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 = \frac{1}{2} m l (ẋ_{1}^{2} + ẏ_{1}^{2}) + \frac{1}{2} m l (ẋ_{2}^{2} + ẏ_{2}^{2}) − m g l (z_{1} + z_{2}) + const
= \frac{1}{2} m l ϕ̇^{2} + \frac{1}{2} m l (ϕ̇^{2} + ψ̇^{2} + 2 ϕ̇ ψ̇ cos(ϕ − ψ)) + m g l (2 cos ϕ + cos ψ)
= m l (ϕ̇^{2} + \frac{1}{2} ψ̇^{2} + ϕ̇ ψ̇ cos(ϕ − ψ)) + m g l (2 cos ϕ + cos ψ) .

Taylor expansion to second order in ϕ, ψ, ϕ̇ and ψ̇ gives

L = m l (ϕ̇^{2} + \frac{1}{2} ψ̇^{2} + ϕ̇ ψ̇) − m g l (ϕ^{2} + \frac{1}{2} ψ^{2}) + 3 m g l .

The Euler-Lagrange equation are then

0 = \frac{d}{dt} \frac{∂L}{∂ϕ̇} − \frac{∂L}{∂ϕ} = 2 ϕ̈ + ψ̈ + 2 \frac{g}{l} ϕ
0 = \frac{d}{dt} \frac{∂L}{∂ψ̇} − \frac{∂L}{∂ψ} = ψ̈ + ϕ̈ + \frac{g}{l} ψ

and in matrix notation

( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) ( ϕ̈ ; ψ̈ ) + ( \begin{matrix}2g/l & 0 \\ 0 & g/l \end{matrix} ) ( \begin{matrix}ϕ \\ ψ \end{matrix} ) = 0 .