# s1110 - ψ ´-m g l ³ φ 2 1 2 ψ 2 ´ 3 m g l The...

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ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 2 Solution for assignment 10: Double Pendulum. Derivation of the Lagrangian: x 1 = l sin φ y 1 = l cos φ x 2 = l sin φ + l sin ψ y 2 = l cos φ + l cos ψ L = 1 2 m l ± ˙ x 1 2 + ˙ y 1 2 ² + 1 2 m l ± ˙ x 2 2 + ˙ y 2 2 ² - m g l ( z 1 + z 2 ) + const = 1 2 m l ˙ φ 2 + 1 2 m l ± ˙ φ 2 + ˙ ψ 2 + 2 ˙ φ ˙ ψ cos( φ - ψ ) ² + m g l (2 cos φ + cos ψ ) = m l ³ ˙ φ 2 + 1 2 ˙ ψ 2 + ˙ φ ˙ ψ cos( φ - ψ ) ´ + m g l (2 cos φ + cos ψ ) . Taylor expansion to second order in φ, ψ, ˙ φ and ˙ ψ gives L = m l ³ ˙ φ 2 + 1 2 ˙ ψ 2 + ˙ φ ˙
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Unformatted text preview: ψ ´-m g l ³ φ 2 + 1 2 ψ 2 ´ + 3 m g l . The Euler-Lagrange equation are then = d dt ∂L ∂ ˙ φ-∂L ∂φ = 2 ¨ φ + ¨ ψ + 2 g l φ = d dt ∂L ∂ ˙ ψ-∂L ∂ψ = ¨ ψ + ¨ φ + g l ψ and in matrix notation ³ 2 1 1 1 ´ ³ ¨ φ ¨ ψ ´ + ³ 2 g/l g/l ´ ³ φ ψ ´ = ....
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## This note was uploaded on 11/10/2011 for the course PHY 4241 taught by Professor Berg during the Spring '11 term at University of Florida.

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