I'm currently working on this question:
A particle moves on the smooth inside surface of the hemisphere
z = -(a^2 - r^2)^(1/2), r <= a,
where (r, theta, z) denote cylindrical polar coordinates, with the z-axis vertically upward.
Initially the particle is at z = 0, and it is projected with speed V in the theta-direction.
1. Show that the particle moves between two heights in the subsequent motion, and find them.
2. Show, too, that if the parameter b = (V^2)/4ga is very large then the difference between the two heights is approximately a/(2b).
Conservation of energy 1/2 m(r'^2 + r^2 theta'^2 + z'^2) + mgz = constant
rtheta' = V
r^2 theta = aV
theta ' = aV/r^2
( ' = dot, differentiation)
The Attempt at a Solution
I'm really stuck on question 1, I got to an answer using conservation of energy that ended with solving a quadratic, this is the particle moves between z=0 and z= (V^2 + sqrt(V^4 + 16g^2a^2))/4g but when I sub in the parameter in (2) I get to the wrong answer, ie 2ab. So I guess I've done 1 wrong. Any ideas guys?