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I have this paragraph in a book:

The vectors $T,U$ and $V=T\times U$ define a moving frame along the path $X$. Let $\Omega$ denote the angular velocity vector describing the rate of rotation of the frame with respect to arclength $s$ so, that $T'=\Omega\times T$, $U'=\Omega\times U$ and $V'=\Omega\times V$. Let $\omega_1,\omega_2,\omega_3$ be the components of $\Omega$ referred to the moving frame, i.e. $\Omega=\omega_1T+\omega_2U+\omega_3V$. Then $\omega_1$ represents the angular rate at which $U$ revolves about $X$.

I understand from wikipedia that the angular velocity vector for a frame is the angular velocity for each vector that composes the frame, but I don't understand how they calculate the angular velocity for each one.

What I understood was that the angular velocity vector of $r$ is $\frac{r\times r'}{|r|^2}$, where the dividend is the cross product of the vector $r$ with its derivative, and the divisor is the square of the norm of $r$. But then why if I will calculate the angular velocity of each vector $\Omega$ is a $3$-dimensional vector if I will get three vectors of dimension $3$?

P.S. I will be really thankful if your explanation is as clear as possible, since I'm learning this by myself.

EDIT:

To make it clear, I don't get how is obtained in a basic way the angular velocity of a frame and I want to understand (if it's possible with an example) how is obtained for a given frame (and if it's related, for only a vector too). What I don't understand from wikipedia is the section of Particle in three dimensions and the part of Spin angular from a frame, because the paragraph I mention states about ONE vector but what I understand from the wikipedia article is that I must calculate THREE vectors, so I need an explanation.

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  • $\begingroup$ "But then why if I will calculate the angular velocity of each vector Ω is a 3-dimensional vector if I will get three vectors of dimension 3?" - Are you asking how come Ω is 3D if you get 3 different vectors (one for each basis vector of the frame)? (Hint - they are all the same.) $\endgroup$ Commented Oct 2, 2020 at 17:20
  • $\begingroup$ Well, yes, it's part of my question. Basically I want to understand how does the angular velocity vector is obtained and why if $\Omega$ is the result, in the wikipedia description I get that I need to calculate three vectors that give me three diferent entries. $\endgroup$
    – iam_agf
    Commented Oct 2, 2020 at 17:46
  • $\begingroup$ Could you just update your question to cite the Wikipedia paragraph that made you confused about that? I think you know what you're referring to, and I think you sort of understood it backwards, but I'm not sure. Including that might help make the answers better. $\endgroup$ Commented Oct 2, 2020 at 18:02
  • $\begingroup$ I added the two paragraphs I just don't understand $\endgroup$
    – iam_agf
    Commented Oct 2, 2020 at 18:10

1 Answer 1

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The formula $${\boldsymbol \omega}\stackrel{?}{=} \frac{{\bf r}\times \dot {\bf r}}{|{\bf r}|^2} $$ is only true if the angular velocity ${\boldsymbol \omega}$ is pependicular to ${\bf r}$. There was another question involving this wrong expression recently. Where did you find this misleading formula?

The correct definition of the angular velocity about some point O is $$ \dot {\bf r}= {\boldsymbol \omega}\times {\bf r}, $$ where ${\bf r}$ is the position vector measured from O.

Note added: I found the misleading formula in the Wikipedia article on angular velocity. This article contains the horribly wrong equation $$ {\boldsymbol \omega}\stackrel{?}{=} \frac{{\bf r}\times \dot {\bf r}}{|{\bf r}|^2}= {\bf e}_1\times \dot {\bf e}_1= {\bf e}_2\times \dot {\bf e}_2= {\bf e}_1\times \dot {\bf e}_2 $$ In which the last three equalities cannot possibly be true because they imply that ${\boldsymbol \omega}$ is perpendicular to all of ${\bf e}_1, {\bf e}_2,{\bf e}_3$, and so must be zero!

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  • $\begingroup$ I read the part of en.wikipedia.org/wiki/… and I understood what I wrote, so that's why I wrote that. Probably I just understood wrong. But then, from what you wrote, how do you get $\omega$ from $r$ and $\dot {r}$? You need to obtain the vector that transforms $r$ to $\dot{r}$ throught the cross product? $\endgroup$
    – iam_agf
    Commented Oct 2, 2020 at 17:43
  • $\begingroup$ There is no unique ${\boldsymbol \omega}$ from single ${\bf r}$ and its velocity $\dot {\bf r}$. You need at least three vectors. Lets take them as an orthonormal triad ${\bf e}_i$. Then one defines $\omega_{ij}= {\bf e}_i\cdot\dot {\bf e}_j$. Orthonormality says that $\omega_{ij}=-\omega_{ji}$ and then the components of the angular velocity are $\omega_i= \epsilon^{ijk}\omega_{jk}/2$. $\endgroup$
    – mike stone
    Commented Oct 2, 2020 at 18:13
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    $\begingroup$ I looked at the wikipedia article and it is just wrong for the general case that it claims to describe. I may edit it if I have time. $\endgroup$
    – mike stone
    Commented Oct 2, 2020 at 18:17
  • $\begingroup$ Why thinking about the orthonormal vectors? Also, could you mention me good references to read about this topic? $\endgroup$
    – iam_agf
    Commented Oct 2, 2020 at 18:20
  • $\begingroup$ Orthonormal because the algebra is much easier! This is described in any book on the mechanics of rigid bodies. $\endgroup$
    – mike stone
    Commented Oct 2, 2020 at 18:25

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