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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.

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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.

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    $\begingroup$ 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? $\endgroup$ – Joshua Benabou Jan 11 '15 at 23:52
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    $\begingroup$ 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. $\endgroup$ – mmesser314 Jan 12 '15 at 1:10
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    $\begingroup$ 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?!?!?! $\endgroup$ – Joshua Benabou Jan 12 '15 at 1:57
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    $\begingroup$ 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. $\endgroup$ – mmesser314 Jan 12 '15 at 2:01
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    $\begingroup$ A torque is two equal and anti parallel forces that are displaced sideways from each other. Here is more on it. physics.stackexchange.com/questions/95234/… $\endgroup$ – mmesser314 Jan 12 '15 at 2:06
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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.

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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.

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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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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.

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