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In space of horizontal $EF$ ($E = (mg)/q$) 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
$\sqrt {\frac{g}{l}} $
$\sqrt {\frac{{2g}}{l}} $
$\sqrt {\frac{{3g}}{l}} $
$\sqrt {\frac{{5g}}{l}} $
Solution
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
$W_{e}=q E l \sin \theta$
and work done by the gravitational force as
$W_{g}=m g(l-l \cos \theta)$
Thus we get
$q E l \sin \theta+m g(l-l \cos \theta)=\frac{1}{2} m v^{2}$
Thus we get
$m g \sin \theta+m g l-m g l \cos \theta=\frac{1}{2} m v^{2}$
as $\theta=45^{\circ},$ we get
$m g l=\frac{1}{2} m v^{2}$
also as $v=\omega l$
we get
$\omega=\sqrt{\frac{2 g}{l}}$
