Question

A ball of mass $m$ with a charge $q$ can rotate in a vertical plane at the end of a string of length $\ell$ in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity must be imparted to the ball in uppermost position so that the tension in the string in the lowermost position of the ball is $15$ times the weight of the ball?

Answer 1

gain in kinetic Energy $=$ Work done by external forces
$⟹$ gain in $KE=$ work done by gravity $+$ work done by electrostatic force
$⟹\frac{1}{2}m\left(v_2^2{-}_1^2\right)=\left(mg\right)\left(2\ell \right)-\left(qE\right)\left(2\ell \right)$ ...(1)
For the ball at lowermost position $Q$:
$qE+T-mg=\frac{m{g}_2^2}{\ell }$ ...(2)
Putting $T=$ Tension $=15mg$
$qE+15mg-mg=\frac{m{v}_2^2}{\ell }$ ...(3)
From (1) and (3):
$⟹\frac{1}{2}(v_2^2-v_1^2)=2mg\ell -2qE\ell$
$⟹m{v}_2^2-m{v}_1^2=4mg\ell -4qE\ell$
$⟹\frac{m{v}_2^2}{\ell }=\frac{m{v}_1^2}{\ell }+4mg-4qE$
Hence, $15mg+qE-mg=\frac{m{v}_1^2}{\ell }+4mg-4qE$
$v_1=\sqrt{\frac{\ell }{m}(10mg+5qE)}$.