Damped Pendulum (generalised) 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". 
 A: If the damping coefficient approaches zero, the differential equation we are looking for needs to approach the 2nd equation you wrote (the one for the physical pendulum). Therefore, we can conclude that the only thing we have to modify in the last equation to get the equation for a damped physical pendulum is to change g/L into mgL/(I_CM+mL^2). 
If we want to find the equation of motion, I am afraid that we have to make a small-angle approximation (or solve the equation numerically which is quite tedious). 
A: 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.
