Question

A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $\omega$. Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega/2$. Then, the ratio $m/M$ is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. $\sqrt{2}$
• E. $\frac{1}{\sqrt{2}}$

Hint:

Whenever a problem mentions "attaching gently" or "dropping" onto a rotating object, assume angular momentum is conserved.

Answer 1

The Correct Option is B

Concept:
Since the masses are attached "gently," no external torque acts on the system. Therefore, the total angular momentum ($L$) is conserved: \[ I_1\omega_1 = I_2\omega_2 \]

Step 1: Calculate Initial and Final Moment of Inertia.

• Initial ($I_1$): Only the ring. $I_1 = mR^2$.
• Final ($I_2$): Ring + two point masses $M$ at distance $R$ from the center. \[ I_2 = mR^2 + 2(MR^2) = (m + 2M)R^2 \]

Step 2: Apply Conservation of Angular Momentum.
Given $\omega_1 = \omega$ and $\omega_2 = \omega/2$: \[ mR^2(\omega) = (m + 2M)R^2 \left( \frac{\omega}{2} \right) \] \[ m = \frac{m + 2M}{2} \] \[ 2m = m + 2M \implies m = 2M \] Therefore, the ratio $\frac{m}{M} = 2$.