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

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Question

(C) The initial moment of inertia of the ring about the central vertical axis is $I = M R^2$.The initial angular velocity is $\omega = 2 \; rad \; s^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The initial moment of inertia of the ring about the central vertical axis is $I = M R^2$.The initial angular velocity is $\omega = 2 \; rad \; s^{-1}$.Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.Initial angular momentum $L_i = I \omega = (M R^2) 	imes 2 = 2 M R^2$.When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia becomes $I' = M R^2 + m R^2 + m R^2 = (M + 2m) R^2$.Let the new angular velocity be $\omega'$. By conservation of angular momentum:$I \omega = I' \omega'$$2 M R^2 = (M + 2m) R^2 \omega'$$\omega' = \frac{2 M}{M + 2m} \; rad \; s^{-1}$.

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