Question

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

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

Answer 1

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



FBD of the particle is as shown above,

Since, particle is rotating in horizontal circle, the vertical force on the particle should be zero.

Balancing the forces in vertical and horizontal direction we get,

$\therefore N\sin \theta =\frac{m{v}^2}{r}$....(1)

$\therefore N\cos \theta =mg$.......(2)

On dividing eq $\left(1\right)÷\left(2\right)$ we get,

$\tan \theta =\frac{v^2}{rg}$.......(3)

Now from the figure we get,

$\tan \theta =\frac{h}{r}......\left(4\right)$

From eq (3) and (4) we get,

$\Rightarrow v=\sqrt{gh}$

Hence, option (c) is the correct answer.

Why this question ? Concept : Horizontal component of the normal force is responsible for the circular motion. For this type of question take the component normal along the plane of circular motion and perpendicular to it.