A bar of mass $m$ resting on a smooth horizontal plane starts moving due to force $F=mg/3$ of constant magnitude. In the process of its rectilinear motion the angle $\alpha$ between the direction of this force and the horizontal varies as $\alpha$ $= as$ , where $a$ is a constant, and $s$ is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle $\alpha$ .

Let at any distance $s$ , the $\alpha$ is the inclination of force. From Newton’s second law, we have

                     $F \cos \alpha = m\frac{dv}{dt}$

or                 $\frac{mg}{3}\cos as = m\left(v\frac{dv}{ds}\right)$

or                              $vdv = \frac{8}{3}\cos \left(as\right)ds$            …(i)

Integrating equation (i), we get

                   ${\int }_0^{v}vdv = \frac{g}{3}{\int }_0^s\cos \left(as\right)ds$

or                ${\left|\frac{v^2}{2}\right|}_0^v = \frac{g}{3a}{\left|\sin as\right|}_0^s$

or               $\left(\frac{v^2}{2}-0\right) = \frac{g}{3a}(\sin as - \sin 0)$

                                   $v = {\left[\frac{2g}{3a}\sin as\right]}^{1/2}$

                                    $= {\left[\frac{2g}{3a}\sin \alpha \right]}^{1/2}$ .