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

Answer

4

Explanation

It is clear from the statement of the problem that the particle leaves the circular path before it crosses the line AB second time, during its upward motion.
Let us assume that the particle leaves 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 :
`1/2 m u^2 = 1/2 m v^2 + mgL(1 + sin theta) " "....(i)`
At point `P : T = mg sin theta = (m v^2)/(L)`. If T becomes zero. `v^2 = gL sin theta " ".....(ii)`
After leaving 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.
`implies 1/2xx "Range" = L cos theta - L/8`
`(v^2 sin 2(90 - theta))/(2g) = L cos theta - L/8 " "....(iii)`
From (ii) and (iii), `cos theta = 1/2 implies theta = 60^@`
Using this in (i) , we get
`u = sqrt([(4 + 3sqrt(3))/(2) gL])`