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

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(A) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.When four particles of mass $m$ are placed on the ring,the new moment of in

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

Vedclass · Answer

(A) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.When four particles of mass $m$ are placed on the ring,the new moment of inertia of the system becomes $I' = Mr^2 + 4(mr^2) = (M + 4m)r^2$.Since there is no external torque acting on the system,the angular momentum remains conserved.Therefore,$L_{initial} = L_{final}$.$Mr^2\omega = (M + 4m)r^2\omega'$.Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{Mr^2\omega}{(M + 4m)r^2} = \frac{M\omega}{M + 4m}$.

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

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