Q. Consider a uniform rod of mass $M = 4m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $\nu$ making angle $\theta = \frac{\pi}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is :
Solution:
Let angular velocity of the system after collision be $\omega$.
By conservation of angular momentum about the hinge :
$m\left(\frac{v}{\sqrt{2}}\right)\left(\frac{\ell}{2}\right)=\left[\frac{4m\ell^{2}}{12}+\frac{m\ell^{2}}{4}\right]\omega$
On solving
$\omega=\frac{3\sqrt{2}}{7}\left(\frac{v}{\ell}\right)$
