Question

A small ball of mass $m$ and charge $+q$ tied with a string of length $l$ is 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. $\theta ={\tan }^{-1}\left(\frac{qE}{mg}\right)$
• B. $\theta =\pi$
• C. $\theta =\pi -{\tan }^{-1}\left(\frac{qE}{mg}\right)$
• D. $\theta =\pi +{\tan }^{-1}\left(\frac{qE}{mg}\right)$

Answer 1

The correct option is A $\theta ={\tan }^{-1}\left(\frac{qE}{mg}\right)$

mass $=m$

charge $=q$

length $=l$

Electric field $=E$

Tension in the string will be minimum $f_0$. Let, the velocity at bottom most point is $V_0$. Then an angle $\theta$ let the velocity of particle is $V$.

$T-mg\cos \theta -qF\sin \theta =\frac{{mV}^2}{l}⟶\left(1\right)$

Using energy conservation,

$\frac{1}{2}{mV}^2-\frac{1}{2}{mV}_0^2=-mgl(1-\cos \theta )+qEl\sin \theta$

$\frac{{mV}^2}{l}=\frac{{mV}_0^2}{l}=-2mg(1-\cos \theta )+2qE\sin \theta$

Putting in equation $\left(1\right)$ we get,

$T=mg\cos \theta +qE\sin \theta +\frac{{mV}_0^2}{l}-2mg(1-\cos \theta )+2qE\sin \theta$

$T=3mg\cos \theta +3qE\sin \theta +\frac{{mV}_0^2}{l}-2mg$

For $T$ to be minimum

$\frac{dT}{d\theta }=0$ $\frac{d^{2}T}{{d\theta }^2}<0$

Thus, $\frac{dT}{d\theta }=-3mg\sin \theta +3qE\cos \theta =0$

$\tan \theta =\frac{qE}{mg}$

$\theta ={\tan }^{-1}\left(\frac{qE}{mg}\right)$ or, $\pi +{\tan }^{-1}\left(\frac{qE}{mg}\right)$