Simple application of euler-lagrange equation

  • #1
740
15
Suppose I have a particle of mass m in a uniform, downward gravitational field g, constrained to move on a frictionless parabola

[tex]
y = x^2
[/tex]

I get

[tex]

L = KE - PE = \frac {1}{2} m (\dot x^2 + \dot y^2) - mgy = \frac {1}{2}m \dot x^2 (1+4x^2) - mgx^2

[/tex]

[tex]
\frac {\partial L}{\partial x} = (\frac {1}{2}m \dot x^2)(8x) - 2mgx [/tex]

[tex]

\frac {\partial L}{\partial \dot x} = (m \dot x)(1+4x^2)

[/tex]

From here, no matter how I go about differentiating, I get a velocity-dependent force ( [itex] m \ddot x [/itex] ), which does not seem reasonable to me.

Where exactly is my error?
 

Answers and Replies

  • #2
vanhees71
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I don't see any error so far. Why shouldn't the force be velocity dependent? You have a constraint here, and the centripetal part of the constraining forces is velocity dependent. Just write down the Euler-Lagrange equation.
 
  • #3
vanhees71
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PS: Of course, you can use energy conservation here, since the Lagrangian is not explicitly dependent on time, i.e.,

[tex]H=\dot{x} \frac{\partial L}{\partial \dot{x}}-L=E=\text{const}.[/tex]
 
  • #4
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hmmm...thinking...
 
  • #5
740
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Ok, I get

[tex]

m \ddot x = \frac {-2mg}{1+4x^2} - \frac {4 x \dot x ^2 m}{1+4x^2}

[/tex]

I was expecting only the first term, which is the one caused by the gravitational field, but yes, what you say makes sense to me. Thank you.

(Constraints are new to me.)
 
Last edited:
  • #6
740
15
Yes, it makes even more sense to me now. For - if there were no gravitational field, the particle would move along the parabola at a constant speed, and [itex] \ddot x [/itex] would still be non-zero, since the path is curved.

Thanks again.
 

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