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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
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Verified Answer
The correct answer is:
$\frac{\omega M}{M+2 m}$
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
$$
\omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega
$$
$$
\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2
$$
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
$$
\omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega
$$
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