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

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(B) 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

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

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(B) 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 $L_1 = L_2$.$Mr^2 \omega = (M + 2m)r^2 \omega'$Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{M \omega}{M + 2m}$.

$A$ 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 will now rotate with an angular velocity:

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