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Lagrangian of simple pendulum

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data
    2 masses, m[itex]_{1}[/itex] and m[itex]_{2}[/itex] are fixed at the endpoints of a rigid rod of length l. mass m[itex]_{1}[/itex] is attached to a horizontal bar so that it may move in the x direction freely, but not in the y direction. let θ be the angle the rod makes with the vertical, what is the corresponding Lagrangian of the system if it is assumed to be in the uniform gravitational field g?


    2. Relevant equations
    Euler Lagrange eqns


    3. The attempt at a solution
    The only issue I am running into with this problem is in the kinetic energy terms. My kinetic terms are : [itex]\frac{1}{2}[/itex](m[itex]_{1}[/itex]+m[itex]_{2}[/itex])[itex]\dot{x}[/itex][itex]^{2}[/itex]+[itex]\frac{1}{2}[/itex]m[itex]_{2}[/itex]l[itex]^{2}[/itex][itex]\dot{θ}[/itex][itex]^{2}[/itex] but the book proposes a solution the same as mine except with the added term [itex]\frac{1}{2}[/itex]m[itex]_{2}[/itex](2l[itex]\dot{x}[/itex][itex]\dot{θ}[/itex]cosθ). I am not understanding where this term comes from, I thought I took care of the x velocity dependence in the first term.
     
  2. jcsd
  3. Jan 14, 2013 #2

    TSny

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    Find expressions for the x and y coordinates of m2 in terms of the x coordinate of m1 and θ.
     
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