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I came across this question where Question

A solid cylindrical body of mass m and radius r rolls without slipping over a smooth inclined plane. Its initial velocity is v. What would be the maximum height attained by the body?

The question seems simple, and so it is, but my answer didn't match. When I checked the solution, I got more confused.

First, let me tell you my approach, and then the approach given in the solution. Let body climb till position B, at height h from initial position, on inclined plane from initial position A. At position A, body will have rolling kinetic energy and by conservation of energy, this rolling kinetic energy is converted into potential energy at position B.

$$\frac{1}{2}mv^2(1+\frac{k^2}{r^2})=mgh$$ where k is radius of gyration.

I solved this equation for conservation of energy. But it turns out this equation isn't correct.

Here's approach given in solution. Since there's no friction, rotational kinetic energy won't change while just translational kinetic energy will change. So, $$\frac{1}{2}mv^2=mgh$$ And this is where I am confused. Theoretically, everything seems alright. There's no external torque from friction, therefore rotational kinetic energy won't change. But then will the body keep on rotating even at B? That seems unrealistic. Am I visualising this solution incorrectly or is it by any chance very much realistic? Can anybody also elaborate why rotational kinetic energy does not change when external torque is absent but presence of external force does not matter in change of kinetic energy?

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    $\begingroup$ rolling without slipping means $~v=\omega~ r$ thus if the translation velocity equal zero , the angular velocity is also zero $\endgroup$
    – Eli
    Commented Apr 29 at 7:07
  • $\begingroup$ This is the JEE session 1 question if I am not mistaken. If so, your wording of the question is unfortunately wrong. 'The body is initially rolling horizontally without slipping with speed v. It then moves up an inclined smooth plane.' So it will not slip while moving horizontally but it will slip while moving along the incline. I suspect it to be the JEE question because the answer given would be more appropriate for the case I have stated. $\endgroup$ Commented May 1 at 1:53
  • $\begingroup$ @wonderingwhy, there is a similar question in JEE, but this one isn't that question. The JEE question's wording is consistent with all the arguments that are stated in the below answers. This one's different. $\endgroup$
    – aayush
    Commented May 1 at 8:54
  • $\begingroup$ Perhaps the answer given was in the context of the former. $\endgroup$ Commented May 1 at 9:36

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“Rolling without slipping” is a way to describe the particular motion of a rolling body and requires there to be a frictional force (of static friction). If there were frictional losses it would be “rolling with slipping” with a force of kinetic friction, if there were no friction there would be no rolling (1), and the question would have to stipulate “frictionless”.
As the question is posed, your answer is correct. You have to take into account both linear and rotational kinetic energy.

The answer provided for this problem is incorrect!

That happens.

(1) Technically the surface could be frictionless and the body rolling but it would have to specifically stated that that was the case.

EDIT: Final Conclusion
After an informative discussion with BobD in comments I have come to this conclusion:
The question poses two contradictory assertions and is therefore invalid;
(1) The object rolls without slipping over an inclined plane
(2) The inclined plane is smooth, meaning frictionless
As the object rolls up the inclined plane its linear velocity decreases. If it rolls without slipping then its rotational speed decreases proportionally. This can only occur if the surface has friction.
If there is no friction on the slope, then the rotational speed is constant as the linear velocity decreases, which implies slipping.
The two assertions cannot both be true. The question is invalid.

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  • $\begingroup$ "'Rolling without slipping' is a way to describe the particular motion of a rolling body and requires there to be a frictional force (of static friction)." If you are saying rolling without slipping requires a static friction force, that is not true. A static friction force is only needed to initiate rolling without slipping, not to maintain it. $\endgroup$
    – Bob D
    Commented Apr 30 at 0:28
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    $\begingroup$ @BobD How would a cylinder slow down on an incline, while maintaining the same frequency of rotation without slipping? In this question, either the slope is without friction, which should be specifically mentioned, or the cylinder will slip. Saying it is smooth, does not imply without friction. As I thought I made clear in my comment on your reference and, in agreement with what you are saying here, friction is required to start rolling without friction, but is not necessary to maintain it. $\endgroup$
    – Rich
    Commented Apr 30 at 3:33
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    $\begingroup$ @BobD That is not always the case. In the case of a cylinder rolling up an inclined plane against gravity, there does need to be a continual frictional force to maintain rolling without slipping. Just the initial frictional force is not enough. Rolling without slipping means $v=r\omega$. As the cylinder slows down, if there is no friction then $r\omega$ is constant, not decreasing with the velocity. Then there is sliding. $\endgroup$
    – Rich
    Commented Apr 30 at 5:19
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    $\begingroup$ For me smooth is a math term; a unique tangent/derivative exists at every point. $\endgroup$
    – Rich
    Commented Apr 30 at 14:01
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    $\begingroup$ @BobD: can you find a more reliable source for the smooth=frictionless claim? I have never encountered it outside this question. $\endgroup$
    – RLH
    Commented Apr 30 at 14:19
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The quoted question is badly worded. It should say the object is initially rolling without slipping, because as its vertical velocity slows down it has to roll with slipping. I think frictionless would be less ambiguous than smooth. The rolling without slipping part is a red hearing to throw you off because for a frictionless surface it is irrelevant whether it is rolling or not.

will the body keep on rotating even at B? That seems unrealistic.

Yes, that is what happens, even if it seems unrealistic.

Can anybody also elaborate why rotational kinetic energy does not change when external torque is absent but presence of external force does not matter in change of kinetic energy?

A linear external force does matter for a change in kinetic energy. Here the linear external force is the force of gravity acting on the centre of the object. A torque is required to change the rotational kinetic energy.

EDIT: While superficially there appears to be an anticlockwise torque acting on the object, due to a component of the surface reaction force being parallel to the incline, this component cannot apply a torque to the object due to the surface being frictionless.

If it helps, remove the inclined plane and imagine launching the cylinder straight up with a vertical velocity of v. It will reach a maximum height and the kinetic energy will go to zero, but the cylinder will continue rotating even at the top and the rotation rate has no effect on the height reached (in a vacuum).

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    $\begingroup$ The problem states rolling without slipping, not frictionless conditions. $\endgroup$
    – RLH
    Commented Apr 29 at 23:11
  • $\begingroup$ @RLH The answer in the book by the people that wrote the problem, states the situation is frictionless, so that must be what they mean by "smooth" surface. They should know what they meant. $\endgroup$
    – KDP
    Commented Apr 29 at 23:23
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    $\begingroup$ They clearly don’t know what they meant, because the problem and answer are inconsistent. The provided question is a canonical question in mechanics that ties into several principles of energetically; the provided answer is for a contrived didactic counterexample problem. $\endgroup$
    – RLH
    Commented Apr 29 at 23:44
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    $\begingroup$ I have never encountered this “smooth=frictionless” assumption. And I’ll say it again: the “rolling without slipping” problem is a canonical, interesting, and useful problem to have asked, and rolling without slipping is stated in the problem; the frictionless problem is at best a counter-example of “what would have happened if there hadn’t been friction” $\endgroup$
    – RLH
    Commented Apr 30 at 14:11
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    $\begingroup$ Not to do an extended conversation, I just want to say to me, the answer in the book is wrong. As said, that happens and I don't see a need to twist things around to try and interpret it as correct. They may not want to publish a new errata, but that's what happened. If they meant frictionless, there is a word for that: frictionless. Every example when I searched this site as suggested clarifies "frictionless". If the writers meant you to guess it was frictionless based on an unspoken code, that's a different test and seems deliberately misleading (which makes me doubt that's what they meant). $\endgroup$
    – Vectorjohn
    Commented Apr 30 at 22:40
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From energy conservation, translational KE, rotational KE and potential energy are conserved:

$$K_{{rot}_{initial}} + K_{{trans}_{initial}} +U_{initial} = $$ $$K_{{rot}_{final}} + K_{{trans}_{final}}+U_{final}$$

Using the center of mass : $$\frac 1 2 mv_{cm}^2 = mgh - \frac 1 2 I(\omega_i^2 - \omega_f^2)$$ where I is the moment of inertia and $\omega$ is the rotational speed. For pure rolling (rolling friction applies)

$$F_{friction}R = -I\alpha$$

$$\omega = \frac{v_{cm}} R$$ The rolling friction assures that when $v_{cm} = 0$, $\omega = 0$. The friction provides the torque and deceleration $\alpha$ that slows the rotational velocity and

$$\frac 1 2 mv_{cm}^2 = mgh - \frac 1 2 I(\omega_i^2)$$

If friction doesn't apply then there is no torque to slow $\omega$ so $\omega_i = \omega_f$ and

$$\frac 1 2 mv_{cm}^2 = mgh$$

$v_{cm}$ will reach 0 but the body will still rotate with its initial rotational velocity: $$\omega_f = \frac{v_{{cm}_i}} R$$

So what does rolling without slipping mean?

It's just a constraint that $v_{cm} = \frac{\omega}{R}$ which can be accomplished on a frictionless surface as long as $v_{cm}$ is constant.

In your problem, $v_{cm}$ is not constant. The only way that rolling without slipping can occur is if the angular velocity changes as well. That requires a rolling friction force.

Your answer is correct.

The question should have read:

A solid cylindrical body of mass m and radius r rolls without slipping to the bottom of a frictionless inclined plane. Its initial velocity is v. What would be the maximum height attained by the body?

The book answer is correct with this wording.

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The problem is poorly worded. It should not say “no slipping”. Rolling without slipping generally means pure rolling. The condition for pure rolling is $v(t)_{com}=R\omega(t)$. That cannot occur going up the incline without static friction, yet the term “smooth surface” implies a frictionless surface. Moreover with no slipping the angular velocity would need to decrease along with a decrease in linear velocity, for a loss of rotational kinetic energy, which is not accounted for in the answer.

Given the above, since there is no friction and thus no torque, the only external force acting parallel to the incline is $mg\sin\theta$ acting through the COM. Thus the cylinder spins at constant angular velocity all the way up the incline (maintaining its rotational kinetic energy) while losing translational kinetic energy due to the negative acceleration of the COM.

Bottom line: If, in fact, the cylinder loses no rotational KE moving up the incline, the surface must be frictionless. But we shouldn’t be talking about “rolling” up the incline. The cylinder is spinning at constant angular velocity while sliding up the incline due to its inertia. Since the surface is frictionless, there is no energy loss in the form of heat. The end result is simply the conversion of translational kinetic energy to gravitational potential energy.

Hope this helps.

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    $\begingroup$ Wait. Slipping to me means the cylinder neither slides nor spins in place. Looks like they only mean sliding Along the surface $\endgroup$
    – Bob D
    Commented Apr 28 at 19:31
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    $\begingroup$ @aayush they should not say there’s no slipping. If they didn’t say that and just said it is going up a frictionless surface it would be OK $\endgroup$
    – Bob D
    Commented Apr 29 at 11:54
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    $\begingroup$ @aayush This exact problem is discussed in the following video: docvphysics.blogspot.com/2014/11/… $\endgroup$
    – Bob D
    Commented Apr 29 at 13:00
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    $\begingroup$ I wouldn't say the problem is poorly worded. As written it's pretty unambiguous, but the answer given is wrong for the question. You have to count the rotational kinetic energy because if the cylinder rolls without slipping then the rotation rate has to be slowed by something (friction). It's very clearly not frictionless, the answer is just wrong. $\endgroup$
    – Vectorjohn
    Commented Apr 29 at 16:20
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    $\begingroup$ @BobD you say "but slipping is occurring". The question states very plainly "rolls without slipping over a smooth inclined plane". Smooth here cannot be interpreted as frictionless, smooth just means there are no bumps or curves. Your assertion is that the answer is correct for a different question, which, sure. But that's not the question. The answer is wrong for the question that was asked. $\endgroup$
    – Vectorjohn
    Commented Apr 30 at 18:48
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To me, smooth means no friction, so it will continue to spin at the same rate whether or not it goes up and incline. Rolling without slipping over an inclined plane means no relative velocity between point of contact of cylinder and the plane at any point as it moves up the plane. So I think the question proposes an impossible situation.

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