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I am confused by rolling friction.

Suppose you have a cylinder rolling which starts at rest at the top of an incline plane and begins to roll down the plane without slipping. Is work done by the incline on the cylinder?

I know from doing some problems that the total kinetic energy (translational and rotational) is $mgh$, which is true only if the only work done on the cylinder is by gravity. But also, the cylinder must have nonzero angular acceleration, so friction must be exerting torque on the cylinder, so work must be done by friction. One of these statements is wrong.

Qmechanic
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math_lover
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    This is a really subtle point and i would have a read of this paper as it can explain it much better than i can http://www4.ncsu.edu/~basherwo/docs/Pseudowork1983.pdf – ChrisM Jan 11 '15 at 23:34
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    The link in the comment above, from Chris2807, is currently broken. Right now this works: https://brucesherwood.net/wp-content/uploads/2017/06/Pseudowork1983.pdf – stafusa Jul 21 '17 at 10:02

6 Answers6

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Work is force times distance. If there is no slip, the force of friction acts over a distance of 0. There is no work.

Gravity does work. As the cylinder rolls down the hill, it accelerates. It gains kinetic energy in two forms: translation and rotation.

Gravity would do the same work on an identical cylinder that slide down the same slope without friction. The kinetic energy of the two would be the same at each position.

The rolling cylinder would travel more slowly than the sliding cylinder. But it would also spin.

mmesser314
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    But there is torque exerted on the cylinder (else it would not increase angular speed, which it does), so you are saying the torque is due to gravity. How is that so? – math_lover Jan 11 '15 at 23:52
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    The torque is due to both forces, but only one of them adds kinetic energy to the cylinder. Friction holds the bottom of the cylinder still, which does not make the bottom go faster. Gravity pulls the center forward and does make it go faster. – mmesser314 Jan 12 '15 at 01:10
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    Take the pivot point point to be the center of mass of the cylinder. No torque is exerted by gravity by definition. Then torque MUST be exerted by the force of friction. Thus work is done by the friction torque. What is wrong with this line of reasoning?!?!?! – math_lover Jan 12 '15 at 01:57
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    Making the center be the pivot point is saying the center is fixed. This would be like a massive pulley with a string wound around it being spun up by a mass mass hanging on the string. In this case, the force holding the pulley up would do no work. The work would be done by the tension of the string. – mmesser314 Jan 12 '15 at 02:01
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    A torque is two equal and anti parallel forces that are displaced sideways from each other. Here is more on it. http://physics.stackexchange.com/questions/95234/toppling-of-a-cylinder-on-a-block/95301#95301 – mmesser314 Jan 12 '15 at 02:06
  • Will normal force do any work? – Kashmiri Jan 09 '21 at 12:49
  • @YasirSadiq - Suppose a cylinder was rolling on a frictionless horizontal table. The normal force would be upward. The bottom point in contact with the table does not move. Suppose it was spinning in place on the table. The displacement of the bottom point is horizontal. Either way $\vec{F} \cdot \vec{d} = \vec{0}$. – mmesser314 Jan 09 '21 at 15:16
  • In rolling it does move eventually. – Kashmiri Jan 09 '21 at 15:21
  • @YasirSadiq - You are right. It has slowed down, and stops just for the instant it is in contact with the bottom, before accelerating upward. But for that instant, it is motionless. $dW = \vec{F} \cdot d \vec{d} = 0$. $dW = 0$ for all instants, so $W = 0$. The internal forces that accelerate the point do work on it and change its kinetic energy. But those forces are not the normal force. – mmesser314 Jan 09 '21 at 15:49
  • I'd like to say'when the force acts, it doesn't move. When it moves, the force doesn't act' – Kashmiri Jan 09 '21 at 15:51
  • Dear mmsers314 here is a related one I'm looking an answer for, I'd be grateful if you could clarify https://physics.stackexchange.com/questions/606389/work-done-by-friction-if-a-body-is-not-purely-rolling?noredirect=1#comment1364909_606389 – Kashmiri Jan 09 '21 at 15:52
  • You are not really answering OP's question. OP was examining the situation by considering torque about the COM. You took his comment and spun it into a completely different scenario without resolving his doubt. – Alpha Delta May 16 '22 at 07:31
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In the case of rolling down the slope without slipping, static friction force holds the contact point between the incline and the cylinder. It does rotational work. We can view this work as a medium between gravitational potential energy and rotational kinetic energy. So when it accelerates down the incline, friction force takes some of the gravitational potential energy and turns it into rotational kinetic energy for the cylinder. And in the end, the amount of gravitational potential energy decrease is equal to the amount of translational kinetic energy plus rotational kinetic energy increase.

physicsguy19
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The work is done by gravity. Friction simply holds the instantaneous contact point stationary to the ramp, so it doesn't do any work. The rotation is around the contact point, and is cause by gravity acting through the centre of mass of the roller.

Marco Ocram
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If you take the perspective that "work is the increase in total mechanical energy of a system", then there really no work being done. The potential energy is just being converted to kinetic energy (of some sort). It's not really correct to say that " total kinetic energy (translational and rotational) is mgh". It would be more correct to say that any increase in kinetic energy:

$$E[total] = E[translationa]+E[rotational] = \frac{1}{2}mv^2+\frac{1}{2} I \omega^2 $$

is associated with an equal decrease in $$mgh =E[potential]$$.

So it's the exchange of potential energy for kinetic energy that is doing hte work. The incline is just a constraint that limits the path over which this exchange takes place in position. Friction is actually causing an increase in heat energy. It is a "dissapative" exchange of potential energy for thermal energy.

DWin
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Here it is more appropriate to say that the net work done by friction is zero, because friction here support rotation and at the same time opposes translation motion.so amount of work done by friction in rotation is cancelled out by work done in translation.So the loss of potential energy is equal to the Gain in total kinetic energy.and here friction acting at lowest point is static.

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Devendra Kumar
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Pure rolling friction does no work because there is no relative motion between the rolling object and the surface it rolls on. Sliding friction does work due to relative motion.

This is similar to fluid flowing in a pipe assuming the fluid sticks to the pipe walls; the pipe does no work on the fluid since there is no relative motion of the fluid relative to the pipe surface.

The constraint from rolling friction does provide a torque that causes rotation of the body as it moves. Force and Torque, yes: work, no.

John Darby
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