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

• B
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## Main Question or Discussion Point

In a previous thread it was concluded that both torque and angular momentum are taken about a chosen origin, and these quantities are generally not invariant under translations of the origin (since $\vec{r}$ changes but $\vec{v}$ does not, etc.). The moment of inertia tensor doesn't implicitly refer to an origin, but it's value depends on the position in space (i.e. the MOI tensor at the centre of mass is different to that at a point on the edge of the body).

However, I did some more reading and found that the angular velocity vector $\vec{\omega}$ as well as the angular acceleration $\vec{\alpha}$ are invariant under translations of the origin. This doesn't make much sense, because if for instance $\vec{v} = \vec{\omega} \times \vec{r}$, and $\vec{v}$ is invariant under origin translations but $\vec{r}$ definitely is not, then $\vec{\omega}$ has to change also!

I tried looking in Morin but he just mentions that $\vec{\omega}$ is parallel to the axis of rotation. On the one hand, it makes sense to speak of the $\vec{\omega}$ of a rigid body about it's centre of mass, which is surely origin invariant (as long as the basis vectors are still oriented in the same directions). However, if now choose an origin in the corner of the lab in which the body is spinning, wouldn't $\vec{\omega}$ be different for every point on the body?

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THEOREM. There exists a unique (pseudo) vector $\boldsymbol \omega$ such that for any two points $A,B$ of a rigid body the following formula holds
$$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol\omega\times\boldsymbol{BA}.$$

DEFINITION. The vector $\boldsymbol\omega$ is referred to as the angular velocity of the rigid body

• vanhees71, PeroK and etotheipi
The moment of inertia tensor doesn't implicitly refer to an origin,
it does

• etotheipi
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THEOREM. There exists a unique (pseudo) vector $\boldsymbol \omega$ such that for any two points $A,B$ of a rigid body the following formula holds
$$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol\omega\times\boldsymbol{BA}.$$

DEFINITION. The vector $\boldsymbol\omega$ is referred to as the angular velocity of the rigid body
Right, so is there a difference between the angular velocity of the rigid body $\vec{\omega}$, and the angular velocity of a point on the rigid body wrt some other fixed origin?

it does
I think we're on the same page here, but I might have used the wrong terminology. The moment of inertia of a body is definitely wrt a point, which we might call the origin. But this point could be anywhere in space, as in we could fix $O$ to be the corner of the room, spin something in the middle of the room, and measure the angular momentum, torque and moment of inertia of the spinning body wrt this point.

A.T.
Right, so is there a difference between the angular velocity of the rigid body $\vec{\omega}$, and the angular velocity of a point on the rigid body wrt some other fixed origin?
Yes. The angular velocity of the rigid body refers to changing orientation, while the angular velocity of a point on the rigid body wrt some origin can be non-zero, even if the body just translates with fixed orientation.

• etotheipi
Right, so is there a difference between the angular velocity of the rigid body →ω\vec{\omega}, and the angular velocity of a point on the rigid body wrt some other fixed origin?
I prefer not to speak about angular velocity of a point especially when the point is in the space. I do not understand what this is
The moment of inertia of a body is definitely wrt a point, which we might call the origin. But this point could be anywhere in space, as in we could fix OO to be the corner of the room, spin something in the middle of the ro
ok we understood each other

• PeroK
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I've seen $\vec{\omega}$ and $\vec{\Omega}$ used; I think the first generally refers to the spin angular momentum mentioned here
DEFINITION. The vector $\boldsymbol\omega$ is referred to as the angular velocity of the rigid body
whilst $\vec{\Omega}$ perhaps refers to the orbital angular velocity, i.e. that of the centre of mass wrt another point.

Since we know that the total angular momentum equals the sum of the orbital and spin angular momenta $$\vec{L} = \vec{L}_{\text{COM}} + \vec{L}_{\text{wrt. COM}}$$ I'd imagine it would make sense to define something like $$\vec{\omega}_{\text{total}} = \vec{\Omega} + \vec{\omega}$$ but I'd need to play around with it for a little while to check. Furthermore, I don't know whether $\omega_{total}$ would be a physically useful quantity.

Also, for angular acceleration we know that $$\vec{\Gamma}_{O}= \boldsymbol{I_{O}}\vec{\alpha}$$ Does this equation only then hold if $O$ is a point on the rigid body? Since I can't imagine $\vec{\alpha}$ has any interpretation if $O$ is fixed elsewhere!

From $$\vec{\Gamma}_{O}= \frac{d\vec{L_{O}}}{dt} = \boldsymbol{I_{O}} \frac{d\vec{\omega}}{dt} = \boldsymbol{I_{O}} \vec{\alpha}$$ since $\vec{\omega}$ is only defined for the rotation of the rigid body as a whole?

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sophiecentaur
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Furthermore, I don't know whether ωtotalωtotal\omega_{total} would be a physically useful quantity.
In the context of the Solar System, the Total (or, for you , the total of totals) is of interest. If you assume that the Sun and its planets all started of as a single nebula then the angular momentum of the nebula would be the same as the total, now and it is divided up between the whole family, including the Kuiper belt etc.). This link makes interesting reading - pay particular attention to the exponents in the two figures quoted for the Sun and the whole of the rest of the SS. Not an intuitive fact and it certainly shows the influence of the x2 term in the calculation of MI.

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• etotheipi
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In the context of the Solar System, the Total (or, for you , the total of totals) is of interest. If you assume that the Sun and its planets all started of as a single nebula then the angular momentum of the nebula would be the same as the total, now and it is divided up between the whole family, including the Kuiper belt etc.). This link makes interesting reading - pay particular attention to the exponents in the two figures quoted for the Sun and the whole of the rest of the SS. Not an intuitive fact and it certainly shows the influence of the x2 term in the calculation of MI.
That's pretty cool, I hadn't come across this but I can sort of understand the underlying idea. Which link were you referring to?

vanhees71
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THEOREM. There exists a unique (pseudo) vector $\boldsymbol \omega$ such that for any two points $A,B$ of a rigid body the following formula holds
$$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol\omega\times\boldsymbol{BA}.$$

DEFINITION. The vector $\boldsymbol\omega$ is referred to as the angular velocity of the rigid body
This is WRONG!!!! [see #13]

One should note though, that the angular velocity of the body (as its spin-angular momentum and torque) depends on the arbitrarily chosen origin of the body-fixed (non-inertial rotating) reference frame. The clever choice of this origin is utmost important for the analysis of the rigid body's motion. E.g., for a free falling rigid body you better choose the center of mass as this body-fixed reference point than any other point.

In other words, your $\vec{\omega}$ depends on $\overrightarrow{BA}$! The choice of one body-fixed origin $O$ then makes $\vec{\omega}$ the same for any body-fixed point $P$, but dependent on the choice of $O$!

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• etotheipi
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I wonder, does anyone know of a resource which explains all of these little subtleties well? I usually like Morin's explanations but I'm finding his section on rotational dynamics to be a little jarring.

Especially with regards to the spin/orbital distinction, and choices of origins, etc?

Also, if we have a rod rotating about a hinge, and we calculate its angular momentum by considering the moment of inertia about the hinge and its angular velocity about the hinge, which type of angular momentum are we computing? Considering that it can't be spin, since the hinge is not at the centre of mass of the rod...

vanhees71
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In this case it's of course the angular momentum with respect to some point on the line of rotation defined by the hinge. Any other choice of a body-fixed reference point would unnecessarily complicate the analysis!

In other words, your →ω\vec{\omega} depends on −−→BA\overrightarrow{BA}!
that is wrong
read the theorem carefully: there exists a unique vector $\boldsymbol\omega$ such that for any two points $A,B$.....
$\exists !.....\forall$

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In this case it's of course the angular momentum with respect to some point on the line of rotation defined by the hinge. Any other choice of a body-fixed reference point would unnecessarily complicate the analysis!
So we pick the origin $O$ to be, for instance, at the hinge. The moment of inertia of the rod wrt this origin, $I_{O}$ is then $\frac{ml^2}{3}$. We say $$L_{O} = I_{O}\omega$$ Though apparently there are two flavours of $\vec{\omega}$, orbital and spin. Orbital refers to $\sum m\vec{r}_{\text{COM}}\times \vec{v}_{\text{COM}}$, and spin angular momentum - apparently dubbed the angular momentum of the rigid body - refers to the same form but instead computed at the centre of mass.

The $\omega$ in the first equation doesn't fit either of these definitions.

vanhees71
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that is wrong
read the theorem carefully: there exists a unique vector $\boldsymbol\omega$ such that for any two points $A,B$.....
$\exists !.....\forall$
You are right! If $A$ and $B$ are arbitrary body-fixed points for all pairs you get the same $\vec{\omega}$.

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I'm getting a little lost ! I wonder if I'm understanding this correctly.

Suppose a rigid body is rotating in space, and we pick two origins $O'$ and $O''$ in the same reference frame, separated by only a translation. The moments of inertia of the rigid body and torques acting on it as measured from $O'$ and $O''$ are $I'$ and $I''$, and $\Gamma'$ and $\Gamma''$ respectively. We write two relations $$\Gamma' = I' \alpha'$$ $$\Gamma'' = I'' \alpha''$$ We also know that, if $\vec{r}_c$ is the vector from O'' to O', $$\Gamma'' = \vec{r}_c \times \vec{F} + \Gamma'$$ Is it the case that $\alpha' = \alpha''$? What does $\alpha$ even refer to in this context?

sophiecentaur
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That's pretty cool, I hadn't come across this but I can sort of understand the underlying idea. Which link were you referring to?
Sorry = I forgot to link it in. Here it is.

• etotheipi
jbriggs444
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You are right! If $A$ and $B$ are arbitrary body-fixed points for all pairs you get the same $\vec{\omega}$.
To be a bit picky, any given pair of body-fixed points gives you only a partial constraint on $\vec{\omega}$. They nail down only two degrees of freedom out of three.

Which means that the theorem quoted in #2 is not correct. The rigid body must consist of points which are not all colinear. [The angular velocity of a line segment rotating around its own axis is undefined]

• vanhees71
Which means that the theorem quoted in #2 is not correct. The rigid body must consist of points which are not all colinear.
Sure, there is such a degenerate case. I like to think that a rigid body is by definition a three dimensional object

• jbriggs444
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Alright, I think I've gotten a bit further. We know $\vec{\Gamma}_{ext} = \dot{\vec{L}}$, since internal force pairs will produce equal and opposite torques that cancel each other when evaluated from any origin chosen. This applies so long as the origin is one of an inertial frame.

We also have from König's theorem, $\vec{L} = \vec{R}\times\vec{P} + \vec{L}_{CM}$, and we can derive wrt time to obtain $$\dot{\vec{L}}_{CM} = \dot{\vec{L}} - \vec{R} \times M\ddot{\vec{R}}$$ since $\dot{\vec{R}}\times \dot{\vec{R}}$ is the zero vector in the second term of the chain rule! This becomes $$\dot{\vec{L}}_{CM} = \sum_{i} (\vec{r}_i - \vec{R}) \times \vec{F}_i = \vec{\Gamma}_{CM}$$ So essentially, as a special case, the total torque on a rigid body relative to the centre of mass equals the rate of change of angular momentum evaluated at the centre of mass - even if the centre of mass is at rest in an accelerated frame. This is not true for other points on the rigid body, since they are in accelerated frames.

As far as I am aware, the quantity $\vec{L}_{CM}$ is the spin angular momentum, and $\vec{L}_{CM} = I_{CM} \vec{\omega}$ where $\vec{\omega}$ is the quantity @wrobel defined all the way back in post #2.

And consequently $\frac{\vec{L}_{CM}}{dt} = I_{CM} \frac{d\vec{\omega}}{dt} = I_{CM}\vec{\alpha} = \vec{\Gamma}_{CM}$, where $\vec{\alpha}$ is again a property of the rigid body (I'm not sure if it has a specific name - "spin angular acceleration?").

Now we might step back and consider a rigid body performing some arbitrary motion relative to an inertial frame, in which we've chosen an origin. We can still evaluate torques relative to the centre of mass and equate these to the rate of change of angular momentum wrt the centre of mass.

However, now we also have to take into account the orbital angular momentum, $\vec{R} \times \vec{P}$. I start with $$\vec{\Gamma}_{ext} = \dot{\vec{L}}$$ I assume that the $\vec{L}$ in this expression refers to the total angular momentum evaluated from the origin. Now $\vec{\Omega}$ is the the orbital angular velocity vector. I had a few queries about this part
1. Can we write $\vec{v}_{CM} = \vec{\Omega} \times \vec{r}$ relative to the origin, (i.e. as well as $\vec{v} = \vec{\omega} \times \vec{r'}$ relative to the centre of mass)
2. How do we relate the external torque to the rate of change of orbital angular velocity?

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There are two most frequently used forms of the angular momentum theorem for a rigid body.

$$J_A\boldsymbol{\dot\omega}+\boldsymbol\omega\times J_A\boldsymbol\omega=\boldsymbol\tau_A.$$
Here $\tau_A$ stands for a total torque which is applied to the rigid body about a point $A$.
The above formula is valid either when $A$ is a center of mass of the rigid body or when $A$ is a fixed point of the rigid body.
$\boldsymbol\omega, \boldsymbol{\dot\omega}$ are considered relative to an inertial frame

• etotheipi
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$$J_A\boldsymbol{\dot\omega}+\boldsymbol\omega\times J_A\boldsymbol\omega=\boldsymbol\tau_A.$$
Thanks; just one point about this, how does one show that $\vec{\omega} \times I_{A} \vec{\omega} = \vec{0}$ if A is the centre of mass?:

how does one show that →ω×IA→ω=→0\vec{\omega} \times I_{A} \vec{\omega} = \vec{0} if A is the centre of mass?:
this term is not obliged to be zero if $A$ is the center of mass

vanhees71
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$$\newcommand{\uvec}{\underline{#1}}.$$
You must be careful. You have to distinguish between vectors and their components. For the vectors the formula reads
$$\dot{\vec{J}} = \vec{\tau}.$$
For the components in the space-fixed (inertial) frame the equation reads the same
$$\dot{\underline{J}}=\underline{\tau},$$
where the underlined symbol are column vectors with the space-fixed components of the corresponding vector.

The body-fixed frame is a rotating frame and thus the time derivative of a vector translates into the formula
$$\dot{\underline{J}}' + \underline{\omega}' \times \underline{J}=\underline{\tau}',$$
where an underlined symbol with prime means column vectors with the body-fixed components of the corresponding vector.

Both the angular momentum and the torque depend on the body-fixed origin, the angular velocity does not. I was wrong about this in #13!!!

The most general motion of a rigid body is described by first choosing an inertial reference frame $(O,\vec{e}_j)$ ($j \in \{1,2,3 \}$), where $O$ is the origin of this "space-fixed" inertial reference frame and the $\vec{e}_j$ are an arbitrary righthanded Cartesian basis. Then you have to also define a "body-fixed" reference frame, which usually is a non-inertial rotating frame. To that end you define an arbitrary point $O'$ fixed relative to the body and a righthanded body-fixed Cartesian basis $\vec{e}_k'$.

Then any point $P_k$ of the rigid body (which for simplicity we assume to consist of a discrete set of point masses) is described by the position vector $\vec{x}_k=\overrightarrow{OP_k}$. Now we write
$$\vec{x}_k=\vec{R} + \vec{r}_k \quad \text{with} \quad \vec{R}=\overrightarrow{OO'}, \quad \vec{r}=\overrightarrow{O'P_k}.$$
These are all basis-independent vectors.

A rigid body is now by definition a body, where for all of its points $P_k$
$$\uvec{r}_k'=\text{const}.$$
Thus for body-fixed relative vectors $\vec{r}_k$ it's convenient to use body-fixed components, which are time-independent.

Now for any vector $\vec{V}$ you have, introducing the transformation matrix between the space-fixed and the body-fixed Cartesian bases (Cartesian Einstein summation convention used):
$$\vec{e}_k'=\vec{e}_j D_{jk},$$
which is $(D_{jk})\in \mathrm{SO}(3)$, i.e., it's a rotation matrix fulfilling $\hat{D}^{\text{T}}=\hat{D}^{-1}$ and $\mathrm{det} \hat{D}=+1$,
$$\vec{V}=V_k' \vec{e}_k'=\vec{e}_j V_k' D_{jk} = \vec{e}_j V_j \; \Rightarrow \; \uvec{V}'=\hat{D} \uvec{V}.$$
Then on the one hand
$$\dot{\vec{V}} = \vec{e}_j \dot{V}_j,$$
because by definition the space-fixed basis is time independent but on the other hand
$$\dot{\vec{V}}=\dot{V}_k' \vec{e}_k' + V_{k}' \dot{\vec{e}}_k',$$
because the body-fixed basis is time-dependent.

To get the body-fixed components of $\dot{\vec{V}}$ it's more convenient to use
$$\uvec{V}=\hat{D}^{\text{T}} \uvec{V}' \; \Rightarrow \; \dot{\uvec{V}}=\hat{D}^{\text{T}} \dot{\uvec{V}}' + \dot{\hat{D}}^{\text{T}} \uvec{V}'.$$
Thus the body-fixed components of $\dot{\vec{V}}$, which I denote as $\mathrm{D}_t \uvec{V}'$, are given by
$$\mathrm{D}_t \uvec{V}'=\hat{D} \dot{\uvec{V}}=\dot{\uvec{V}}' + \hat{D} \dot{\hat{D}}^{\text{T}} \uvec{V}'.$$
Now the matrix in the 2nd term is antisymmetric, because
$$\hat{D} \hat{D}^{\text{T}}=\hat{1} \; \rightarrow \; \mathrm{d}_t (\hat{D} \hat{D}^{\text{T}}) = \dot{\hat{D}} \hat{D}^{\text{T}} + \hat{D} \dot{\hat{D}}^{\text{T}}=0$$
and from that
$$\hat{\Omega}'=\hat{D} \dot{\hat{D}}^{\text{T}}=-\dot{\hat{D}} \hat{D}^{\text{T}}=-\hat{\Omega}^{\prime \text{T}}.$$
Thus one can write
$$\Omega_{kl}'=-\epsilon_{klm} \omega_m',$$
where
$$\omega_m'=-\frac{1}{2} \epsilon_{klm} \Omega_{kl}$$
is the Hodge dual of $\Omega_{kl}'$ (which is a one-to-one mapping between antisymmetric matrices and vector components). Thus we finally have
$$(\mathrm{D}_t \uvec{V}')_k=\dot{V}_k' + \hat{\Omega}_{kl}' V_l'=\dot{V}_k' - \epsilon_{klm} \omega_m' V_l' = \dot{V}_k'+\epsilon_{kml} \omega_m' V_l',$$
i.e., in column-vector notation
$$\mathrm{D}_t \uvec{V}'=\dot{\uvec{V}}' + \uvec{\omega}' \times \uvec{V}'.$$
Supposed you have a spinning top, i.e., if you fix the body at one of its points (to make the thing simpler) and choose $O=O'$, i.e., you make the origin of the space-fixed reference the same as the origin of the body-fixed origin $O'$ which you choose as the point of the body which is fixed, then you have
$$\mathrm{D}_t \uvec{J}' = \dot{\uvec{J}}'+\uvec{\omega}' \times \uvec{V}' = \uvec{\tau}'.$$
This is an equation for body-fixed components, which is convenient, because as the further analysis of this case shows there's a symmetric tensor with time-independent body-fixed components $\uvec{\Theta}'$ such that
$$\uvec{J}'=\hat{\Theta}' \uvec{\omega}',$$
there $\overleftrightarrow{\Theta}=\Theta_{kl}' \vec{e}_k' \otimes \vec{e}_l'$ is the tensor of inertia. Since it's symmetric, it can be diagonalized by choosing the body-fixed basis as oriented along the corresponding principle axes, i.e., you make the body-fixed basis vectors the eigenvectors of the symmetric matrix. A well-known theorem tells you that you can always make these eigenvectors to form a right-handed Cartesian basis, so that the above considerations apply to this most convenient choice of the body-fixed basis. With $\hat{\Theta}'=\mathrm{diag}(A,B,C)$ you have
$$\uvec{J}'=\begin{pmatrix} J_1' \\ J_2' \\ J_3' \end{pmatrix} = \begin{pmatrix} A \omega_1' \\ B \omega_2' \\ C \omega_3' \end{pmatrix},$$
and plugging this in the above equation for $\mathrm{D}_t \uvec{J}'$ you get Euler's equation of motion for $\uvec{\omega}'$ of the spinning top.

See

https://en.wikipedia.org/wiki/Euler's_equations_(rigid_body_dynamics)
where they use of course somewhat different notation.

• etotheipi
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Goodness, thanks a bunch @vanhees71 for writing that all out! I can understand roughly the first half but it's going to take me a while to get to grips with the second half (I'm not great with indices!).

`I'll see how far I can get!