Is centripetal acceleration the same as angular acceleration? I know that the centripetal acceleration changes the direction of the tangential speed. But can I calculate it as the derivative of the angular speed with respect to time? Or are these different things?
 A: They cannot be the same thing because they have different units.  Centripetal acceleration $a_c = v^2/R = \omega^2 R$ has units of $\rm m/s^{2}$, while angular acceleration $\alpha = d\omega/dt$ has units of $\rm \text{radian}/s^2$.
Centripetal acceleration is the component of the acceleration vector that's perpendicular to the velocity, and responsible for changing the direction of the motion.  The component of acceleration parallel (or antiparallel) to the velocity vector, $a_\parallel$, changes the speed but not the direction.  If you're moving in a circle, you can prove pretty easily that $a_\parallel = \alpha R$ relates the angular acceleration $\alpha$ to the tangential acceleration $a_\parallel$.  So $a_\parallel$ and $a_c$ are two orthogonal components of the vector acceleration.
A: No, these are not the same. Say we have an object moving in the cartesian plane. We can parameterize it's motion using polar coordinates: 
$$r\langle\cos \theta, \sin \theta\rangle$$
To be clear, $r$ and $\theta$ should be thought of as functions of time. It would be more accurate to write $r(t)$ and $\theta(t)$, but this would take up too much space. 
Consider the velocity of the object:
$$\frac{d}{dt}\big[r\langle\cos \theta, \sin \theta\rangle\big]=\dot{r}\langle\cos\theta,\sin\theta\rangle+r\omega\langle-\sin\theta,\cos\theta\rangle$$
Where $\omega = \dot{\theta}$. Usually, $\omega$ is referred to as the object's angular velocity. Notice that the component of the velocity vector that is tangent to the object's motion has magnitude $r\omega$. This is referred to as tangential velocity. 
Now, consider the acceleration of the object: 
$$\frac{d}{dt}\big[\dot{r}\langle\cos\theta,\sin\theta\rangle+r\omega\langle-\sin\theta,\cos\theta\rangle\big]$$$$=\ddot{r}\langle\cos\theta,\sin\theta\rangle+\dot{r}\omega\langle-\sin\theta,\cos\theta\rangle+(\dot{r}\omega+r\alpha)\langle-\sin\theta,\cos\theta\rangle-r\omega^2\langle\cos\theta,\sin\theta\rangle$$
Where $\alpha=\dot{\omega}=\ddot{\theta}$. Usually, $\alpha$ is referred to as the object's angular acceleration. Let's simplify the above expression: 
$$\left(\ddot{r}-r\omega^2\right)\langle\cos\theta,\sin\theta\rangle+\left(2\dot{r}\omega+r\alpha\right)\langle-\sin\theta,\cos\theta\rangle$$
The component of the acceleration vector perpendicular to the object's motion is called its centripetal acceleration. Notice it has magnitude $\ddot{r}-r\omega^2$.
The component of the acceleration vector tangent to the object's motion is called its tangential acceleration. Notice it has magnitude $2\dot{r}\omega+r\alpha$. If we called the object's angular momentum $L$, then the tangential acceleration is also equal to the following (assuming that the object's mass does not change):
$$\dot{L}\cdot\frac{1}{rm}$$
A: They're not the same, but you can calculate the angular momentum with respect to orbital momentum (centripetal acceleration) and visa versa if you know the rigidness of the system and the radius to the point you are measuring. Any sort of angular momentum always involves some level of centripetal acceleration but centripetal acceleration doesn't always involve angular acceleration because in rigid systems coordinate transformations aren't subject to the same definitions as non-rigid systems for whatever reason.
