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One twirls a circular ring ( of mass \( M \) and radius \( R \) ) near the tip of one's finger as shown in Figure-1. In the process the finger
\( \mathrm{P} \) never loses contact with the inner rim of the ring. The finger

W traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is \( r \). The finger rotates with an angular velocity \( \omega_{0} \). The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure-2). The coefficient of friction between the ring and the finger is \( \mu \) and the acceleration due to gravity is g.
[JEEAdvanced-2017]
Figure-1
The total kinetic energy of the ring is :
(a) \( \frac{1}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2} \)
(b) \( \frac{3}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2} \)
(c) \( \mathrm{M} \omega_{0}^{2} \mathrm{R}^{2} \)
(d) \( \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2} \)
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