A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to…
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
$\frac{\omega(M-2 m)}{M+2 m}$
$\frac{\omega M}{M+2 m}$
$\frac{\omega(M+2 m)}{M}$
$\frac{\omega M}{M+m}$
Solution
Key Idea Apply law of conservation of angular momentum.
$\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2$
In the given case
$\begin{aligned}
\mathrm{I}_1 & =\mathrm{MR}^2 \\
\mathrm{I}_2 & =\mathrm{MR}^2+2 \mathrm{mR}^2 \\
\omega_1 & =\omega
\end{aligned}$
Then
$\omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega$
~