Minimum Angular Velocity to Prevent Ring Slipped Off Finger

To find the minimum threshold angular speed $\omega_0$ required to prevent the ring from slipping down, we analyze the motion and forces acting on the ring.

1. Motion of the Center of Mass of the Ring

Let the origin $O$ be the center of the circular path traced by the contact point (having radius $r$).

Since the contact point between the finger and the inner rim of the ring is always at a distance $r$ from $O$ and $R$ from the center of mass $C$ of the ring, the distance between $O$ and $C$ is: $OC = R - r$

The line $OC$ rotates with an angular speed $\omega_0$. Therefore, the center of mass $C$ of the ring undergoes uniform circular motion of radius $(R - r)$ with angular speed $\omega_0$.

The centripetal acceleration of the ring's center of mass is directed towards $O$ and is given by: $a_c = (R - r)\omega_0^2$

2. Normal Force Analysis

The only force providing the necessary horizontal centripetal force to the ring is the normal force $N$ exerted by the finger at the contact point. By Newton's second law in the horizontal radial direction: $N = M a_c = M(R - r)\omega_0^2$

3. Friction and Vertical Equilibrium

In the vertical direction, two forces act on the ring:

1. Downward gravitational force $Mg$.
2. Upward static friction force $f_s$ exerted by the finger.

For the ring not to slip vertically, the vertical forces must balance: $f_s = M g$

The condition to prevent the ring from slipping down is that the required static friction force $f_s$ must be less than or equal to the maximum static friction force $f_{s,\text{max}}$: $f_s \le \mu N$

Substituting $f_s = Mg$ and $N = M(R - r)\omega_0^2$: $M g \le \mu M(R - r)\omega_0^2$

Dividing both sides by $M$: $g \le \mu(R - r)\omega_0^2$

Solving for $\omega_0$: $\omega_0^2 \ge \frac{g}{\mu(R - r)}$

$\omega_0 \ge \sqrt{\frac{g}{\mu(R - r)}}$

Thus, the minimum value of $\omega_0$ below which the ring will drop down is: $\omega_0 = \sqrt{\frac{g}{\mu(R - r)}}$

Correct Answer: A