Question

A particle is suspended vertically from a point O by an inextensible massless spring of length $L$. A vertical line $AB$ is at a distance $L/8$ from O as shown. The object 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 circular motion ceases at point $P$. At this point tension in the string vanishes. Let $v$ be the velocity at point $P$. Let $OP$ make angle $\theta$ with horizontal. Beyond $P$, the motion of the object is parabolic. The highest point of the parabolic path lies on $AB$. Obviously $PQ$ is equal to the half range of the projectile.

Equation of motion of the particle at point $P$

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

as $T=0$

$v^2=gh\sin \theta ....\left(1\right)$

Applying law of conservation of mechanical energy during the motion from the lowest point to $P$ we have $\frac{1}{2}m{u}^2-\frac{1}{2}m{v}^2=mg(L+L\sin \theta )$

$u^2=v^2+2gL(1+\sin \theta )$ [substituting value of $v^2$]

$u^2=gL\sin theta+2gL(1+\sin \theta )=gL(2+3\sin \theta ).....\left(ii\right)$

From the equation of projectile

Half range $=\frac{v^2\sin \theta \cos \theta }{g}$

From geometry $PQ=\frac{v^2\sin \theta \cos \theta }{g}=L\cos \theta -\frac{L}{8}$

$v^2=\frac{gL}{\sin \theta \cos \theta }(\cos \theta -\frac{1}{8}).....\left(iii\right)$

From eq (i) and (iii)

$gL\sin \theta =\frac{gL}{\sin \theta \cos \theta }(\cos \theta -\frac{1}{8})\to \cos \theta =\frac{1}{2}$

$\theta ={60}^{\circ }$

Substituting in eq. (ii)

$u^2=gL(2+\frac{3\sqrt{3}}{2})$

$u=\sqrt{gL(2+\frac{3\sqrt{3}}{2})}$.