Question

A bar of mass $m$ resting on a smooth horizontal plane starts moving due to the force $F=\frac{mg}{3}$ 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 $\alpha$

• A. $v=\sqrt{\left(\frac{g}{a}\right)\sin \alpha }$
• B. $v=\sqrt{\left(\frac{g}{3a}\right)\sin \alpha }$
• C. $v=\sqrt{\left(\frac{2g}{3a}\right)\sin \alpha }$
• D. $v=\sqrt{\left(\frac{5g}{3a}\right)\sin \alpha }$

Answer 1

Answer: (C)

$v=\sqrt{\left(\frac{2g}{3a}\right)\sin \alpha }$

Newton's second law of motion in projection form, along horizontal or x-axis i.e. $F_x=m{w}_x$ gives,
$F\cos \left(as\right)=mv\frac{dv}{ds}$ (as $\alpha =as$)
or, $F\cos \left(as\right)ds=mvdv$
Integrating over the limits for $v\left(s\right)$
$\frac{F}{m}{\int }_0^{\infty }\cos \left(as\right)ds=\frac{v^2}{2}$
or $v=\sqrt{\frac{2F\sin \alpha }{ma}}$
$=\sqrt{\frac{2g\sin \alpha }{3a}}$ (using $F=\frac{mg}{3}$)
which is the sought relationship.