A small ball if mass `2 xx 10^(-3) Kg` having a charges of ` 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 which should be imparted to the lower ball so that it can make 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`