Simple Lagrangian question, not getting right answer

  • Thread starter Thread starter Clever-Name
  • Start date Start date
  • Tags Tags
    Lagrangian
Click For Summary

Homework Help Overview

The problem involves a particle of mass 'm' sliding on a smooth surface defined by the equation y = Ax². The task is to derive the equation of motion for the particle after it is displaced from equilibrium and released, using Lagrangian mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of the kinetic energy in terms of the correct coordinates and the application of the Euler-Lagrange equations. There is a focus on ensuring the kinetic energy is expressed accurately based on the particle's motion along the surface.

Discussion Status

Some participants have provided guidance on correcting the kinetic energy expression and have pointed out potential algebraic errors. There is ongoing exploration of the contributions to the terms in the equation of motion, with participants working through their calculations and seeking clarification on discrepancies with the expected solution.

Contextual Notes

There appears to be confusion regarding the representation of kinetic energy and the resulting terms in the equation of motion, leading to differing results. Participants are encouraged to show more detailed work to facilitate further assistance.

Clever-Name
Messages
378
Reaction score
1

Homework Statement


A particle of mass 'm' slides on a smooth surface, the shape of which is given by [itex]y = Ax^{2}[/itex] 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
[tex]T = \frac{1}{2}m(\dot{y})^{2}[/tex]
[tex]V = mgy[/tex]

The Attempt at a Solution


Using the chain rule for my derivatives and plugging in the functions for the lagrangian I got:

[tex]L = \frac{1}{2}m(2Ax\dot{x})^{2} - mgAx^{2}[/tex]

Applying this to the Euler-Lagrange equations:

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

[tex]\frac{\partial L}{\partial x} = -2mgAx[/tex]

End up with, after some cancelling,:

[tex]2A^{2}x^{2}\ddot{x} + 4A^{2}x\dot{x}^{2} + gAx = 0[/tex]

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

[tex](1 + 4A^{2}x^{2})\ddot{x} + 4A^{2}x\dot{x}^{2} + 2Agx = 0[/tex]


I can show more detail in my work if needed, I just didn't want to type it all out.
 
Physics news on Phys.org
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.
 
Of course! Can't believe I missed that. Alright well by applying it that way it boils down to:

[tex](1+4A^{2}x^{2})\ddot{x} + 8A^{2}x\dot{x}^{2} + 2Agx = 0[/tex]

There's still that factor of 8 infront of the first derivative term that isn't matching with the solution.
 
Clever-Name said:
Of course! Can't believe I missed that. Alright well by applying it that way it boils down to:

[tex](1+4A^{2}x^{2})\ddot{x} + 8A^{2}x\dot{x}^{2} + 2Agx = 0[/tex]

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

Similar threads

Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
4K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K