A thin circular ring of mass M and radius R i | vedclass

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(C) The initial moment of inertia of the ring about its axis is $I_1 = MR^2$.The initial angular momentum of the system is $L_1 = I_1 \omega = MR^2 \o

Vedclass · Accepted Answer

$\frac{\omega M}{M + 2m}$

Vedclass · Answer

(C) The initial moment of inertia of the ring about its axis is $I_1 = MR^2$.The initial angular momentum of the system is $L_1 = I_1 \omega = MR^2 \omega$.When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia of the system becomes $I_2 = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.Since no external torque acts on the system,the angular momentum is conserved,so $L_1 = L_2$.$MR^2 \omega = (M + 2m)R^2 \omega^{\prime}$.Dividing both sides by $R^2$,we get $M \omega = (M + 2m) \omega^{\prime}$.Therefore,the new angular velocity is $\omega^{\prime} = \frac{\omega M}{M + 2m}$.

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

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