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Lagrange with friction

  1. Nov 17, 2007 #1

    Dale

    Staff: Mentor

    Hi Everyone,

    I want to use the Lagrange approach (which I am not terribly familiar with) to model a system with friction. I was thinking of modeling the losses due to friction as a simple constant dissipation of energy over time. Can I simply add a term of the form -Ft to the potential energy? (F is the amount of energy lost to friction in a unit time)

    If not, what is the easiest way to add an energy-dissapation term?

    -Thanks
    Dale
     
  2. jcsd
  3. Nov 17, 2007 #2
    If you add a term -Ft it will not appear then it will not appear in the equations of motion!

    Better to use the Euler-Lagrange equations to find the equations of motion, and then insert your friction directly into the equations of motion.

    Edit: Some treatments due this by making the RHS of the Euler-Lagrange equations non-zero.
     
  4. Nov 17, 2007 #3

    Ben Niehoff

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    Science Advisor
    Gold Member

    There is a standard way to do this, by means of a "dissipation function" [itex]\mathcal F[/itex]. The dissipation function represents the power lost to friction, so it is often a quadratic function of the generalized velocities [itex]\dot q_i[/itex]. Lagrange's equations of motion then become:

    [tex]\frac d{dt} \left( \frac{\partial L}{\partial \dot q_i} \right) - \frac{\partial L}{\partial q_i} = - \frac{\partial {\mathcal F}}{\partial \dot q_i}[/tex]

    Note that [itex]\mathcal F[/itex] is quadratic in the velocities only when the frictional forces are linear in the velocities; that is, for ordinary friction. For viscous cases when the forces are proportional to the square of the velocity, [itex]\mathcal F[/itex] would take on a different form.
     
    Last edited: Nov 17, 2007
  5. Nov 17, 2007 #4

    Dale

    Staff: Mentor

    Thanks, to both of you, this is exactly what I need.
    By frictional forces being linear in velocity I assume that you mean something to the effect that the energy dissipated by ordinary friction over a unit time is proportional to the velocity, because the force is constant. Is that correct? I am just trying to figure out how to construct this function. It looks like it should be in units of energy/time, perhaps representing the rate at which energy enters the system (negative numbers for dissipation, and 0 for purely conservative).

    Edit: never mind, I just noticed that you already said it was power (energy/time) lost to friction (positive for dissipation). But if you have any references that discuss this in depth I would be appreciative!
     
    Last edited: Nov 17, 2007
  6. Nov 17, 2007 #5

    Ben Niehoff

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    Gold Member

    Ack, I forgot a minus sign! I've fixed the original post. The revised equation is:

    [tex]\frac d{dt} \left( \frac{\partial L}{\partial \dot q_i} \right) - \frac{\partial L}{\partial q_i} = - \frac{\partial {\mathcal F}}{\partial \dot q_i}[/tex]

    [itex]\mathcal F[/itex] is then a positive definite quadratic form in the generalized velocities. Specifically, if the generalized forces of friction are

    [tex]Q_i = -k_i \dot q_i[/tex]

    then [itex]\mathcal F[/itex] is given by

    [tex]{\mathcal F} = \frac 12 \sum_i k_i \dot q_i^2[/tex]

    such that

    [tex]Q_i = - \frac{\partial {\mathcal F}}{\partial \dot q_i}[/tex]

    The reference I'm using is Classical Mechanics, by Goldstein, Poole, & Safko. I looked in Marion & Thornton also, but I couldn't find it (it may be in there; I didn't look very hard).
     
    Last edited: Nov 17, 2007
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