Question

A particle describes a horizontal circle of radius $r$ on the smooth surface of an inverted cone as shown. The height of plane of circle above vertex is $h$. The speed of particle should be

• A. $\sqrt{r g}$
• B. $\sqrt{2 r g}$
• C. $\sqrt{g h}$
• D. $\sqrt{2 g h}$

Answer 1

Answer: (C)

$N \cos \left(90^{\circ}-\theta\right)=\frac{m v^{2}}{r}$
$N \sin \left(90^{\circ}-\theta\right)=m g$
$N=\frac{m g}{\cos \theta}$ ...(i)
and $N \sin \theta=\frac{m v^{2}}{r}$ ...(ii)
dividing (ii) and (i)
$m g \tan \theta=\frac{m v^{2}}{r}$
$g\left(\frac{h}{r}\right)=\frac{v^{2}}{r}$
$\Rightarrow v=\sqrt{g h}$