Physics - Rotation Question with Solution | TestHub

Angular momentum is conserved during the inelastic collision. The initial angular momentum of the disk is $L_i=I_{\text{disk}}{\omega }_0=\frac{1}{2}M{R}^2{\omega }_0$. The small object is dropped gently, so its initial angular momentum about the center is zero. The final moment of inertia of the disk-object system is $I_f=I_{\text{disk}}+I_{\text{object}}=\frac{1}{2}M{R}^2+m{R}^2=R^2(\frac{M}{2}+m)$. Equating $L_i=I_f{\omega }_f$, we have $\frac{1}{2}M{R}^2{\omega }_0=R^2(\frac{M}{2}+m){\omega }_f$. Solving for ${\omega }_f$: ${\omega }_f=\frac{\frac{1}{2}M{\omega }_0}{\frac{M}{2}+m}=\frac{M{\omega }_0}{M+2m}$.