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In Goldstein it is said "It is intuitively obvious that the rotation angle of a rigid body displacement, as also the instantaneous angular velocity vector, is independent of the choice of origin of the body system of axis."

Unfortunately, not intuitively obvious for me.

Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference:

$$ \left(\frac{d\mathbf r}{dt}\right)_s = \left(\frac{d\mathbf r}{dt}\right)_r +\mathbf \omega\times\mathbf r$$

where $s$ denotes the space fixed frame and $r$ denotes the rotating frame.

How can $\omega$ given as vector components in a space-fixed frame, fully describe the rotation of of a rotating frame whose axis are fixed in, say, a rigid body? It seems to me that the choice of origin of the body fixed axis must also be included and is crucial to determining the subsequent motion. The origin dictates which point(s) the axis of rotation goes through. If I rotate the rigid body about origin point $P_1$ with angular momentum $\omega_1$, the motion will certainly be different than the rotation about origin point $P_2$ also with $\omega_1$. I've gone through the proof on why the angular velocity vector is the same in either choice of body origins, but still remain confused. Just about the only thing I can reconcile is that a given the same $\omega$ and two different body origins, the rotational orientation of a rigid body will be the same, but the body will be translated very differently from the origin of the space-fixed axis. Yet this leads me to think it would complicate the nice separation of translation and rotational motion we seek when solving equations of motion.

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Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference:

$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$

where $s$ denotes the space fixed frame and $r$ denotes the rotating frame.

That is the relevant equation for the instantaneous co-moving inertial frame. What about a non-comoving space frame? Generalizing this to a space frame in which the origin of the rotating frame is moving results in

$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r_0}{dt}\right)_s + \left(\frac{d(\boldsymbol r - \boldsymbol r_0)}{dt}\right)_r +\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_0)$$

where $\boldsymbol r_0$ is the displacement vector from the origin of the space frame to the origin of the rotating frame.

Suppose you use some other point $\boldsymbol r_1$ that is fixed from the perspective of the rotation frame (i.e., $\left(\frac{d(\boldsymbol r_1-\boldsymbol r_0)}{dt}\right)_r \equiv 0$. Go through the math (an exercise I'll leave up to you) and you'll find that

$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r_1}{dt}\right)_s + \left(\frac{d(\boldsymbol r - \boldsymbol r_1)}{dt}\right)_r +\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_1)$$

In other words, the angular velocity $\boldsymbol \omega$ is independent of the choice of origin.

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  • $\begingroup$ By co-moving space frame, you mean that the space frame origin and rotating frame origin coincide at all times, correct? $\endgroup$ – Lone Wolf Jul 25 '15 at 15:26
  • $\begingroup$ @LoneWolf - No. What if the object in question is accelerating? At any point in time, there exists an inertial frame of reference that is instantaneously coaligned with the body frame, has the same origin as that of the body frame, and in which the instantaneous velocity of the origin of the body frame is zero. This is the instantaneous co-moving inertial frame. $\endgroup$ – David Hammen Jul 25 '15 at 15:59
  • $\begingroup$ If the object had translational acceleration and a body axis was fixed at a single point within the object, it sounds like the co-moving inertial frame must also move with the same acceleration (origins coincide at all times) in order for the instantaneous velocity of the body frame to be zero relative to the co-moving inertial space frame. Right? How come we call such a co-moving space frame an "inertial" frame when it clearly accelerates relative to a non-comoving inertial space frame? $\endgroup$ – Lone Wolf Jul 25 '15 at 16:21
  • $\begingroup$ @Lone Wolf: "If the object had translational acceleration and a body axis was fixed at a single point within the object, it sounds like the co-moving inertial frame must also move with the same acceleration (origins coincide at all times)" -- Such a frame/system would be (permanently) co-moving, but necessarily not inertial. However, in a flat region, it is possible to identify a family of instantaneously co-moving inertial systems, where the members of any one such system determine zero speed of the single accelerating material point "within the object" at only just one event. $\endgroup$ – user12262 Oct 22 '15 at 5:47

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