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Uniform Circular Motion in Lagrangian Formalism

  1. Mar 4, 2008 #1
    Problem: Consider a particle of mass m, constrained to move in a circle of radius r. Find the Lagrangian:

    Relevant Equations: L = T - V

    Where L is the Lagrangian, T is the kinetic energy, and V is the potential energy.

    My questions is this. T is the kinetic energy and would simply equal mV^2/2 or mr^2w^2/2 depending on the coordinate system chosen.

    What about V? There has to be a force on the particle holding it in its circular trajectory or it would simply fly off. However, no central force is mentioned in the problem. For all I know the particle may be held in place by a string, or maybe it rides in a circular track. Anyhow, does it make sense to talk about a potential energy associated with centripetal force?

    Is it possible that L = T - V doesn't hold in this case since the forces involved are velocity dependent (centripetal force)? I know for the EM Lagrangian L is not T - V.

  2. jcsd
  3. Mar 4, 2008 #2

    Andy Resnick

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    It's been a while since I've done anything like this, and I don't have my copy of Goldstein handy, but doesn't Lagrange's equation also contain a term for a contraint?

    That is, the full equation is:

    [tex] Q = \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial q}[/tex]

    Where Q is the constraint force...?
  4. Mar 7, 2008 #3
    That's news to me. Where did you get this formula from? Forces of constraint are not explicitly given in Lagrange's equations. That's the beauty of them. It looks to me that the Q above refers to non-monogentic external forces (i.e. forces for which there is no associated potential energy function) such as friction.

    Best wishes

  5. Mar 7, 2008 #4
    well you're lagrangian coordinates are (\theta,r) do the transformation on velocities and you'l find T..... you have a rotational symmetry so V=V(r)....
    if F=mV^2/r..... what is the V??? remebere that F=-grad(V)....

  6. Mar 7, 2008 #5

    Andy Resnick

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    It's entirely possible I am confused. I do recall there's a way to introduce constraint forces into the Lagrange formalism (we had to do a problem from Symon- a ball rolling off another ball), but can't remember exactly what we did- my class notes are at home, buried in the basement IIRC.
  7. Mar 7, 2008 #6


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    [tex] Q = \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}[/tex]

    Sorry to be pedantic. A dot was missing.
  8. Mar 7, 2008 #7
    [QUOTE[tex] Q = \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}[/tex]
    [/QUOTE]I still don't see where you got this from. Perhaps yo just made slight mistake. There is a similar expression in Lagrangian mechanics which is reminiscent of your equation. It differs from yours by a negative sign and the interpretation of Q. The equation I speak of is

    [tex] Q_i = \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} [/tex]

    where L = Lagrangian of system = T - U and Qi are the components of the generalized force. U = generalized potential aka velocity-dependant potential.

    See - http://electron6.phys.utk.edu/phys594/Tools/mechanics/summary/lagrangian/lagrangian.htm [Broken]

    For a single charged particle moving in an electromagnetic field U = q[itex]\phi[/tex] - qv*A where q = charge of particle, v particle's velocity and A is the magnetic vector potential.

    But these generalized forces are not forces of constraint. They are the total force acting on the ith particle. And yes, I was a bit wrong in my statement :redface:. That's why I love posting!! I learn new things when I least expect it :smile:. They part of this force is the force of contstraint. In fact this force can be thought of as representing the sum of the external force and the .

    This is found in Classical Mechanics - Third Ed., by Goldstein, Safko and Poole (2002), page 23.

    In the example you gave U = V = 0.

    I hope this has helped some?

    Best wishes

    Last edited by a moderator: May 3, 2017
  9. Mar 10, 2008 #8
    Thanks for the replies.

    I believe Pete is correct. The constraining forces are not conservative and therefore are not associated with a potential in the Lagrangian. Constraining forces can be added to the Euler-Lagrange equation as Lagrange multipliers.

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