Question

A thin horizontal uniform rod $AB$ of mass $m$ and length $l$ can rotate freely about a vertical axis passing through its end $A$. At a certain moment, the end $B$ starts experiencing a constant force $F$ which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane. Find the angular velocity of the rod as a function of its rotation angle $\theta$ measured relative to the initial position.

Answer 1

Work done by the torque
$△W=\int \tau d\theta ={\int }_0^{\theta }Fl\cos \theta d\theta$
$△W=Fl\sin \theta$
Now using work energy theorem,
$△W=△k$
$\therefore Fl\sin \theta =[\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2-0]$
Which gives, $\omega =\sqrt{\frac{6F\sin \theta }{ml}}$.