# A Mechanics Problem Involving Equilibrium Rotational Motion

The following is a useful Mechanics 3 level problem involving a particle in equilibrium whilst undergoing rotational motion. I give my suggested solution below.

The Problem

Suggested solution

I sketch the relevant forces in the following diagram:

Since the particle is in equilibrium, the centripetal and centrifugal forces must be equal so resolving towards the central axis of the cone we must have

$R(\leftarrow)$, $\frac{mv^2}{\frac{1}{3}a} - F_C = 0$                      (1)

Since there is no motion up or down the inner surface of the cone we must also have (resolving downwards along the inner surface)

$R(\swarrow)$, $mg\cos(90^{\circ} - \theta) - F_C\cos\theta = 0$        (2)

But

$\cos(90^{\circ} - \theta) = \sin\theta$                  (3)

Using (1) and (3) in (2) we get

$mg\sin\theta = \frac{mv^2}{\frac{1}{3}a}\cos\theta$

and therefore

$g\tan\theta = \frac{v^2}{\frac{1}{3}a}$                     (4)

But

$\tan\theta = \frac{\frac{1}{3}h}{\frac{1}{3}a}$             (5)

Using (5) in (4) gives

$g\frac{\frac{1}{3}h}{\frac{1}{3}a} = \frac{v^2}{\frac{1}{3}a} \Longleftrightarrow v = \sqrt{\frac{1}{3}hg}$ as required.