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

[wrobel]

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

 

[wrobel]

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

 

[vanhees71]

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
$$\dot{\vec{e}}_k' = -\vec{e}_l' \epsilon_{lkm} \omega_m',$$
leading to
$$\dot{\vec{V}}=\vec{e}_k' \dot{V}_k' -\vec{e}_l' \epsilon_{lkm} \omega_m' V_k'$$
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.

 

[wrobel]

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

 

[A.T.]

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.

 

[vanhees71]

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!

 

[etotheipi]

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?