A uniform rod of mass 3 kg and length 1.2 m lies at rest on a smooth horizontal table. A putty ball of mass 1 kg moving at 8 m s^-1 perpendicular to…

The table is smooth, so linear momentum and angular momentum about any fixed point are conserved. The centre of mass of the rod plus ball is displaced from the rod centre towards the ball by $x_c=\dfrac{1\times0.6}{4}=0.15\ \mathrm{m}$, and it moves with $V=8/4=2\ \mathrm{m\,s^{-1}}$. Take angular momentum about the final centre of mass, which lies $0.45\ \mathrm{m}$ from the ball's line of motion: $L=mv(0.6-0.15)=1\times8\times0.45=3.6\ \mathrm{kg\,m^2s^{-1}}$ (the rod has none initially). The moment of inertia about this point is $I=\dfrac{3(1.2)^2}{12}+3(0.15)^2+1(0.45)^2=0.36+0.0675+0.2025=0.63\ \mathrm{kg\,m^2}$. Hence $\omega=L/I=3.6/0.63=5.71\ \mathrm{rad\,s^{-1}}$.