# Euler-Lagrange equation for paraboloid plane

1. Jun 5, 2006

### wavemaster

I have a classical mechanics question I couldn't conclude. The reason seems to be mathematical. It's this:
There's a paraboloid shaped plane of mass M, which is standing on a frictionless surface and can slide freely. It's surface is $$y=ax^2$$. A point mass m is place on the plane. Solve the Euler-Lagrange equations of the system for little mass.

My (unsuccessful) solution is as follows:

I chose $$x_m$$ as the x (horizontal) coordiante of point mass, and for sliding, plane it's $$x_M$$. I defined another coordinate for point mass: $$x$$ is the horizontal distance of point mass from the center (or bottom) of the parabol. So, coordinates of m are
$$x_m = x_M + x$$
$$y_m = ax^2$$
so velocities are
$$\dot{x_m} = \dot{x_M} + \dot{x}$$
$$y_m = a2x\dot{x}$$

So kinetic energy of the system is:
$$T = \frac{m\dot{x_m}^2}{2} + \frac{M\dot{x_M}^2}{2}$$
$$T = \frac{m ( \dot{x_M}^2 + \dot{x}^2 + 2\dot{x}\dot{x_M} + 4a^2x^2\dot{x}^2 )}{2} + \frac{M\dot{x_M}^2}{2}$$
and potential is
$$V = mgy = mgax^2$$

and
$${\cal L} = T-V$$

Now, Euler-Lagrange equations are:
(1) $$\frac{d}{dt} \frac{\partial {\cal L}}{\partial \dot{x_M}} - \frac{\partial {\cal L}}{\partial x_M}$$
(2) $$\frac{d}{dt} \frac{\partial {\cal L}}{\partial \dot{x}} - \frac{\partial {\cal L}}{\partial x}$$

The solution of the first one yields (of course, if I done the math correctly):
$$\ddot{x_M}(m+M) + \ddot{x}m = 0$$
and second is
$$\ddot{x} + \ddot{x_M} + 4a^2x\dot{x}^2 4a^2x^2\ddot{x} + 2gax = 0$$

I tried combining two solutions and got this non-linear differential equation:

$$\ddot{x}(1- \frac{m}{m+M} + 4a^2x^2) + 4a^2x\dot{x}^2 + 2gax = 0$$

Maybe there was a mathematical mistake in my solution, maybe this differential equation could be solved with some tricks, or maybe I'm downright wrong by choosing such coordinates. I don't know.

Is there someone who can solve this question?

Last edited: Jun 5, 2006