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A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $\omega$. Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be
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Verified Answer
The correct answer is:
$\left(\frac{M}{M+4 m}\right) \omega$
According to conservation of angular momentum,
$I \omega=\text { constant }$
i.e. we can write,
$\begin{array}{rlrl}
I_1 \omega_1 & =I_2 \omega_2 & \\
\text { or } & M R^2 \omega =(M+4 m) R^2 \omega_2 & \\
\text { or } & \omega_2 =\left(\frac{M}{M+4 m}\right) \omega &
\end{array}$
$I \omega=\text { constant }$
i.e. we can write,
$\begin{array}{rlrl}
I_1 \omega_1 & =I_2 \omega_2 & \\
\text { or } & M R^2 \omega =(M+4 m) R^2 \omega_2 & \\
\text { or } & \omega_2 =\left(\frac{M}{M+4 m}\right) \omega &
\end{array}$
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