:: force/math question :: - Page 1
Forum section: electronics / beginners

author: 3roomlab
author_rank: Frequent Contributor
timestamp: August 03, 2014, 11:26:16 pm
content: hi folks,its a calculation question. if i have a solid metal cylinder of say 100kg mass, 200cm diameter. assuming friction free condition. and if i apply a force at the edge to turn it, if i need to (example) accelerate the cylinder from zero to 180degrees in 1 second (half a turn). how do i calculate the force required (kgm/s ?). an approximate math or an exact math tip if any input ... much appreciated.must 1 make a calculation to assume approximated mass at the edge of block?... tyvm yall

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 03, 2014, 11:41:45 pm
content: your question need to revise back to the college book.your question can be complex since involving rotational motion.your question is contradicting itself. if there is no friction, the cylinder wont roll, it just goes straight linear motion if you mean "edge" as the side of the cylinder and pushing right toward the center of gravity.maybe for a start... F = ma coupled with kinematics...http://www.schoolphysics.co.uk/age16-19/Mechanics/Kinematics/text/Equations_of_motion_proof/index.html

author: ejeffrey
author_rank: Super Contributor
timestamp: August 04, 2014, 12:02:48 am
content: For rotational motion you need to talk about torque and moment of inertia. Basically all the equations of linear motion apply to rotational with the following equivalences: mass -> moment of inertia. force -> torque, x -> theta, velocity -> rotation frequency, acceleration -> angular acceleration.Themoment of inertiaof a solid cylinder is 0.5 * m * r^2. So the moment of inertia of your cylinder is 50 kg * m^2.You want a 180 degree rotation in 1 second, so 0.5 * a * t^2 = pi. Set t = 1 s, and solve to get a = 2*pi radian/ s^2.torque = moment of inertia * angular acceleration = 50*pi * kg * m^2 / (s^2) ~ 157 N*m.If you apply a force of 150 N at a distance 1 m from the axis of rotation, that should do the trick. Of course you need a bearing at the axis of rotation, or you need to apply half the force on opposite points of the cylinder to get it to rotate but not translate. If you apply the force closer or farther from the axis of rotation you need more or less force, but the torque is the same: 157 N*m.

author: westfw
author_rank: Super Contributor
timestamp: August 04, 2014, 12:19:36 am
content: Do you want it to stop after 180 degrees as well? If so, you need to apply force to decelerate it (opposite and equal to the force you applied to get it going.)I'm pretty sure you get multiple solutions, depending on how much you want to "coast" between acceleration and deceleration. Rather like spaceflight: all the SF spaceships accelerate at 1G for half the trip, then "flip" and decelerate at 1G for the 2nd half, conveniently providing "normal gravity" for the passengers. But most of the real ships kick themselves in the right direction and then coast most of the distance.

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 12:23:15 am
content: Quote from: 3roomlab on August 03, 2014, 11:26:16 pmFor this question to be reasonable, you must assume the cylinder to be pivoted on a frictionless axis running through the center. You must also assume the cylinder to be of uniform density. Given those assumptions, you can use moment of inertia as mentioned above.

author: 3roomlab
author_rank: Frequent Contributor
timestamp: August 04, 2014, 12:28:25 am
content: thanks for the inputs folks, yea i think my question is missing bits n pieces.OK got it .. moment of inertia !

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 04:15:16 am
content: I think you are getting a bit confused about force, work, energy and equations of motion.Let's see if you can understand motion in a straight line before looking at rotation.Suppose you had a stationary mass of 100 kg. What force would you need to apply to move it over a distance of 1 metre in 1 second?How much work would be done on the object and how much energy would be expended?

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 04:39:33 am
content: Hint: you would use the following two equations:1. Equation of distance moved under acceleration:s=ut+ ½at2(wheres= distance moved,u= initial velocity,a= acceleration andt= time)2. Newton's second law of motion:F=ma(whereF= force applied,m= mass,a= acceleration)You can later solve the rotational problem using analogs of these equations.

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 04:50:11 am
content: See my post above.The equation is:s=ut+ ½at2Now the distances= 1 m, the initial velocityu= 0 m/s, the timet= 1 s.So:1 = ½athereforea = 2 m/s2Next, use:F=maWe havem= 100 kg anda= 2 m/s2ThereforeF= 100 x 2 = 200 NSince the force is applied over a distance of 1 m, the work done is force x distance, orW= 200 x 1 = 200 JThe work done is the same as the energy expended since no energy is lost to friction or other causes.

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 04:51:47 am
content: Quote from: 3roomlab on August 04, 2014, 04:41:29 amYes, that's right.

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 05:00:33 am
content: Now the equations are going to work in a similar way for rotational motion.Instead ofs=ut+ ½at2we are going to have(theta) = (omega)t+ ½ (alpha)t2Where (theta) is angular position in radians, (omega) is angular velocity in rad/s, and (alpha) is angular acceleration in rad/s2.The equivalent of Newton's second law is going to be(tau) =I(alpha)where (tau) is the torque in Nm,Iis the moment of inertia in kg m2and (alpha) is the angular acceleration in rad/s2(Unfortunately this forum can't do Greek letters)Also the work done when a torque moves through a given angle is going to be from the similar equationW= (tau)(theta)

author: 3roomlab
author_rank: Frequent Contributor
timestamp: August 04, 2014, 05:11:18 am
content: ahhhh so this is what i missed, i think i confused myself cos i totally forgot the base stuff-- i will be back after playing with more newtons

author: ejeffrey
author_rank: Super Contributor
timestamp: August 04, 2014, 05:21:47 am
content: Also be careful with units. Torque has units of newton-meter, but it is not a measure of energy. Torque is the rotational equivalent of force. With linear motion we have the formula work = force * distance. For angular motion that is work = torque * angle.

author: 3roomlab
author_rank: Frequent Contributor
timestamp: August 04, 2014, 05:28:52 am
content: okay, based on the rectangular rotating blockMoment of inertia (rectangular) I = 1/12 x m(X^2+Y^2)Assume X=2, Y=3. m = 180kg. Therefore I = 1/12 x 100(13) = 150kg/m2To move/accelerate the object 1 complete rotation in 1 sec on its CG pivot (assume lossless), from position of rest. Angular distance moved = 360 x pi / 180 = 2(pi)Angular velocity = 2(pi)Angular acceleration = A (unknown)Solving for A, we get 2(pi) = 0.5 * A * 1A = 4(pi) rad/s/sSolving for “F=ma”T = I * A = 600pi NmAnd so work done, energy expendedW = 600(pi) x 2(pi) = 1200(pi)^2 joulesso this 1200pi.pi joules being expended in 1 second, also expressable as 1200pi.pi watts? yes ? (=11.8kNm ... 11.8kJ)

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 05:47:22 am
content: Quote from: 3roomlab on August 04, 2014, 05:28:52 amBe careful with your arithmetic and with your units of measure.You haveI = 1/12 m (x2+ y2)wherem = 180 kg, x = 2 m, y = 3 m.SoI = 1/12 x (180 kg) x [(2 m)2+ (3 m)2]thereforeI = 1/12 x 180 kg x (4 m2+ 9 m2) = 195 kg m2Note: this isnot"kg/m2", it is "kg x m2".QuoteIt's best to forget the degrees, they are only confusing. Everything is radians in physics and there are 2 pi radians in a complete revolution. (If you wonder why there are two pies in a single turn, you may go look up some YouTube videos about why "pi is wrong")

author: IanB
author_rank: Super Contributor
timestamp: August 04, 2014, 06:12:27 am
content: Quote from: 3roomlab on August 04, 2014, 06:02:20 amRadians happen because they are a natural way of expressing turning.If you have a point on the edge of a disk 1 m in diameter and you turn the disk through 1 radian the point will move through a distance of 1 m.So with radians you can convert radius to distance moved just by multiplying by the angle turned.

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 04, 2014, 04:42:47 pm
content: the fun part about pushing something in pure freefall vaccum not in the line of its CG is that it will also moves linearly as it rotates "angularly"and then you need to split your exerted force into linear force and torque at the CG in somesort of superposition method, but thats the easy part. then you have 2 branches of formulation for linear and angular, redefine what you want (the angular and linear motion), and backtrace it to get your required force. you want angular motion alone? dont exert a force, exert a pure "twist", then thats the pure torque.