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

`u = sqrt(5 g l - (q^2)/(4 pi epsilon_0 m l)) = ((275)/(8))^(1//2)m//s`

Explanation

If the ball has just complete the circle, then the tension must vanish at the topmost point, i.e., A. From Newton's second law
`T_2 + m g - (q^2)/(4 pi epsilon_0 l^2) = (m v^2)/(l)` ...(i)
At the topmost point, `T_2 = 0`. So
`mg -(q^2)/(4 pi epsilon_0 l^2) = (m v^2)/(l)` ... (ii)
From energy conservation,
Energy at lowest point = Energy at topmost point
or `(1)/(2) m u^2 = (1)/(2) m v^2 + m g 2 l` ...(iii)
or `v^2 - u^2 = -4 g l` ...(iv)
From Eq. (ii), `v^2 = g l -(q^2)/(4 pi epsilon_0 m l)` ...V
From Eqs. (iv) and V,
`u = sqrt(5 g l-(q^2)/(4 pi epsilon_0 m l)) = ((275)/(8))^(1//2) = 5.86 ms^-1`.
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