Physics - Rotation Question with Solution | TestHub

The initial angular momentum of the bullet about the pivot is $L_i=mv(L/2)$. After the collision, the bullet gets embedded, and the system (rod + bullet) rotates with angular velocity $\omega$. The moment of inertia of the rod about the pivot is $I_{rod}=\frac{1}{3}M{L}^2$. The moment of inertia of the bullet about the pivot (located at $L/2$ from the pivot) is $I_{bullet}=m(L/2{)}^2=\frac{m{L}^2}{4}$. The total moment of inertia of the system is $I_{sys}=I_{rod}+I_{bullet}=\frac{1}{3}M{L}^2+\frac{1}{4}m{L}^2=L^2(\frac{4M+3m}{12})$. By conservation of angular momentum, $L_i=L_f$, so $mv(L/2)=L^2(\frac{4M+3m}{12})\omega$. Solving for $\omega$, we get $\omega =\frac{mv(L/2)}{L^2(\frac{4M+3m}{12})}=\frac{mv}{2L(\frac{4M+3m}{12})}=\frac{6mv}{L(4M+3m)}$.