Physics - Rotation Question with Solution | TestHub

During the collision, angular momentum about the pivot $O$ is conserved as there are no external torques about $O$. The initial angular momentum is due to the bullet: $L_i=mv(L/2)$. The final angular momentum of the rod-bullet system is $L_f=I_{total}\omega$, where $I_{total}$ is the moment of inertia of the rod with the embedded bullet about the center. The moment of inertia of the rod about its center is $I_{rod}=\frac{1}{12}M{L}^2$. The moment of inertia of the embedded bullet (as a point mass at distance $L/2$ from the pivot) is $I_{bullet}=m(L/2{)}^2=\frac{m{L}^2}{4}$. Thus, $I_{total}=I_{rod}+I_{bullet}=\frac{1}{12}M{L}^2+\frac{1}{4}m{L}^2=\frac{(M+3m)L^2}{12}$. Equating initial and final angular momenta: $mv(L/2)=\frac{(M+3m)L^2}{12}\omega$. Solving for $\omega$, we get $\omega =\frac{mvL/2}{(M+3m)L^2/12}=\frac{6mv}{(M+3m)L}$.