# Dynamics - Pendulum Motion Derivation

Where a mass, $$m$$ is swinging a distance $$l$$ away from point $$O$$, with angular displacement of $$\Psi$$ from the vertical. The mass moment of inertia of the pendulum about the pivot $$O$$ is defined as

$$I _o = ml^2$$

Equalising the torques on the pendulum, with the torque from the swing

$$\tau = I_o \ddot{\Psi}$$

and the weight of the pendulum against the swing

$$\tau = -mg l \sin \Psi$$

Substituting in the inertia and simplifying:

$$ml^2 \ddot{\Psi} + mg l \sin \Psi = 0$$

$$\ddot{\Psi} + \frac{g}{l} \sin \Psi = 0$$

For a pendulum bar with mass spread evenly along the length, the mass moment of inertia is given by parallel axis theorem as

$$I_o = I_G + m (\frac{l}{2})^2$$

$$I_G = \frac{ml^2}{12}$$

Equating torques:

$$I_o \ddot{\Psi} + mg \frac{l}{2} \sin \Psi = 0$$

Substituting out via parallel axis theorem:

$$(I_G + \frac{ml^2}{4}) \ddot{\Psi} + mg \frac{l}{2} \sin \Psi = 0$$

Since

$$I_G = mk^2 = \frac{ml^2}{12}$$

Substituting and simplifying:

$$\frac{l}{3} \ddot{\Psi} + \frac{g}{2} \sin \Psi = 0$$