Physics - Rotation Question with Solution | TestHub

Angular momentum is conserved about the pivot during the inelastic collision. The initial angular momentum of the particle about the pivot is $L_i=mvL$. The final moment of inertia of the rod-particle system about the pivot is $I_f=I_{\text{rod}}+I_{\text{particle}}=\frac{1}{3}M{L}^2+m{L}^2=L^2(\frac{M}{3}+m)$. Equating $L_i=I_f\omega$, we get $mvL=L^2(\frac{M}{3}+m)\omega$. Solving for $\omega$: $\omega =\frac{mvL}{L^2(M/3+m)}=\frac{mv}{L(M/3+m)}=\frac{3mv}{L(M+3m)}$.