Q.
A rod of length $50\,cm$ is pivoted at one end. It
is raised such that if makes an angle of $30^\circ$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\, s^{-1}$) will be $(g = 10\,ms^{-2})$
Solution:
Work done by gravity from initial to final position is,
$W = mg \frac{\ell}{2} \sin30^{\circ} $
$ = \frac{mg \ell}{4} $
According to work energy theorem
$W = \frac{1}{2} I \omega^{2} $
$ \Rightarrow \frac{1}{2} \frac{m \ell^{2}}{3} \omega^{2} = \frac{mg \ell}{4} $
$ \omega = \sqrt{\frac{3g}{2\ell}} = \sqrt{\frac{3\times10}{2 \times0.5}} $
$\omega = \sqrt{30} $ rad / sec
