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Lagrangian for a Spherical Pendulum (Goldstein 1.19)

  1. Sep 9, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the Lagrangian and equations of motion for a spherical pendulum

    2. Relevant equations

    L=T-U and Lagrange's Equation

    3. The attempt at a solution

    I found the Lagrangian to be L = 0.5*m*l222sin2(θ)) - mgl*cos(θ) where l is the length of the rod, ω is (theta dot) and Ω is (phi dot). Here, the angle θ is measured vertically down from the z-axis and Φ is measured in the xy-plane.

    My question comes when solving the Euler-Lagrange equation for Φ, namely the term: (d/dt)(∂L/∂Ω).
    The inner term, ∂L/∂Ω is easy enough: ∂L/∂Ω = ml2Ωsin2(θ). The trick for me is coming when finding the total time derivative of that. I've seen two sources online that give different values, but what I did was:

    d/dt(∂L/∂Ω) = d/dt(ml2Ωsin2(θ)) = ∅*ml2*sin2(theta) + 2ml2Ωsin(θ)cos(θ)ω

    Here, ∅ = (phi double dot). Is this right? A lot of things I have seen online leave out the ω = (theta dot) factor in the second term. This has to be there for a total time derivative, right?
  2. jcsd
  3. Sep 9, 2016 #2


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    If you think about the value of ##\frac{\partial L}{\partial \Phi}##, you should see that there is no need to worry about carrying out the total time derivative ##\frac{d}{dt} \frac{\partial L}{\partial \Omega}##.

    Also, I'm not sure how you are defining ##\theta##. Does ##\theta = 0## correspond to the pendulum hanging vertically? That is, does the potential energy increase if ##\theta## increases? If so, I think you should check the sign of your potential energy function in the Lagrangian.
  4. Sep 9, 2016 #3
    Yeah, I just caught the potential energy problem myself and fixed it. However, I'm not quite sure what you mean about not needing to carry out the total time derivative. In the equation for Φ, ∂L/∂Φ = 0, so this would mean that the physical quantity described by (d/dt)(∂L/∂Ω) is conserved. Am I misunderstanding this? That is, would it just be the value corresponding to ∂L/∂Ω that is conserved rather than (d/dt)(∂L/∂Ω)?
  5. Sep 10, 2016 #4


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    Yes. The equation (d/dt)(∂L/∂Ω) = 0 implies that ∂L/∂Ω is conserved.
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