Question

A mass m is released from the top of a vertical circular track of radius r with a horizontal speed $v_0$. Find angle $\theta$ where it leaves the contact with circular track.

Answer 1

Given,

Initial velocity = $v_0$

Using the third equation of motion,

$v^2=v_0^2+2gh=v_0^2+2gr(1-cos\theta )$

The centripetal acceleration is given by,

$\frac{{mv}^2}{r}=\frac{{mv}_0^2}{r}+2mg(1-cos\theta )$............................(i)

Condition of leaving: $\frac{{mv}^2}{r}=mg\cos \theta$....................(ii)

From equation (i) and (ii) we get,
$mg\cos \theta =\frac{{mv}_0^2}{r}+2mg(1-cos\theta )$

$\cos \theta =[\frac{v_0^2}{3rg}+\frac{2}{3}]$

Thus, $\theta =co{s}^{-1}[\frac{v_0^2}{3rg}+\frac{2}{3}]$