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(C) Before impact,the bullet is moving with velocity $v$. The initial angular momentum of the system about the hinge point $O$ is $J = m v L$.After th

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$\frac{3 m v}{(M+3 m) L}$

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(C) Before impact,the bullet is moving with velocity $v$. The initial angular momentum of the system about the hinge point $O$ is $J = m v L$.After the bullet gets embedded in the rod,suppose the system attains angular velocity $\omega$. The moment of inertia of the bullet-rod system about the axis through $O$ is:$I = I_{	ext{bullet}} + I_{	ext{rod}} = m L^2 + \frac{1}{3} M L^2 = \left( \frac{M + 3m}{3} ight) L^2$.By the principle of conservation of angular momentum,the initial angular momentum equals the final angular momentum:$J = J' \implies m v L = I \omega$$m v L = \left( \frac{M + 3m}{3} ight) L^2 \omega$Solving for $\omega$:$\omega = \frac{3 m v}{(M + 3m) L}$.

$A$ uniform rod of mass $M$ and length $L$ is suspended from a rigid support. $A$ small bullet of mass $m$ hits the rod with velocity $v$ and gets embedded into the rod. The angular velocity of the system just after impact is

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