Explanation

(c) Centrifugal force on rod, $F = \frac{mv^{2}}{r}$ along $BF.$ Let $\theta$ be the angle which the rod makes with the vertical. Forces acting on the rod are shown in figure
[image:4_image139.emf]
Resolving $mg$ and $F$ into two rectangular components, we have,
Forces parallel to rod,
$mg\cos{\theta + \frac{mv^{2}}{r}}\sin{\theta = T}$
Force perpendicular rod
$= mg\sin{\theta - \frac{mv^{2}}{r}}\ \cos\theta$
The rod will be balanced if
$mg\sin{\theta = \frac{mv^{2}}{r}}\cos{\theta = 0}$
or $mg\sin{\theta = \frac{mv^{2}}{r}\cos\theta}$
or tan $\theta = \frac{v^{2}}{rg} = \frac{(10)^{2}}{10 \times 10} = 1 = \tan{45{^\circ}}\ $or $\theta = 45{^\circ}$