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Lagrangian for a Spherical Pendulum (Goldstein 1.19)
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[QUOTE="Yosty22, post: 5562784, member: 460241"] [h2]Homework Statement [/h2] Find the Lagrangian and equations of motion for a spherical pendulum [h2]Homework Equations[/h2] L=T-U and Lagrange's Equation [h2]The Attempt at a Solution[/h2] [/B] I found the Lagrangian to be L = 0.5*m*l[SUP]2[/SUP](ω[SUP]2[/SUP]+Ω[SUP]2[/SUP]sin[SUP]2[/SUP](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 = ml[SUP]2[/SUP]Ωsin[SUP]2[/SUP](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(ml[SUP]2[/SUP]Ωsin[SUP]2[/SUP](θ)) = ∅*ml[SUP]2[/SUP]*sin[SUP]2[/SUP](theta) + 2ml[SUP]2[/SUP]Ωsin(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? [/QUOTE]
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Lagrangian for a Spherical Pendulum (Goldstein 1.19)
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