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.

Answer 1

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.
$mu\frac{L}{2}+mu\frac{L}{2}=[2m\frac{L^2}{12}+m{\left(\frac{L}{2}\right)}^2+{\left(\frac{L}{2}\right)}^2]\omega$
$muL=[\frac{m{L}^2}{6}+\frac{m{L}^2}{4}+\frac{m{L}^2}{4}]\omega =\frac{2m{L}^2}{3}\omega$ or $\omega =\frac{3u}{2L}=\frac{3\times 6}{2\times 1}=9rad/s$