Question

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

Answer 1

Answer:

u=5gl−q24πε0ml−−−−−−−−−−−√=(2758)1/2m/s

Explanation:

Solution :

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

T2+mg−q24πε0l2=mv2l ...(i)

At the topmost point, T2=0. So

mg−q24πε0l2=mv2l ... (ii)

From energy conservation,

Energy at lowest point = Energy at topmost point

or 12mu2=12mv2+mg2l ...(iii)

or v2−u2=−4gl ...(iv)

From Eq. (ii), v2=gl−q24πε0ml ...V

From Eqs. (iv) and V,

u=5gl−q24πε0ml−−−−−−−−−−−√=(2758)1/2=5.86ms−1