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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
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Verified Answer
The correct answer is:
$\frac{M \omega}{M+2 m}$
In the absence of external torque, angular momentum remain constant
$$
\begin{aligned}
\mathrm{L} & =\mathrm{I} \omega=\mathrm{I}^{\prime} \omega^{\prime} \\
\therefore \quad \mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\
\omega^{\prime} & =\frac{\mathrm{M} \omega}{(\mathrm{M}+2 \mathrm{~m})}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{L} & =\mathrm{I} \omega=\mathrm{I}^{\prime} \omega^{\prime} \\
\therefore \quad \mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\
\omega^{\prime} & =\frac{\mathrm{M} \omega}{(\mathrm{M}+2 \mathrm{~m})}
\end{aligned}
$$
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