How does conservation of mechanical energy in a Pendulum work? How does conservation of mechanical energy in a pendulum work? 
I understand that gravitational potential energy is converted, as an object falls, into kinetic energy, and is otherwise given by $mgh$, for an object at a height of $h$. $ \ $ So, if the object falls a distance of $x$, from rest, then it gains $mgx$ kinetic energy at the height of $ \ h-x$. 
But the bob of a pendulum isn't simply falling.  Rather it follows a circular-arc trajectory, so the resultant force causing its motion is different than the weight of the bob.  Yet, the potential energy in a bob at a height of $h$, is also $mgh$.  So, as the bob reaches a height of $ \ h-x \ $, in its circular-arc motion, it will also have a kinetic energy of $mgx. \ \ $  
This then must imply that the bob will be moving at the same speed as the object described above, when it reaches the height of $ \ h-x \ $, but in a very different direction. Is this accurate?
I'm having a hard time understanding how this makes sense intuitively.  
On one hand, the bob has the same speed as the object even though a different resultant force than its weight got it to that speed. On another hand, it's not even moving in the same direction. But, still, both phenomena are referred to as gravitational potential energy, and no distinction seems to be made between the two when solving problems. 
Why is this (assuming I got everything right in the above)? 
 A: 
This is a surprisingly difficult problem to solve, actually. Looking at the figure, we can surmise that:
$$
\frac{dv}{dt} = \ddot{x} = g\sin(\theta)
$$
It's natural to switch to angular coordinates here ($x = r\theta, \dot{x} = r\dot{\theta}, \ddot{x} = r\ddot{\theta}$):
$$
r\ddot{\theta} = g\sin(\theta)
$$
Unfortunately this is a non-linear differential equation and my recollection (and a cursory search supports this) is that there is no analytical solution.
We suspect however, that there should be a conserved quantity here: energy. The kinetic energy is given by $\frac{1}{2}mr^2\dot{\theta}^2$ and the potential energy is given by (above the lowest point) $mgr\left(\cos(\theta) - 1\right)$. Therefore we expect the following to be true:
$$
mr\left(\frac{1}{2}r\dot{\theta}^2 + g\cos(\theta) - g\right) = C
$$
Which really means that the following is constant:
$$
\frac{1}{2}r\dot{\theta}^2 + g\cos(\theta) = C
$$
You can verify, through inspection that, sure enough, if you differentiate the above equation with respect to time, you get:
$$
r\dot{\theta}\ddot{\theta} - g\sin(\theta)\dot{\theta} = 0 \leadsto \dot{\theta}\left(r\ddot{\theta} - g\sin(\theta)\right) = 0
$$
Finally we recover the original acceleration equation:
$$
r\ddot{\theta} = g\sin(\theta)\text{, q.e.d.}
$$
Comments
I'll be the first admit this isn't a complete answer. 1) I worked backwards from what we already assumed (conservation of energy) and 2) at no point did I show what that value of C should be or how it can be used to calculate the actual speeds.
Super late edit
It's worth pointing out that our differentiation of energy gave us a weird (yet valid) solution which was $\dot{\theta} = 0$, corresponding to the case that we hold the bob at any angle we want--and never ever! move it!
Note that this solution isn't possible from the original assumptions because this extraneous solution requires introducing another force (one that cancels out gravity exactly). In the original, $r\ddot{\theta} = g\sin\left(\theta\right)$, certainly $\theta = 0$ (and thus $\dot{\theta} = 0$) is a solution but $\theta = A$ (a constant) does not satisfy this equation.
