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Consider a rotating object with mass $m$, moment of inertia $I$, along an inclined plane of vertical height $h$. Then simply speaking the following conservation law holds.

$$\frac{1}{2}(mv_{CM}^2+I\omega^2) = mgh$$

Recognise that all other letters than $v_{CM}$ and $\omega$ are given constants. Thus, solving for $\omega$,

$$\omega = \sqrt{\frac{2mgh-mv_{CM}^2}{I}}$$

The result above means that we get the angular velocity as a function of the velocity of the centre of mass.

But I personally feel that this contradicts with my intuition, because experimentally if there are given inclination $\theta$, height $h$, moment of inertia $I$, and initial velocity $v_0=0$, then there should be a unique set of $(v_{CM}, \omega)$. What's the problem?

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  • $\begingroup$ Am I missing something? I dont see anything about angular momentum. Do you mean angular velocity of the rigid body in your title? $\endgroup$
    – Kksen
    Aug 25, 2021 at 4:13
  • $\begingroup$ @Kksen Someone has changed the title. I also think it's inappropriate so I will change it again. $\endgroup$ Aug 25, 2021 at 5:06
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    $\begingroup$ You are forgetting the kinematics, where $v_{\rm CM}$ and $\omega$ are linked together, $\endgroup$ Aug 26, 2021 at 1:32
  • $\begingroup$ You are missing the CM subscript on the mass moment of inertia. As kinetic energy is invariant to the location it is measured if the parallel axis theorem is used $$KE = \frac{1}{2}(mv_{CM}^2+I_{CM} \omega^2) = \frac{1}{2}(mv_{A}^2+I_A\omega^2) $$ $\endgroup$ Aug 26, 2021 at 12:23

3 Answers 3

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$$\frac{1}{2}(mv_{CM}^2+I\omega^2) = mgh$$

You have only changed the subject to $\omega$, considering only the term $I\omega^2$. But $\omega$ is hiding inside $\frac{1}{2}mv_{CM}^2$ too, because $v_{CM}=\omega r$, in rolling without slipping motion. So you have to present $v_{CM}$ in terms of $\omega$ before quoting $\omega$.

(It is obvious that you are considering rolling without slipping because you have used "initial PE=final KE". If you consider rolling with slipping you have to consider energy loss due to friction)

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There are a unique set of $\omega$ and $v_{cm}$.

They are in fact both related to each other by the equation $$\omega=\frac{v_{cm}}{r}$$

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  • $\begingroup$ I guess the reason why we can consider the formula $v_{CM} = R \times \omega$ is that we approximate every infinitesimal movement as the centre of mass undergoing a very short rotation with the axis being at the intersection between the object and the incline. But what if the object - despite having symmetry along the axis of rotation - has different radius? I mean like, represented by a typical polar coordinate system $r=r(\theta)$. $\endgroup$ Aug 25, 2021 at 14:29
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Assuming the rotating object is a wheel and is rolling along the inclined plane, then the angular velocity $\omega$ is related to the velocity of COM, $v_{CM}$ by $v_{CM}=\omega R$, where $R$ is the radius of the wheel. So they are interrelated that once you solve one of them by energy conservation, you get the other one as well.

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