A small bead of mass m moving with velocity v | vedclass

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(C) The angular momentum of the system about the center $O$ is conserved because there is no external torque acting on the system.Initial angular mome

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$v/2R$

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(C) The angular momentum of the system about the center $O$ is conserved because there is no external torque acting on the system.Initial angular momentum of the bead about $O$ is $L_i = mvR$.The moment of inertia of the semicircular ring about its center is $I_{ring} = mR^2$.The moment of inertia of the bead at the rim of the ring is $I_{bead} = mR^2$.When the bead comes to rest relative to the ring,the total moment of inertia of the system is $I = I_{ring} + I_{bead} = mR^2 + mR^2 = 2mR^2$.By the conservation of angular momentum,$L_i = L_f$,so $mvR = I\omega$.$mvR = (2mR^2)\omega$.Solving for $\omega$,we get $\omega = \frac{mvR}{2mR^2} = \frac{v}{2R}$.

$A$ small bead of mass $m$ moving with velocity $v$ gets threaded on a stationary semicircular ring of mass $m$ and radius $R$ kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

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