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.

  • $\begingroup$ I'm still somewhat new to the whole concept of angular momentum so I'm not surprised I'm missing things; but for instance, the $\vec{J}=I\vec{\omega}$ bit is something I've derived myself multiple times to make sure I feel comfortable enough with it; and yet, I don't at all feel comfortable with it yet... and it would appear that with good reason, since I'm at a loss right now! $\endgroup$
    – Shay
    Oct 3, 2021 at 9:30

1 Answer 1


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:

  1. When $r_{CM}=r_0$. In other words, we calculate angular momentum with respect to centre of mass.
  2. When $v_0=0$. In other words, we calculate the angular momentum with respect to fixed point.

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.

  • $\begingroup$ Thanks a lot for your helpful & detailed answer! What I meant when I said I'm interested in changing the frame of reference is that I'll be changing it but not the point(/axis) w.r.t which I calculate the angular momentum. That is, I make a certain (misguided, perhaps?) distinction in my mind between the origin point of my reference/coordinate frame and the point w.r.t which I calculate angular momentum; so the position vector is invariant. Is this where my error lies? That is: if I calculate ang. momentum w.r.t CoM, then must the velocity also be calculated within the CoM reference frame? $\endgroup$
    – Shay
    Oct 4, 2021 at 8:17
  • $\begingroup$ To clarify: if I calculate angular momentum with respect the CoM, then as you've shown, $J=I\omega$; but then if I change my reference frame but still calculate angular momentum with respect to the center of mass, then the velocity changes but the position of each body in the system with respect to the CoM remains unchanged. At the same time, we're still calculating with respect to the center of mass, and so $J=I\omega$ should still hold. This is the bit I'm confused about. $\endgroup$
    – Shay
    Oct 4, 2021 at 8:20
  • 1
    $\begingroup$ Yes, in non-relativistic mechanics the position of the CoM is the same in all frames, Thus $J_{CM}$ will be the same in all inertial frames. But importantly, when we calculate $J_CM$ we calculate it with respect to moving CoM with whatever the velocity it has in the given FoR. It's not a fixed point coinciding with the CoM at the given time. In other words, $J_CM$ is the angular momentum in the CoM FoR with respect to CoM. $\endgroup$ Oct 5, 2021 at 11:17
  • $\begingroup$ Thank you very much :) This definitely answers my question! $\endgroup$
    – Shay
    Oct 7, 2021 at 7:36

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