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Physics Question on Gravitation

One twirls a circular ring (of mass MM and radius RR ) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger 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 rr. The finger rotates with an angular velocity ω0\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 μ\mu and the acceleration due to gravity is gg. The total kinetic energy of the ring is

A

Mω02R2M \omega_{0}^{2} R^{2}

B

12Mω02(R−r)2\frac{1}{2} M \omega_{0}^{2}(R-r)^{2}

C

Mω02(R−r)2M \omega_{0}^{2}(R-r)^{2}

D

32Mω02(R−r)2\frac{3}{2} M \omega_{0}^{2}(R-r)^{2}

Answer

Mω02(R−r)2M \omega_{0}^{2}(R-r)^{2}

Explanation

Solution

Answer (c) Mω02(R−r)2M \omega_{0}^{2}(R-r)^{2}