A small ball of mass `2 xx 10^-3 kg`, having a charge `1 mu C`, is suspended by a string of length `0.8 m`. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity that should be imparted to the lower ball so that it can make a complete revolution.

Answer

5.86

Explanation

Given `q = 1 mu C = 10^(-6) C " " :" " m = 2 xx 10^(-3) kg " " and " " l=0.8` m
Let u be the speed of the particle at its lowest point and v its speed at highest point.
At highest point three forces are acting on the particle
(a) electrostaic replusion
`F_(e) = (1)/(4 pi epsi_(0) )(q^(2))/(l^(2))` (outwards)
(b) weight v = mg ( inwards)
(c) tension T (inwards)
T = 0 , if the particle has just to complete the circle and the necessary centripetal force is provided by ` v - F_(e) ` i .e
`(mv^(2))/(l) = w - F_(e) or " "v^(2) = (l)/(m) (mg - (1)/(4 pi epsi_(0)) (q^(2))/(l^(2)))`
`v^(2) = (0.8 )/( 2 xx 10^(3)) ( 2 xx 10^(3) xx 10 - (9.0 xx 10^(9) xx (10^(-6))^(2))/((0.8)^(2))m^(2) //s^(2)`
or `v^(2) = 2.4 m^(2) // s^(2)`
Now, the electrostatic potential energy at the lowest and highest points are equal. Hence, form conservation of mechanical energy.
Increase in gravitational potential energy decrease in kinetic energy
or ` m g (21) = (1)/(2) m (u^(2) - v^(2)) or u^(2) = v^(2) + 4 gl`
Substituting the values of `v^(2)` from Eq.(i) we get
`u^(2) =2.4 A ( 10)(0.8) = 34.4 m^(2) // s^(2) " " therefore " "u = 5.86 m //s`
Therefore, minimum horizontal velocity imparted to the lower ball, so that it can make complete revolution, is `5.86 m//s`.