A uniform rod of mass \(M\) and length \(L\) is suspended from the 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

Before impact, the bullet is moving with velocity \(v\). The initial impact, angular momentum of the system about \(O\) the hinge point is \(J=L \times m v=m v L\)
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,
\(I=\) (M.I Of bullet + M.I of rod)
\(=m L^2+\frac{1}{3} M L^2\)
\(I=\left(\frac{M+3 m}{3}\right) L^2\)
Final angular momentum of the system is

By conservation of angular momentum,
\(J=J^{\prime}\left(\frac{M+3 m}{3}\right) L^2 \omega=m v L\left(\frac{M+3 m}{3}\right) L^2 \omega=m v L\)
So, the angular velocity of the system just after impact is
\(\omega=\frac{3 m v}{(M+3 m) L}\)