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A double pendulum has two degrees of freedom. Produce a reasonable Lagrangian L(01, 02, 01, 02) for describing a double
pendulum free to move in a plane in gravity g? Assume the connecting rods are rigid.
4₁
m₁
L₂
m₂²₂
Hint: Let's just assume L₁
2
2
L = ¼m²³² (0₂² +401² +30₁02 cos(0₁ - 0₂)) + ½ mgl (3 cos 0₁ + cos 0₂).
L₂ = 1 and that m₁ = m₂ = m and show that the Lagrangian reduces to :

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