A particle is suspended vertically from a point O by an inextensible mass less string of length L. A vertical line AB is at a distance of `L/8` from O as shown in figure. The particle is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u.

Explanation

Let the string slacks when the particle is at point P as shown
At point P, `T+mgcosalpha=(mV^(2))/(L)`
where T = 0 (as string slacks)
`therefore mgcosalpha=(mV^(2))/(L)`

`impliesV^(2)=gLcosalpha" "......(i)`
After this it undergoes parabolic path. When it passes through line AB its velocity is horizontal which implies that `(L sin theta – L//8)` is half of horizontal range.
i.e., `Lsinalpha-(L)/(8)=(1)/(2)[(V^(2)sin(2alpha))/(g)]`
`therefore` from equation (i)
`Lsinalpha-(L)/(8)=(1)/(2)[(gLcosalpha(2sinalphacosalpha))/(g)]`
`therefore sinalpha-(1)/(8)=sinalphacos^(2)alpha`
`impliessinalpha=(1)/(2)impliesalpha=30^(@)`
`therefore V^(2)=(gLsqrt(3))/(2)`
Also by applying work energy theorem
`-mgL(1+cosalpha)=(1)/(2)m(V^(2)-u^(2))`
`=-gL(1+(sqrt(3))/(2))=(1)/(2)((gLsqrt(3))/(2)-u^(2))`
`therefore` On solving, we get `u=sqrt((gL)/(2)(4+3sqrt(3)))`