Question

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

• A. M / (M + m)
• B. M / (M + 2m)
• C. (M - 2m) / (M + 2m)
• D. (M + 2m) / M

Hint:

When mass is added to a rotating system, the angular speed decreases because the moment of inertia increases while angular momentum stays constant.

Answer 1

The Correct Option is B

Step 1: By Conservation of Angular Momentum: $I_1 \omega_1 = I_2 \omega_2$.
Step 2: Initial moment of inertia $I_1 = Mr^2$.
Step 3: Final moment of inertia $I_2 = Mr^2 + m(r^2) + m(r^2) = (M + 2m)r^2$.
Step 4: $Mr^2 \omega = (M + 2m)r^2 \omega_2 \implies \omega_2 = \frac{M}{M + 2m} \omega$.