Physics - Circular Motion Question with Solution | TestHub

Let the semi-vertical angle of the conical funnel be $\theta$.

From the geometry of the cone, the height $h$ of the circular plane from the vertex and its radius $R$ are related by:

$\tan \theta =\frac{R}{h}$

The forces acting on the particle are its weight $mg$ acting vertically downwards and the normal reaction $N$ acting perpendicular to the inner surface of the funnel.

The normal reaction $N$ makes an angle $\theta$ with the horizontal. Resolving $N$ into vertical and horizontal components:

The vertical component balances the weight:

$N\sin \theta =mg$

The horizontal component provides the necessary centripetal force for circular motion:

$N\cos \theta =\frac{m{V}^2}{R}$

Dividing the vertical equation by the horizontal equation, we get:

$\tan \theta =\frac{Rg}{V^2}$

Equating the two expressions for $\tan \theta$:

$\frac{R}{h}=\frac{Rg}{V^2}$

$h=\frac{V^2}{g}$

Answer: $\frac{{\mathrm{V}}^2}{\mathrm{g}}$