Question

A weightless rod of length 2l carries two equal masses 'm', one secured at lower end A and the other at the middle of the rod at B. The rod can rotate in verticle plane about a fixed horizontal axis passing through C. What horizontal velocity must be imparted to the mass at A so that it just completes the vertical circle.

Answer 1

Here,
$\frac{1}{2}m{v}_a^2+\frac{1}{2}m{v}_b^2=2lmg+4lmg$
$\frac{v_a}{2l}=\frac{v_b}{l}$
$v_a=2v_b$
And substituting valve of $v_b$-
$\frac{1}{2}m{v}_a^2+\frac{1}{2}m{\left(\frac{v_a}{2}\right)}^2=6lmg$
So, finally,
$v_a=4\sqrt{\frac{3}{5}lg}$