Question

A uniform rod of mass $200$ gm and length $1$ m is initially at rest in a 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 $m=100$ gm each having horizontal velocity of equal magnitude $u=6$ m/s strike the rod at the top and the bottom simultaneously as shown and stick to the rod. Find the angular speed (in rad/s) of the rod when it becomes horizontal.

Answer 1

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