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Question
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become
Options
\[\frac{\omega M}{M + m}\]
\[\frac{\omega M}{M + 2 m}\]
\[\frac{\omega\left( M - 2 m \right)}{M + 2 m}\]
\[\frac{\omega\left( M + 2 m \right)}{M}\]
Solution
\[\frac{\omega M}{M + 2 m}\]
No external torque is applied on the ring; therefore, the angular momentum will be conserved.
\[I\omega = I'\omega'\]
\[ \Rightarrow \omega' = \frac{I\omega}{I'} ..........(1)\]
\[I = M r^2 \]
\[I' = M r^2 + 2m r^2\]
On putting these values in equation (1), we get
\[\omega' = \frac{\omega M}{M + 2 m}\]
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