Question

A Rod of mass M and length L stands vertically on a smooth horizontal surface. A particle of mass m travelling at speed u strikes the top of the rod and comes to rest. The angular velocity of the rod after the impact is

Answer 1

Answer on Question #49921-Physics-Mechanics-Kinematics-Dynamics

A Rod of mass M and length L stands vertically on a smooth horizontal surface. A particle of mass m travelling at speed u strikes the top of the rod and comes to rest. The angular velocity of the rod after the impact is

Solution

Conservation of angular momentum:

$$m u L = I \omega = \left(\frac {M L ^ {2}}{3} + m L ^ {2}\right) \omega .$$

The angular velocity of the rod after the impact is

$$\omega = \frac {m u L}{\frac {M L ^ {2}}{3} + m L ^ {2}} = \frac {m u}{L} \left(\frac {1}{1 + \frac {M}{3 m}}\right).$$

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