Bead on a ring fixed at a point on the ring A ring of mass $M$ and radius $R$ is supported from a pivot located at one point of the ring about which it is free to rotate in its own vertical plane. A bead of mass $m$ slides without friction about the ring.  Problems And Solutions On Mechanics, Page 571, Question 2048
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Now, from the picture above I have the coordinates of the bead to be $(R\sin\theta)+R\sin\varphi, R\cos\theta+R\cos\varphi)$ and coordinates of the center to be $(R\sin\theta, R\cos\theta)$. Moreover the velocities of the bead and the center are $(R\dot{\theta}\cos\theta + R\dot{\varphi}\cos\varphi, -R\dot{\theta}\sin\theta+R\dot{\varphi}\sin\varphi$) and $(R\dot{\theta}\cos\theta,-R\dot{\theta}\sin\theta)$ respectively.
In the book that I have mentioned above the author considers only the rotational kinetic energy to be the kinetic energy of the ring (not of the system as a whole), which is $\dfrac{I\dot{\theta^2}}{2} =  MR^2\dot{\theta^2}$. Where as in case of the bead both the rotational and the translational kinetic energy is considered.
Shouldn't the kinetic energy of the ring be $\dfrac{Mv^2}{2} + \dfrac{I\dot{\theta^2}}{2}$ which is $2MR^2\dot{\theta^2}$.
I am confused. Kindly help me out.
Thanks.
 A: You’re trying to apply the parallel axis theorem twice.
If the ring were rotating about its center, its moment of inertia would be $I_\text{center} = MR^2$, and its kinetic energy would be $T_\text{center} = \frac12 I_\text{center}\dot\theta{}^2 = \frac12 M v^2$, where $v=R\dot\theta$ would be the speed of the edge of the ring.
Your ring is rotating around a point on its edge. So in addition to the rotational energy $\frac12 I_\text{center}\dot\theta{}^2$, the center is moving with speed $v=R\dot\theta$ relative to the pivot, and there is a translational kinetic energy $\frac12 Mv^2$ associated with this motion. This gives a total kinetic energy $T_\text{edge} = \frac12 I_\text{center}\dot\theta{}^2 + \frac12 Mv^2 = MR^2\dot\theta{}^2$.
The parallel axis theorem says that, if your pivot is some distance $D$ from the center of mass, the effective moment of inertia is
$$
I_\text{effective} = I_\text{center} + MD^2
$$
and that using this effective moment of inertia accounts for the energy and angular momentum of the center of mass about the pivot.
In that picture, the rotational energy of the ring about its edge should be
\begin{align}
T_\text{edge} &= \frac12 I_\text{edge} \dot\theta{}^2
\\ &= \frac12 \left(
I_\text{center} + MR^2
\right) \dot\theta{}^2
\end{align}
These two expressions for $T_\text{edge}$ are equivalent.
