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

MI.In summary, angular velocity and angular acceleration are both invariant under translations of the origin, but the moment of inertia tensor is not. The total angular momentum can be divided into spin and orbital components, with the total angular velocity being the sum of the spin and orbital angular velocities. The concept of angular acceleration only applies if the origin is fixed on the rigid body. The total angular momentum of a system can be divided among its components, as seen in the case of the Solar System.
  • #36
vanhees71 said:
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
 
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  • #37
in my version (1977) it is formula (2.24) at page 27
 
  • #38
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.
 
  • #39
have you taken into account that in your last formula ##V'=\omega'## for the case under consideration?
 
  • #40
etotheipi said:
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.
 
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  • #41
A.T. said:
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.
 
  • #42
wrobel said:
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!
 
  • #43
wrobel said:
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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