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 its circular motion cases at upper half circle at an angle `'theta'` with horzontal as shown in figure. Also it cannot cases in lower half circle, because in lower half circle, its can velocity cannot be horzontal while passing through AB, F.B.D at 'P' w.r.t. non-inertial from

`because At"p"` circuilar motion cases, its tension becomes zero.
`(mv^(2))/(L)=mg sintheta" "...(1)`
`v=sqrt(gLsintheta)`
Now, applying coservation of mechanical energy between bottom most point and at P. `K_(i)+U_(i)=H_(f)+U_(f)` [Assumign bottom most point as fererence]
`1/2m u^(2)+0=1/2mv^(2)+mgL(1+sin0)`
From equation (1),
`u^(2)=g Lsintheta + 2gL(1+ sin theta)" "...(2)`
The motion of particle after P is projectile and when it passes through line AB, then its becomes horzontal, it means at AB line it will be at its maximum height, from the given diagram.
`PQ=1/2"Range"`
`L cos 0-L/8=1/2(v^(2)sin2alpha)/(g)`
`2L(costheta)1/8=9Lsin theta(2 sin alpha cosalpha)/(g)`
`costheta-1/8=sinthetasin (90-theta)con(90-theta)`
`=sin^(2)thetacostheta`
`=(1-cos^(2)theta)costheta`
`costheta-1/8=costheta-cos^(3)theta`

`cos theta-1/8=costheta-cos^(3)theta`
`impliescos^(3)theta=1/8`
`impliescostheta=1/2`
`impliestheta=60^(@)`
From equation (2),
`u^(2)=2gL+3gL(sqrt3)/(2)`
`u=sqrt((gL)/(2)(4+3sqrt3))`