A circular ring of mass m and radius R is rot | vedclass

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(D) According to the principle of conservation of angular momentum,since no external torque acts on the system,the angular momentum remains constant:

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) According to the principle of conservation of angular momentum,since no external torque acts on the system,the angular momentum remains constant: $L_1 = L_2$.The initial moment of inertia of the ring is $I_1 = mR^2$.The initial angular momentum is $L_1 = I_1 \omega = mR^2 \omega$.When two particles of mass $M$ are attached at the ends of a diameter,the new moment of inertia $I_2$ is the sum of the ring's moment of inertia and the moment of inertia of the two particles: $I_2 = mR^2 + M R^2 + M R^2 = (m + 2M)R^2$.Equating the angular momenta: $I_1 \omega = I_2 \omega'$.$mR^2 \omega = (m + 2M)R^2 \omega'$.Solving for $\omega'$,we get: $\omega' = \frac{m \omega}{m + 2M}$.

$A$ circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two particles of mass $M$ are gently attached to the ring at the ends of a diameter. The ring will now rotate with an angular velocity $\omega'$ equal to:

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