Skip to content
dronacharya.ai
← JEE Advanced 2017 · Paper 2
PhysicsSingle correct+3Question 18 of 54

JEE Advanced 2017 · Paper 2, Physics question 18

Passage

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 minimum value of ω0\omega_0 below which the ring will drop down is

Choose one option
Answer and worked solution

Official answer: A

Normal force. The finger's normal force on the ring supplies the centripetal force for the centre's circle of radius R−rR-r:
N=Mω02(R−r).N=M\omega_0^2(R-r).

Friction. It must support the weight, so μN≥Mg\mu N\ge Mg:
ω0≥gμ(R−r).\omega_0\ge\sqrt{\frac g{\mu(R-r)}}.

(A)

Topics:FrictionCircular motion dynamics
← Previous questionNext question →