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I am a bit confused about the kinetic energy of a rotating object when the axis of rotation does not pass through the center of mass. For example, consider a thin hoop of mass $m$ and radius $R$ that is hanging from a pivot and is free to rotate about that pivot. If the hoop were rotating about it's center of mass, the kinetic energy would just be $$K = \frac{1}{2}I\omega^2 = \frac{1}{2} m R^2 \omega^2,$$ because the center of mass of the hoop is not moving (i.e. the translational kinetic energy is zero). However, when the hoop is rotating about the pivot, the center of mass is no longer stationary. Is this compensated for in the change in the moment of inertia, or does this translational piece need to be added? In other words, do we have $$ K_{tot}= K_{trans}+K_{rot}=\frac{1}{2} m v^2+\frac{1}{2}I\omega^2=\frac{1}{2}m(R\omega)^2+\frac{1}{2}(2mR^2)\omega^2 = \frac{3}{2} m R^2 \omega^2 $$ or $$ K_{tot}= K_{trans}+K_{rot}=0+\frac{1}{2}I\omega^2=\frac{1}{2}(2mR^2)\omega^2 = m R^2 \omega^2 $$ I am fairly certain that the latter is the correct answer but I just wanted to make sure I am thinking about this correctly.

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  • $\begingroup$ The error is in your 1st eqn for $K_{tot}$ : you used the moment of inertia about a point on the rim ($2mR^2$), instead you should have used the moment of inertia about the centre ($mR^2$). $\endgroup$ – sammy gerbil Jul 10 '18 at 3:51
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The kinetic energy of a rigid body of mass $m$ rotating about a fixed point $O'$ with angular velocity $\vec\omega=\omega\hat n$ and translating with velocity $\vec V$ is given by $$T=\frac 12mV^2+\frac 12\omega^2I_n+m\vec R_{cm}'\cdot(\vec V\times\vec\omega),$$ where $I_n$ the moment of inertia about the axis along $\hat n$ and $\vec R_{cm}'$ is the centre of mass vector with respect to the point $O'$. If $O'$ is the centre of mass, then $\vec R_{cm}'=0$.

In your example, the pivot is the point $O'$ and since it is fixed, $\vec V=0$. On the other hand, $I_n=2mR^2$ is the moment of inertia relative to the axis through the pivot and perpendicular to the plane of the hoop. Thus $$T=mR^2\omega^2.$$

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You get the same result by both methods if you use them correctly. If you choose to decompose the motion into motion of the COM and rotation around the COM, then of course the rotational part should use the moment on inertia around the COM. So there is no difference. You can use either way, just use the right terms.

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