Question

A rigid rod of mass $m$ & length $\ell$ is pivoted at one of its ends. If it is released from its horizontal position, find the speed of the centre of mass of the rod when it becomes vertical.

Answer 1

Suppose that in the vertical position the speed of the centre of mass of the rod is $v$
$\omega =\frac{v}{OP}\Rightarrow \omega =\frac{v}{\ell /2}=2\frac{v}{\ell }$ $\left(1\right)$
The total $K.E.$ of the rod about the point $P$ is given by,
$K.E.=\frac{1}{2}{I}_p{\omega }^2$ $\left(2\right)$ where $I_p=I_o+m(OP{)}^2$
$\Rightarrow I_p=\frac{m{\ell }^2}{12}+m(\ell /2{)}^2$
$\Rightarrow I_p\frac{m{\ell }^2}{3}$ $\left(3\right)$
Using $\left(2\right)$ & $\left(3\right)$
$K.E.=\frac{m{\ell }^2{\omega }^2}{6}$ $\left(4\right)$
From $\left(1\right)$,
$K.E.=\frac{m{\ell }^2}{6}{\left(\frac{2v}{\ell }\right)}^2=\frac{2}{3}m{v}^2$ $\left(5\right)$
Since the rod is released from rest, its $K.E.$ at the position $1$
$I=K.E.,=0$
The change in $K.E.$, between the position $1$ and $2$; $\Delta K.E.=K.E_2-K.E_1$
$\Rightarrow \Delta K.E.=\frac{2}{3}m{v}^2-0=\frac{2}{3}m{v}^2$ $\left(6\right)$
Since the rod falls through a vertical distance of $\ell /2$, its gravitational potential energy decreases.
Change in $P.E.=mg\left(\ell /2\right)$
$\Rightarrow 2m{v}^2/3=mg\left(\ell /2\right)\Rightarrow v=\sqrt{\frac{3g\ell }{4}}=\frac{\sqrt{3g\ell }}{2}$