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 $L8$ 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

The particle has circular path before it passes $AB$ second time during its upward motion. Let the particle have circular path at ${}^{\prime }{P}^{\prime }$ and $V$ be its velocity at $P$.

Applying conservation of energy at $P$ and at the lowest point gives:

$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mgL(1+\sin \theta )$

At point $P$

$T+mg\sin \theta =\frac{m{v}^2}{2}$

and when $T=0$

$2g\sin \theta =v^2$

After moving in a circular path, the particle moves along a parabolic trajectory. Since the velocity of the particle when it crosses $AB$ is horizontal. Therefore, it is passing through its highest point in the parabolic path. Thus, half of the range of this parabolic path $Lcos\theta =\frac{L}{8}$

Projection angle of the particle at $P$

$=90°-\theta$

$\therefore$ minimum range

$=\frac{v^2\sin 2(90-\theta )}{g}\therefore \frac{1}{2}\frac{v^2\sin 2(90-\theta )}{g}=L\cos \theta -\frac{L}{8}$

$\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60°$

Using the above equations:

$u=\sqrt{\left(\frac{4+3\sqrt{3}}{2}\right)gL}$