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Question

# A uniform rod of mass 200 grams and length L = 1 m is initially at rest in vertical position. The rod is hinged at the centre such that it can rotate freely without friction about a fixed horizontal axis passing through its centre. Two particles of mass m = 100 grams, each having horizontal velocity of equal magnitude u = 6 m/s strike the rod at the top and bottom simultaneously as shown, and stick to the rod. The angular speed (in rad/sec.) of the rod when it becomes horizontal is . Write upto two digits after the decimal point.

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Solution

## Assuming that there is no loss of energy and the rod - mass system keeps rotating as the masses stick to it, we can say that it will rotate with a constant angular velocity. So, from conservation of angular momentum. muL2+muL2=[2mL212+m(L2)2+(L2)2]ω muL=[mL26+mL24+mL24]ω=2mL23ω or ω=3u2L=3×62×1=9 rad/s

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