Question

A ring which can slide along the rod is kept at the mid point of a smooth rod of length L. The rod is rotated with a constant angular velocity $\omega$ about vertical axis passing through one of its ends. Ring is released from mid point. Find the velocity of the ring when it just leave the rod.

Answer 1

Centrifugal force
$m{\omega }^{2}x=ma$
${\omega }^{2}x=\frac{vdv}{dx}$
${\int }_{L/2}^L{\omega }^{2}xdx={\int }_0^{v}vdv$ (integrate both side).
${\omega }^2{\left(\frac{x^2}{2}\right)}_{L/2}^L={\left(\frac{v^2}{2}\right)}_0^v{\omega }^2\left(\frac{L^2}{2}-\frac{L^2}{8}\right)=\frac{v^2}{2}v=\frac{\sqrt{3}}{2}\omega L$
Velocity at time of leaving the rod
$v^{\prime }=\sqrt{(\omega L{)}^2+{\left(\frac{\sqrt{3}}{2}\omega L\right)}^2}=\frac{\sqrt{7}}{2}\omega L$