Question

In space of horizontal Electric field $(E=(mg\left)/q\right)$ exist as shown in figure and a mass $m$ attached at the end of a light rod. If mass $m$ is released from the position shown in figure, find the angular velocity of the rod when it passes through the bottom most position:

• A. $\sqrt{\frac{g}{l}}$
• B. $\sqrt{\frac{2g}{l}}$
• C. $\sqrt{\frac{3g}{l}}$
• D. $\sqrt{\frac{5g}{l}}$

Answer 1

The correct option is B $\sqrt{\frac{2g}{l}}$

According to the work energy theorem we have
$W_e+W_g=\frac{1}{2}m{v}^2$
We have work done by electrostatic force as
$W_e=qElsin\theta$
and work done by the gravitational force as
$W_g=mg(l-lcos\theta )$
Thus we get
$qElsin\theta +mg(l-lcos\theta )=\frac{1}{2}m{v}^2$
Thus we get
$mgsin\theta +mgl-mglcos\theta =\frac{1}{2}m{v}^2$
as $\theta ={45}^o$, we get
$mgl=\frac{1}{2}m{v}^2$
also as $v=\omega l$
we get
$\omega =\sqrt{\frac{2g}{l}}$