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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?

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  • $\begingroup$ In the first paragraph in Wikipedia, you can read that what you ask for is called tangential acceleration. See: en.wikipedia.org/wiki/Angular_acceleration $\endgroup$ Apr 25, 2021 at 2:03
  • $\begingroup$ @Deschele Schilder I precisely stated in my question that tangential acceleration is the one that I don’t have any problems with, but it’s with angular acceleration. It’s fine though, I have arrived at the conclusion that the value I’m asking doesn’t have its own standard name. $\endgroup$
    – mikeeei
    Apr 25, 2021 at 11:57
  • $\begingroup$ In the article one is called angular acceleration and the other tangential acceleration. $\endgroup$ Apr 25, 2021 at 12:07
  • $\begingroup$ I don't think you understood my question. $\endgroup$
    – mikeeei
    Apr 26, 2021 at 8:33
  • $\begingroup$ I think I get the question now (I was even thinking about it in bed!). The tangential acceleration of a particle is the acceleration for r is constant. If the particle is accelerated only tangential (r constant, circular motion), then it obviously gains tangential speed. Because of this, the centripetal force is increased. So there is a change in force directed towards the center. Not a change in distance. Is it the change in centripetal acceleration you are interested in? $\endgroup$ Apr 26, 2021 at 12:15

4 Answers 4

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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.

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  • $\begingroup$ The could probably express the tangential acceleration in terms of the curvature of the path and thus obtain a "standard expression". $\endgroup$
    – Semoi
    Apr 24, 2021 at 18:41
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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.

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  • $\begingroup$ I'm sorry, but I don't understand how this answers the question. $\endgroup$
    – mikeeei
    Apr 24, 2021 at 14:43
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I think it's the angular acceleration; its formula is $\alpha = \frac{d \omega}{dt} = \frac{d^2 \theta}{dt^2}$

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    $\begingroup$ I said in my question why I couldn't simply call it that. $\endgroup$
    – mikeeei
    Apr 24, 2021 at 14:21
  • $\begingroup$ Sorry, I don't think I have quite understood your question then. You say you have a nonuniform circular motion so the radius doesn't change, does it? $\endgroup$
    – Jones
    Apr 24, 2021 at 14:31
  • $\begingroup$ The reason I can't call this angular acceleration is because that term is reserved for the time derivative of angular velocity. I want to know how the time derivative of angular speed is called. $\endgroup$
    – mikeeei
    Apr 24, 2021 at 14:35
  • $\begingroup$ Differentiating the usual tangential speed doesn't create any ambiguity but doing so with angular speed does. $\endgroup$
    – mikeeei
    Apr 24, 2021 at 14:40
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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.

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