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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.
Solution
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
\[I\omega = I'\omega'\]
\[\left[ m \left( \frac{L}{2} \right)^2 + m \left( \frac{L}{2} \right)^2 \right]\omega = \left[ 2m \left( \frac{L}{2} \right)^2 + m \left( \frac{L}{2} \right)^2 \right]\omega'\]
\[2m L^2 \omega = 3m L^2 \omega'\]
\[ \Rightarrow \omega' = \frac{2}{3}\omega\]
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