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A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity
A
$\frac{\omega M}{M + m}$
B
$\frac{\omega(M - 2m)}{M + 2m}$
C
$\frac{\omega M}{M + 2m}$
D
$\frac{\omega(M + 2m)}{M}$
Detailed Solution
The objects are attached gently, so no external torque acts and angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$
Initial moment of inertia of the ring: $I_1 = MR^2$
After attaching two masses m at distance R from the axis: $I_2 = MR^2 + 2mR^2 = (M + 2m)R^2$
$MR^2\omega = (M + 2m)R^2\omega'$
$\omega' = \frac{\omega M}{M + 2m}$
Initial moment of inertia of the ring: $I_1 = MR^2$
After attaching two masses m at distance R from the axis: $I_2 = MR^2 + 2mR^2 = (M + 2m)R^2$
$MR^2\omega = (M + 2m)R^2\omega'$
$\omega' = \frac{\omega M}{M + 2m}$