Mass on a sphere problem.

1. Sep 20, 2010

lylos

1. The problem statement, all variables and given/known data
Consider a smooth hemisphere of radius a placed in the Earth's magnetic field. Place a small point mass on the top of the sphere and provide an initial small displacement as to allow the mass to slide down the sphere. Calculate the point where it falls off the sphere.

This is from chapter 2.4 of Goldstein.

2. Relevant equations
$$L=T-V$$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}$$

3. The attempt at a solution
First, I followed Goldstein in using a coordinate axis that is centered at the bast of the hemisphere with z pointing to top of sphere. The resulting motion can be contained in the xz plane if we consider the initial velocity in y to be zero.

$$L=\frac{1}{2}m(\dot{x}^2+\dot{z}^2)-mgz+\lambda(\sqrt{x^2+z^2}-a)$$

When I transform this to spherical coordinates, keeping in mind that R is constant, I have:

$$L=\frac{1}{2}m(R^2\dot{\theta}^2)-mgRCos(\theta)+\lambda(R-a)$$

Which yields the following equations:

$$mR\dot{\theta}^2-mgCos(\theta)+\lambda=0$$
$$mR^2\ddot{\theta}=mgRSin(\theta)$$
$$R-a=0$$

Goldstein states that you would solve the 2nd then solve the 1st and you can then solve for lambda. I am wondering what trick you must use to solve for the 2nd equation. I feel that the small angle approximation won't work here. Please enlighten?

2. Sep 20, 2010

Mindscrape

Hmm, good question. The taylor expansion may need to go a bit further for something analytical. If it were me, I'd just go to numerics! Let me see if I can find something about it and post back later. My compy is about to run out of juice.

3. Sep 21, 2010

lylos

We have:
$$mR^2\ddot{\theta}=mgRSin(\theta)$$

Which can be rewritten:
$$mR^2\frac{d\dot{\theta}}{dt}=mgRSin(\theta)$$

We make the substitution:
$$\omega=\dot{\theta}$$

Such that:
$$mR^2\frac{d\omega}{dt}=mgRSin(\theta)$$

By chain rule:
$$\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\frac{d\omega}{d\theta}\omega$$

Finally:
$$mR^2\frac{d\omega}{d\theta}\omega=mgRSin(\theta)$$

Upon integrating and replacing $$\omega$$ with $$\dot{\theta}$$
You finally have:
$$\dot{\theta}^2=\frac{-2g}{r}Cos(\theta)+\frac{2g}{r}$$

Took me a while to find this trick. I wish I could have figured it out myself without having to resort to someone else's work.