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 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. Take $r\ll R$

• A ring of mass $M$ and radius $R$.

• Contact point with finger moves on a smaller circle of radius $r$ ($r\ll R$).

• Finger rotates with angular velocity $\omega_0$.

• Ring rolls without slipping on this path.

Step 1: Kinematics condition

For rolling without slipping:

vcm​=Rω=rω0​

where $\omega$ is angular velocity of the ring about its center.

So,

ω=Rrω0​​ 

Step 2: Forces on the ring

• Normal reaction at contact: $N = Mg$.

• Friction $f \leq \mu Mg$ provides centripetal force.

Centripetal force needed:

Fc​=RMvcm2​​=RM(r2ω02​)​

This must be supplied by friction:

RMr2ω02​​≤μMg 

Step 3: Condition on $\omega_0$

ω02​≤r2μgR​

So maximum possible angular velocity of the finger is:

ω0,max​=r2μgR​​