Editing Inverted pendulum (section)

===Stationary pivot point===
In a configuration where the pivot point of the pendulum is fixed in space, the equation of motion is similar to that for an [[Pendulum (mathematics)|uninverted pendulum]].  The equation of motion below assumes no friction or any other resistance to movement, a rigid massless rod, and the restriction to [[2-dimensional]] movement.

:<math>\ddot \theta - {g \over \ell} \sin \theta = 0</math>

Where <math>\ddot \theta </math> is the [[angular acceleration]] of the pendulum, <math>g </math> is the [[standard gravity]] on the surface of the Earth, <math>\ell</math> is the length of the pendulum, and <math>\theta</math> is the angular displacement measured from the equilibrium position.

When <math>\ddot \theta </math> added to both sides, it has the same sign as the angular acceleration term:

:<math>\ddot \theta = {g \over \ell} \sin \theta</math>

Thus, the inverted pendulum accelerates away from the vertical unstable equilibrium in the direction initially displaced, and the acceleration is inversely proportional to the length. Tall pendulums fall more slowly than short ones.

'''Derivation using torque and moment of inertia:'''
[[Image:cart-pendulum.svg|thumb|300px| A schematic drawing of the inverted pendulum on a cart. The rod is considered massless. The mass of the cart and the point mass at the end of the rod are denoted by M and m. The rod has a length l.]]

The pendulum is assumed to consist of a point mass, of mass <math>m </math>, affixed to the end of a massless rigid rod, of length <math>\ell</math>, attached to a pivot point at the end opposite the point mass.

The net [[torque]] of the system must equal the [[moment of inertia]] times the angular acceleration:
:<math>\boldsymbol{\tau}_{\mathrm{net}}=I \ddot \theta</math>
The torque due to gravity providing the net torque:
:<math>\boldsymbol{\tau}_{\mathrm{net}}= m g \ell \sin \theta\,\!</math>
Where <math> \theta\ </math> is the angle measured from the inverted equilibrium position.

The resulting equation:
:<math> I \ddot \theta= m g \ell \sin \theta\,\!</math>

The moment of inertia for a point mass:
:<math>I = m R^2</math>
In the case of the inverted pendulum the radius is the length of the rod, <math> \ell </math>.

Substituting in <math>I = m \ell ^2</math>
:<math> m \ell ^2 \ddot \theta= m g \ell \sin \theta\,\!</math>

Mass and <math>\ell^2</math> is divided from each side resulting in:
:<math>\ddot \theta = {g \over \ell} \sin \theta</math>