A particle moves on the smooth inside surface of the hemisphere
z = −sqrt(a^2− r^2)
where (r, θ, 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 θ-direction.
Show that the particle moves between two heights in the subsequent motion, and find
them.
Show, too, that if the parameter β = V^2/4ga is very large then the difference between
the two heights is approximately a/2β.
Ok
I've spent days trying to do this and I cant seem to get it!
I end up with z^2 = z^2.v^2 - 2.g.z.a^2 - 2gz^3
where bold denotes 'z dot' (derivative of z wrt time)
factorising this I get z((z - ((v^2 + sqrt(v^2 - 4ga))/4g)(z - ((v^2 - sqrt(v^2 - 4ga))/2z^2 + a^2
getting the particle lying between((v^2 + sqrt(v^2 - 4ga)) and ((v^2 - sqrt(v^2 - 4ga))
is this correct?
how should I go about the next bit?
many thanks