Hamiltonian on Simple Pendulum Motion

∟Hamiltonian on Simple Pendulum Motion

This section provides an example of calculating the Hamiltonian on a mechanical system of an single object in simple pendulum motion and applying the Law of Conservation of Energy.

What Is Simple Pendulum Motion? A Simple Pendulum Motion is an object of mass m hanging on a string from a pivot point so that it is constrained to move on a circle of a fixed radius.

If we map the space in Cartesian coordinates, the Simple Pendulum Motion is a 2-dimensional problem, both the position, r, and the speed, v, have 2 components:

r = (x, y)
v = (x', y')

If we introduce an extra variable θ as the angular position of the object from the vertical line, r and v can be expressed as:

r = (l*sin(θ), -l*cos(θ))
v = (l*cos(θ)*θ', l*sin(θ)*θ')

Now the object's kinetic energy, T, can be expressed as:

T = m*|v|**2

or:
  T = 0.5*m*l*l*(cos(θ)**2)+sin(θ)**2)*θ'*θ'
    # Pythagorean Theorem applied

or:
  T = 0.5*m*l*l*θ'*θ'
    # Since cos(θ)**2)+sin(θ)**2 = 1

The potential energy, V, can be expressed as:

V = m*g*y

or:
  V = m*g*(-l*cos(θ))

or:
  V = -m*g*l*cos(θ)

So the Hamiltonian, H, can be expressed as:

H = T + V                          (H.1)

or:
  H = 0.5*m*l*l*θ'*θ' - m*g*l*cos(θ)
    # Replaced T and V with their expressions

Since this simple pendulum motion can be considered as an isolated conservative system, we can apply the Law of Conservation of Energy:

H = constant

or:
  0.5*m*l*l*θ'*θ' - m*g*l*cos(θ) = constant

or:
  d(0.5*m*l*l*θ'*θ' - m*g*l*cos(θ))/dt = 0
    # Since d(constant)/dt = 0

or:
   m*l*l*θ'*θ" + m*g*l*sin(θ)*θ' = 0
   l*θ" + g*sin(θ) = 0
   θ" = - g*sin(θ)/l