In classical mechanics:
Logically, it appears to me that if I draw a mark on a ball and let it roll, the amount of time that will pass before the mark reaches the same position (in terms of angles: for instance, the very top of the ball) should be constant across different reference frames. Of course, this isn't only true for $2\pi$ radians specifically, but for any angle I choose. This goes to say that the angular velocity of a body should be constant across reference frames, if I'm not making any silly, false assumptions here.
To take a more concrete example: say I have a sphere rolling on a catwalk. Its linear velocity may be different between the floor's frame of reference and the catwalk's; but it should still complete a rotation about its own center within the exact same amount of time: an observer standing on the floor and an observer standing on the catwalk will agree on the time this takes. Hence: $\omega$ should be constant across reference frames.
However… I've been attempting to "prove" this using angular momentum and the relation that claims that, at least about the axis of rotation of such an object, we have: $\vec{J}=I\vec{\omega}$ (in which $I$ is the moment of inertia with respect to its axis of rotation).
This is where things got wonky for me:
If I look at even a simple case of, say, a particle rotating about some center. Then its angular momentum is $\vec{J}=R\hat{r}\times m\vec{v}$. However, while $R$ and $m$ remain consistent across reference frames, $\vec{v}$ does not. This means that the angular momentum isn't constant across various reference frames even about the same point — which is already weird to me. But to make things worse, we have that it is equal to $I\vec{\omega}$. $I$ is the same since we're still measuring it about the same point. This means that $\vec{\omega}$ is supposedly changing in magnitude across different reference frames. This… feels quite off (at least in classical mechanics), and so I'm sure I'm making some silly false assumption here. Would appreciate any and all help with spotting it.
Best Answer
A. Angular momentum “knows” nothing about the axis of rotation. Angular momentum is a physical quantity $J=\int (r-r_0)\times v dm$, where $r_0$ is the coordinates of the centre in respect to which you calculate it. It's easy to see that angular momentum changes when you just change the centre, you don't even need to change the velocity of the reference frame.
B. Now, let's speak about formula $J=I\omega$. In case of flat motion: $$ J = \int (r-r_0)\times v dm=\int (r-r_0)\times \big(v_0+\omega\times(r-r_0)\big) dm =\\ \int (r-r_0)\times v_0 dm + \omega\int (r-r_0)^2 dm = M(r_{CM}-r_0)\times v_0+I\omega. $$ We can see that $J=I\omega$ works only in two cases:
When you change the frame of reference, the centre is no longer fixed and it wasn't CoM in the first place, so the formula is not valid.
C. Now let's speak about what is $\omega$? If you are given a particle moving with the speed $v$, what is its angular velocity? Of course, the answer will depend on the centre you choose (same arguments as with angular momentum applies). However, if you have many points coming from a rigid body like wheel, then you have an obvious choice of a centre: the point with zero velocity. Then the velocity of every point can be calculated as $v=\omega\times(r-r_0)$. Now if you change the frame of reference, the old centre will gain the speed and can no longer be considered a natural centre. However, in case of flat motion we can always find such $\Delta r$, so $$ v'=v+v_{FR}=\omega\times(r-r_0) + v_{FR} = \omega\times(r-(r_0+\Delta r)) $$ so the velocity field can be considered rotational with the same angular velocity around some new centre $r_0+\Delta r$. I leave the proof of this fact to you.