Simple Lagrangian question, not getting right answer

[Clever-Name]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[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.

 

[Dick]

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.

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.

 

[capandbells]

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

 

[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