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Let $\vec \omega = (\omega_1, \omega_2, \omega_3)$ be the angular velocity of a rigid body with respect to the body frame, where the body frame is right-handed orthonormal.

I have gathered 2 definitions of $\vec \omega$ from different sources and I am confused at how they connect to one another. One is that the rigid body rotates with $\vec \omega$ through its Center of Mass at rate $abs(\vec \omega)$. The other is that each component of $\vec \omega$ represents the rate at which the rigid body rotates about that particular basis axis of the body frame.

Does this mean we can somehow add the 3 rotations (which are about different axes) and get an equivalent rotation about some other (single) axis?

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Does this mean we can somehow add the 3 rotations (which are about different axes) and get an equivalent rotation about some other (single) axis?

Yes. It's one of many consequences of Euler's rotation theorem. This neat little trick of finding the Euler rotation axis works in three dimensional space, and in three dimensional space only. The Lie group SO(3) is a very special space.

A more generic way of looking at rotation is that rotation in any Euclidean space comprises a combinations of rotations parallel to a two dimensional plane rather than about an axis. There's only one plane in two dimensional space, so rotation in 2D space requires but one parameter. There are three pairs of planes of rotation in three dimensional space (the YZ, ZX, and XY planes), so rotation in 3D space requires three parameters. You can think of those planar rotations as being about an axis (with an arbitrary decision as to what qualifies as positive or negative rotation). There are six planes of rotation in four dimensional space, so rotations in 4D space requires six parameters. This can result in something very weird, something you don't see in 2D or 3D space, a Clifford rotation.


Source: http://eusebeia.dyndns.org/4d/vis/10-rot-1

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  • $\begingroup$ thanks. What I dont get is how Euler's angles connect with w(t). Say w(t)=[w1(t),w2(t),w3(t)] and each component represents how w(t) change along that direction (of the basis axis) and they certainly do NOT tell us about the rotation around those axes. That is, the rotation is always around w(t), which itself is moving around. But then Euler's angles come in and w now depends on 3 other angles (a,b,c) which are themselves some other rotations(so functions of time too). Why then does w(t) = w(a,b,c)? Conceptually they seem completely irrelevant to me. $\endgroup$
    – Martin Cheung
    Jan 13, 2015 at 15:33
  • $\begingroup$ @MartinCheung I am afraid that none of these answers is correct. I am not sure that the three projections of $\vec \omega$ mean that there is a rotation with $\omega_x$ around the axis $x$, with $\omega_y$ around the axis $y$, and with $\omega_z$ around the axis $z$. $\endgroup$
    – Sofia
    Jan 13, 2015 at 16:06
  • $\begingroup$ @Sofia -- Your comment is completely wrong, at least in three dimensional space. One can look at rotations in 3D space as intrinsic or extrinsic. The intrinsic point of view leads to Euler angles, Tait–Bryan angles, Cardan angles, and other evils. The extrinsic point of view is also evil, but it is a lesser evil. This point of view that results in viewing angular velocity as a vector (better, a pseudovector), with distinct components in each orthogonal direction. $\endgroup$
    – David Hammen
    Jan 13, 2015 at 17:20
  • $\begingroup$ @MartinCheung - You have to go the math to see why this works in 3D space (and it most certainly does). Humans can and do build devices that detect those components of angular velocity. We wouldn't be able to operate spacecraft if those devices didn't exist. $\endgroup$
    – David Hammen
    Jan 13, 2015 at 17:24
  • $\begingroup$ @DavidHammen : Is that true that performing a rotation by an angle $\omega_x$ around the axis $x$, then by an angle $\omega_y$ around the axis $y$, and then by an angle $\omega_z$ around the axis $z$, we get in fact a rotation by $|\vec \omega|$ around the axis $\vec \omega$ ? Is that true? With the Euler angles we work differently, and as far as I know, one doesn't get the same result if one performs the rotations in a different order. $\endgroup$
    – Sofia
    Jan 13, 2015 at 18:40

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