1

When a ball rolls without slipping down a track, it seems like static friction from the track does rotational work on the ball. As explained in this post: Is work done in rolling friction?, this work is exactly the same as the work done by gravity around the pivot point. But shouldn't the track also do linear (i.e. translational, not rotational) work on the ball? After all, the ball is moving.

(The fact that the pivot point is not moving does not seem to be a sufficient explanation because $F=ma$ holds and therefore so should the work-energy theorem for the displacement of the ball.)

Eric David Kramer
  • 1,627
  • 8
  • 21
  • Greetings! Please don’t work around the question-closing system by deleting and re-posting. See the [help] for advice on getting a closed question re-opened. – rob Nov 10 '21 at 23:24

3 Answers3

4

But shouldn't the track also do linear (i.e. translational, not rotational) work on the ball? After all, the ball is moving.

when a block is sliding down a slope, the friction force $F$ does work and slows the block down. Work is done on the block because the force is acting parallel to the velocity.

For the ball, it's true that it's moving down the slope, $v_1$, but the point $P$ where the friction is acting is moving perpendicular to the force, in the direction of $v_3$.

enter image description here

The point $P$ moves in the shape of a cycloid, and moves perpendicular to the surface when in contact with it (e.g. at $2\pi a$)

enter image description here

The formula for work done is $W= Fd\cos\theta$, where $\theta$ is the angle between the force and the distance moved, so (ideally) no work is done on a rolling ball by friction.

John Hunter
  • 13,700
  • Nice answer! But shouldn't the work-energy theorem be valid for the ball as a whole, not just the contact point? – Eric David Kramer Nov 10 '21 at 15:44
  • Just to mess things up: the rolling ball accelerates less than it would if there were no friction, because of course it wouldn't roll. That's because the potential energy (from height) doesn't get siphoned off into rotational energy, etc. – Carl Witthoft Nov 10 '21 at 16:07
  • @ Eric David Kramer Yes, as Carl says, the overall force down the slope is reduced due to friction, so the acceleration is less than a sliding ball. The 'work done' just comes from the loss in gravitational potential energy, less translational speed/energy can occur as there is also rotational kinetic energy created. – John Hunter Nov 10 '21 at 19:52
3

This is a very (needlessly) confusing topic that, for whatever reason, is frequently taught incorrectly.

The power of a specific force is given by $P=\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of application of $\vec F$. This simple definition works for any mechanical force in any classical mechanical scenario. Then the work done by that force is simply $W=\int P \ dt$.

Now, specifically for this scenario, a ball rolling without slipping down an incline, there are three forces: the normal force, the frictional force, and the gravitational force.

The normal force and the frictional force are both applied at the contact point. The contact point is moving, but the velocity of the material at the contact point is 0. The contact point is not an object, so its motion is not important. What is important is the motion of the material at the contact point, which is $\vec v=0$. So then $P=0$ and $W=0$ for both the normal force and the frictional force.

The gravitational force, in contrast, is applied at the center of mass. The center of mass is moving at a nonzero $\vec v$ and $\vec F$ is not perpendicular, so $P=\vec F \cdot \vec v = F v \ \sin(\theta)$. So all of the work is done by gravity. The only source of kinetic energy, both linear and rotational, is the decreased PE from gravity.

Although the frictional force does provide torque it does not provide energy. That makes sense because the frictional force does not have an associated potential energy. The frictional force does convert some of the gravitational PE into rotational KE, but it does not do so by doing any work. The energy comes only from gravity.

This does not contradict the work energy theorem in any way. The net force is still related to the change in translational KE as specified by the theorem. The work energy theorem only tells you about the net force and the change in translational KE. It does not tell you anything about the work done by any individual force, even when there is only one force. Any attempt to use the work energy theorem to deduce the work done by an individual force is a misuse of the theorem.

Dale
  • 99,825
  • You say at the end of your post that the work-energy theorem should still hold. For a ball rolling without slipping down an incline, the torque from gravity around the pivot point is $mgR\sin\theta$. I think that this should be equal to the torque of the friction force around the center of the sphere? If that's the case, then the friction force should be equal to $mg\sin\theta$. If the ball rolls down a length $L$, then the work done by friction should be $-mg\sin\theta L=-mgh$. If the work done by gravity is $+mgh$, then that means the total work is zero. That doesn't seem right... – Eric David Kramer Nov 10 '21 at 21:54
  • Wait I'm wrong! The torques are not equal. One of them must be equal to $I\alpha$ and the other to $(mR^2+I)\alpha$ by the parallel axis theorem. Thus the friction force must be less than $mg\sin\theta$. The friction force will do negative work and the total work will be the final (linear) kinetic energy of the ball and that's the answer! The work energy theorem from $F=ma$ is only for linear motion! – Eric David Kramer Nov 10 '21 at 22:11
  • @EricDavidKramer as explained above very clearly the work done by friction is 0. There is no ambiguity or ambivalence. It is 0. That it produces a torque does not imply that it does work. It does not do negative work or positive work. It does 0 work. You are correct that the work energy theorem only describes translational KE, – Dale Nov 11 '21 at 00:04
  • Sorry, I disagree with what you're saying. – Eric David Kramer Nov 11 '21 at 11:51
  • $\sum \vec{F}{\rm ball} = m \vec{a}{\rm CM, ball}$. Integrate wrt $\vec{r}{\rm ball}$ and get $\int!\sum \vec{F}{\rm ball} \cdot d\vec{r}{\rm CM, ball} = \frac12 m \vec{v}{\rm CM, ball, f}^2-\frac12 m \vec{v}^2_{\rm CM, ball, i}$. That's a mathematical fact. And $\sum \vec{F}{\rm ball}$ includes the static friction, and the center of mass displacement $d\vec{r}{\rm CM, ball}$ is non-zero. Maybe we are just arguing about nomenclature. – Eric David Kramer Nov 11 '21 at 11:59
  • It isn’t a matter of opinion. What you are saying is factually incorrect. The frictional force does no work, and it is a misuse of the work energy theorem to ever try to use it to try to determine the work done by an individual force. The net force does include the frictional force, but the work energy theorem never tells you anything about the work of an individual force. Not even in the case where there is only a single force acting on the system. – Dale Nov 11 '21 at 12:10
  • Look, it is not your fault that the work energy theorem is taught so poorly. And it is also not your fault that the mechanical power formula is not emphasized in the curriculum. But as a physicist since the early 2000’s and a physics instructor for the last several years, I am telling you straight up that you have this wrong. – Dale Nov 11 '21 at 12:21
  • I am happy for you that you're a physics instructor but I would like you to tell me what's wrong with my calculation. I did a work integral on $\sum F = ma_{CM}$. The work done by gravity is $mgh$, and the $\Delta \frac12 m v^2_{CM}$ is less than $mgh$. Why are $\int! F_g dx_{CM}$ not equal to $\Delta \frac12 m v^2_{CM}$? – Eric David Kramer Nov 11 '21 at 15:40
  • Your calculations are fine. Physics is about more than just calculations. You need to also understand the meaning of the calculation. In this case, the integral of the net force over distance, sometimes confusingly called “net work”, ONLY gives you the change in translational KE. It provides no additional information. In particular, it does not tell you anything about the work done by gravity. The reason that $\int F_g dx_{CM} \ne \Delta \frac{1}{2}mv^2_{CM}$ is because they are completely unrelated quantities that have nothing to do with each other – Dale Nov 11 '21 at 18:09
  • I see. So is it correct to say $\int F_g dx_{CM} = \Delta \frac12 m v_{CM}^2 + \int f_s dx_{CM}$, where $f_s$ is the static friction? – Eric David Kramer Nov 11 '21 at 20:06
  • Yes that is correct. However, $\int f_s dx_{CM}$ is not the work done by the frictional force. That is $\int \vec f_s \cdot \vec v \ dt =0$. The frictional force is not applied at the center of mass – Dale Nov 11 '21 at 22:26
  • OK I think I get it now, thanks. Your explanations were helpful. What I really meant to ask was whether $\int \vec{f}s\cdot \vec{v}{CM} dt$ is non-zero. And it is. Sorry for using the word "work". Maybe we should give it a new name. The "shmerk". The shmerk done by the static friction is non-zero. I like that. – Eric David Kramer Nov 12 '21 at 08:51
  • I just found an explanation to this in Kleppner & Kolenkow, Example 6.17 (p.268). Their point of view seems to be somewhere between yours and mine. – Eric David Kramer Dec 22 '21 at 12:08
-1

Yes, the track does work. You can verify that the final velocity of the ball is less than $\sqrt{2gh}$. This is because the friction force did negative work.

This is exactly what is the work-energy theorem says: The sum of the forces on an object times the displacement of the center of mass of the object is equal to the change in $\frac12 mv^2$, where $v$ is the velocity of the center of mass of the object. This is precisely what is happening here. The static friction force acts against the motion of the ball and reduces its final linear kinetic energy.

As for conservation of energy, the work done by the friction force is exactly equal to the final rotational kinetic energy of the ball. The total change in kinetic energy (linear plus rotational) is indeed equal to $mgh$. So everyone goes home happy :).

Eric David Kramer
  • 1,627
  • 8
  • 21
  • Yes and no, it's true that the resultant linear force down the slope is $mg\sin\theta - F$ so the linear Kinetic energy is $(mg\sin\theta - F)d = mgh - Fd$, so in this sense the work-energy theorem holds true. However it's doesn't seem valid to say the friction force does negative work. The rotational K.E. created is like Fd, but $T\alpha$ where $T$ is torque of the part of the weight not causing linear acceleration i.e. $T = (mg\sin\theta - (mg\sin\theta - F)) \times r = Fr$, so $T\alpha$ rotational K.E. due to the weight is $Fr\times d/r = Fd$, numerically the same, but from gravity. – John Hunter Nov 10 '21 at 23:13
  • The track does not do work. The energy of the track does not change. If the track did negative work without gaining energy then energy would not be conserved. It would magically disappear. – Dale Nov 11 '21 at 00:36
  • The track certainly does gain energy, but it's infinitely massive so the change in velocity is zero. – Eric David Kramer Nov 11 '21 at 10:19
  • @EricDavidKramer it does not. To see this explicitly, let the track have mass $M$. Do the calculation, and then take the limit as M goes to infinity. The energy is zero. However, even with finite M the energy gained by the track is not equal to the amount you falsely claim it is. So your explanation fails even for a finite mass ramp – Dale Nov 11 '21 at 12:07
  • @ Dale, Eric, Dale is right. You could imagine a reflection of the track joined to the original and two balls rolling down, one each side. Any forces on the tracks would cancel out, so if energy is gained by the tracks from the balls it must be as heat. A sliding mass heats up a track, but a rolling ball doesn't. The K.E. isn't as high simply because the loss of GPE is shared between translational movement and rotational K.E. The rotational K.E. has to be created for the ball to roll, that's the role friction plays, it makes the GPE unable to turn completely into translational K.E. – John Hunter Nov 11 '21 at 13:15
  • @Dale If you use conservation of momentum, the Earth (including the inclined plane) does move in reverse to match the ball's momentum. But I agree it's not frictional force doing that work. – Carl Witthoft Nov 11 '21 at 14:37
  • By conservation of energy: mgh = (1/2)m$v^2$ + (1/2)I$ω^2$ where ω = v/r. – R.W. Bird Nov 11 '21 at 15:22
  • OK I take it back the track doesn't gain energy, sorry. But can't the friction force still do negative work on the center of mass (CM)? It has to, because the friction force accelerates the CM according to Newton's 2nd law. Integrate Newton's second law w.r.t $\vec{r}{CM}$ and you get: $\int m a{CM} dx_{CM} = \int m \frac{dv_{CM}}{dt}v_{CM}dt = \Delta \frac12 m v_{CM}^2$. Maybe it's wrong to say the track does work. The force does work. The friction force is moving, but not the track. – Eric David Kramer Nov 11 '21 at 15:48
  • I see that John Hunter answered my question in his comment. For some reason I missed that. @JohnHunter Thanks! – Eric David Kramer Nov 11 '21 at 17:10