Question

Two small balls A and B, each of mass m, are joined rigidly by a light horizontal rod of length L. The rod is clamped at the centre in such a way that it can rotate freely about a vertical axis through its centre. the system is rotated with an angular speed $\omega$ about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod.

Answer 1

Since external torque $=0$Therefore $l_1{\omega }_1=l_2{\omega }_2$$I_1=\frac{m{l}^2}{4}+\frac{m{l}^2}{4}=\frac{m{l}^2}{2}$${\omega }_1=\omega$$I_2=\frac{2m{l}^2}{4}+\frac{m{l}^2}{4}=\frac{3m{l}^2}{4}$Therefore ${\omega }_2=\frac{l_1{\omega }_1}{l_2}=\frac{\left(\frac{m{l}^2}{2}\right)\times \omega }{\frac{3m{l}^2}{4}}=\frac{2\omega }{3}$