About the time derivative of tangential speed Let an object have nonuniform circular motion, with increasing tangential speed. The object would still have a velocity vector tangent to its circular trajectory, but its acceleration vector would not be directly towards the rotation axis, and a little bit towards outside of the trajectory.
I understand that this acceleration vector can be decomposed into one component tangent to the trajectory, and the other directly towards the rotation axis. The latter would represent centripetal acceleration, and the former (since it's parallel with the tangential velocity vector) would change the magnitude of the velocity vector (and not affecting its direction), therefore changing tangential speed.
This tangential acceleration component must be then a time derivative of tangential speed, which is defined by $v_{t} = \omega r$ where $\omega$ stands for angular speed. Differentiating $v_{t}$ simply gives $a_t$, but differentiating $\omega$ gives a concept for which I don't even know how to call it. I had no problems differentiating speed when it came to tangential speed because there was no ambiguity (differentiating velocity gives acceleration, doing so with tangential speed gives tangential acceleration) but for angular speed, I can't call its time derivative simply angular acceleration because angular acceleration is already defined as the time derivative for angular velocity.
How is this value named? Or does this have no specific name?
 A: There's some vector/scalar confusion here.  Consider an object undergoing (not necessarily uniform) circular motion centered at the coordinate origin in the $(x,y)$-plane.  The angular velocity vector is given by $\boldsymbol \omega = \mathbf r \times \mathbf v$, which can be seen to be pointing in the $z$-direction; also, since $\mathbf r \cdot \mathbf v = 0$, we have that $\mathbf v = \boldsymbol \omega\times \mathbf r$.
Differentiating the velocity $\mathbf v$, we find
$$\frac{d}{dt} \mathbf v = \boldsymbol \alpha \times \mathbf r+\boldsymbol \omega \times \mathbf v \equiv \mathbf a_t + \mathbf a_c$$
where the first term is the tangential acceleration and the second is the radial acceleration - i.e. the centripetal acceleration.

Up to this point, we've worked with vector quantities.  Taking magnitudes gives the corresponding scalar expressions.  For example,
$$ \mathbf v = \boldsymbol \omega \times \mathbf r\implies v = \omega r$$
where we've used that $\mathbf r \cdot \boldsymbol\omega = 0$.
If we want to take the time derivative of the speed $v$ (as opposed to the velocity $\mathbf v$), we can do so:
$$\frac{d}{dt} v = \frac{d}{dt}\sqrt{\mathbf v \cdot \mathbf v}= \frac{\mathbf v \cdot \mathbf a}{\sqrt{\mathbf v \cdot \mathbf v}} = \hat v \cdot \mathbf a$$
where $\hat v$ is the unit vector in the direction of the velocity. This quantity doesn't have a standard name, other than "tangential acceleration" or something like that.
Similarly,
$$\frac{d}{dt}\omega = \frac{d}{dt}\left(\frac{v}{r}\right) = \frac{1}{r} \frac{d}{dt} v = \frac{\hat v \cdot \mathbf a_t}{r} = \hat\omega \cdot \boldsymbol \alpha$$
Once again, this object doesn't have a standard name in general.  It is the component of the angular acceleration along $\hat \omega$.  Of course, as long as the object is moving in the $(x,y)$-plane, all of $\boldsymbol \alpha$ is directed along $\hat \omega$, but in general this is not true.
A: The vector representing angular velocity is defined as being along the axis of rotation. If the direction of the axis is not changing, then dω/dt does give the magnitude of the angular acceleration vector (which is also along the axis).  To get the tangential acceleration, you assume that, r, is constant.  Then: $a_t$ = rα.
A: I think it's the angular acceleration; its formula is $\alpha = \frac{d \omega}{dt} = \frac{d^2 \theta}{dt^2}$
A: The acceleration of a particle, in general, has two components. One that lies along the direction of motion and one that is tangent to this direction. The last is the one you ask for, i.e., the tangential acceleration.
