# A small ball of mass m and charge +q tied with a string of length l , rotating in a vertical circle under gravity and a uniform horizontal electric field E as shown . The tension in the string will be minimum for

$(a)\;\theta=tan^{-1} (\large\frac{qE}{mg})\qquad(b)\;\theta=\pi\qquad(c)\;\theta=0^{0}\qquad(d)\;\theta=\pi+tan^{-1} (\large\frac{qE}{mg})$

Answer : (d) $\;\theta=\pi+tan^{-1} (\large\frac{qE}{mg})$
Explanation :
Let the velocity at bottom most point is $\;V_{0}$
Then at angle $\;\theta\;$ let the velocity of particle is V
$T-mg cos \theta-qE sin \theta=\large\frac{mV^2}{l}--(1)$
Using energy conservation
$\large\frac{1}{2}\;mV^2-\large\frac{1}{2}\;mV_{0}^2=-mgl (1-cos \theta) + qEl sin \theta$
$\large\frac{mV^2}{l}=\large\frac{mV_{0}^2}{l}-2mg(1-cos \theta) + 2qE sin \theta$
Putting in equation (1) we get
$T=mg cos \theta +qE sin \theta+\large\frac{mV_{0}^2}{l}-2mg(1-cos \theta) +2qE sin \theta$
$T=3mg cos \theta +3qE sin \theta+\large\frac{mV_{0}^2}{l}-2mg$
For T to be minimum
$\large\frac{d T}{d \theta}=0$ & $\;\large\frac{d^2T}{d \theta ^2} > 0$
Thus $\;\large\frac{dT}{d \theta}=-3 mg sin \theta + 3 qE cos \theta =0$
$tan \theta= \large\frac{qE}{mg}$
$\theta=tan^{-1}\;(\large\frac{qE}{mg})\quad \; or \quad \; \pi+tan^{-1}\;(\large\frac{qE}{mg})$
$\large\frac{d^2 T}{d \theta^2}=-3\;(mg cos \theta+qE sin\theta) > 0$
At $\;\theta=tan^{-1}(\large\frac{qE}{mg})\quad\;\large\frac{d^2T}{d \theta^2} < 0$
At $\;\theta=\pi+tan^{-1}(\large\frac{qE}{mg})\quad\;\large\frac{d^2T}{d \theta^2} > 0$
Thus the desired $\;\theta \;$ at which tension is minimum is $\;\pi+tan^{-1} (\large\frac{qE}{mg})\;.$
answered Feb 17, 2014 by
edited Feb 17, 2014 by yamini.v