Angular position vector?

Question

I'm a mathematician, so I like my angular velocities to be vectors. It makes my angular momenta and torques vectors as well, and so they have nice operations I can do on them.

Because of that, I pick my definition of $\omega$ to be $\frac{\vec{r} \times \vec{v}}{\vec{r} \cdot \vec{r}}$, instead of $\frac{d\theta}{dt}$.

But if I use that definition, what is $\theta$ now? I tried to integrate $\omega$ with respect to $t$, and all I'm getting is $\int \frac{\vec{r}}{\vec{r} \cdot \vec{r}} \times d\vec{r}$, and I can't go any further.

Is there some way of interpreting $\theta$ so that I don't have to use this stubborn integral as a definition?

EDIT: I've worked out some examples, and it's path dependent. So it won't simplify past an integral. But (at least in the 2D case), it works out to be $2\pi$ times the winding number around $0$. I'm having trouble extending it to three though.

Answer 1

You won't be able to understand this solely by manipulating symbols algebraically on a page.

To understand what's going on here, you need to actually draw a rotating coordinate system (presumably attached to a rotating body) superimposed over a fixed 3D inertial coordinate system, and then interpret $\theta$ itself as a vector. In this interpretation, each of the three components of $\overrightarrow{\theta}$ represents the infinitesimal angular change experienced by the two axes perpendicular to that component. For example, if $(x_{1},x_{2},x_{3})$ represents one coordinate system, and $(x^{'}_{1},x^{'}_{2},x^{'}_{3})$ represents a coordinate system which has been rotated by an infinitesimal amount, then $d\theta_{3}$ represents the angular rotation about the $x_{3}$ axis required in order to transform $x_{1} \rightarrow x^{'}_{1}$ and $x_{2} \rightarrow x^{'}_{2}$.

Performing a truly complete derivation is out of scope for a venue such as this, but fortunately for you, most dedicated textbooks on classical mechanics will have a discussion of exactly this issue, including a sketch of $\overrightarrow{\theta}$ as it relates to the two coordinate systems, somewhere within the first chapter or two. In my copy of Fetter and Walecka's "Theoretical Mechanics of Particles and Continua" for example, (which is still in print) the relevant material can be found in Chapter 2, "Accelerated Coordinate Systems", Section 7, "Infinitesimal Rotations". And BTW, I would definitely recommend attempting to look this up in a text which is dedicated specifically to classical mechanics; a generic text such as you would use in a first year introductory physics class won't cover it, as it's a little too advanced for that level.

Answer 2

In a general sense $\vec{\omega}$ is NOT $\frac{{\rm d}\theta}{{\rm d}t}$ since it is a vector quantity and rotations are not. When rotation is constrained along a fixed axis, then the two are equivalent. You could say $\vec{\omega} = \frac{{\rm d}\theta}{{\rm d}t} \hat{z}(t)$ but all you are doing is separating magnitude and direction. Unfortunately you cannot integrate rotational velocity to get angles.

It works the other way around. Given two coordinate frames, defined by a relative 3×3 rotation matrix $R={\rm Rot}(\hat{z},\theta)$, then the coordinate frame orientations are

$$ \mbox{orientation of (1)} = E_1 \\ \mbox{orientation of (2)} =E_2 = E_1 R $$

The derivative of any (non-constant) vector defined on a rotating frame is done with the operator $$\frac{{\rm d}}{{\rm d}t} = \frac{\partial }{\partial t} + \vec{\omega}\times$$

Since the frame coordinates (columns of $E_1$) are fixed to the frame, then $\frac{\partial E_1}{\partial t}=0$ and the derivative of $E_1$ is

$$ \dot{E_1} = \frac{\partial }{\partial t} E_1 + \vec{\omega}_1\times E_1 =\vec{\omega}_1\times E_1 $$

Similarly

$$ \dot{E_2} = \vec{\omega}_2 \times E_2 \\ \dot{R} = \dot\theta \hat{z}\times R $$

Now the derivative of $E_2$ will give us the kinematics of the rotating frames

$$ \begin{aligned} \dot{E_2} &= \frac{{\rm d}}{{\rm d}t} ( E_1 R) \\ \vec{\omega}_2 \times E_2 &= \dot{E_1} R + E_1 \dot{R} \\ & = \vec{\omega}_1\times E_1 R + E_1 \left( \dot\theta \hat{z}\times R \right) \\ & = \vec{\omega}_1\times E_2 + \left(E_1 \hat{z} \dot\theta\right)\times \left(E_1 R \right) \\ & = \vec{\omega}_1\times E_2 + \left(E_1 \hat{z} \dot\theta\right)\times E_2 \\ \vec{\omega}_2 \times E_2 &= \left(\vec{\omega}_1 + E_1 \hat{z} \dot\theta\right)\times E_2 \end{aligned} $$

$$\boxed{ \vec{\omega}_2 = \vec{\omega}_1 + E_1 \hat{z} \dot\theta } $$

The same methodology can be applied for multiple consequent rotations but the math gets more complex. So rotational velocities are defined from the sequence of (n) joint angles $\theta_1, \theta_2, \ldots, \theta_n$, their axes $\hat{z}_1,\hat{z}_2,\ldots,\hat{z}_n$ and the angle speeds $\dot\theta_1, \dot\theta_2, \ldots, \dot\theta_n$. Combined you have

$$\vec{\omega}_n = \hat{z}_1 \dot\theta_1 + R_1 \left(\hat{z}_2 \dot\theta_2 + R_2 \left( \ldots R_{n-1} \hat{z}_n \dot{\theta}_n\right)\right) $$

A must read is the vector notation of rigid body motion, and then the spatial notation using screw theory.

For 3D bodies $\vec{\omega}$ must be considered together with the linear velocity $\vec{v}_A$ measured at some point A on the rigid body. This give us the motion screw of the rigid body.