Lagrange Equations on Simple Pendulum

∟Introduction of Generalized Coordinates

∟Lagrange Equations on Simple Pendulum

This section provides an example of applying the Lagrange Equations on an object in simple pendulum motion using generalized coordinates.

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 radius L.

Let's introduce 2 generalized coordinates to describe this motion, θ and l:

q = (θ, l)

where:
  # θ is the angular position from the vertical line
  # l is the length of the string

The transformation functions are:

r = (x, y)

r = (r₁(), r₂())

r = (l*sin(θ), -l*cos(θ))

Then the velocity is:

r' = dr/dt

or:
  r' = (l*cos(θ)*θ', l*sin(θ)*θ')

Then the kinetic energy is:

T = m*v*v/2

or:
  T = 0.5*m*r'*r'

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

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

Then the potential energy is:

V = m*g*y

or:
  V = m*g*(-l*cos(θ))

or:
  V = -m*g*l*cos(θ)

The Lagrangian function in generalized coordinates becomes:

L = T - V                          (G.1)

or:
  L = 0.5*m*l*l*θ'*θ' + m*g*l*cos(θ)

Now take the Lagrange Equations in generalized coordinates:

d(∂L/∂q')/dt = ∂L/∂q               (C.3)

or:
  d(∂L/∂θ')/dt = ∂L/∂θ
  d(∂L/∂l')/dt = ∂L/∂l

or:
  d(m*l*l*θ')/dt = -m*g*l*sin(θ)
  0 = 0

or:
  m*l*l*θ" = -m*g*l*sin(θ)

or:
  l*θ" = -g*sin(θ)