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A hollow cylinder is rolling down an inclined plane. We have to get the speed of the center of mass of the cylinder at the end of the inclined plane.

Energy Conservation tells us:

$$ mg(h + R) = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 + mgR $$

Cancelling $mgR$: $$ mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 $$

If we assume frictionless rolling, we can now substitute $\omega = \frac{v_{cm}}{R}$.

Which yields, with $I_{cm} = mR^2$:

$$ v_{cm} = \sqrt{gh} $$

Is this right? If it is indeed right, why do we not have to take the moment of inertia about the point of contact with the incline, using the parallel-axis theorem? Is it because the point of contact with the incline is stationary, and we would have to consider the speed of the cylinder about the point of contact with the ground, which would be $v_{g}=0$?

Some clarification would be very much appreciated.

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closed as off-topic by stafusa, Jon Custer, sammy gerbil, Mitchell, JMac Dec 25 '17 at 15:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – stafusa, Jon Custer, sammy gerbil, Mitchell, JMac
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. You might want to consider posting on some other Physics website, such as the PhysicsForums. $\endgroup$ – stafusa Dec 22 '17 at 19:42
  • $\begingroup$ It's not a checked work type example, I'm specifically asking if the idea here is the right one. See: Is this right? If it is indeed right, why do we not have to take the moment of inertia about the point of contact with the incline, using the parallel-axis theorem? Is it because the point of contact with the incline is stationary, and we would have to consider the speed of the cylinder about the point of contact with the ground, which would be vg=0vg=0? Which is in my opinion an understanding type question, which should be following your rules. $\endgroup$ – Christopher Dec 22 '17 at 19:51
  • $\begingroup$ First, did you mean $\sqrt{gh}$? As for which axis to take, it's as arbitrary as the choice of an origin. And, with respect to closing the question, it's just my interpretation of this policy that the question is off-topic $-$ but it's good you defended your question: it might swing some people and, since a total of five votes are necessary to close, if I'm mistaken, my vote shouldn't have any effect. $\endgroup$ – stafusa Dec 22 '17 at 20:03
  • $\begingroup$ I did mean $\sqrt{gh}$, thanks for pointing that out. I understand that the choice is arbitrary, but my intuition would have had me choose the moment of inertia about the point of contact with the inclined plane, even when using the speed of the center of mass, which would obviously influence the result. Specifically because $\omega=\frac{v}{R}$, which would be the angular speed using the speed of a point on the outside of the circle. I'm specifically asking why we can assume $\omega=\frac{v_{cm}}{R}$ and still use the moment of inertia about the center of mass. This does not make sense to me. $\endgroup$ – Christopher Dec 22 '17 at 20:15
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    $\begingroup$ Very helpful. I was indeed wrong! Physics is very fun and surprising.. $\endgroup$ – Christopher Dec 22 '17 at 21:02
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The calculation is correct.

Let's see what happens if we take as reference point the point of contact ${P}$ between the cylinder and the plane:

  • We can use the parallel axis theorem to obtain the moment of inertia: $$I = I_{cm}+md^2 = mR^2 + mR^2 = 2mR^2. $$
  • The angular speed is the same, as we can see considering the speed of the point opposite to $P$, $v_P=2v_{cm}$, which is at a distance $r=2R$ from $P$, so $\omega = v/r = 2v_{cm}/(2R) = v_{cm}/R.$
  • The kinetic energy is only rotational now, as $P$ is immovable: $$ \Delta K = \frac{1}{2}I\omega^2 = \frac{1}{2} 2m R^2 \frac{v_{cm}^2}{R^2} = m v_{cm}^2. $$
  • Then, using that $\Delta K = \Delta U = mgh$, we have: $$ m v_{cm}^2 = mgh \Longrightarrow v_{cm}=\sqrt{gh}. $$

That is, there's no change $-$ just as expected, since the choice of axis is arbitrary.

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  • $\begingroup$ Now I see what you mean by arbitrary! I couldn't think that far ahead. Very enlightening! I would upvote if I could.. $\endgroup$ – Christopher Dec 22 '17 at 20:55

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