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

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(B) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the

Vedclass · Accepted Answer

$\frac{M\omega}{M + 4m}$

Vedclass · Answer

(B) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the final angular momentum.$L_1 = L_2$$I_1 \omega_1 = I_2 \omega_2$Initially,the moment of inertia of the ring is $I_1 = MR^2$ and angular velocity is $\omega$.After placing four masses $m$ at the rim,the new moment of inertia is $I_2 = MR^2 + 4(mR^2) = (M + 4m)R^2$.Substituting these into the conservation equation:$MR^2 \omega = (M + 4m)R^2 \omega_2$$\omega_2 = \frac{MR^2 \omega}{(M + 4m)R^2} = \frac{M\omega}{M + 4m}$.

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Four objects,each of mass $m$,are placed gently at the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be:

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