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Lagrangian Constraint Problem

  1. Feb 22, 2004 #1
    So a bead slides down a frictionless parabolic wire of shape y=ax^2. I have to express the Lagrangian in terms of x and y. Then I have to use the constraint equation to express this solely in terms of x. Then I have to find the equations of motion, and simplify them for small oscillations.

    How do I get from

    L=1/2m(x')^2 + 1/2m(y')^2 - mgy


    L=1/2m(x')^2(1 + 4a^2x^2) -mgax^2 ???

    And in this specific example I am having trouble seeing what the constraint force or constraint potential is. It seems to that we know that the force of gravity is present and acting in the negative y direction, and we are told there is no force of friction, so what is the constraint?
  2. jcsd
  3. Feb 22, 2004 #2
    You have two degrees of freedom, x and y, and one constraint, so you are left with one degree of freedom. Your bead has to roll on a parabola with equation y=ax^2, so (surprise, surprise), the constraint is y=ax^2. Simply substitute y and you'll get the correct lagrangrian.
  4. Feb 22, 2004 #3
    Yes, but since y'=2ax, how do you come up with the term 1/2m(x')^2 *4ax^2? In other words, how does a term of x' end up in there? And I am not talking about the term 1/2m(x')^2 *1 which is also present in the Lagrangian.
  5. Feb 22, 2004 #4


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    I don't think that's true...
  6. Feb 23, 2004 #5
    You've made an error in taking the derivative. Yo didn't take the derivative with respect to time, you took the derivative with respect to x. The correct derivative is obtained using the chaing rule. The result is

    y' = dy/dt = 2ax dx/dt = 2ax x'
  7. Feb 23, 2004 #6
    Stupid stupid me
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