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 ω 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

Let the new angular speed of the rod be ω'.

Here, net torque on the system is zero.

So, the angular momentum is conserved.

Therefore, we get

\[I\omega = I'\omega'\]

\[\left[ m \left( \frac{L}{2} \right)^2 + m \left( \frac{L}{2} \right)^2 \right]\omega = \left[ 2m \left( \frac{L}{2} \right)^2 + m \left( \frac{L}{2} \right)^2 \right]\omega'\]

\[2m L^2 \omega = 3m L^2 \omega'\]

\[ \Rightarrow \omega'   = \frac{2}{3}\omega\]