Timeline for Velocity in a turning reference frame
Current License: CC BY-SA 4.0
12 events
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| Sep 6, 2023 at 21:20 | comment | added | EE18 | (5) Last question (sorry for all these). You write "It must be noticed that the rotation operator $\;\Bbb{S}(t)\;$ depends generally on time as all other quantities $\mathbf{R},\mathbf{U},\mathbf{A},\mathbf{r},\mathbf{u},\mathbf{a}, \ldots$ ". I am confused for, if this is so, doesn't this mean that the angular velocity $\omega$ of the second frame with respect to the first will depend on which vector $\mathbf{R}$ we're interested in? That is, on which particle we look at? | |
| Sep 6, 2023 at 21:18 | comment | added | EE18 | Regarding my question 3, I think I may understand but perhaps you can clarify. In your equation (01), it's important to note that the vectors on each side of the equality are from a fixed origin, either $O$ or $O'$, so we can interpret the vector used to describe the given quantity in the given frame as simply rotated from one another? (My initial concern was that we were measuring from different origins, and we would therefore not have an orthogonal transformation preserving norms. I wasn't reading your notation closely enough.) | |
| Sep 6, 2023 at 21:06 | comment | added | EE18 | Regarding my question 4, it seems like from this (physics.stackexchange.com/questions/757624/…) answer, one actually defines $\omega$ as the angular velocity based on this manipulation? | |
| Sep 6, 2023 at 20:56 | comment | added | EE18 | (4) Below (14), you write "The vector $\;\boldsymbol{\Omega}\;$ is nothing else than the instantaneous angular velocity of the system $\;O'x'y'z'\;$ relative to the system $\;Oxyz$." Can I ask why this should be so? That is, I can see that we can represent the effect of the antisymmetric matrix as a cross product with some vector, but why should it be the angular velocity vector of the other frame specifically? If you have a reference here I would be glad to read it. | |
| Sep 6, 2023 at 20:49 | comment | added | EE18 | (3) When you say "Since the two systems are orthonormal any accented vector (upper-case $\mathbf{V}^{\prime}$ or lower-case $\mathbf{v}^{\prime}$) that is expressed with $O'x'y'z'$-coordinates may be expressed with $Oxyz$-coordinates via an orthonormal transformation", I don't follow. I am familiar with the notion of change of basis, but all this means in general is that the coordinates we use to label a given fixed vector change. Here, we seem to say that there are two different vectors -- should I understand $\Bbb{S}(t)\$ as a map between orthonormal bases of different vector spaces? | |
| Sep 6, 2023 at 20:37 | comment | added | EE18 | (2) When you write “This transformation represents also the rotation which when applied to system $Oxyz$ brings it in coincidence with $O'x'y'z'$", are you using that the active transformation which takes $Oxyz$ and brings it in coincidence with $O'x'y'z'$ is the inverse of the passive transformation which tells us how coordinates change between these? | |
| Sep 6, 2023 at 20:37 | comment | added | EE18 | Thank you so much for this lovely answer. If it’s OK, I’d like to ask a couple questions to clarify some things I didn’t understand. (1) When you write, for example, in your point 3 "of a particle with respect to $Oxyz$ expressed by coordinates of the other system $O'x'y'z'$", does this mean the vector (as expanded in the basis of $O'x'y'z'$) from the origin $O$ (of $Oxyz$) to the particle? | |
| Jul 12, 2018 at 22:05 | history | edited | Voulkos | CC BY-SA 4.0 |
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| Apr 28, 2016 at 21:39 | history | edited | Voulkos | CC BY-SA 3.0 |
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| Apr 28, 2016 at 7:48 | history | edited | Voulkos | CC BY-SA 3.0 |
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| Apr 26, 2016 at 20:46 | history | edited | Voulkos | CC BY-SA 3.0 |
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| Apr 26, 2016 at 20:38 | history | answered | Voulkos | CC BY-SA 3.0 |