Question

A particle describes a horizontal circle of radius $r$ at a uniform speed, on the inside of the smooth surface of an inverted cone as shown in figure. The height of the plane of the circle above the vertex is $h$, then the speed of particle should be:

• A. $\sqrt{rg}$
• B. $\sqrt{2rg}$
• C. $\sqrt{gh}$
• D. $\sqrt{2gh}$

Answer 1

The correct option is C $\sqrt{gh}$

Let $N$ is the normal reaction from the wall of the cone on the particle. It makes angle $\theta$ with the vertical , as shown in figure.


From the frame of reference of the revolving particle, a centrifugal force $F_{\text{centrifugal}}=\frac{m{v}^2}{r}$ will be acting along radially outward direction.
$\therefore$Applying equilibrium condition in horizontal and vertical direction from the particle's frame of reference:

$N\sin \theta =\frac{m{v}^2}{r}...\left(i\right)$
$N\cos \theta =mg...\left(ii\right)$
where $v=$ speed of the particle)

Dividing Eq. $\left(i\right)$ by $\left(ii\right)$,
$\tan \theta =\frac{\left(\frac{m{v}^2}{r}\right)}{mg}$
$\Rightarrow v=\sqrt{rg\tan \theta }$

From the geometry in figure,
$\tan \theta =\frac{h}{r}$
$\Rightarrow v=\sqrt{rg\times \frac{h}{r}}$
$\therefore v=\sqrt{hg}$