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(C) Initial moment of inertia of the ring is $I_i = MR^2$. Initial angular velocity is $\omega_i = \omega$. Initial kinetic energy is $K_i = \frac{1}{

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$\frac{Mm}{(M + 2m)} \omega^2 R^2$

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(C) Initial moment of inertia of the ring is $I_i = MR^2$. Initial angular velocity is $\omega_i = \omega$. Initial kinetic energy is $K_i = \frac{1}{2} I_i \omega_i^2 = \frac{1}{2} MR^2 \omega^2$.Since no external torque acts on the system,angular momentum is conserved: $L_i = L_f$.$I_i \omega_i = I_f \omega_f$.The new moment of inertia after attaching two masses $m$ at distance $R$ is $I_f = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.Thus,$\omega_f = \frac{I_i \omega_i}{I_f} = \frac{MR^2 \omega}{(M + 2m)R^2} = \frac{M \omega}{M + 2m}$.Final kinetic energy is $K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} (M + 2m)R^2 \left( \frac{M \omega}{M + 2m} \right)^2 = \frac{1}{2} \frac{M^2 R^2 \omega^2}{M + 2m}$.Loss in kinetic energy $\Delta K = K_i - K_f = \frac{1}{2} MR^2 \omega^2 - \frac{1}{2} \frac{M^2 R^2 \omega^2}{M + 2m}$.$\Delta K = \frac{1}{2} MR^2 \omega^2 \left( 1 - \frac{M}{M + 2m} \right) = \frac{1}{2} MR^2 \omega^2 \left( \frac{M + 2m - M}{M + 2m} \right) = \frac{1}{2} MR^2 \omega^2 \left( \frac{2m}{M + 2m} \right) = \frac{Mm}{(M + 2m)} \omega^2 R^2$.

$A$ ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this,the kinetic energy loss will be

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