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 m//s`

Explanation

Given : `q=1 muC=10^(-6)C`

`m=2xx10^(-3)` kg and `l=0.8 m`
Let u be the speed of the particle at its lowest point and v its speed at high point. At highest point three forces are acting on the particle.
(i) Electrostatic repulsion
`F_(e)-1/(4pi epsilon_(0))=q^(2)/l^(2)` (outwards)
(ii) Weight W=mg (inwards)
(iii) Tension T (inwards)
`T=0`, if the particle has just to complate the circle and the necessary centripetal force provided by
`W-F_(e)` i.e., `(mv^(2))/l=W-F_(e)`
`rArr v^(2)=l/m(mg-1/(4pi epsilon_(0))xxq^(2)/l^(2))`
`v^(2)=0.8/(2xx10^(3))(2xx10^(3)xx10-(9.0xx10^(9)xx(10^(-6))^(2))/((0.8)^(2)))`
`rArr v^(2)=2.4 m^(2)//s^(2)` ...(i)
Now, the electrostatic potential energy at the lowest and highest points are equal. Hence, from conservation of mechanical energy.
Increase in gravitational potential energy
`=` Decrease in kinetic energy
`rArr mg(2l)=1/2 m (u^(2)-v^(2))rArr u^(2)-v^(2)=4gl`
Substituting the values of `v^(2)` from Eq (i), we get
`u^(2)=2.4+4(10)(0.8)=34.4 m^(2)//s^(2)`
`:. 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`