The bob of a simple pendulum of length l is given a sharp hit to impart it a horizontal speed of ` sqrt(3gl)` . When it was at its lowermost position. Find
angle `alpha` shown of the string from upside of vertical and speed of the particle when the string becomes slack.

Explanation

Since `sqrt(2gl)ltu ltsqrt(5gl)`, the string slacks somewhere between horizontal point and the topmost point.

Let string slack at P, where speed is say v. At point P,
`impliesT+mgcosalpha=(mv^(2))/(l)`

As the string slacks, `T = 0`
`impliesmgcosalpha=(mv^(2))/(l)`
`impliesv=sqrt(glcosalpha)" "....(i)`
Applying work energy theorem for motion from A to P
`-mgh_(1)=(1)/(2)m(v^(2)-u^(2))`
`therefore` from equation (i)
`-mgl(1+cosalpha)=(1)/(2)m[g(lcosalpha)-3gl]`
`impliescosalpha=(1)/(3)`
`implies alpha=cos^(-1)((1)/(3))`
`therefore` equation (i), `u=sqrt((gl)/(3))`