Modeling Friction with Lagrange: Easiest Method?

[Dale]

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

 

[Crosson]

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.

 

[Ben Niehoff]

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

\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}

Note that \mathcal F 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, \mathcal F would take on a different form.

 

[Dale]

Edit: Some treatments due this by making the RHS of the Euler-Lagrange equations non-zero.

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

\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}

Thanks, to both of you, this is exactly what I need.

Note that \mathcal F 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, \mathcal F would take on a different form.

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!

 

[Ben Niehoff]

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

\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}

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

Q_i = -k_i \dot q_i

then \mathcal F is given by

{\mathcal F} = \frac 12 \sum_i k_i \dot q_i^2

such that

Q_i = - \frac{\partial {\mathcal F}}{\partial \dot q_i}

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).