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We knew that a after a bowling ball is threw out with a certain velocity to a non smooth track, it first rolls and skids as the translational velocity (decelerates due to friction) of the center of mass is greater than tangential velocity of the point of contact of ball with the floor which has a opposite direction. But after some time torque due to friction causes the angular velocity to increase and eventually the tangential velocity of point of contact achieve same value with translational velocity and have v=0, it starts rolling without slipping, and eventually come to stop.

My question here is after rolling without slipping is achieve how does the translational kinetic energy and rotational kinetic energy change? How does translational kinetic energy decrease while rotational kinetic energy is increase by torque due to friction?

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    $\begingroup$ Exemplary question. I think you may wish to specify that you are asking a physical / molecular explanation as to what's going on the section of the body in contact with surface $\endgroup$
    – 666User666
    Aug 21 at 15:50
  • $\begingroup$ I don't get it. In a ball that is rolling without slipping, translational and rotational energy are directly coupled to each other, they both depend on the square of the rolling velocity (the angular velocity is coupled to the rolling velocity by the radius of the ball). So once a ball is rolling without slipping, that conversion you mention isn't happening anymore. $\endgroup$
    – MaxD
    Aug 21 at 23:53
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In rolling without slipping motion, velocity of the centre of mass is related to angular velocity by $v=\omega r$ [1]. The term $\omega r$ represents the tangential velocity of a particular point on the object.

There is a maximum friction force that the floor can exert on the ball. Since the rotation of rolling motion is helped by this friction force, there is a pre-defined maximum angular velocity which the ball can have. Therefore there is a maximum tangential velocity too that one point can obtain.

Let's take the tangential velocity of the point of contact between ball and the track as $v'$. Then, if the initial velocity of the centre of the mass ($v$) is greater than $v'$, there cannot be a rolling without slipping motion, because $v\neq v'$ (or $v\neq \omega R$). Therefore the friction force acts against the translational motion, which results in slowing down the ball (because friction forces acts against the translational motion of a body most of the times causing deceleration) . When $v$ decreases to $v'$, the necessity for rolling without slipping motion is obtained. Thus it starts to roll with the aid of the torque provided by the friction force (quite ironical, because the same force caused the deceleration of the translational motion). Note that this rolling motion cannot happen if you place the ball on the track without an initial velocity. That is because, then there is no motion and therefore no resistance to motion by friction and therefore no torque provided by the friction. Thus it is obvious that rolling is helped by friction.

If the ball is perfectly rigid, it will roll forever due to that torque. Although in real life we cannot find perfectly rigid bodies, hence the ball will slow down due to phenomenon called rolling resistance [2], [3].

Hope this helps.

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    $\begingroup$ You have described the scenario but you haven't answered the question. OP is asking 'how' the 'what' happens rather than 'what happens'. $\endgroup$
    – 666User666
    Aug 21 at 15:55
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    $\begingroup$ @Buraian I think the second to last paragraph answers the question: "If the ball is perfectly rigid, it will roll forever due to that torque. Although in real life we cannot find perfectly rigid bodies, hence the ball will slow down due to phenomenon called rolling resistance[2], [3]." Energy is lost to rolling resistance due to deformation of the ball by the surface it is rolling on. $\endgroup$
    – Andrew
    Aug 21 at 16:12
  • $\begingroup$ In the model we take, it is idealized not rigid. Using the real life arguement does not explain this ideal case. $\endgroup$
    – 666User666
    Aug 22 at 2:45
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If you treat the ball as a rigid body there is no dissipation of energy as an increase in internal energy- "heating"- of the ball due to friction. So you can apply the laws of classical mechanics: sum of forces is change in acceleration of the CM and friction provides a torque about the center of mass. This explains the decrease in translational and rotational kinetic energy. This earlier discussion may help: Consistent Approach for Calculating Work By Friction for Rigid Body in Planar Motion

Consider the situation once the ball rolls without slipping on a long track. For "pure rolling" on a flat track the force of friction is zero! The assumption of pure rolling is usually made for a relatively short track, and the translational and rotational kinetic energy remain constant. But in reality there is not pure rolling and the force of friction- although relatively small compared to pure sliding- is not zero and will eventually slow the ball down over a long track. The de-acceleration of the center of mass (CM) is $a_{cm} =-f_{fric}/m$ where $f_{fric}$ is the force of friction for the ball of mass $m$. For no slipping, the angular acceleration of the ball $\alpha = a_{cm}/R$ where $R$ is the radius of the ball; since $a_{cm}$is negative, $\alpha$ is also negative. Both translation and translational kinetic energy decrease. In summary, it is the force of friction that causes the ball to eventually stop over a long track, even though that force is relatively small for rolling without slipping.

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    $\begingroup$ I thought there is a torque due to friction that accelerates angular velocity? $\endgroup$
    – Eugene Ng
    Aug 22 at 1:54
  • $\begingroup$ Yes, while rolling with slipping, that is true. But, once there is rolling without slipping the force of friction both slows the linear acceleration and the angular acceleration about the center of mass as discussed above. Friction is a phenomena that is poorly discussed in most physics texts, in my opinion. You can search this site for "friction" and find many good- and some not so good- discussions of friction. Hope this helps. $\endgroup$
    – John Darby
    Aug 22 at 2:24
  • $\begingroup$ "If you treat the ball as a rigid body there is no dissipation of energy as an increase in internal energy- "heating"- of the ball due to friction. So you can apply the laws of classical mechanics: sum of forces is change in acceleration of the CM and friction provides a torque about the center of mass. " There is no disspation of energy as what $\endgroup$
    – 666User666
    Aug 22 at 2:47
  • $\begingroup$ No dissipation of energy as change in internal energy of the rigid body. $\endgroup$
    – John Darby
    Aug 22 at 2:57
  • $\begingroup$ Ohh I get it, but it may also be that the dissipated energy goes into heating environment. $\endgroup$
    – 666User666
    Aug 22 at 3:00
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There are two forces causing a real rolling ball to lose energy. One is friction with the air. The other results from the slight deformation of the two surfaces at the point of contact. The ball is effectively rolling “uphill”.

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  • $\begingroup$ "The ball is effectively rolling “uphill”." -- I found this to be a very insightful comment, thank you. $\endgroup$
    – Andrew
    Aug 21 at 17:56
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My question here is after rolling without slipping is achieve how does the translational kinetic energy and rotational kinetic energy change? How does translational kinetic energy decrease while rotational kinetic energy is increase by torque due to friction?

Once the ball is rolling without slip, it will roll forever, with its translational and rotational kinetic energy unchanged. There will be no friction force acting at the point of contact, so long as the track is flat and horizontal.

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    $\begingroup$ I am wondering how does the frictional force vanish? $\endgroup$
    – Eugene Ng
    Aug 22 at 2:00
  • $\begingroup$ It’s not necessary anymore. Once the velocity of the point of contact is zero, it stays zero. The motion is a combination of translation plus rotation. $\endgroup$
    – Evan
    Aug 22 at 2:10
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Are you considering an idealised model? This would have:

  1. There is no air resistance.
  2. The ball and the track are perfectly smooth; the track is perfectly flat and the ball is a perfect sphere.
  3. The ball and the track are perfectly rigid.

In that case, there is only a single point of contact between the ball and the track, and the tangential velocity of that point of the ball is zero relative to the track, once the ball is rolling without slipping. If the translational velocity of the centre of mass wasn't a perfect match for the tangential velocity of the point of contact to be exactly zero relative to the track there would be slipping; in that case friction would slow down the translation and/or speed up the rotation until they were perfectly matched. Once there is no relative motion of the point of contact between ball and track there can also be no friction between ball and track. So at that point there is no net force or torque on the ball, so the ball will continue with its current velocity and rotation forever.

In the real world, none of the assumptions I detailed in my first paragraph are true:

  1. The ball will lose energy to air resistance.
  2. There are imperfections in the surface of both the ball and the track which make the "single point of contact with exactly zero relative velocity" idea false, so some energy will be lost to friction after all.
  3. Even if the ball and track had perfect idealised shapes, they are not perfectly rigid, so the weight of the ball slightly deforms the surface of both. This leads to rolling resistance, that constantly saps energy from the rolling motion.

There are probably more effects I haven't covered. They are all fairly small, which is why "rolling has no friction" is a decent first-order approximation (and why wheeled vehicles are efficient). But they are non-zero.

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    $\begingroup$ How does the frictional force vanish when rolling without slipping had achieved? $\endgroup$
    – Eugene Ng
    Aug 22 at 2:03
  • $\begingroup$ @EugeneNg I sortof explained that in the big paragraph in the middle. The friction involved is of the two surfaces resisting slipping past one another. This acts to oppose the relative velocity of the two surfaces, (which means it's pushing "back" on the ball). Eventually the relative velocity of the point of contact is zero, because it's been gradually reduced by friction. Since the surfaces aren't slipping past each other anymore, there's no friction (if there were, which direction should it push, forward, back, or off to the side? there's no longer a relative velocity to tell you which). $\endgroup$
    – Ben
    Aug 22 at 4:03
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How does translational kinetic energy decrease while rotational kinetic energy is increase by torque due to friction?

Since, you are asking about the situation AFTER rolling without slipping has been achieved, i think it is incorrect to say that rotational KE increases.

Once rolling without slipping has been achieved, the rotational KE will no longer increase.
Both rotational KE and translational KE will start to decrease.

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    $\begingroup$ But how and why they decrease ? $\endgroup$
    – Eugene Ng
    Aug 21 at 13:29
  • $\begingroup$ This also does not answer the question due to same reasons I commented on ACB's post $\endgroup$
    – 666User666
    Aug 21 at 15:56
  • $\begingroup$ Please see my answer. $\endgroup$
    – John Darby
    Aug 21 at 16:14
  • $\begingroup$ @Buraian This answer was written by me, BEFORE the question was edited. If the original question was what it is now, i would have written a different answer . $\endgroup$
    – user196948
    Aug 21 at 16:41
  • $\begingroup$ @physics_newb I made that edit. But it was just a tag edit. I didn't change the body. $\endgroup$
    – ACB
    Aug 21 at 17:17

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