Question

The height at which body leaves the vertical circle is:

• A. $\frac{2}{3}r$
• B. $\frac{3}{2}r$
• C. $3r$
• D. r

Answer 1

Answer: (A)

When body leaves the vertical circle its reaction is zero.
The various forces acting on the body are as shown.
Let body leave the path at point $P$, then
$R=0=mg\sin \theta -\frac{m{v}^2}{r}$
where $\frac{m{v}^2}{r}$ is the centripetal force.
$\Rightarrow v^2=rg\sin \theta ?$ ... (i)
Also, from law of conservation of energy Potential energy of
fall from $=$ Kinetic energy $Q$ to $P$.
$\therefore mgr-mgr\sin \theta =\frac{1}{2}m{V}^2$... (ii)
From Eqs. (i) and (ii), we get
$2-2\sin \theta =\sin \theta$
$\Rightarrow \sin \theta =\frac{2}{3}$
Also, $\sin \theta =\frac{h}{r}=\frac{2}{3}$
$\Rightarrow h=\frac{2}{3}r$