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Langrange Equation of Motion

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A charged particle of mass m and charge q is free to move in the horizontal (x, y) plane, under
    the influence of the Coulomb potential due to another charge Q that is fixed at the origin.

    Find the Lagrangian and the differential equations of motion of the mass m, in terms of
    Cartesian coordinates (x, y).

    2. Relevant equations

    dL/dqi - d/dt * dL/dq'i = 0
    L = T - V
    T = .5mv^2
    V= KQq/r

    3. The attempt at a solution

    L = T - V

    Langrangian = 0.5mx'^2 - KQq/x

    dL/dx - d/dt * dL/dx' = 0
    d/dx(-KQq/x) - d/dt * d/dx'(0.5mv^2)=0

    -dV/dt - d/dt(mx')=0
    Fx = mx''

    Is this the right answer? The question asks for it to be in terms of the cartesian coordinates.
    However, my answer isn't. Please Help.
  2. jcsd
  3. Nov 10, 2008 #2


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    Your solution is schematically ok, if you are only working in one variable. Except you mean dV/dx, right? Now just do the same thing in two variables q1=x(t), q2=y(t). T=(1/2)mv^2=(1/2)m(x'(t)^2+y'(t)^2), V=KqQ/(x^2+y^2)^(1/2).
  4. Nov 10, 2008 #3
    The question says that the charge can only move horizontally, so is it necessary to have the y component?
  5. Nov 10, 2008 #4


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    You said it's constrained to move in the (x,y) plane, right? Not just along a line.
  6. Nov 10, 2008 #5
    It says the charge is free to move in the horizontal (x,y) plane. Does that not mean it is only allowed to move horizontally?

  7. Nov 10, 2008 #6


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    The entire xy plane is horizontal, the z-direction would be the vertical....think of it in terms of standing on a flat earth...you can move North, West, 22 degree East by South-East and In all of these cases you are moving in the horizontal plane....it is only if you were to jump up and down that you would be moving vertically....so yes, you need to consider both x and y in your Lagrangian
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