# How can angular velocity be independent of the choice of origin?

No! The basis vectors of the non-inertial reference frame are (in general) time dependent
sure, they are time dependent. You can study this formula by the following textbook
https://www.springer.com/us/book/9783322909435

in my version (1977) it is formula (2.24) at page 27

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I have no access to this book, but it can be found in any textbook on classical mechanics, e.g., Sommerfeld vol. 1.

If ##\vec{e}_j## is the righthanded Cartesian basis of the inertial frame and ##\vec{e}_k'## the one in a rotating frame, then for any vector ##\vec{V}## you have
$$\vec{V}=\vec{e}_j V_j = \vec{e}_k' V_k',$$
and by definition the ##\vec{e}_j## are time-independent, while then ##\vec{e}_k'## are necessarily time dependent, because the non-inertial frame is rotating. Thus you have
$$\dot{\vec{V}}=\vec{e}_j \dot{V}_j=\vec{e}_k' \dot{V}_k' + \dot{\vec{e}}_k' V_k'.$$
Now you can write
$$\dot{\vec{e}}_k' = \vec{e}_l' \Omega_{lk}',$$
and the matrix transforming between ##\vec{e}_j## and ##\vec{e}_k'## is just a rotation matrix you have ##\Omega_{lk}'=-\Omega_{kl}'## and thus you can write
$$\dot{\vec{e}}_k' = -\vec{e}_l' \epsilon_{lkm} \omega_m',$$
$$\dot{\vec{V}}=\vec{e}_k' \dot{V}_k' -\vec{e}_l' \epsilon_{lkm} \omega_m' V_k'$$
or
$$\dot{\vec{V}}=\vec{e}_k' (\dot{V}_k' + \epsilon_{klm} \omega_l' V_k').$$
Thus in matrix-vector notation (with ##\underline{V}'=(V_1',V_2',V_3')^{\text{T}}##)
$$\dot{\vec{V}}=:\vec{e}_k' (\mathrm{D}_t V_k') \quad \text{with} \quad \mathrm{D}_t \underline{V}_k' = \dot{\underline{V}}_k' + \underline{\omega}' \times \underline{V}'.$$
So there is this characteristic additional term with the cross product, defining the momentaneous angular velocity of the non-inertial basis wrt. the inertial basis.

have you taken into account that in your last formula ##V'=\omega'## for the case under consideration?

Why does it not hold in this case?
If think in your exchange with @wrobel two meanings of angular velocity get mixed up again. See post #5.

etotheipi
If think in your exchange with @wrobel two meanings of angular velocity get mixed up again. See post #5.

I think that's right, it doesn't help that we use the same letter for everything 😅 ! I think others have used ##\vec{\omega}## to refer to the rigid body/spin angular velocity. I.e. the angular velocity measured wrt an inertial coordinate system of any two chosen points on a rigid body. So it could be the angular velocity of a point on the rigid body about the centre of mass, or just two other arbitrarily chosen points on the body.

This is as opposed to an orbital angular velocity of a centre of mass wrt an inertial coordinate system. But I'm not sure if this second type is too important.

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have you taken into account that in your last formula ##V'=\omega'## for the case under consideration?
Ok, I've overlooked this. Then you are right of course, because trivially ##\underline{\omega}' \times \underline{\omega}'=0##. Sorry for that!

Assume we have two coordinate frames: the frame I and the frame II, and we also have a rigid body.
Let the angular velocity of the frame II relative to the frame I be ##\boldsymbol \omega_1##
let the angular velocity of the rigid body relative to the frame II be ##\boldsymbol \omega_2##
let the angular velocity of the rigid body relative to the frame I be ##\boldsymbol \omega##

THEOREM: ##\boldsymbol \omega=\boldsymbol \omega_1+\boldsymbol \omega_2.##

Ah, okay I see now.

For the gyroscope, the axle frame rotates at ##\vec{\Omega} = \Omega_z \hat{z}## wrt the lab frame and the angular velocity of the flywheel wrt the rotating frame is ##\vec{\omega} = \omega_{x'} \hat{x'}##, so then the angular velocity of the flywheel wrt the inertial frame is ##\vec{\Omega} + \vec{\omega}##. So if we let ##I_o## be the moment of inertia matrix computed at the origin of the lab frame, the total angular momentum is $$\vec{L} = I_o(\vec{\Omega} + \vec{\omega})$$ which I'm assuming (!) is equivalent to $$\vec{L} = I_1 \vec{\Omega} + I_2\vec{\omega}$$ if ##I_1## is the moment of inertia of the COM wrt the origin of the inertial frame and ##I_2## is the moment of inertia of the body wrt the COM.

Whilst this doesn't seem to work for the Earth scenario. Assuming both rotations occur about the ##z## direction, ##\vec{\Omega} = \omega_{year} \hat {z}## and ##\vec{\omega} = \omega_{day} \hat {z}##. But we can't add these now because the second frame is translating. Is this sort of correct?

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