The spherical pendulum

In this section, we are going to look at the more general case of the simple pendulum. The mass will no longer be constrained to a plane. Since the only restriction on the mass is that its distance to the origin is L, the system has two degrees of freedom. To describe the system, we choose spherical coordinates. The hamiltonian of the system is then:

$$H = \frac{p_{\theta}^2}{2mL^2} + \frac{p_{\varphi}^2}{2mL^2\sin^2\theta}+mgL\cos\theta$$

Since we have two degrees of freedom, we also have two canonical momenta. The corresponding equations are the following:

$$\begin{align} \dot{\theta} &= \frac{p_{\theta}}{mL^2}, \nonumber \\ \dot{p_{\theta}} &= mgL\sin\theta + \frac{p_{\varphi}^2}{mL^2\sin^2\theta \tan\theta},\nonumber \\ \dot{\varphi} &= \frac{p_{\varphi}}{mL^2\sin^2\theta},\nonumber \\ \dot{p_{\varphi}} &= 0.\nonumber \end{align}$$

By looking at the equations, we can see that the axial angular momentum is a conserved quantity. This is a consequence of the axial symmetry that the system has. If we set the system to the following initial conditions:

L = 1.0  # Length of the pendulum arm (meters)
g = 9.81 # Gravitational acceleration (m/s^2)
m = 1.0  # Mass (kg)

theta_0   = 2.6  # In radians
p_theta_0 = 0.0  # Theta momentum
phi_0     = 0.0  # In radians 
p_phi_0 = 0.5    # Axial momentum
Nt = 401               # Number of time steps
dt = 0.025             # Time resolution

and evolve it with the Runge-Kutta method, we can obtain the first of these animations:

Precession of the pendulum.

Another precession movement but this time, the pendulum has enough energy to flip over.

We can see that the pendulum follows a very peculiar movement, it is not cahotic, but it is definitely not intuitive.