Radial/centripetal vs. tangential/linear vs. angular acceleration I am thoroughly confused as to how exactly radial, tangential, and linear accelerations of points on a rotating wheel relate to each other. Every time I think I know what they mean individually, I find a source that tells me that no, that is wrong. I was actually writing down in this post what I think is correct, only to realize that I was writing the same thing for two obviously different concepts.
First, is radial acceleration the same as centripetal and tangential the same as linear?
Second, how do these different accelerations relate to each other on a spinning disk? I was reading through my textbook, and I think I understand angular velocity perfectly in this context. However, I do not understand, for example:

If a disk rotates at a constant angular velocity, why does a point on the rim have radial but not tangential acceleration? Mathematically, this makes sense (since $a=r\alpha$, so if $\alpha=0$, then so must $a$), but not conceptually. Does it have radial/centripetal acceleration simply because it is still exhibiting circular motion? Would the linear acceleration change––is there any linear acceleration?

Basically, I do not really understand how radial, tangential, and linear accelerations relate to each other? I have just a big salad of greek letters in my head.
 A: I think I understand your confusion.  It might be worth pointing out that when it comes to points on the edges of rotating disks, these points can have many different kinds of acceleration.


*

*Rotational or angular acceleration.  The point was rotating at 25 rev/min, and has increased to 45 rev/min over the last 18 seconds.  This is rotational acceleration.

*Centripetal acceleration (also known as radial acceleration) - if the "point" on the disk has mass then there has to be some kind of force that points to the center of the disk "keeping" the point in its circular motion.  And any time you have a force of any kind acting on a mass, there is an acceleration.

*Tangential acceleration:  You state in your post that this makes mathematical sense, but not conceptual sense.  I basically feel the same way.  However, if you were viewing a rotating point "edge on" you would see the point oscillating back and forth, and there's a certain "acceleration" to that oscillation.  Furthermore, you could move around and look at the rotating point "edge on" from some other axis, and continue to see this "acceleration".  Putting these two edge-on-view accelerations together by summing them together as vectors gives a rather peculiar acceleration that we call tangential acceleration.  This may offend your ordinary sense of acceleration (as it does mine) -- one might have to just place more faith in the math than in your own instinct about what acceleration is.


Note that with this "centripetal acceleration", you can still have centripetal acceleration even if the point is rotating at a constant angular speed (e.g. 32 revolutions/minute, with no angular acceleration).
Lastly, you mention linear acceleration.  The way I think of linear acceleration, there really is none for an object going in a circle.  However, if you're one of those people who think that tangential acceleration is a peculiar form of linear acceleration, then I suppose there is.  In my mind, this is somewhat of a "if the tree falls in the forest . . ." type of semantic dilemmas.
I'll stop my rant here to avoid belaboring the point.  I hope that this has helped.
A: 
First, is radial acceleration the same as centripetal [...]

Yes.

[...] and tangential the same as linear?

Well, the tangential acceleration is a linear acceleration, yes. But so is the radial (centripetal) acceleration. "Linear motion" is a category. Two categories of motion exist:


*

*linear  (or translational) measured in metres-per-second or metres-per-second-squared etc. and

*angular (or rotational) measured in radians-per-second or radians-per-second-squared etc.



Second, how do these different accelerations relate to each other on a spinning disk?

They are perpendicular to each other. Actually, the tangential and radial accelerations are the two components of the total acceleration:
$$\vec a=\vec a_{tan} + \vec a_{rad}$$
The tangential is... tangential to the motion. The radial is perpendicular to the motion. This is how they are related. But they have no influence on each other in any way. The size of one is can be anything regardless of what the other is.

Basically, I do not really understand how radial, tangential, and linear accelerations relate to each other? I have just a big salad of greek letters in my head

As I just described above, "linear acceleration" is the overall category. Both the radial and tangential accelerations are linear. So split those words apart in your head.
And regarding Greek letters; you have not mentioned anything with Greek letters anywhere in you question (only in the quote). You have only mentioned linear but not angular terms.
Now, to give the overview, let's have a quick look at the relevant formulas:
$$a_{rad}=\frac{v^2}{r} \tag{1}$$
$$s=r\theta \tag{2}$$
$$v=r\omega \tag{3}$$
$$a=r\alpha \tag{4}$$


*

*$(1)$ describes a circular motion but only involves linear terms, namely speed $v$ and $a_{rad}$. The reason is that the $a_{rad}$ is perpendicular to $v$, as mentioned. $a_{tan}$ is parallel and speeds up the motion since it pulls it forward, but $a_{rad}$ pulls sideways and thus turns the motion. Keep turning the motion by continuously pulling perpendicular and you get a curve - keep the pull (the $a_{rad}$) constant and you soon complete that curve to a circle.


So, those two linear terms $a_{rad}$ and $v$ cause constant turning and thus rotational motion, even though no angular terms are involved. Of course that rotational motion does have some angular terms (an angular speed $\omega$, changes in angular position $\theta$, maybe angular acceleration $\alpha$ etc.), but those terms are just not involved in that formula.
But they are involved in equations $(2)$, $(3)$ and $(4)$. These are sometimes called geometric bonds, since they couple linear and angular terms together.


*

*$(2)$ couples linear position $s$ to angular position $\theta$. Think of tetherball (a tennis ball in a string swung around a pole). If the ball moves 2 metres then maybe that is a quarter of the whole round. Then those 2 metres correspond to $90^\circ$. When the string winds up and the ball comes closer while stilling swinging around the pole, then the new circle is smaller. Moving 2 metres is now more than a quarter. The angular position is now more than the $90^\circ$. The radius $r$ clearly has some influence. It turns out that the formula that combines the linear and angular positions looks like $(2)$.

*Similarly for $(3)$, where moving a certain number of metres-per-second $v$ corresponds to turning a number of degrees-(or radians)-per-second $\omega$.

*And similarly for $(4)$, where speeding up a certian number of metres-per-second every second $a$ corresponds to speeding up the turning with a number of degrees-(radians)-per-seconds every second.


I hope this clears it out. $\omega$ is the rotational version of $v$, because you can speed up linearly but you can also speed up a rotation. And $\theta$ is the rotational version of $s$ and $\alpha$ the equivalent version of $a$.
