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Let's suppose that we have a cylinder of moment of inertia $I$ rolling on the floor without sliding, moving with linear velocity $v$ and rotating around an axis passing through the center of mass with angular velocity $\omega = v / R$. (a composition of translation and rotation)

Now if we consider the motion as a rotation around the instant rotation axis, will the angular velocity with respect to that axis be $\omega$ too or will it be different? Will it be the same if the cylinder were sliding?

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    $\begingroup$ Check your math. It should be $$\omega = \frac{v}{R}$$ $\endgroup$ – ja72 Mar 25 '15 at 0:06
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The fastest way is to compare kinetic energies in the two cases: \begin{align*} KE &= \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} So $\omega_{\text{inst}}=\omega_{\text{cm}}$. The first equation is rotational KE plus translational KE (remember that $\omega = v/R$) for rotation about the center of mass, while the second equation is only rotational energy around the instantaneous axis, where we used the parallel axis theorem.

Note: if the cylinder was sliding then the translational kinetic energy in the first equation will be less than $\tfrac{1}{2}M(R\omega)^2_{\text{cm}} $, so you have to replace the the above system with \begin{align*} KE &< \tfrac{1}{2}I_{\text{cm}}\omega^2_{\text{cm}} + \tfrac{1}{2}M(R\omega)^2_{\text{cm}} \\ KE &=\tfrac{1}{2}I_{\text{inst}}\omega_{\text{inst}}^2 = \tfrac{1}{2} (I_{\text{cm}} + MR^2)\omega^2_{\text{inst}} \end{align*} there for $\omega_{\text{inst}}<\omega_{\text{cm}}$, consistent with intuition.

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