# Equilibrium - Particle in Bowl, looking for guidance

1. Dec 6, 2007

### m0nk3y

Equilibrium - Particle in Bowl, looking for guidance!!

1. The problem statement, all variables and given/known data
A point particle of mass m is confines to the frictionless surface of a sphere cal bowl. There are 2 degrees of freedom. Prove that the equilibrium point is the bottom of the bowl. Near the bottom of the bowl, what is the most general form possible for the shape of the bowl in order to maintain the stability of the equilibrium point at the bottom?

2. Relevant equations
spherical coordinates: rsin$$\thetacos\varphi + rsin\thetasin\varphi + rcos\theta$$
Lagrange = Kinetic Energy(T)- Potential (V)

3. The attempt at a solution
I started off by getting the Lagrange and got:
$$\stackrel{1}{2}$$r^2 (theta)'^2 - mgrcos(theta)
Then I got the E.O.M
(mr^2 (theta)'')/2 - mgsin(theta)

I have to prove that it is at equilibrium at $$\theta$$=0
But when I plug in 0 I am still left with (mr^2 (theta)'')/2
What am I doing wrong

[NOTATION::(theta)' is theta dot or velocity and (theta)'' is theta double dot or acceleration]

Any help is appreciated!!
Thanks

2. Dec 7, 2007

### Mr.Brown

i guess it´s not useful to go with the lagrangian just build up your potential in terms of spherical coordinates.
I guess it is V=m*g*r*sin(theta) as you said. And as in your case theta is between 0 and Pi/2
the minimum is at theta =0 because this is the lowest value sine take on [0,Pi/2]
how about that ?