Thread: Particle moving on a sphere mechanics help required

1. Particle moving on a sphere mechanics help required

A particle P is projected horizontally with speed u, where u^2 < 2ag, from the highest point A of a fixed smooth sphere of radius a and whose centre is at O. The particle slides on the outer surface of the sphere. Show that it leaves the sphere when PO makes an angle arccos ( $\frac{2}{3}+\frac{u^2}{3ag}$) with the vertical.

I figured out that when speed=2ag, reaction will be 0 and the particle will leave the sphere. And the change in speed have something to do with the energy change mgh=.5 mv^2. But all my workings to find out h lead to very complicated expressions. So how do i prove the angle is arccos ( $\frac{2}{3}+\frac{u^2}{3ag}$) ?
(vertical circular motion)

A particle P is projected horizontally with speed u, where u^2 < 2ag, from the highest point A of a fixed smooth sphere of radius a and whose centre is at O. The particle slides on the outer surface of the sphere. Show that it leaves the sphere when PO makes an angle arccos ( $\frac{2}{3}+\frac{u^2}{3ag}$) with the vertical.

I figured out that when speed=2ag, reaction will be 0 and the particle will leave the sphere. And the change in speed have something to do with the energy change mgh=.5 mv^2. But all my workings to find out h lead to very complicated expressions. So how do i prove the angle is arccos ( $\frac{2}{3}+\frac{u^2}{3ag}$) ?
(vertical circular motion)
let O be the zero point for gravitational potential energy.

$\theta$ = vertical angle between O and P

total initial mechanical energy ...

$mga + \frac{1}{2}mu^2$

at point P ...

$mga\cos{\theta} + \frac{1}{2}mv^2$ , $v > u$

since energy is conserved ...

$
mga + \frac{1}{2}mu^2 = mga\cos{\theta} + \frac{1}{2}mv^2
$

$2ga + u^2 = 2ga\cos{\theta} + v^2$

$v^2 = u^2 + 2ga(1 - \cos{\theta})$

at point P, the normal force of the sphere is zero. centripetal acceleration ...

$\frac{v^2}{a} = g\cos{\theta}$

$\frac{u^2}{a} + 2g - 2g\cos{\theta} = g\cos{\theta}$

$\frac{u^2}{a} + 2g = 3g\cos{\theta}$

$
\frac{u^2}{3ga} + \frac{2}{3} = \cos{\theta}
$