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

• B
Another little comment.
Let $\boldsymbol e_1 \boldsymbol e_2 \boldsymbol e_3$ be a basis of an inertial frame and let
$\boldsymbol u_1\boldsymbol u_2\boldsymbol u_3$ be a basis of a body-fixed frame.
We can present $\boldsymbol \omega$ in two forms
$$\boldsymbol \omega=\omega^i\boldsymbol e_i=\Omega^i\boldsymbol u_i.$$
Introduce a notation
$$\boldsymbol {\dot\omega}=\dot\omega^i\boldsymbol e_i,\quad \frac{\delta\boldsymbol \omega}{\delta t}=\dot\Omega^i\boldsymbol u_i.$$

THEOREM. $$\boldsymbol {\dot\omega}=\frac{\delta\boldsymbol \omega}{\delta t}.$$

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Consider the Earth rotating around the Sun. The Earth rotates about its axis with angular velocity $\vec{\omega}_{1}$, and the Earth about the sun with angular velocity $\vec{\omega}_{2}$. Then it would seem the total angular velocity of the Earth is $\vec{\omega}_{1} + \vec{\omega}_{2}$.

But now consider a rod rotating about a hinge. The angular velocities of the centre of mass and that of the rod about the centre of mass are both equal, $\omega$. Their sum definitely does not equal the angular velocity of the rod about the hinge, which is also $\omega$. So what gives?

jbriggs444
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Consider the Earth rotating around the Sun. The Earth rotates about its axis with angular velocity $\vec{\omega}_{1}$, and the Earth about the sun with angular velocity $\vec{\omega}_{2}$. Then it would seem the total angular velocity of the Earth is $\vec{\omega}_{1} + \vec{\omega}_{2}$.
Why would you add those two? The Earth is not a rigid body with a fixed point anchored to the sun. Its revolution rate around the sun (or any other form of fancy translation it might undergo) is irrelevant to its rotation rate.

Possibly you are thinking of adding the rotation rate of the Earth relative to a noontime marker pointing directly at the sun (the solar day) to the similarly-directed revolution rate of the Earth about the sun (the year) to get the rotation rate of the Earth relative to a fixed background (the sidereal day).

But that's silly. If you are making up a thought experiment, why would you pretend-measure a rotation rate against a rotating reference when it is just as easy to pretend-measure a rotation rate against a non-rotating reference.
But now consider a rod rotating about a hinge. The angular velocities of the centre of mass and that of the rod about the centre of mass are both equal, $\omega$.
The rod is a rigid object. For a rigid object, the rotation rate of any pair of fixed points on the object about one another will be the same as for any other pair of fixed points. The hinge and the center of mass are one pair of fixed points. The center of mass and an arbitrary other point on the rod are another pair of fixed points. Of course, the two rotation rates will match.

Their sum definitely does not equal the angular velocity of the rod about the hinge, which is also $\omega$.
Yes, of course not. It gives double the rotation rate of the rod. $\omega + \omega = 2\omega$

A.T.
Consider the Earth rotating around the Sun. The Earth rotates about its axis with angular velocity $\vec{\omega}_{1}$, and the Earth about the sun with angular velocity $\vec{\omega}_{2}$. Then it would seem the total angular velocity of the Earth is $\vec{\omega}_{1} + \vec{\omega}_{2}$.
As @jbriggs444 points out, simply adding angular velocites around different centers is not of much use, unless you combine it with the radius vectors:

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Why would you add those two? The Earth is not a rigid body with a fixed point anchored to the sun. Its revolution rate around the sun (or any other form of fancy translation it might undergo) is irrelevant to its rotation rate.
`I agree it makes little sense, however I was reading through some lecture notes about gyroscopes and came across stuff like the following (a flywheel rotating about its axis with angular speed $\omega_{s}$, and this axle is rotating about a vertical axis with angular speed $\Omega$).

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: $\boldsymbol \omega=\boldsymbol \omega_1+\boldsymbol \omega_2.$

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So let's suppose the rod is rotating about its hinge, about the $z$ axis, with speed $\omega$ with respect to one frame. The angular velocities of the centre of mass relative to the hinge, the rod relative to the centre of mass and the rod relative to the hinge are all equal, namely $$\vec{\omega}_H = \vec{\omega}_{CM} = \vec{\omega}_R = \begin{pmatrix} 0 \\ 0 \\ \omega \end{pmatrix}$$ Why does it not hold in this case?

the rod relative to the centre of mass
you can not rotate relative a point you can rotate relative a frame

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you can not rotate relative a point you can rotate relative a frame
Is that to say the origins of the two frames must coincide?

vanhees71
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Another little comment.
Let $\boldsymbol e_1 \boldsymbol e_2 \boldsymbol e_3$ be a basis of an inertial frame and let
$\boldsymbol u_1\boldsymbol u_2\boldsymbol u_3$ be a basis of a body-fixed frame.
We can present $\boldsymbol \omega$ in two forms
$$\boldsymbol \omega=\omega^i\boldsymbol e_i=\Omega^i\boldsymbol u_i.$$
Introduce a notation
$$\boldsymbol {\dot\omega}=\dot\omega^i\boldsymbol e_i,\quad \frac{\delta\boldsymbol \omega}{\delta t}=\dot\Omega^i\boldsymbol u_i.$$

THEOREM. $$\boldsymbol {\dot\omega}=\frac{\delta\boldsymbol \omega}{\delta t}.$$
No! The basis vectors of the non-inertial reference frame are (in general) time dependent (particularly if the non-inertial frame is rotating against an inertial frame). For details see my previous long posting.

That's where the usual "inertial forces", among them Coriolis and centrifugal forces in frames rotating against the inertial frames, are coming from!

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

vanhees71
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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?

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.

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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
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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!

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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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