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A ball rolls without slipping on a table. It rolls onto a piece of paper. You slide the paper around in an arbitrary (horizontal) manner. (It’s fine if there are abrupt, jerky motions, so that the ball slips with respect to the paper.) After you allow the ball to come off the paper, it will eventually resume rolling without slipping on the table. Show that the final velocity equals the initial velocity.

This was a problem I came across in David Morin's Introduction to Classical Mechanics. I don't understand how the final velocity will be equal to the initial velocity since evidently, there is negative work being done by kinetic friction of the paper when slipping occurs. So, its kinetic energy is being lost in the process and suppose, even if it starts rolling without slipping again (I understand that pure rolling is bound to happen since kinetic friction will reduce translational velocity till the point that the v=rw condition will hold and thereafter only static friction will act on the contact point of the ball), wouldn't it have lesser velocity than its initial velocity? Where is my reasoning going wrong here?

As far as I've thought about it, the only way this is possible is if friction does positive work later or there's some other external force I haven't taken into account which gives the ball kinetic energy. And even if that is the case, how can I prove that this will always happen?

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One needs to understand that the purpose of friction is to ensure no relative slipping between the contact points of the bodies in contact. When the ball was pure rolling on the table (its point of contact was stationary) relative to the table, friction was not acting and it was going with some velocity. When we introduced the paper and gave the same velocity, there was slipping between the contact point of the ball and the paper because the contact point of the ball was initially at rest. When the paper has some velocity relative to the ball, the kinetic friction of the paper starts acting on the system of the ball and paper. It retards the paper and It increases the translatory velocity of the ball in the direction towards the motion of the paper which also results in a negative torque about the center of gravity i.e., angular torque decreasing its angular velocity. the force of friction will act until the contact point of the ball and paper (which is translating) have the same velocity. So the kinetic energy which was initially KE = 1/2(Iω²+mv²) has its individual terms changed now due to friction resulting in increased translational velocity and decreased angular velocity. So there is not necessarily a decrease in the kinetic energy of the ball. When the ball leaves contact with the paper, kinetic friction will act again until the contact point is stationary relative to the table which itself is stationary. It will decrease its translating velocity and increase its angular velocity. We can conserve the angular momentum of the ball about the contact point because there is no external torque about it. Initially, it was rotating about that point. so we can write: (I+mr²)ω = (I +mr²)ω' (After some time the ball rolls purely again) Clearly, ω = ω' Here, I : MOMENT OF INERTIA OF THE BALL ABOUT ITS CENTER OF GRAVITY; ω = INITIAL ANGULAR VELOCITY; ω' = FINAL ANGULAR VELOCITY; v = initial velocity; m = mass of the ball In this case, we are assuming that the paper is moving in the direction where the ball was moving initially, that is why velocity increased and angular velocity decreased. If we assume the opposite the velocity will decrease and angular velocity will increase. If we move the paper in any horizontal way such that it has velocity then the ball will have either its velocity or angular velocity increased and the other increased. (while it is on the paper) We are taking the frame of reference as the table which is inertial.

If we had initially given the ball a translational velocity, then kinetic friction would have only acted for a limited time until the ball rolled. However, friction would act continuously if we were applying a constant force that did not lead to pure rolling. I was assuming the first scenario . While on the paper, kinetic friction acts until both the contact points move with the same velocity. You need to imagine the force of friction acting on the contact point of the ball parallel to the paper in either direction of motion of paper. If the paper was moving forward, then kinetic friction would act forward for the ball which increases its translational velocity but decreases its angular velocity because it has a torque about the center of gravity. So, work is being done by kinetic friction. And, since frictional force passes through the contact point, it has no torque about it, so we can conserve angular momentum about that point. (comment for more clarifications)

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Commented Nov 23, 2023 at 7:58
  • $\begingroup$ I'm a little confused here; firstly, when u said that no friction acts on the ball when it is performing pure rolling, that's wrong; static friction continuously acts on the ball, but it does not perform any work. Secondly, how is the work being done by kinetic friction increasing the translational kinetic energy? Won't it simply dissipate in the form of heat? For example, if the ball is rolling forward and we just yank the page backward once and let the ball continue, then yes agreed its translational velocity increases in the direction of the paper but will it again attain its initial speed? $\endgroup$ Commented Nov 25, 2023 at 9:09
  • $\begingroup$ see edit (last paragraph in the answer) and comment for more clarifications. $\endgroup$ Commented Nov 25, 2023 at 10:31
  • $\begingroup$ Oh okay, now I understand $\endgroup$ Commented Nov 25, 2023 at 16:56

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