

EXPERIMENT NO. 15

THE DENSITY OF A LIQUID

OBJECT:

In this experiment the density of alcohol (or some other liquid as directed by the instructor) is to be found by three different methods.

By the Specific Gravity Bottle.

METHOD.

Several different forms of specific gravity bottles are in use, one of the most common forms being that shown in the Figure 22. It consists of a glass bottle with a glass stopper through which runs a narrow hole. In order to fill the bottle with exactly the same volume of liquid each time sufficient liquid is poured in the bottle to cause a little of it to run out through the hole of the stopper when the latter is inserted in the bottle. The number of cubic centimeters of volume of the bottle is generally marked upon it but it is never wise to depend upon the accuracy of such markings. The actual volume of each bottle should be determined by finding the weight it holds of a liquid whose density is known, the most convenient liquid for this purpose being water at a known temperature. After the exact volume of the bottle has been determined by weighing with water, the weight of the liquid just filling it, whose density is to be determined, is found and its density calculated.

FIGURE 22

DIRECTIONS.

If the bottle appears clean and dry on the inside you may use it without cleaning; but if there is any suspicion of liquid or moisture on the inside it should be rinsed with a little alcohol and dried by drawing air thru it with a suction pump.

Weigh the empty dry bottle with its stopper to the nearest 0.01 gram.

Fill the bottle with distilled water of known temperature, wiping off any that may run out of the top of the hole with filter paper, and weigh.

Empty the water and fill with the liquid whose density is to be found, and weigh to the same accuracy as before.

Divide the weight of the water (without weight of bottle) by its density as obtained from a table of contents, to obtain the volume of the bottle.

Calculate the density of the liquid from its weight and volume.

By the Mohr (Westphal) Balance.

METHOD.

This method consists in comparing the buoyant force of the liquid whose density is to be measured with the buoyant force of distilled water on the same glass plummet, F of Figure 23.

The glass plummet and the weights are supposed to be adjusted by the maker of the instrument so that with the plummet immersed in water at 15 degrees Cent. the buoyant force is exactly balanced when the largest of the hook-shaped weights is hung from the hook on the end of the beam of the balance. Other hooks just one-tenth, one-hundredth, and one-thousandth as heavy as the largest hook come with the instrument, and the beam is divided into ten equal parts. Any one of the weights can be hung from any division of the beam. The buoyant force of any liquid, relative to that of water, can therefore be read directly to four places, as soon as the four weights are put on such divisions of the beam as to give a balance.

Such a balance, however, should not be used without testing for the accuracy of its readings by taking observations on the density reading of some liquid of known density. The wire which supports the plummet is probably not the same as originally came with the instrument, and the readings are affected by the weight and size of this wire. A balance correction which is easily applied can be found by taking the reading with distilled water at a known temperature. Suppose for example that the balance reading is found to be 0.9994 with water whose density as given by physical constant tables is 0.9962. This would show that the balance readings have an error of 0.0032. To get the true density of any liquid it will be necessary to subtract this amount from the reading of the balance in that liquid. In the figure the heaviest and the third lightest of the hook weights are both on the 7 mark, and the middle weight on the 4 mark. The balance reading is therefore 0.7470. If the correction term is that given in the above example the density is 0.7438.

DIRECTIONS.

If the upright of the balance is not approximately vertical, make it so by adjustment of the screw S. Turn the balance beam so that its arm points in the direction of the screw S.

Fill the jar with distilled water to about the level indicated in the figure, immerse the plummet in the water, at such height that it will not touch the bottom of the glass and so that the twisted part of the wire will not touch the surface of the water as the beam swings up and down. Adjust the weights on the beam divisions until P swings equally on either side of p. Measure the temperature of the water, look up and record the density of water, to the fourth decimal place, for this temperature.

Take the plummet out of the jar without changing the location of balance frame on the table, and mark the height of the water in jar.

Replace the water in jar with the liquid whose density is to be determined, filling to the same height as with water.

Immerse the plummet in this liquid and again adjust weights for equal swings of P. Subtract or add correction term to get the density.

Measure and record the temperature of the liquid.

FIGURE 23

Experiment No. 11.

ROTARY HARMONIC MOTION

MEASUREMENT OF MOMENT OF INERTIA

The motion, backward and forward in a straight line, of a mass which has an acceleration always proportional to its distance from its equilibrium position and always directed toward that position is said to be simple harmonic motion. Similarly the motion of a body supported so that it rotates backward and forward through an angle about its equilibrium position with an angular acceleration always proportional to the angular displacement from the equilibrium position, and directed toward the equilibrium position, is called rotary harmonic motion. Such a motion is that of a body hung on a steel wire, twisted through a slight angle and then released. The frequency of such motion is given by an exactly analogous to that for S.H.M.

$$ V_{\text{iz}} \qquad N^2 = \frac{980}{4\pi^2} \frac{(\text{M of F})}{\theta} \qquad \text{Equation 1} $$

Where N is the frequency in oscillations per second, I is the moment of inertia of the body in gram x cm² and M of F is the moment of force required to twist the wire through the angle θ, M of F being measured in gram x cm. and θ in radians. (M of F)/θ evidently depends on the supporting wire and it may be shown to be equal to $\frac{m\pi r^4}{2L}$.

Where r is the radius of the wire in cm. L is the length of the wire in cm. and n is the "modulus of rigidity" of the wire which may be found in tables of constants? The value of $m\pi r^4$ will be given by the instructor and the student can then easily calculate the value of (M of F)/θ which corresponds to the experimentally determined F/x in experiment No. 9.

Determine by experiment the frequency of an iron disc oscillating about the wire by finding the time for 20 oscillations. Make three observations of this time and take the average. From this value of N calculate from Equation 1 the value of I, the moment of inertia of the disc. Check this result from the formula for a disc rotating about its axis.

$$ I = \frac{MR^2}{2} \qquad \text{Equation 2} $$

and calculate the per cent error.

Repeat for a rectangular block and check by the equation for the moment of inertia of a rectangular block rotating about an axis through its center and perpendicular to one of its faces.

$$ I = \frac{M(a^2 + b^2)}{12} \qquad \text{Equation 3} $$

where M is the mass of the block in grams and a and b are length and breadth respectively in cm. (measured perpendicular to the axis)

Repeat for an irregular solid. Here there is no method for computing the moment of inertia from dimensions and mass and this experimental method is most valuable for finding such moments of inertia.

In your report show all calculations and draw a diagram of the apparatus.

EXPERIMENT NO. 13

ACCURATE WEIGHING AND DENSITY OF A SOLID.

METHOD OF WEIGHING.

Balances used for accurate weighings are enclosed in glass cases to avoid the effect of dust and of air currents. The beam and pans of such a balance swing from knife edges resting on agate planes. In order to prevent the roughening of the planes and knife edges, and the displacement of the pan hangers from their proper position the balance must be handled with much care.

Never put weights or objects to be weighed either on or off the pans unless the beam is lifted from its knife edges.

The beam is lifted from its knife edges by turning the large screw head on the outside of the balance case to the right. To put the beam and pans on the knife edges the screwhead should be turned *slowly* to the left.

Before starting to weigh always dust off the pans with a camel's hair brush. Dust on the pans would cause no error if it remained from the beginning to the end of the operation, but it is likely to be knocked off during the weighing.

After dusting, let beam rest on knife edges, by turning the screw slowly to the left. If the pointer stops on the scale but does not swing, set it swinging by fanning one of the pans gently with air from the hand. If the pointer swings on either side beyond the scale-divisions, arrest the beam and put on the knife edges again more slowly. If the pointer should come to rest at one side, or if it does not swing so that the middle of its swing is within three scale divisions of the center, the balance needs adjustment. Either the beam or one of the pan hangers is out of proper position. In either case ask the instructor to show you how to correct the trouble, so that you can get the pointer to swing with the mean of its swing within three scale divisions of the center. It would be a waste of time to try to make the adjustment so that middle of swing would be exactly at center of scale.

1. Find Zero-Point of Balance with Pans Empty.

After the adjustment has been tested by taking the average of two successive swings one on either side, the mean of the swings is to be found accurately by getting the mean of five successive swings of the pointer, the last being on the same side as the first. This method is quicker and more accurate than letting the pointer come to rest. In the latter case the resting point would probably not be the true position of equilibrium on account of the effects of friction. See example of observations which are to be made on last page.

The middle of the scale is always to be called 10, not 0, the zero being taken ten divisions to the left of the center. The turning points are to be estimated to the tenths of the smallest scale division and the eye to be moved with the pointer so as to avoid parallax. The equilibrium position with nothing on the pans is to be computed as in the example given below from five successive turning points. We will call this the *zero-point* of the balance. The equilibrium positions when loaded will be called *rest-points*.

Observations for each equilibrium position must be taken and calculated at least twice. If they do not agree within 0.3 of a scale division they must be repeated until they do, and the mean of the two closest results taken.

2. Find Rest-Points with Body to be Weighed and Weights on Pans.

After the zero-point of the balance has been ascertained, place the cylinder to be weighed on the left pan. Place weights estimated to balance it on the right pan and turn the arrestment screw to the left only far enough to see whether the weight on the pan is too much or too little. Weights must always be picked up by the pincers and

not with the fingers. Dirt from the latter might easily cause errors. Always being careful to have beam arrested when taking off or putting on weights, find by repeated trials the weight of body to the nearest 0.01 gram.

The final balancing is to be made by means of a wire "rider" on the graduated arm attached to the beam or by adjusting length of chain hanging from balance arm if you are using a chainomatic balance. In either case you are to set rider or chain only on milligram marks, and get the tenth-milligrams by calculation as is done in the example on next page. The rider may be moved any distance desired from the central knife edge by a hook which is operated from the outside of the balance case. The actual weight of the rider is 0.01 gram and it would balance this much weight of the body if placed over the knife edge of the weight pan. When placed on the beam nearer the center its effect will be in proportion to its distance, making an equivalent weight of 0.001 gram (one milligram) when placed at the mark 1, two milligrams at the mark 2, and so on.

After you have found the milligram division which brings the equilibrium position of the pointer within a division or two of the zero-point, find the accurate rest-point for this weight, by reading as accurately as possible five successive turning points of the pointer, and calculating their mean as shown in the example under "Weights 64.343". Then increase or decrease the weight on arm by one milligram, whichever will bring the rest-point nearer the zero-point you obtained with pans empty. Find the new rest-point. The difference between these last two rest points is a measure of the sensitivity of the balance, that is the motion of the pointer for a difference of one milligram weight. Then calculate, as shown in the example, the weight, to the nearest tenth-milligram, which would be required to bring the pointer to the zero-point you found with empty pans.

## TO FIND THE VOLUME OF CYLINDER.

The volume of the cylinder is to be calculated, both in c. c. and cu. ins., by getting the product of its length by the area of its end.

Use the vernier calipers to get the length and diameter. Take at least four measurements of each about equally spaced over the cylinder and use the mean in calculation. Take measurements both in cms. and in inches.

## TO FIND DENSITY.

The density of a substance is the mass of one unit volume of it.

Divide the weight of the cylinder as found in grams by 454 to get its weight in lbs., and calculate its density both in grams per cubic centimeter and in pounds per cubic inch. Calculate per cent error of first result from value found in some table of physical constants.

## RECORD.

<table><thead><tr><th></th><th>Length</th><th>Diameter</th></tr><tr><th></th><th>cms.</th><th>in.</th></tr></thead><tbody><tr><td>Cylinder No.</td><td></td><td></td></tr><tr><td>Material</td><td></td><td></td></tr><tr><td>Mass in grams</td><td>in lbs.</td><td>mean</td></tr><tr><td>Volume</td><td>c. c.</td><td>cu. ins.</td></tr><tr><td>Density</td><td>gr/c. c.</td><td>lbs./cu. in.</td></tr><tr><td>True Density</td><td>gr/c. c.</td><td></td></tr><tr><td>%Error</td><td></td><td></td></tr></tbody></table>

-4-

V. A. The molecular weight of CO₂ = 44. If one liter of CO₂ gas at S.T.P. weighs 1.964 grams what is the gram molecular volume of CO₂? 22.4 liters (4)

gram molecular volume: 44
liters = 1.964 gm

gram molecular weight: 44

C 1.964

What is the volume occupied by one gram molecular weight of any gas at S.T.P. 22.4 liters (2)

2

B. What is the weight of 1 liter of SO₂ gas at S.T.P.?

At. wt. of S = 32. At. wt. of O = 16.

$$
\begin{array}{c} 5 \\ 32 \uparrow \uparrow \\ * \, O_2 \, * \, 64 \, \text{g} \end{array}
$$

gram molecular weight of SO₂: 32

volume of 9; 22.4liters

1 liter weight $\frac{64}{22.4}$ = 2.86 grams

C. Set up the following problem so that if the arithmetic were done the correct answer would be obtained. The answer to the first step of this problem is 175.4 ml. The volume measured was 265.4 ml.

A sample of a gas measured at 75°C and 640 mm pressure weighs 0.454 grams. What is its molecular weight? (10)