A thin ring of mass M and radius R is rotatin | vedclass

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(A) The initial moment of inertia of the ring is $I = MR^2$. The initial angular momentum is $L = I\omega = MR^2\omega$.When $4$ point masses of mass

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$\left( {\frac{M}{{M + 4m}}} \right)\omega $

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(A) The initial moment of inertia of the ring is $I = MR^2$. The initial angular momentum is $L = I\omega = MR^2\omega$.When $4$ point masses of mass $m$ are placed at the ends of two mutually perpendicular diameters,they are all at a distance $R$ from the axis of rotation.The new moment of inertia of the system becomes $I' = MR^2 + 4(mR^2) = (M + 4m)R^2$.Since there is no external torque acting on the system,the angular momentum remains conserved $(L_{initial} = L_{final})$.$MR^2\omega = (M + 4m)R^2\omega'$Dividing both sides by $R^2$,we get $M\omega = (M + 4m)\omega'$.Therefore,the new angular velocity is $\omega' = \left( \frac{M}{M + 4m} \right)\omega$.

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