A small sphere tied to the string of length `0.8m` is describing a vertical circle so that the maximum and minimum tensions in the string are in the ratio `3:1`. The fixed end of the string is at a height of `5.8m` above ground.
(a) Find the velocity of the sphere at the lowest position.

(b) If the string suddenly breaks at the lowest position, when and where will the sphere hit the ground?

Explanation

(a) Let u and v be the speed of sphere at the bottom and the top positions and m be its mass.
Radius of circle = length of string = r = 0.8 m`
`T_(b) = tension at the bottom or the maximum tension`
`T_(t) = tension at the top or the minimum tension

`T_(b) - mg = (mu^(2))/(r)`
`T_(t) + mg = (mv^(2))/(r) `
`Rightarrow T_(b) = 3 T_(t)`
`((mu^(2))/(r) + mg) = 3((mv^(2))/(r) - mg )`
`Rightarrow (3v^(2) - u^(2)) = 4rg ...(i) `
Using conservation of energy. Loss in KE from bottom to the top = gain in GPE
`1/2mu^(2) - 1/2(mv)^(2) = mg (2r)`
`Rightarrow v^(2) = u^(2) - 4rg ...(ii)`
Using (i) and (ii), we get
`3(u^(2) - 4gr) - u^(2) = 4rg` `Rightarrow 2u^(2) = 16 rg`
`Rightarrow u = sqrt(8(0.8)10) = 8 m/s`
(b)After breaking away from the string the sphere moves along a parabolic path, and strikes the ground at G.
Vertical displacement of sphere
` (##NAR_NEET_PHY_XI_P2_C06_SLV_039_S02.png" width="80%">
`T = sqrt(2S_(y))/(g) = sqrt(2xx5)/(10) = 1 sec`
So `x = u_(x) T = uT = 8 xx 1 = 8 m`
The sphere hits the ground after 1 second at a horizontal distance 8m from the fixed end of the string.