Swinging Pendulum ODE The problem is a pendulum with the ability to swing freely
I have a system of first order differential equations of the following form:
$\dfrac {d\theta} {dt}$=$\omega$
$\dfrac {d\omega} {dt}$=$-\dfrac glsin\theta-\dfrac{r(w)} {lm}$
we also know that $r(0)=0$ and $r'(0)>0$
where $\theta$ is the angle between the string and the vertical position and $\theta=0$ is the pendulums natural resting position, $l$ is the length of the string, $m$ is the mass of the object at the end of the string
I am trying to find the equilibrium points of the ODE system above and to describe what position of the pendulum they correspond to. 
I imagine that to find the equilibrium points you set both equations equal to 0 and solve. And just from an intuitive standpoint I would guess that the equilibrium points are the exact moment when the pendulum reaches it's max peak height through each swing before it starts to fall again but I am having trouble getting that from the equations
 A: Hint 1: You are on the right track for the equilibrium points (EPs).
We have the EPs when we simultaneously set each equation to $0$ and get the critical points:
$$(n \pi, 0)$$
where $n \in \mathbb{Z}$ which of course includes the point $(0,0)$.
Now, what is the $\pi$ term telling you?
Hint 2:
Linearize around those critical points and analyze the resulting system.
Hint 3:
Draw a phase portrait and analyze the behaviors.
A: The equilibrium position would correspond to the location for which your angular acceleration is zero and the angular velocity is zero. The angular velocity being zero allows $\theta$ to remain constant. The angular acceleration being zero allows $\omega$ to stay zero. 
So we know $\omega=0$ and $\frac{d \omega}{dt}=0$ which makes the second equation,
$$ 0=-\frac{g}{l} \sin(\theta) -\frac{r(0)}{ml}$$
Since $r(\omega)$ is the frictional force the pendulum experiences we know this must be $0$ when $\omega=0$ (friction doesn't cause motion it impedes it). This gives the equation,
$$ 0 = -\frac{g}{l} \sin(\theta)  \quad \Rightarrow \quad \theta=0,\pi$$
Therefore the equilibrium position is at the bottom of the swing. The object will stay in that equilibrium provided that its velocity is zero.
There is also an equilibrium at $180^\circ$ from the bottom (i.e, the top), but this is only physically realizable if the pendulum is a rigid rod. A ball on a string couldn't stay there in equilibrium.
