Simple Pendulum Why Generalized Coordinate Always Angle?  
When writing the equations of motion for the simple pendulum, why do textbooks always choose $\theta$ to be the generalized coordinate? The force of gravity is in the y-direction so wouldn't it be natural to write everything in terms of y instead of $\theta$? Since the string is of fixed length $l$, we can write $x=\sqrt{l^2-y^2}$ and so shouldn't we be able to write the Equations of Motion completely in terms of $y$? 
 A: The force of gravity is in the $\hat y$ direction, but that's not the only force in the problem. There's also tension in the string, which points along the string. The total force then points tangentially to the circle.
Another way to say this is: You can work in coordinates $x,y$ with basis vectors $\hat x$, $\hat y$, but then the force points in both the $\hat x$ and $\hat y$ directions. If you instead work in the coordinates $r,\theta$, with unit vectors $\hat{r}$, $\hat{\theta}$, you'll find that the force points only in the $\hat \theta$ direction, with no component along $\hat r$. So these are nicer coordinates to use!
A: 
Since the string is of fixed length l, we can write $x=\sqrt(l^2-y^2)$

Umm no
$x=\pm\sqrt{l^2-y^2}$
So there are two different positions of the system with the same $y$ value, one with positive $x$ and one with negative $x$, a little bit of thought about how a pendulum swings shows that both positive and negative $x$ values are part of the normal operating region of the pendulum.
A: You can use any coordinate system you like.  Some of them make it a lot easier to solve the equations of motion however.  In particular if you choose $\theta$ then you end up with a system which manifestly has one degree of freedom, while if you choose $x$ & $y$ you need to express it as being in two dimensions with a constraint between them: $y = -\sqrt{l^2 - x^2}$.
People generally like to choose the coordinate system which makes the solution easiest.
A: Peter Green's answer already showed you the error ($x=\sqrt{l^2-y^2}$ isn't generally true), but you can also directly see that $y$ isn't a sufficient coordinate:
No matter how fast the pendulum moves, at the bottom we always have $y=-l$ and $\dot{y}=0$. Therefore, you can't describe the state of the system just by $y$ and $\dot{y}$.

Edit: It's also worth pointing out that the other answers are indeed correct that $\theta$ is used instead of cartesian coordinates also because it actually is the choice that gives the simplest (and, subjectively, most natural) equations.
A: The coordinates $x$, $y$ or even (arc length) $s$ are as good as $\theta$ as generalized coordinate. The difference among them can only be a matter of convenience. 
As a short exercise let us see what happens when we choose either $\theta$ or $y$ as generalized coordinate.
First $q=\theta$. Then the lagrangian is
$$L=\frac{1}{2}l^2\dot\theta^2+gl\cos\theta,$$
where we choose unit mass for simplicity. The equation of motion reads
$$\ddot \theta+\frac gl\sin\theta=0,$$
and small oscillations means $\sin\theta\approx\theta$, hence
$$\ddot \theta+\frac gl\theta=0.$$
Now let $q=y$. Then
$$L=\frac 12\left(1+\frac{y^2}{l^2-y^2}\right)\dot y^2+gy=\frac 12\frac{l^2\dot y^2}{l^2-y^2}+gy,$$
since we eliminate $x$ using the constraint $x^2+y^2=l^2$. The equation of motion looks much worse,
$$\frac{d}{dt}\left(\frac{l^2\dot y}{l^2-y^2}\right)-\frac{l^2y\dot y^2}{(l^2-y^2)^2}-g=0.$$
since time derivative must act both on $\dot y$ and $y$. Moreover, small oscillations approximation is not straightforward as in the previous case. You shall consider $l-y\ll l$ and Taylor expand.
Also notice that since the constraint is holonomic, you can consider a dual problem which has actually three degrees of freedom ($x$, $y$ and $\lambda$)
$$L=\frac{1}{2}(\dot x^2+\dot y^2)+gy+\lambda(x^2+y^2-l^2).$$
Taking Euler-Lagrange equations for $q_i=x,\ y,\ \lambda$, 
$$\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=0,$$
we obtain the equations of motion for $x$, $y$ and $\lambda$, the later being just the constraint. Yet it is just much easier to keep with $q=\theta$.
A: In lagrangian mechanics, the tension of the string is a constraint (unknown) force so it's "replaced" by a holonomic constraint:
$$x^2+y^2=\ell^2 \tag 1\\$$ 
You're free to choose $x$ or $y$ as the generalised coordinate, but you can choose only one of them because there is one constraint and two cartesian coordinates $(x,y)$, so the number of degrees of freedom (which is the same as the number of generalised coordiantes) is simply: $2-1=1$ degree of freedom ($1$ generalised coordiante). Further more, the generalised coordiantes must be independent i.e there is no relation or formula that gather them which is not the case as it clear from the constaint $(1)$ (e.g if I know the value of $x$ at time $t_1$ and substitute it in eq$(1)$ I can reduce the value of $y$). 
However, there exists another equivalent constraint:
$$S=\ell \theta \tag 2\\$$
where $S$ is the arc  length of the circular path of the bob.
These two constrains are basically the same (and they count as one) because they contain the same information about the geometry of the system: the path of motion is circular and they tell us that the string is unstrechable (with a constant length). 
Following the same reasoning, you can choose $S$ or $\theta$ as the generalised coordinate. Since we have a rotational motion it's convenient to use the angle $\theta$ as the generalised coordinate.  
It's important to note that every choose of the generalised coordiantes gives the same differential equation of motion (hence the same solution for the coordinate and the same natural frequency of vibration $\omega_0$).
