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 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 r. The finger rotates with an angular velocity ω0. 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 and the acceleration due to gravity is g.
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 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 r. The finger rotates with an angular velocity ω0. 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 and the acceleration due to gravity is g.
The total kinetic energy of the ring isFigure 1Figure 2
(A)Mω02R2
(B)21Mω02(R−r)2
(C)Mω02(R−r)2✓ Correct
(D)23Mω02(R−r)2
Answer: (C)
Solution
The ring has both translational and rotational kinetic energy. The center of the ring moves in a circle of radius (R−r) with angular velocity ω0. Translational KE: 21Mω02(R−r)2. The ring also rotates about its own axis with angular velocity ω=Rω0(R−r) (rolling condition). Rotational KE: 21Iω2=21MR2⋅R2ω02(R−r)2=21Mω02(R−r)2. Total: KE=Mω02(R−r)2.
JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option
