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A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity $\omega$. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
A
$\frac{(M + 2m)\omega}{2m}$
B
$\frac{2M\omega}{M + 2m}$
C
$\frac{(M + 2m)\omega}{M}$
D
$\frac{M\omega}{M + 2m}$
Detailed Solution
The objects are attached gently, so no external torque acts; 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 the two masses at distance r from the axis: $I_2 = Mr^2 + 2mr^2 = (M + 2m)r^2$
$Mr^2\omega = (M + 2m)r^2\omega'$
$\omega' = \frac{M\omega}{M + 2m}$
Initial moment of inertia of the ring: $I_1 = Mr^2$
After attaching the two masses at distance r from the axis: $I_2 = Mr^2 + 2mr^2 = (M + 2m)r^2$
$Mr^2\omega = (M + 2m)r^2\omega'$
$\omega' = \frac{M\omega}{M + 2m}$