A simple pendulum with a bob of mass `m = 1` kg, charge `q = 5muC` and string length imis given a horizontal velocity a in a uniform electric field `E = 2 xx 10^6 V/m` at its bottommost point A, as shown in figure. It is given that the speed a is such that the particle leaves the circle at point C. Find the speed a (Take `g = 10 m/ s^2`)

Explanation

From work energy theorem,

`1/2m(v^2-u^2)=-mgl(1+sin60^@)+qElcos60^@`
Substituting the values, we get
`u^2-v^2=32.32`………..i
Further at C tension in the string is zero.
Hence, `(mv^2)/l=mgsin60^@-qEcos60^@`
`or v^2=3.66`....ii
From eqn i and ii we get
`u=6m//s`