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

It is clear from the statement of the problem that the particle has the circular path before it crosses the line AB second time, during its upward motion.

Let us assume that the particle have the circular path at P and v is the velocity of the particle at P. Then, conservation of mechanical energy at P and at the lowest point yields:
`1/2u^2=1/2mv^2+mgL(1+sintheta)`
`mu^2=2gL(1+sintheta)+mv^2` (i)
At point `T+mg sin theta=(mv^2)/(L)`.
If T becomes zero:
`v^2=gLsintheta` (ii)
After having the circular path, the particle moves along a parabolic trajectory as a projectile.
As it is given that velocity of the particle when it crosses AB is horizontal, it is clear that, when it is crossing AB, it is passing through its highest point in the parabolic path.
Hence, half of the range of this projectile path `=Lcostheta-L/8`
As projection angle of the particle with respect to horizontal time at
`P=(90^@-theta)`, therefore, maximum range `=(v^2sin2(90^@-theta))/(8)`
Hence, `1/2xx(v^2sin2(90^@-theta))/(g)=Lcostheta-L/8` (iii)
From Eq. (ii) and (iii): `costheta=1/2impliestheta=60^@`
Using this information and (I) and (ii),
`u=sqrt([(4+3sqrt3)/(2)gL])`