Classical mechanics - Lagrange multipliers

[LiorE]

Homework Statement

A disk moves on an inclined plane, with the constraint that it's velocity is always at the same direction as it's plane (similar to an ice skate, maybe). In other words: If \hat{n} is a vector normal to the disk's plane, we have at all times: \hat{n} \cdot \vec{v} = 0. Also, it's free to move without friction, and always perpendicular to the plane. (as seen in the figure.)

I need to get and solve the equations of motion for certain initial conditions that I'll write promptly. We set an x-y coordinate system at the top-right corner of the plane with the y-axis going downwards, and denote that angle between \hat{n} as \varphi.

Homework Equations

The constraint is:

c_1 = \dot{x}\cos\varphi + \dot{y}\sin\varphi

and accordingly the Lagrangian is:

L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}I\dot{\varphi}^2 + mgy + \lambda(\dot{x}\cos\varphi + \dot{y}\sin\varphi)

The initial conditions that were given are that at t=0:

x=0, y=0, \dot{x}=0, \dot{y}=0, \varphi = 0, \dot{\varphi} = \omega_0

The Attempt at a Solution

The obvious way of solving is to use Euler-lagrange and get the equations of motion. The problem is that I can't solve them! They're too damn complicated. There is a hint that I should try to find constants of motion by setting t=0 in the equations, but I can't seem to find them.

I would appreciate any help...

Thanks in advance!

 

[latentcorpse]

ok well if y=0 [/atex] and \dot y=0 then y=0 \forall t which will simplfy your Lagrangian

 

[latentcorpse]

The same would happen for x so I get L = \frac{1}{2} I {\omega_0}^{2} Then solving the Euler-Lagrange I get I \ddot \psi = 0. This could be massively wrong but I though I'd give it a shot anyway.

 

[LiorE]

Thanks, I solved it by variating the action integral - the TA said that Lagrange multipliers give here the wrong answer, and he doesn't know why. I guess it's a question for mathematicians to answer.