# The relationship between velocity of centre of mass and angular velocity of a rigid body

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?

• Am I missing something? I dont see anything about angular momentum. Do you mean angular velocity of the rigid body in your title? Aug 25, 2021 at 4:13
• @Kksen Someone has changed the title. I also think it's inappropriate so I will change it again. Aug 25, 2021 at 5:06
• You are forgetting the kinematics, where $v_{\rm CM}$ and $\omega$ are linked together, Aug 26, 2021 at 1:32
• 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 $$KE = \frac{1}{2}(mv_{CM}^2+I_{CM} \omega^2) = \frac{1}{2}(mv_{A}^2+I_A\omega^2)$$ Aug 26, 2021 at 12:23

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

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)

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

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

• 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 $r=r(\theta)$. Aug 25, 2021 at 14:29

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.