# Simple Lagrangian question, not getting right answer

1. Oct 16, 2011

### Clever-Name

1. The problem statement, all variables and given/known data
A particle of mass 'm' slides on a smooth surface, the shape of which is given by $y = Ax^{2}$ where A is a positive constant of suitable dimensions and y is measured along the vertical direction. The particle is moved slightly away from the position of equilibrium and then released. Write down the equation of motion of the particle.

2. Relevant equations
Lagrangian
$$T = \frac{1}{2}m(\dot{y})^{2}$$
$$V = mgy$$

3. The attempt at a solution
Using the chain rule for my derivatives and plugging in the functions for the lagrangian I got:

$$L = \frac{1}{2}m(2Ax\dot{x})^{2} - mgAx^{2}$$

Applying this to the Euler-Lagrange equations:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 8mA^{2}x(\dot{x})^{2} + 4mA^2x^{2}\ddot{x}$$

$$\frac{\partial L}{\partial x} = -2mgAx$$

End up with, after some cancelling,:

$$2A^{2}x^{2}\ddot{x} + 4A^{2}x\dot{x}^{2} + gAx = 0$$

However the solution I have says the final answer should be:

$$(1 + 4A^{2}x^{2})\ddot{x} + 4A^{2}x\dot{x}^{2} + 2Agx = 0$$

I can show more detail in my work if needed, I just didn't want to type it all out.

2. Oct 16, 2011

### capandbells

Your kinetic energy is wrong. You were right to assume you only need one coordinate, but that coordinate needs to represent position along the surface, so it can't be just y. First write your kinetic energy in terms of x-dot and y-dot, then write x and y in terms of this other coordinate and substitute those into the kinetic energy.

3. Oct 16, 2011

### Clever-Name

Of course! Can't believe I missed that. Alright well by applying it that way it boils down to:

$$(1+4A^{2}x^{2})\ddot{x} + 8A^{2}x\dot{x}^{2} + 2Agx = 0$$

There's still that factor of 8 infront of the first derivative term that isn't matching with the solution.

4. Oct 16, 2011

### Dick

I think you need to show more of your work before we can help with that one. There's a contribution to the x*x-dot^2 term from both the dL/dx and dL/dx-dot parts.

5. Oct 16, 2011

### capandbells

Double check your algebra. Your dL/dq should have a term that turns that 8 into a 4.

6. Oct 16, 2011

### Clever-Name

Oh.. wow.. Yea I found it. I completely skipped over the x in the velocity term when taking dL/dx.

Thanks for your help guys, it's all fixed now

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook