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Lagrangian/Hamilton's principle when no holonomic forces

  1. Apr 20, 2010 #1


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    My professor (that I appreciate each time less for several reasons) told us that there are many contradictions and few resources in physics literature when it comes to express an equivalent form to Euler-Lagrange equations of a system when we're dealing with no-holonomic systems. Or he said that the Euler-Lagrange equations needed to be changed into another form (that I'm not sure of). For instance I see in my notes that [tex]\frac{\partial L}{\partial q_i}-\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q_i} \right ) = \sum _{p=1} ^s \lambda _p (t) \frac{\partial \Phi _p}{\partial \dot q_i}[/tex] and I don't see the definition of the terms in the pages I have my hand over.
    The matter is that after all this "mess", he took an example: an annulus rolling over an inclined surface. He said it was a bad example of no-holonomic system because it is possible to "integrate directly" the equations of motions or something like that, that I do not remember well.
    Anyway he said the problem could be treated as if the system was no-holonomic. The annulus doesn't slide. He said that we could take the same problem with friction but that it would be more complicated and that we would have to include the friction force into the modified E-L's equations. I then raised my hand to ask him that in the case of an annulus not sliding, there must be some static friction (although I agree the force doesn't depend on the velocity of the sliding annulus). He got pissed off by my question and said "NO, NO, NO. I said there is NO friction in the problem and that including friction requires another information". I just said "Oh ok, thanks". He introduced in me a very little doubt, is there a static friction force in the case of a purely rolling (without slipping) annulus? I'm 99.99999999% sure there is thanks to my professor of my introductory course to mechanics. Otherwise how does the annulus starts to rotate? Where does its torque come from? Obviously from the contact point, but it isn't right according to my professor. I just want to be sure.
    The funny part was he solved the problem by writing the Lagrangian. Then he wrote "[tex]\Phi (x,\dot x , \theta , \dot \theta)=r \dot \theta - \dot x =0[/tex].
    Thus [tex]\frac{\partial L}{\partial x}-\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot x} \right ) =- \lambda =mg \sin \varphi - m\dot x[/tex] where [tex]\varphi[/tex] is the angle the inclined surface makes with the horizontal floor.
    [tex]\frac{\partial L}{\partial \theta} - \frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \theta} \right ) = \lambda r =-mr^2 \ddot \theta[/tex].
    After this, he wrote [tex]r\ddot \theta = \ddot x[/tex], [tex]\lambda =-mr \ddot \theta =-m \ddot x[/tex] and [tex]m\ddot x =\frac{1}{2}mg\sin \varphi[/tex].
    [tex]\Rightarrow x=\frac{1}{2}at^2+ v_0t+x_0[/tex], [tex]\theta =...[/tex]." Where does this last line come from? To me it looks like using Newtonian mechanics instead of the Lagrangian one. It seems that instead of solving the 2 differential equations, he used the methods we've been introduced to in the first year... Strange.
    By the way, do you know any book that "correctly" deal with forces depending on say velocity, in Analytical Mechanics?
    Thanks for any comment.

    Edit: By the way, do you know what's called a "generalized coordinates"? My friend keep telling me that they must be independent coordinates while from what I understand from wikipedia, they are just coordinates and kept this name from Lagrange's time or so (if I remember well... how bad is my memory?!).
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  3. Apr 20, 2010 #2
    I think this is completely analogous to using Lagrange multipliers to take into account constraints. In this case you include a no-slip constraint, which automatically implement friction and normal forces that makes sure that the annulus rolls without slipping.
  4. Apr 20, 2010 #3


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    Thanks for the reply. I knew that the non-slip constraint implied friction. However I never thought about Lagrange multipliers. What exactly do you want to minimize in the applied case of the rolling annulus?
  5. Apr 20, 2010 #4
    You minimize the action: Integral of L dt

    You can derive the EL equations by setting up the usual arguemnt involvoing changing the trajectory qi(t) infinitessimally to qi(t) + delta qi(t). If qi(t) minimizes the action, then that means that the first order contribution in deltaqi(t) should vanish (otherwise you could change the qi(t) to make the action even smaller, contradicting that
    qi(t) minimizes the action).

    However, if you cannot vary all the qi(t) for all t independently, then the argument is not valid. You can cure this problem by introducing Lagrange multipliers, completely analogous to how you do this in ordinary calculus.

    Instead of a single lagrange multiplier, you get a function lambda(t), which usually can be interpreted as a force (like the normal force or some static friction force) that in the physical situation makes sure that the system satisfies the contraints.
  6. Apr 21, 2010 #5


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    Ok thank you very much. I'll try to do an example alone.
    I'm just wondering. If there are only holonomic constraints, should I ALWAYS be able to derive Euler-Lagrange equations? The case only gets complicated when dealing with non-holonomic constraints, right? I just read on wikipedia
    which is analogous to the example of my professor I believe (hence his words "bad example as non-holonomic system").
  7. Apr 21, 2010 #6
    Yes, in general you cannot handle constraints within the Euler-Lagrange formalism. If in the physical situation you would have friction forces that would do work leading to dissipation of energy, then that would be off limits to E-L.
  8. Apr 21, 2010 #7


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    Ok perfect, I get it.
    My question was more about if there is no dissipative forces or more precisely, the system is holonomic. Will I always be able to "vary all the qi(t) for all t independently"? (making allusion to your words
    ). I believe that yes, if the system is holonomic then I could always derive Euler-Lagrange equations. Is this right?
  9. Apr 21, 2010 #8
    Yes, but you then need to use the Lagrange multipliers. They can then compensate for the fact that the qi satisfy the constraints.
  10. Apr 21, 2010 #9


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    Thanks once again.
    I don't really see where are the Lagrange multipliers for example in this case http://en.wikipedia.org/wiki/Lagrangian_mechanics#Pendulum_on_a_movable_support. There are 2 holonomic constraints (according to me!): The suspended mass m has to be at a constant distance from the other mass M and the mass M cannot move vertically.
    I've even found the equations of motion (though didn't solve them), but I didn't realize where I used the Lagrange multipliers.
    Are there included into the Lagrangian? For example when I write the position of M: [tex]\vec r_1 = x \hat i + 0 \hat j[/tex]. Then I derivate to get the velocity and then I square it in order to get a part of the kinetic energy needed for the Lagrangian.
  11. Apr 21, 2010 #10
    What they do on the Wiki-page is to express the Lagrangian in terms of the "free" variables that are not constrained. You can also write down the Lagrangian that ignores the constraints of the system. That is then not the correct Largranian, of course. However, instead of solving the contraint equations for the variables and working with only the free variables, you can add lamba(t) times constraint to that Lagrangian.

    Yopu can actually see how this works better by studying some very simple examples, like the simple pendulum. If we completely ignore the fact that the mass is always a distance L away from the support point of the pendulum, we would have the Lagrangian:

    L = 1/2 m xdot^2 - m g z

    Instead of switching to polar coordinates and imposing the constraint x^2 + z^2 = R^2 by noting thast then only the polar angle is a free vaiable, you can use Lagrange multipliers as follows.

    If we define:

    f(x, z) = x^2 + z^2 - R^2

    then the constraint is f(x,z) = 0

    Then we consider the Lagrangian:

    L_new = L - lambda f

    Where lambda is to be treated as a dynamical variable on equal footing with x and z.

    So, you now get an extra equation by varying lambda:

    d/dt dL_new/dlamba-dot - dL_new/dlambda = 0

    Now L does not depend on lambda-dot, so we get:

    f(x,z) = 0

    Which is indeed the constraint equation we want.

    However, you can now also consider the variation of L_new w.r.t. x and z as if they are not constrained. It is the new dynamical variable lamba appearing in L_new which will make sure that the x and z will end up on a trajectory that in fact does satisfy the constraint.

    It turns out that lamba is can be identified with the normal force (up to a sign).
  12. Apr 22, 2010 #11


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    Thank you so much Count Iblis for the lesson, too bad I don't have ink anymore to print your reply. I'm going to copy it in my draft-book. It makes perfect sense to me.
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