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:

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