I know the differential equation for the swinging of a simple pendulum:
$\displaystyle\frac{\partial^2\theta}{\partial t^2} + \left(\frac{g}{L}\right)\sin\theta = 0$
where:
- $L$ is the length of the (massless) wire or rod that the mass is attached to
- $g$ is the acceleration due to gravity
- $\theta$ is the angle of the pendulum with the vertical
The corresponding equation for a physical pendulum is:
$\displaystyle\frac{\partial^2\theta}{\partial t^2} + \left(\frac{mgL}{I_\text{C of M} + mL^2}\right)\sin\theta = 0$.
where:
- $L$ is the distance between the pivot point and the body's centre of mass
- $g$ is the acceleration due to gravity
- $\theta$ is the angle of the body with the vertical
- $m$ is the mass of the body
- $I_\text{C of M}$ is the body's moment of inertia about its centre of mass
However, in reality, pendulums will slow down due to frictional damping and air resistance. Neglecting air resistance, bringing in a damping coefficient $\xi$ changes the equation for a simple pendulum to: (correct me if I'm wrong)
$\displaystyle\frac{\partial^2\theta}{\partial t^2} + \left(\frac{\xi}{L}\right)\frac{\partial\theta}{\partial t} + \left(\frac{g}{L}\right)\sin\theta = 0$.
My question is, what would the differential equation above translate to if applied to a physical pendulum? Please do not use "small-angle approximations".
Best Answer
A physical pendulum as described above behaves identically to a simple pendulum with a length $L^\prime=\frac{I_{\rm{CM}}+mL^2}{mL}$. So my inclination would be that you just replace both L's in your equation with the $L^\prime$ I just defined. That is:
$\frac{\partial^2\theta}{\partial t^2}+\left(\frac{\xi m L}{I_{\rm{CM}}+mL^2}\right)\frac{\partial\theta}{\partial t}+\left(\frac{mgL}{I_{\rm{CM}}+mL^2}\right)\sin(\theta)$
Although of course this may be different depending on how your damping coefficient is defined.