8.4.2 The double pendulum

Figure 1 shows a sketch of a double pendulum: two identical rods of length $2L$ and mass $m$ are freely hinged together at one end, and the linked pair is freely pivoted to a fixed support at one end. We can describe the position of the two pendulums at any instant with the two angles $\theta$ and $\phi$ as shown.

Figure 1. Sketch of the double pendulum

We can derive the governing equations for free motion by using Lagrange’s equations. First, we need expressions for the potential and kinetic energies. In terms of Cartesian x-y axes with an origin at the fixed pivot, the centre of mass of the top pendulum is at the point

$$[L \sin \theta, -L \cos \theta] \tag{1}$$

while that of the lower pendulum is at the point

$$[2L \sin \theta + L \sin \phi, -2L \cos \theta – L \cos \phi]. \tag{2}$$

We can immediately write down an expression for the potential energy, which is entirely due to gravity:

$$V=mgL(1 – \cos \theta) +mgL[2(1- \cos \theta) + (1-\cos \phi)] \tag{3}$$

where the reference configuration with $V=0$ has been taken to be the equilibrium position $\theta=\phi=0$.

For the kinetic energy, we first need the velocities of the two centres of mass. For the top pendulum this has components

$$L \dot{\theta} [\cos \theta, \sin \theta] \tag{4}$$

while for the lower pendulum the components are

$$2L \dot{\theta} [\cos \theta, \sin \theta] + L \dot{\phi} [\cos \phi, \sin \phi]. \tag{5}$$

We can now write down an expression for the kinetic energy: for each pendulum separately, we can use the general rigid-body expression $(1/2)m v_G^2 + (1/2) I_G \Omega^2$ where $m$ is the mass, $I_G$ is the moment of inertia about the centre of mass, $v_G$ is the magnitude of the velocity of the centre of mass, and $\Omega$ is the angular velocity.

The result is

$$T=\dfrac{1}{2}mL^2 \dot{\theta}^2 (\cos^2 \theta + \sin^2 \theta) +\dfrac{1}{2}\left(\dfrac{1}{3}mL^2\right) \dot{\theta}^2+\dfrac{1}{2}\left(\dfrac{1}{3}mL^2\right) \dot{\phi}^2$$

$$+ \dfrac{1}{2}mL^2 \left[ \left(2 \dot{\theta}\cos \theta + \dot{\phi}\cos \phi\right)^2 + \left(2 \dot{\theta}\sin \theta + \dot{\phi}\sin \phi\right)^2 \right] \tag{6}$$

which, after multiplying out and simplifying, comes down to

$$T=\dfrac{1}{2}mL^2 \left[ \dfrac{16}{3} \dot{\theta}^2 + \dfrac{4}{3} \dot{\phi}^2 + 4 \dot{\theta} \dot{\phi} \cos (\phi-\theta) \right] . \tag{7}$$

Now we use the Lagrange equations. Applied to the angle $\theta$, the equation is

$$\dfrac{d}{dt}\left[ \dfrac{\partial T}{\partial \dot{\theta}} \right] – \dfrac{\partial T}{\partial \theta} + \dfrac{\partial V}{\partial \theta} = 0 \tag{8}$$

leading to

$$\dfrac{d}{dt}\left[\dfrac{16}{3}mL^2 \dot{\theta} + 2mL^2 \dot{\phi} \cos (\phi – \theta) \right] $$

$$-2mL^2 \dot{\theta} \dot{\phi} \sin (\phi – \theta) + 3mgL \sin \theta = 0 \tag{9}$$

which expands and simplifies to the final result

$$\dfrac{16}{3}\ddot{\theta}+ 2\ddot{\phi}\cos (\phi – \theta) – 2 \dot{\phi}^2 \sin (\phi – \theta) + \dfrac{3g}{L} \sin \theta = 0. \tag{10}$$

The corresponding Lagrange equation for the angle $\phi$ gives

$$\dfrac{d}{dt}\left[\dfrac{4}{3}mL^2 \dot{\phi} + 2mL^2 \dot{\theta} \cos (\phi – \theta) \right] $$

$$+2mL^2 \dot{\theta} \dot{\phi} \sin (\phi – \theta) + mgL \sin \phi = 0 \tag{11}$$

which expands and simplifies to give

$$\dfrac{4}{3}\ddot{\phi}+ 2\ddot{\theta}\cos (\phi – \theta) + 2 \dot{\theta}^2 \sin (\phi – \theta) + \dfrac{g}{L} \sin \phi = 0. \tag{12}$$

Equations (10) and (12) determine the free motion of the pendulum system. Because we have two second-order differential equations, we need four initial values: the most natural things to specify are the values at time $t=0$ of $\theta$, $\dot{\theta}$, $\phi$ and $\dot{\phi}$.