Question

A particle is suspended vertically from a point $O$ by an inextensible massless string of length $L$. A vertical line $A B$ 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 $A B$. At the instant of crossing $A B$, 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, $Q D=$ half the range of the projectile.
Now, we have following equations

(1) $T_{Q}=0$. Therefore, $m g \sin \theta=\frac{m v^{2}}{L}\,\,\, ...(i)$
(2) $v^{2}=u^{2}-2 g h=u^{2}-2 g L(1+\sin \theta)\,\,\, ...(ii)$
(3) $Q D=\frac{1}{2}$ (Range)
$\Rightarrow\left(L \cos \theta-\frac{L}{8}\right)=\frac{v^{2} \sin 2\left(90^{\circ}-\theta\right)}{2 g}=\frac{v^{2} \sin 2 \theta}{2 g}\,\,\, ...(iii)$
Eq. (iii) can be written as $\left(\cos \theta-\frac{1}{8}\right)=\left(\frac{v^{2}}{g L}\right) \sin \theta \cos \theta$
Substituting value of $\left(\frac{v^{2}}{g L}\right)=\sin \theta$ 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), $v^{2}=g L \sin \theta=g L \sin 60^{\circ}$
or $v^{2}=\frac{\sqrt{3}}{2} g L$
$\therefore$ Substituting this value of $v^{2}$ in Eq. (ii)
$u^{2} =v^{2}+2 g L(1+\sin \theta) $
$=\frac{\sqrt{3}}{2} g L+2 g L\left(1+\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3}}{2} g L+2 g L$
$=g L\left(2+\frac{3 \sqrt{3}}{2}\right)$
$u=\sqrt{g L\left(2+\frac{3 \sqrt{3}}{2}\right)}$