This is really cool.
"[MEDIA=youtube[/URL]
I notice the balls change speeds proportionate to their height on the wall as they follow their elliptical paths. Does their speed follow Kepler's 2nd Law?[/QUOTE]
Definitely cool.
But the speed doesn't follow Keppler's law. The equations of motion can be derived using Lagrangian mechanics.
[url]http://en.wikipedia.org/wiki/Lagrangian_mechanics[/url]
To do this, we compute the Lagrangain, which is just the difference of the kinetic and potential energies.
The kinetic energy of the ball is just 1/2 m v^2.
v can be divided up into two components - the horizontal speed, and the vertical speed. The square of the total velocity is the sum of the squares of the two components.
The state of the system can be represented by the height, h of the ball, and the angle, [itex]\phi[/itex] that the ball makes. [itex]2 \alpha[/itex] is the angle at the point of the cone.
Using a dot above a variable to represent taking it's derivative, we can then write the following:
The horizontal speed of the ball is just
[tex]\mathrm{tan} (\alpha)\, h \dot{\phi}[/tex]
The vertical speed of the ball, accounting for the slope it is on, is
[tex]
\frac{\dot{h}}{\mathrm{cos} (\alpha)}
[/tex]
The potential energy of the ball is just mghWe can then write down the Lagrangian, setting the mass of the ball to 1 for simplicity (it won't contribute anything meaningful to the solution)
[tex]
L = T-V = \frac{1}{2}(\mathrm{tan} (\alpha) \, h \dot{\phi})^2 + \frac{1}{2} (\frac{\dot{h}}{\mathrm{cos} (\alpha)})^2 - g h
[/tex]
Lagrange's equations give us the equations of motion
[tex]
d/dt (\partial L / \partial \dot{\phi}) = \partial L / \partial \phi = 0
[/tex]
Thus
[tex]
\dot{\phi} h^2 \mathrm{tan}^2 (\alpha) = \mathrm{constant}
[/tex]
There's another equation for h, but I won't write it down unless someone is interested.