s1110 - -m g l 2 + 1 2 2 + 3 m g l . The Euler-Lagrange...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ˙
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -m g l 2 + 1 2 2 + 3 m g l . The Euler-Lagrange equation are then = d dt L -L = 2 + + 2 g l = d dt L -L = + + g l and in matrix notation 2 1 1 1 + 2 g/l g/l = ....
View Full Document

Ask a homework question - tutors are online