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Lagrangian Method- 1st form? Zwangskraefte?

  1. Jun 10, 2004 #1
    Lagrangian Method-- 1st form? Zwangskraefte?

    Mechanics will be the death of me. For some reason, I can do E&M, and particle stuff, but give me a pendulum, and I'm dumbfounded.

    THat said, I'm working on some problems using Lagrangians, and the only book I have available to me is Goldstein (which is a bit above where I'm at, but helpful, once it clicks). The course I'm taking will count as my uppper level, undergrad mechanics course.

    Anyway, I've got a problem. Supoose you have a particle that's moving frictionless in a gravitational field on the inside of a paraboloid (which is described in cylindrical coordinates as r= az.

    How do I come up with the Lagrangian for this? If given one, I can usually figure out more on how to solve this, but it's stumping me. And another thing, I can't figure out what the english term for 'zwangskraft' would be, forced force? I'm supposed to find the size and direction of these forces, but I'm not really sure what it means. Eeep.


    There's a note saying that I'm supposed to make it a requirement that energy and rotational momentum are conserved.

    THat isn't heling me at all. Suggestions? ANy other suggestions for texts (that aren't too expensive), that might help me? My exam is in 3 weeks, and I'm lost.
     
  2. jcsd
  3. Jun 10, 2004 #2

    turin

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    This seems to hint that the problem has a reduced number of dimensions (from 3). In other words, you aparently have a particle confined on some surface. The frictionless stipulation tells you that the Euler Lagrange expression is equation to zero (which gives you the Euler Lagrange equation).




    This seems to be a description of the submanifold onto which the particle is constrained to move. However, that equation doesn't look like a paraboloid to me? In fact, I am even more confused about the bold vs. non-bold.




    It's hard to say without a better understanding of the physical situation (because Lagrangian are Taylor-made). I will try to give you a general approach:

    If this paraboloid is a 2-D surface, then devise some 2-D coordinate system that is capable of unambiguously describing the position of the particle on this surface. It is usually a good idea to do this so that either the kinetic energy or the potential energy only depends on one of the coordinates and not the other. Then, you need to express the two energies in terms of these new coordinates: T = (1/2)mv2 -> ?, V = mgh -> ?. Finally, of course, L = T - V.

    If this is a 1-D parabolic curve, then the same as above, except that you don't worry so much about the coordinate dependence of T and V. You still need to express both in terms of the 1 coordinate that you chose. For instance, suppose the parabola is y = x2. Then
    V = mgy = mg(x2)
    and
    T = (1/2)m{vx2 + vy2}
    = (1/2)m{vx2 + 2xvx2}
    = (1/2)m{1 + 2x}vx2

    I'm not clear on what your submanifold is, though.




    From the sound of it, I would guess it means "constraint force." This is the same basic idea as the normal force you deal with in 1st semester physics. You don't approach the problem knowing it. In fact, the value doesn't matter to you except that it necessarily counteracts (usually) the weight (and any inertial forces) to balance the force and keep the particle on your submanifold. (In fact, this is just about the most important idea behind the Lagrangian formalism.)

    You might try
    www.dictionary.com/translator




    Well, you have problably have two mechanical energies: kinetic and potential. Since there's no friction, these should be conserved. The rotational momentum is conserved because you're dealing with forces that have a (scalar) potential energy associated with them.
     
  4. Jun 10, 2004 #3

    Gokul43201

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    Must be [tex]r^2=az[/tex]

    Unless it's a conical surface.
     
  5. Jun 10, 2004 #4

    arildno

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    German has such a delightfully(?) dictatorial ring to it..
    "Zwangskräfte" means, (in non-physics), "forces of subjugation"..:wink:
     
  6. Jun 10, 2004 #5
    Yes, r^2= az. Typo that happened when bolding. Sorry :(

    But thank you very much for your response. THat's exactly what I need, more of a step by step thought process on how to solve a problem like this. It all seems a bit clearer now :)
     
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