Question

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. The object 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 $u$.

Answer 1

Let the string slacks at point $Q$ as shown in figure. From $P$ to $Q$ path is circular and beyond $Q$ path is parabolic. At point $C$, velocity of particle becomes horizontal, therefore, $QD=$ half the range of the projectile.
Now, we have following equations

(1) $T_Q=0$. Therefore, $mg\sin \theta =\frac{m{v}^2}{L}...(i)$
(2) $v^2=u^2-2gh=u^2-2gL(1+\sin \theta )...(ii)$
(3) $QD=\frac{1}{2}$ (Range)
$\Rightarrow \left(L\cos \theta -\frac{L}{8}\right)=\frac{v^2\sin 2\left(90^{\circ }-\theta \right)}{2g}=\frac{v^2\sin 2\theta }{2g}...(iii)$
Eq. (iii) can be written as $\left(\cos \theta -\frac{1}{8}\right)=\left(\frac{v^2}{gL}\right)\sin \theta \cos \theta$
Substituting value of $\left(\frac{v^2}{gL}\right)=\sin \theta$ from Eq. (i), we get
$\left(\cos \theta -\frac{1}{8}\right)={\sin }^2\theta \cdot \cos \theta =\left(1-{\cos }^2\theta \right)\cos \theta$
or $\cos \theta -1/8=\cos \theta -{\cos }^3\theta$
$\therefore {\cos }^3\theta =1/8$ or $\cos \theta =1/2$
or $\theta =60^{\circ }$
From Eq. (i), $v^2=gL\sin \theta =gL\sin 60^{\circ }$
or $v^2=\frac{\sqrt{3}}{2}gL$
$\therefore$ Substituting this value of $v^2$ in Eq. (ii)
$u^2=v^2+2gL(1+\sin \theta )$
$=\frac{\sqrt{3}}{2}gL+2gL\left(1+\frac{\sqrt{3}}{2}\right)=\frac{3\sqrt{3}}{2}gL+2gL$
$=gL\left(2+\frac{3\sqrt{3}}{2}\right)$
$u=\sqrt{gL\left(2+\frac{3\sqrt{3}}{2}\right)}$