The bob of a pendulum has mass m = 1 kg and charge q = 40 μC. Length of pendulum is l = 0.9 m. The point of suspension also has the same charge 40 μC. What the minimum speed u should be imparted to the bob so that it can complete vertical circle?

Force of repulsion between the charges,

$\mathrm{F}=\frac{9\times {10}^9\times 40\times 40\times {10}^{-12}}{(0.9{)}^2}=\frac{160}{9} \mathrm{N}$

Since F > mg, so string always remains tight at any position even if velocity of bob is zero. So minimum speed at lowest point will be such that it is sufficient to take particle to highest point, i.e:, velocity becomes zero at highest point. For this

${\mathrm{u}}_{\min }=\sqrt{4\mathrm{gl}}=\sqrt{4\times 10\times 0.9}=6 \mathrm{m}/\mathrm{s}$