# Homework Help: Classical mechanics - Lagrange multipliers

1. Dec 15, 2008

### LiorE

1. The problem statement, all variables and given/known data
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 $$\varphi$$.

2. Relevant 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:

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

3. 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...

#### Attached Files:

• ###### image002.gif
File size:
5 KB
Views:
168
2. Dec 18, 2008

### latentcorpse

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

3. Dec 18, 2008

### 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.

4. Dec 23, 2008

### 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.