Understanding the relationship between the angular velocity and angular acceleration vectors

Question

In this question the people who answered helped me validate my understanding of the $\vec\omega$ vector, the angular velocity of a particle or a rigid object.

I would now like to add the angular acceleration $\vec\alpha$ into the mix, and validate my understanding.

This question assumes that a rigid body, or a particle, is rotating in any arbitrary way around some axis. It may also be moving translationally.

It is now my understanding that $\vec\omega$ is always parallel to the current plane of rotation. So if we define the z axis to be the axis of rotation at all points in time (which my textbook seems to do), $\vec\omega$ only has a component in the z axis. The two other components are always 0.

Thus, $\vec\omega$ can always be written like so for any point in time $t$:

$\vec\omega(t) = (0, 0, \omega_z(t))$

It follows that the angular acceleration $\vec\alpha$, which is the time derivative of $\vec\omega$ can always be described as:

$\vec\alpha(t) = d(\vec\omega)/dt = (0, 0, d(\omega_z)/dt) = (0, 0, \alpha_z(t))$

Meaning - $\vec\alpha$ also has a component only in the z direction.

All of the above implies that for any point in time $t$ the following holds:

1. $\omega$ (the magnitude of $\vec\omega$) = $\omega_z$
2. $\alpha$ (the magnitude of $\vec\omega$) = $\alpha_z$

So my questions would be:

1. Is my understanding so far accurate? I realize that this is basic stuff - and so I would like to build confidence regarding this

2. My textbook often assumes that the derivative of $\omega$ is $\alpha$, and that $\int\alpha dt = \omega$ (note - these are magnitudes, not vectors). Is this always correct? We know that the derivative of a magnitude of a vector isn't necessarily equal to the magnitude of the derivative. However - it seems that all of the above implies that this relationship does hold in the case of $\omega$ and $\alpha$. Am I right?

Answer 1

So my understanding is that you are always finding the direction of the angular velocity and calling that the $z$-direction, such that $\boldsymbol{\omega} = \omega \boldsymbol{k}$ where $\boldsymbol{k}$ is the unit vector in the $z$-direction. The important thing to note is that this $z$-axis rotates in space for a general motion and also rotates with respect to an observer that is rotating with the rigid body. I point this out because in some descriptions of rigid body motion, axes that are attached to and rotating with the rigid body are used. This is not that case. In this case the angular acceleration has to include both the time derivative of the magnitude of the angular velocity, and the time derivative of $\boldsymbol{k}$.

$$\boldsymbol{\alpha}=\frac{d\boldsymbol{\omega}}{dt}=\frac{d\omega}{dt}\boldsymbol{k}+\omega\frac{d\boldsymbol{k}}{dt}$$

Note that if $\boldsymbol{k}$ were attached to the rigid body then we would have $\frac{d\boldsymbol{k}}{dt}=\boldsymbol{\omega}\times\boldsymbol{k}$. But again, $\boldsymbol{k}$ is not attached to the rigid body in your description except for simple cases where there is some reason that axis is fixed, e.g. 2-D problems.

Answer 2

Your textbook is starting you with the 2 dimensional case but using 3 dimensional notation to prepare you for 3 dimensional rotations. In x-y plane rotation, $\vec \omega$ and $\vec \alpha$ are always in the same direction, as you toiced. If $\vec \omega$ were not in the z direction, then the object would be rotating out of the x-y plane. If $\vec \alpha$ was not in the z direction, then $\omega$ would, over time, move off of the z axis, and once again you would have an object rotating out of the x-y plane.

Free yourself from the x-y plane, and they no longer need to be in the same direction.

The calculus is correct, however. Angular acceleration is defined to be the derivative of angular velocity with respect to time ($\frac{d\omega}{dt}$). The only change from what you're being taught right now is that those will be vector quantities, not just magnitudes: $\vec \alpha=\frac{d\vec\omega}{dt}$. You can see why this simplifies to simply integrating magnitudes in the case where $\vec \omega$ and $\vec \alpha$ are in the same direction by decomposing them into a scalar magnitude $\omega$ and a unit vector $\hat \omega$: $\vec \omega = \omega \hat \omega$ and correspondingly $\vec \alpha = \alpha \hat \alpha$. When you do the integration or differentiation in this way, you see that $\hat \omega$ or $\hat \alpha$ is nothing more than a constant, and gets pulled out of the equation. This leaves you with a calculus equation containing only the scalars $\omega$ and $\alpha$, just like your book is doing.