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The angular acceleration is an axial vector when the axis of rotation is fixed. Angular acceleration and angular velocity vectors along that axis.
Is there anybody who can explain this statement, please?

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  • $\begingroup$ Hi Hena. Are you looking for an explanation of how/why angular properties can be considered as axial vectors? $\endgroup$
    – Steeven
    Aug 25, 2020 at 8:12
  • $\begingroup$ Yes that is what I needed $\endgroup$
    – Hena Anvar
    Aug 25, 2020 at 11:34
  • $\begingroup$ I've added an answer that I hope can help to clear it out. $\endgroup$
    – Steeven
    Aug 25, 2020 at 12:24

3 Answers 3

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  • Linearly (called translation) moving backwards or forwards are two different displacements.

  • Angularly (called rotation) something can turn clockwise or counterclockwise, which are two different angular displacements.

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Directionality is clearly involved in both cases. And we typically define directional properties as vectors.

It's easy enough to call a linear directional property a vector since its direction is straight and well-defined. It is an "arrow". But how do we do it with rotation? Should it be a "curved" arrow?

Well, we could define it in that way. But we could also define the rotation in terms of its axis, since the axis is straight and well-defined (and uniquely defines the rotation direction if we employ the right-hand rule). And that is in fact what is typically done: Angular properties are defined as vectors in the way that the vector is pointing not along with the rotation (not clockwise or counterclockwise) but rather along with the axis-of-rotation.

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There are some issues with this approach, though. The angular vector has namely become a so-called pseudo-vector (as opposed to a natural vector or an Euclidian vector) for which usual vector algebra and transformations don't necessarily work. For instance, the commutative law of vector addition ("the order of addends doesn't matter") clearly breaks when you flip a book on a table, first along one sequence and then along the same sequence backwards, where the end-result isn't the same:

enter image description here

By keeping the axis fixed, though, this issue is overcome and the rotation vector works just fine and can be added to any other rotation vector about that same fixed axis. That is why you'll often see the requirement of the axis being fixed in this regard.

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Welcome to the rock and roll physics show!

In this article (good stuff for you to read, I think) you can read:

In equation form, angular acceleration is expressed as follows: α=ΔωΔt α =Δω Δt, where Δω is the change in angular velocity and Δt is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2.

So, in differential form this reads:

$$\vec{\alpha}=d{\vec{\omega}}dt$$

Now,

$$\vec{\omega}=\vec v\times \vec{r},$$

which is the exterior product of $\vec v$ and $\vec r.$

Now, as stated in your question, the acceleration vector is $\vec{\alpha}$ is a pseudovector aligned with the axis of rotation.
Inserting this the expression for $\vec{\alpha}$ above we get:

$$\vec{\alpha}=d(\vec v\times \vec{r})dt,$$

with $d(\vec v\times \vec{r})$ perpendicular to the velocity vector or the axis of rotation i.e. lying on the same line on which lie the former two.

Because $\vec{a}$ is continuously changing, we have to express is in differential form.

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The vectors representing rotational motion are chosen along the axis of rotation because that is the only direction associated with the motion of a rotating body which is constant. The right-hand-rule works well with a vector product (I don't know which came first).

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