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When an object is in pure rolling motion its kinetic energy is equal to $\frac{1}{2}mv^2+\frac{1}{2} I \omega^2$ where $I$ is moment of inertia through centre of mass and $\omega$ is angular velocity about centre of mass. But it's KE is can also be expressed as if it is a pure rotation about axis passing through the point of contact which is equal to $\frac{1}{2}\hat{I} \omega^2$ where $\hat{I}$ is moment of inertia about the axis passing through point of contact . But $\omega$ should also be about axis passing through point of contact, but it's taken about centre of mass . Why is that??

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  • $\begingroup$ use the parallel axis theorem, also called Steiner's theorem $\endgroup$ Commented Jan 8, 2017 at 16:19

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The two are the same.

The angular speed about the point of contact satisfies

$$v=R\omega_1$$

On the other hand, in the reference frame of the CM, the point of contact is moving with velocity $v$, and hence $$v=R\omega_2$$

So $$\omega_1=\omega_2$$

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A rolling body is easier to describe in center of mass system. When assuming the $\omega$ axis on the point of contact you must add a Steiner term

$E_+ = \frac{1}{2}I_+ \omega^2;I_+ = mr^2$

while the computation of the translational kinetic energy would be more difficult.

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For a body to pure roll, the point of contact with the ground should be at rest and hence $v_{CM}=Rw$ where $w$ is the angular velocity about CM and hence for a circular body, From the point of contact, to the top most point the distance is $2R$ and velocity w.r.t ground is $v_{CM}+Rw$ which is equal to $2Rw_{\rm about\ the\ point\ of\ contact}$ and $v_{CM}=Rw$ and hence the conclusion follows....

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