A mass m is tied to a string of length l and is rotated in a vertical circle with centre at the other end of the string.
(a) Find the minimum velocity of the mass at the top of the circle so that it is able to complete the circle.
(b) Find the minimum velocity at the bottom of the circle corresponding to the above condition.

Explanation

At all positions, there are two forces acting on the mass :
its own weight and the tension in the string.
Here radius of the circle = l
(a) At the top : Let `v_(t) =` velocity at the top
Net force towards centre = `(mv_(t)^(2))/(l)` (centripital force)
` T + mg = (mv_(t)^(2))/l Rightarrow T = (mv_(i)^(2))/(l) - mg`

For the movement in the circular path, the string should remain tight i.e. the tension must be positive at all positions. As the tension is minimum at the top, `T_(top) ge 0`
`Rightarrow (mv_(t)^(2)/(l) - mg ge 0` `Rightarrow v_(i) ge sqrt(lg)`
`Rightarrow` minimum or critical velocity at the top
` = sqrt lg`(b) Let `v_(b)` be the velocity at the bottom. As the particle goes up, its KE decreases and GPE increases.
` (##NAR_NEET_PHY_XI_P2_C06_SLV_034_S02.png" width="80%">
`Rightarrow` loss in KE = gain in GPE
`Rightarrow 1/2(mv_(b)^(2) - 1/2 (mv_(t)^(2) = mg(2l)`
`v_(b)^(2) = v_(t)^(2) + 4 gl`
`(v_(b))_(min) = sqrt((v_(t)^(2))_(min) + 4gl ) = sqrt 5gl`