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A thin circular ring of Mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:
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Verified Answer
The correct answer is:
$\frac{M \omega}{M+4 m}$
According to law of conservation of angular momentum
$$
\begin{aligned}
& M r^2 \omega=\left(M r^2+4 m r^2\right) \omega^{\prime} \\
& \Rightarrow \omega^{\prime}=\frac{M \omega}{M+4 m}
\end{aligned}
$$
$$
\begin{aligned}
& M r^2 \omega=\left(M r^2+4 m r^2\right) \omega^{\prime} \\
& \Rightarrow \omega^{\prime}=\frac{M \omega}{M+4 m}
\end{aligned}
$$
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