Question

A light rod of length $L$ can revolve in a vertical circle around point $O$. The rod carries two equal masses of mass $m$ each such that one mass is connected at the end of the rod and the second mass is fixed at the middle of the rod. $u$ is the velocity imparted to the end $P$ to deflect the rod to the horizontal position. Again mass $m$ in the middle of the rod is removed and mass at end $P$ is doubled. Now $v$ is the velocity imparted to end $P$ to deflect it to the horizontal position. Then ${\left(\frac{u}{v}\right)}^2$ is

Answer 1

In first case,
As velocity of mass attached to the middle of rod is $\frac{u}{2}$
Using conservation of energy
$\frac{1}{2}m{u}^2+\frac{1}{2}m{\left(\frac{u}{2}\right)}^2=mgL+mg\frac{L}{2}$
$u=\sqrt{\left(\frac{12}{5}gL\right)}$
In second case,
Using conservation of energy
$\frac{1}{2}\left(2m\right)v^2=2\text{mgL}v=(2gL{)}^{\frac{1}{2}}$
Now,
${\left(\frac{u}{v}\right)}^2=\frac{\left(\frac{12}{5}gL\right)}{\left[2gL\right]}=\left[\frac{6}{5}\right]=1.2$