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JEE Advanced 2017 · Paper 2, Physics question 17

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PARAGRAPH 2

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.

Figure 1: a finger held up, tracing a cone (dotted), with a horizontal ring of radius R hanging on its tip. Figure 2: top view: a large circle of radius R (the ring) touching internally a small circle of radius r (the path of the contact point).

The total kinetic energy of the ring is

Choose one option
Answer and worked solution

Official answer: Bonus: full marks to everyone

Official key: marks to all (after internal review).

The intended working: the ring's centre moves on a circle of radius R−rR-r at speed ω0(R−r)\omega_0(R-r). Rolling without slipping on the smaller circle makes the ring spin with ω=ω0(R−r)/R\omega=\omega_0(R-r)/R about its centre. Then
K=12Mω02(R−r)2+12MR2ω2=Mω02(R−r)2.K=\tfrac12M\omega_0^2(R-r)^2+\tfrac12MR^2\omega^2=M\omega_0^2(R-r)^2.
That is option (C). The description of the motion is open to other readings, which is why marks were given to all.

Answer: (C) (marks to all)

Topics:Rolling motionRotational kinetic energy
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