Confusion about circular motion So today in class, I learnt that the angular quantities are related to the linear quantities by
$$s = r\theta \qquad \qquad v = r\omega \qquad \qquad a = r\alpha$$
where it is assumed that the object travels in a perfect circle.
My teacher then went on to derive the formula
$$a = \frac{v^2}{r}$$
where it is further assumed that the object has constant linear velocity $v$.
Then I got a bit confused. If we rearrange for $\alpha$, we get
$$\alpha = \frac ar = \frac{v^2}{r^2} = \omega^2$$
But if $\omega$ is a constant, then shouldn't we have $\alpha = 0$? What have I done wrong?
 A: $a$ is the tangential acceleration which is equal to zero in uniform circular motion and $\frac{v^2}{r}$ is the centripetal acceleration that's why $a_{tangential} \neq \frac{v^2}{r}$
A: The problem is the distinction between the radial (or centripetal) acceleration and the tangential acceleration (it is key to know the difference between them). 
The radial acceleration (I will call it $a_r$) is always there in circular motion. It is the thing that keeps circular motion, changing the direction of the tangential velocity vector. Its formula is $a_r = \frac{v^2}{r}$, as you said.
Meanwhile, the tangential acceleration (which I will call $a_t$) makes the rotational motion accelerate. Its formula is $a_t = r\alpha$. If you get some intuition here, as the angular acceleration increases, i.e. the body rotates faster and faster, a point in the rim of the body should move faster and faster, right? That's why they are proportional.
So the fact that $\alpha = 0$ doesn't imply $\omega = 0$ because $a_r \neq a_t$ (in most of the cases). 
So, in summary, the basic point is to avoid confusing the radial and tangential components of motion.
A: Both expressions for the acceleration are correct, the problem is that they do not refer to the same thing.
In any kind of motion (of a point particle) you should know that the velocity is always tangent to the trajectory of the object. The case of circular motion is the same.
You may not know this but the right way to express angular velocity is as a vector (just like with regular velocity) that points in the direction of the axis of rotation. The magnitude of the vector gives the instantaneous rate of rotation.

Through a vector relation, we get the vector velocity due to the rotation:
$$\overrightarrow{\omega}\times\overrightarrow{r}=\overrightarrow{v}$$
In the case of circular motion $\overrightarrow{\omega}$ and $\overrightarrow{r}$ are perpendicular, so that 
$$\left|\overrightarrow{\omega}\times\overrightarrow{r}\right|=\left|\overrightarrow{\omega}\right|\left|\overrightarrow{r}\right|\mathrm{sin}(\phi)=\omega r\text{ }\mathrm{sin}(\phi)=\omega r$$
That is the expression you posted in the beginning, but note that it is only the magnitude of the velocity vector. This is important because the acceleration is also a vector that gives you both the magnitude and the direction of change of the velocity.
To get the acceleration you simply differentiate the velocity with respect to time (as always):
$$\overrightarrow{a}=\frac{d}{dt}\left(\overrightarrow{v}\right)=\frac{d}{dt}\left(\overrightarrow{\omega}\times\overrightarrow{r}\right)$$
And you use the fact that the cross product follows Leibniz's rule (careful not to change the order of the products):
$$\overrightarrow{a}=\frac{d\overrightarrow{\omega}}{dt}\times\overrightarrow{r}+\overrightarrow{\omega}\times\frac{d\overrightarrow{r}}{dt}=\boxed{\overrightarrow{\alpha}\times\overrightarrow{r}+\overrightarrow{\omega}\times\overrightarrow{v}}$$
In circular motion, $\overrightarrow{\alpha}$ goes in the same direction as $\overrightarrow{\omega}$ because the axis of rotation doesn't change. The magnitude of the first term is then simply $\alpha r$, your first expression for acceleration. The second term is also simply $\omega v$ but by using $v=\omega r$, the magnitude yields $v^2/r$, your second expression.
So both are valid, but if you look carefully you should realize that the $\overrightarrow{\alpha}\times\overrightarrow{r}$ vector points in the direction of the velocity, it is thus called the tangential acceleration $\overrightarrow{a_{t}}$. On the other hand, the $\overrightarrow{\omega}\times\overrightarrow{v}$ points towards the center of the circle, and is perpendicular to $\overrightarrow{a_{t}}$, it is therefore called the normal acceleration $\overrightarrow{a_{n}}$.
Just a little extra: You can see that the tangential acceleration is generated by the angular acceleration, this simply means that if you spin your particle faster the velocity increases. But the normal acceleration is always there (as long as $\omega\neq 0$) this is because the velocity is always changing its direction along the circular trajectory. The normal acceleration makes it change so that it is always tangent to this circle.
A: As the other answers and comments suggest, you are mixing up different properties by accidentally giving them the same labels.
The actual formulas are:
$$a_\perp = \frac{v^2}r$$
$$a_\parallel = r\alpha$$
A parallel $_\parallel$ acceleration is along with the circular path (also called tangential). 
A perpendicular $_\perp$ acceleration is sideways from the circular path (also called radial or centripetal). 
When you in a circular motion have constant speed, so that $a_\perp = \frac{v^2}r$ applies, then $a_\parallel=0$. If $a_\parallel$ wasn't zero, then $v$ wouldn't be constant - and then  $a_\perp = \frac{v^2}r$ wouldn't apply. 
