Q.
The height at which body leaves the vertical circle is:
J & K CETJ & K CET 2000
Solution:
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
$R=0=m g \sin \theta-\frac{m v^{2}}{r}$
where $\frac{m v^{2}}{r}$ is the centripetal force.
$\Rightarrow v^{2}=r g \sin \theta ?$ ... (i)
Also, from law of conservation of energy Potential energy of
fall from $=$ Kinetic energy $Q$ to $P$.
$\therefore m g r-m g r \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, $\sin \theta=\frac{h}{r}=\frac{2}{3}$
$\Rightarrow h=\frac{2}{3} r$
