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Question about Lagrangian Mechanics.

  1. Apr 11, 2015 #1
    This is not a homework equation at all, however I have devised my own example problem in order to convey my misunderstanding. (My question is at the end of the problem)

    Question that I had come up with:
    A particle's motion is described in the x direction by the equation x = x(t). The particle's motion in the y direction is given by the constraint equation, y = (x(t))2. This particle's motion is on earth and is acted on by the force, -mg. Find the equation of motion.

    Attempted solution:
    because of the constraint equation for y, I did not write the lagrangian in terms of either x-dx/dt-y, but with just the coordinates x-dx/dt. Because of this logic, my Lagrangian was:

    http://www.sciweavers.org/upload/Tex2Img_1428780064/render.png [Broken]

    Plugging into the Euler-Lagrange, I obtained the Equation of Motion to be:

    http://www.sciweavers.org/upload/Tex2Img_1428780484/render.png [Broken]

    This is undoubtedly wrong as the equation of motion is one dimensional while the actual motion of the particle is in two dimensions. What did I do wrong? And more importantly, how would you guys solve this type of problem?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 11, 2015 #2
    When you have constraints there are many methods of taking care of them. I think that in this case, the simplest is to first write the Lagrangian in terms of x and y, then substitute your constraint equation y(x).

    The resulting ODE is for x(t) of your particle. If you want a position function of your particle i.e. r(t) then use your constraints again to generate it. I.e. you know x(t), then y(t) = y( x(t) ) then r(t)= x(t) i + y (x(t)) j

    makes sense?
     
  4. Apr 13, 2015 #3
    What the constraint equation does is it reduces the number of degrees of freedom. While the particle can move in two dimensions the changes in one co-ordinate can be described by the changes in the other co-ordinate. This means that you can completely describe the particle's motion by describing it's motion only in terms of one of these coordinates. This makes solving the problem easier as now you have only one equation to solve, instead of two.

    From your set up you have a horizontal and vertical coordinates x and y and gravity that acts in the negative y-direction. This is all well and good. You have added the constraint that the particle must move on some parabola y=x^2. This is an arbitrary addition to the problem, maybe the particle is stuck on the surface of some parabola shaped bowl and is unable to leave it's surface. That is the sort of situation where constraint equations come into use. Since you know the shape of the bowl and that the particle must be on the bowl's surface, if you know it's horizontal position you know it's vertical position. The exact shape and size of the bowl would be determined by the initial conditions you prescribe.


    You appear to have made a small mistake in setting up your Lagrangian as well.
    The kinetic energies in this case should also be ##\frac{m}{2}(\dot{x}^2 + \dot{y}^2)## So your second term should be positive but other than that your Lagrangian is correct.

    When computing your partial derivatives for the Euler Lagrange equations you have to remember a few maths things. This is especially important when you have an interesting lagrangian which has a term with both ##x## and ##\dot{x}## in it. Remember that all the other variables are treated as constants when taking a partial derivative. This means that this middle term should appear in both your partial with respect to ##x## and your partial with respect to ##\dot{x}##. You should have ##\frac{d}{dt}(m\dot{x}+4m\dot{x}x^2) - 4\dot{x}^2 x-2mgx = 0##. Also, when computing the total time derivative you have to use both the product rule and the chain rule as both ##x## and ##\dot{x}## are functions of t so your final expression is a bit more complicated than you might expect.
     
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