# Radial/centripetal vs. tangential/linear vs. angular acceleration [duplicate]

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.

• Possible duplicate of Types of circular acceleration? – sammy gerbil Feb 22 '18 at 2:13
• -1 No research effort. See related questions on right. – sammy gerbil Feb 22 '18 at 2:15
• @sammygerbil, the answers to that question do not really answer mine because, for example, I do not really understand the significance of linear acceleration. I understand what it is, as in what it is defined as, but I do not understand what it means. – Daniel Feb 22 '18 at 2:34
• Please see related questions on right. Radial acceleration is $\ddot r - r\dot\theta^2$. Centripetal acceleration is $-r\dot\theta^2$. Tangential acceleration is $r\ddot \theta$. Linear acceleration is the vector sum of radial and tangential accelerations. – sammy gerbil Feb 22 '18 at 2:52
• @sammygerbil, I do not know that notation, and I am looking for a conceptual answer like the one by the_photon, not a mathematical one. – Daniel Feb 22 '18 at 4:04

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.

1. 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.
2. 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.
3. 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.

• How can there be tangential acceleration but no angular acceleration? Mathematically, $a=\alpha r$, so if $\alpha=0$, then how can $a$ be anything other than $0$ as well? – Daniel Feb 22 '18 at 4:05
• Sorry, made an editing mistake in my response. I've now changed "notice this tangential acceleration" to "notice this centripetal acceleration". The point you just brought up was quite valid. – the_photon Feb 22 '18 at 4:22

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

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 $a$.

• The answer to a related question that asked to find the linear acceleration of a rigid body rotating with constant angular acceleration indeed could be got only when we solved for the net acceleration. But that confused me alot cause when we derive the tangential velocity 'v', we say that it is the linear velocity of a "say" point on the rigid body. – suiz Apr 28 '18 at 11:18
• Based on which I assumed that the linear acceleration is its derivative but it's wrong. It is confusing why linear is net acceleration and not tangential. – suiz Apr 28 '18 at 11:18
• You have mentioned that net acceleration is the linear acceleration but not mentioned why. Could you please help me to understand this? – suiz Apr 28 '18 at 11:22
• @suiz Linear acceleration is just a term for any acceleration along a path (also along a curved path). Anything measured in meters-per-second-squared is called linear acceleration. It is just a name. So when you mention "net acceleration", then of course it is a linear acceleration, because it is measured in meters-per-second-squared. Many other accelerations are linear as well - for instance, both the tangential and the radial accelerations are also linear. Because they are measured in meters-per-second-squared. – Steeven Apr 28 '18 at 14:04