Question

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 placed gently at the opposite ends of diameter of the ring. The ring now rotates with an angular velocity

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

Answer 1

Answer: (B)

$\omega M/(M+2m)$

Answer 2

As on torque is acting on the system, so from law of conservation of angular momentum, Initial angular momentum = final angular momentum . i.e., $M r^{2} \omega=(M+2 m) r^{2} \omega$ $\Rightarrow \omega'=\frac{M \omega}{(M+2 m)}$