# conservation of energy/cylindrical polar coordinates

• Jan 3rd 2009, 01:09 PM
James0502
conservation of energy/cylindrical polar coordinates
A particle moves on the smooth inside surface of the hemisphere
z = −sqrt(a^2r^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

• Jan 4th 2009, 01:17 AM
DangerMaths
Here's a hint in those questions you need to write down the Lagrangian, and then solve Euler-Lagrange equations.

I leave you to do the calculations.

Cheers, (Evilgrin)
• Jan 4th 2009, 01:31 AM
James0502
we havent been lectured that yet.. I think the way I have given is how we are expected to solve it, but I think I must've made a mistake somewhere? thanks though
• Jan 4th 2009, 04:44 AM
HallsofIvy
One force acting on the particle is gravity: -mg in the z direction, but there is also a "passive force" acting perpendicular to the hemisphere to keep it on the hemisphere. Find the components of -mg perpendicular to and parallel to the hemisphere and then discard the componet perpendicular.
• Jan 4th 2009, 05:25 AM
James0502
Yeh..

I get 1/2(r^2 +r^2.theta^2 + z^2) - gz = 1/2.V^2

eliminating r in favour of z and theta, I get the equation written in my first post..

where bold again denotes differentiation
• Jan 6th 2009, 06:29 AM
DeFacto
Quote:

Originally Posted by James0502
Yeh..

I get 1/2(r^2 +r^2.theta^2 + z^2) - gz = 1/2.V^2

eliminating r in favour of z and theta, I get the equation written in my first post..

where bold again denotes differentiation

I'm not sure how you got the -gz - i get a positive gz. I'm stuck on this question too actually.
• Jan 6th 2009, 06:58 AM
James0502
Quote:

Originally Posted by DeFacto
I'm not sure how you got the -gz - i get a positive gz. I'm stuck on this question too actually.

yeh.. sorry - I got +ve gz also.. The question is really starting to get to me now! like over a week since I first attempted it!