A rigid body motion can be decomposed into translation and rotation. My question is, given a rigid body motion velocities of all points in the body, how to decompose this velocity field into a translation and a rotation? Is this decomposition unique? Or is it unique given an arbitrarily chosen "pivot" point, fixed to the body, which may or may not be the center of mass?

  • $\begingroup$ The second one is correct. Also note that, if the body is rigid, you'll hardly have different velocities along the body, because rigid means that all atoms move together $\endgroup$ – FGSUZ Jul 24 '20 at 20:17
  • $\begingroup$ So if I choose a point fixed to the body as my pivot point, its own motion path is considered the translation path, and rotations around this point is considered rotation? Is this the correct statement? By rotation I mean a 3D rotation tensor (time dependent) rotating the MATERIAL POINTS coordinate vectors through the space, relative to a coordinate system located at the pivot, and does not rotate itself. $\endgroup$ – user138668 Jul 24 '20 at 21:20

The decomposition you are asking about is not unique, it is chosen so that the equations of motion are as simple as possible.

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Here is the general dynamical picture.

Fix an inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z $ with origin $O$. We represent the moving and rotating rigid body as a system of continuum many particles, each particle being represented by the position vector $$\vec{r} = \vec{r}(t) = x(t)\, \vec{e}_x + y(t)\, \vec{e}_y + z(t)\, \vec{e}_z $$ pointing from $O$ to the said particle at time $t$ with respect to the inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z $.

Furthermore, fix a point $Q$ on the rigid body, so that $Q$ moves with the body, firmly attached to it. Denote $\vec{r}_Q(t) = \vec{OQ}(t)$. The important part is that the body is rigid, so we can fix a coordinate system $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$ attached firmly to it. Consequently, $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$ rotates together with the body and the body is at rest with respect to $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$. The position of a point on the body can be expressed as $$\vec{R} = X\, \vec{E}_X + Y\, \vec{E}_Y + Z\, \vec{E}_Z$$ where $\vec{R}$ doesn't change with time with respect to $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$. Then, there is a time dependent rotation matrix $U = U(t)\, \in \, \text{SO}(3)$ such that $$\vec{r}(t) = \vec{r}_Q(t) \, + \, U(t)\,\vec{R}$$ where $$\vec{r}(t) = \begin{bmatrix} x(t)\\y(t)\\z(t)\end{bmatrix} \, \,\, \vec{r}_Q(t) = \begin{bmatrix} x_Q(t)\\y_Q(t)\\z_Q(t)\end{bmatrix} \, \, \text{ and } \,\, \vec{R} = \begin{bmatrix} X\\Y\\Z\end{bmatrix}$$ In particular, the center of mass $G$ of the body is represented by a fixed vector $$\vec{R}_G = X_G\, \vec{E}_X + Y_G\, \vec{E}_Y + Z_G\, \vec{E}_Z$$ in the body fixed frame $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$ and by a time-dependant vector $$\vec{r}_G(t) = \vec{r}_Q(t) \, + \, U(t)\,\vec{R}_G$$ in the inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z $

When we calculate the first derivative (the velocity) of the position vector in the inertial frame of a point from the body, we arrive at the expressions $$\frac{d\vec{r}}{dt} = \frac{d\vec{r}_Q}{dt} \, + \, \frac{d U}{dt}\, \vec{R} = \frac{d\vec{r}_Q}{dt} \, + \, U \big(\vec{\Omega} \times \vec{R}\big)$$ The last equality holds because for any time-dependent orthogonal matrix $U = U(t)$, there exists a time-dependent vector $\vec{\Omega} = \vec{\Omega}(t)$, called angular velocity, such that $$U^{-1}\frac{d U}{dt} \vec{R} = U^T\frac{d U}{dt} \vec{R} = \vec{\Omega} \times \vec{R}$$

To summarize, the position and orientation of the rigid body at any moment of time $t$ is fully determined by $$\text{the position vector }\, \vec{r}_Q = \vec{r}_Q(t) \, \in \, \mathbb{R}^3 \,\, \text{ and the rotation matrix }\,\, U = U(t) \, \in \, \text{SO}(3) $$ in the inertial coordinate frame $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z$

Assume the body may be moving in a force field $\vec{f}\big(\vec{r}, \,t\big)$, and maybe a bunch of forces are acting on the body at specific points fixed on it. Then the forces should be represented by vector-functions (in general) $$\vec{f}_j = U\, \vec{F}_j$$ where $\vec{f}_j$ are the coordinates of the forces with respect to the inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z $ and $\vec{F}_j$ are the coordinates of the forces in the body-fixed frame $Q\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$. The orthogonal matrix $U$ is the transformation matrix defined above between the body-fixed frame and the inertial frame. Assume each force $\vec{f}_j = U\, \vec{F}_j$ is applied to a certain point $\vec{r}_j = \vec{r}_Q + U\,\vec{R}_j$, fixed on the body, which means that while $\vec{r}_j = \vec{r}_j(t)$ changes with time, $\vec{R}_j$ is fixed in time.

The system of differential equations for the unknown vector and matrix functions $$\vec{r}_Q = \vec{r}_Q(t) \, \in \, \mathbb{R}^3, \,\,\, \vec{\Omega} = \, \vec{\Omega}(t)\, \in \, \mathbb{R}^3, \,\,\, U = U(t) \, \in \, \text{SO}(3)$$ defining the dynamics of the rigid body in vector form, are \begin{align*} m\,\frac{d^2\vec{r}_Q}{dt^2} \, + \, m\, U\left(\frac{d\vec{\Omega}}{dt} \times \vec{R}_G\right) \, &+ \, m \, U \left( \vec{\Omega} \times \Big( \, \vec{\Omega} \times \vec{R}_G \, \Big)\right) \, = \, \vec{f}_a\big(\vec{r}_Q, U, t\big) \, + \, \sum_j\, U \vec{F}_j\\ J_Q\frac{d\vec{\Omega}}{dt} \, + \, m\, \vec{R}_G \times \left( U^T\frac{d^2\vec{r}_Q}{dt^2}\right) \, &+ \, \vec{\Omega} \times J_Q\vec{\Omega} \, + \, m\, \vec{\Omega} \times \Big(\, \vec{R}_G \times \left( U^T\frac{d\vec{r}_Q}{dt}\right)\,\Big)\\ &= \, \vec{T}_f\big(\vec{r}_Q, U, t\big) + \sum_j\, \vec{R}_j \times \vec{F}_j \\ & \frac{dU}{dt} \, = \, U\big(\vec{\Omega} \times \cdot\big) \end{align*} where $$\vec{f}_a\big(\vec{r}_Q, U, t\big) \, = \, \int_{\text{Body}}\,\vec{f}\big(\vec{r}_Q + U\vec{R}, \, t\big)\, dR$$ $$\vec{T}_f\big(\vec{r}_Q, U, t\big) \, = \, \int_{\text{Body}}\,\Big\{\,\vec{R} \times \Big[ \, U^T\vec{f}\big(\vec{r}_Q + U\vec{R}, \, t\big) \Big] \,\Big\}\, dR$$ it's important to note that the constant three by three matrix $J_Q$ is the inertia tensor of the body, calculated in the body-fixed frame relative to the point $Q$.

These equations simplify a lot, if we take the point $Q$ to coincide with the center of mass $G$ of the body, i.e. $Q \equiv G$. Then the vector $\vec{R}_G = \vec{QG}$ in the body fixed frame is the zero vector. Hence \begin{align*} & m\,\frac{d^2\vec{r}_G}{dt^2} \, = \, \vec{f}_a\big(\vec{r}_G, U, t\big) \, + \, \sum_j\, U \vec{F}_j\\ & J_G\frac{d\vec{\Omega}}{dt} \, + \, \vec{\Omega} \times J_G\vec{\Omega} \, = \, \vec{T}_f\big(\vec{r}_G, U, t\big) \, + \,\sum_j\, \vec{R}_j \times \vec{F}_j \\ & \frac{dU}{dt} \, = \, U\big(\vec{\Omega} \times \cdot\big) \end{align*}


Edit: If you have the motion path $\vec{r} = \vec{r}(t)$ of each point on the body with respect to the inertial system, you do not need the velocity field at all in order to determine the rotation matrix $U = U(t)$. You simply need to choose the motion path of a reference point $\vec{r}_Q = \vec{r}_Q(t)$ and the motion paths of three more points $\vec{r}_1 = \vec{r}_1(t), \,\, \vec{r}_2 = \vec{r}_2(t)$ and $\vec{r}_3 = \vec{r}_3(t)$ with respect to the inertial system, where all four points are not coplanar and no three of them are collinear (so they form a non-degenerate tetrahedron). Write them as time dependent column vectors, with coordinates with respect to the inertial coordinate system: $$ \vec{r}_Q(t) = \begin{bmatrix} x_Q(t)\\y_Q(t)\\z_Q(t)\end{bmatrix} \,\, \,\, \vec{r}_1(t) = \begin{bmatrix} x_1(t)\\y_1(t)\\z_1(t)\end{bmatrix} \,\, \,\, \vec{r}_2(t) = \begin{bmatrix} x_2(t)\\y_2(t)\\z_2(t)\end{bmatrix} \,\, \text{ and } \, \, \vec{r}_3(t) = \begin{bmatrix} x_3(t)\\y_3(t)\\z_3(t)\end{bmatrix} $$ Form the time dependent 3 by 3 matrix $$M(t) \, = \, \Big[\vec{r}_1(t) - \vec{r}_Q(t) \,\,\,\,\,\,\, \vec{r}_2(t) - \vec{r}_Q(t) \,\,\,\,\,\,\, \vec{r}_3(t) - \vec{r}_Q(t) \Big]$$ or componentwise $$M(t) \, = \, \begin{bmatrix} x_1(t) - x_Q(t) \, & \, x_2(t) - x_Q(t) \, & \, x_3(t) - x_Q(t)\\ y_1(t) - y_Q(t) \, & \, y_2(t) - y_Q(t) \, & \, y_3(t) - y_Q(t)\\ z_1(t) - z_Q(t) \, & \, z_2(t) - z_Q(t) \, & \, z_3(t) - z_Q(t)\end{bmatrix} $$ Now, the orthogonal matrix $U = U(t) \, \in \, \text{SO}(3)$ you are looking for should be such that $$\vec{r}_j(t) \, = \, \vec{r}_Q(t) \, + \, U(t)\,\big(\vec{r}_j(0) - \vec{r}_Q(0) \big) $$ for $j=1,2,3$ or analogously, $$\vec{r}_j(t) \,-\, \vec{r}_Q(t) \, = \, U(t)\,\big(\vec{r}_j(0) - \vec{r}_Q(0) \big) $$ If you put these equations together in matrix form, you get $$M(t) \, = \, U(t)\, M(0)$$ Since the four points are different, not coplanar or not collinear, the matrix $M(t)$ is always invertable, so $$U(t) \, = \, M(t) \, M(0)^{-1} $$

Alternatively, you may simply have the paths of three non-collinear points, say just $\vec{r}_Q = \vec{r}_Q(t), \,\, \vec{r}_1 = \vec{r}_1(t)$ and $\vec{r}_2 = \vec{r}_2(t)$ Then form the normalized cross product vector $$\vec{w}_3(t) \, = \, \frac{\,\big(\vec{r}_1(t) - \vec{r}_Q(t) \big) \times \big(\vec{r}_2(t) - \vec{r}_Q(t) \big)\,} {\big|\big(\,\vec{r}_1(t) - \vec{r}_Q(t) \big) \times \big(\vec{r}_2(t) - \vec{r}_Q(t)\big)\, \big|}$$ then the unit vector $$\vec{w}_1(t) \, = \, \frac{1}{\big|\,\vec{r}_1(t) - \vec{r}_Q(t) \,\big|}\,\big(\vec{r}_1(t) - \vec{r}_Q(t) \big)$$ and finally the unit cross-product vector $$\vec{w}_2(t) \, = \, \vec{w}_3(t) \times \vec{w}_1(t)$$ As before, think of the three pairwise orthogonal unit vectors $\vec{w}_1(t),\,\, \vec{w}_2(t), \,\, \vec{w}_3(t)$ as column vectors. Then arrange them in a matrix $$W(t) \, = \, \Big[ \vec{w}_1(t)\,\,\, \vec{w}_2(t) \,\,\, \vec{w}_3(t) \Big]$$ which by construction is orthogonal matrix. Thus $$U(t) \, = \, W(t) W(0)^T$$

Now, if on the other hand, you have the velocity field of every point on the body with respect to the inertial system at only one moment of time, you can find the angular velocity $\vec{\omega}$ in the inertial system, at that moment of time. The angular velocity $\vec{\omega}$ is uniquely determined by the velocity field at the given time moment.

To find $\vec{\omega}$, assume your vectors are in the inertial coordinate system. Recall that the velocity field on the body comes from the motion $$\vec{r}(t) \, = \, \vec{r}_Q(t) \, + \, U(t) \, \vec{R}$$ as explained before. Inverting this transformation gives us $$\vec{R}\, = \, U(t)^T\big(\vec{r}(t) - \vec{r}_Q(t)\big)$$ Then $$\frac{d\vec{r}}{dt} \, = \, \frac{d\vec{r}_Q}{dt} \, + \, \frac{dU}{dt} \, \vec{R} \, = \, \frac{d\vec{r}_Q}{dt} \, + \, \left(\frac{dU}{dt} \,U^T\right) \big(\vec{r}(t) - \vec{r}_Q(t)\big)$$ There exists a time-dependent vector $\vec{\omega} = \vec{\omega}(t)$ in the inertial coordinate system, called angular velocity, such that $$\frac{d U}{dt}\, U^{-1} \vec{R} = \frac{dU}{dt}\,U^T (\vec{r} - \vec{r}_Q) = \vec{\omega} \times (\vec{r} - \vec{r}_Q)$$ which justifies the formula $$\vec{v} \, = \, \vec{v}_Q \, + \, \vec{\omega} \times (\vec{r} - \vec{r}_Q)$$ For any two points on the body we have $$\vec{v}_1 \, = \, \vec{v}_Q \, + \, \vec{\omega} \times (\vec{r}_1 - \vec{r}_Q)$$ $$\vec{v}_2 \, = \, \vec{v}_Q \, + \, \vec{\omega} \times (\vec{r}_2 - \vec{r}_Q)$$ and if we subtract the two equations and rearrange them, we get $$\vec{v}_2 \, = \, \vec{v}_1 \, + \, \vec{\omega} \times (\vec{r}_2 - \vec{r}_1)$$ since this is true for any two points, the angular velocity $\vec{\omega}$ is the same for any choice of the point $Q$ on the body (being point 1 or point 2 or any other point).

To find $\vec{\omega}$, given information at only one moment of time, assume your vectors are in the inertial coordinate system. If you have a selected point $Q$ on the body with position vector $\vec{r}_Q$ and velocity $\vec{v}_Q$, then knowing the velocities $\vec{v}_1$ and $\vec{v}_2$ at two other points $\vec{r}_1$ and $\vec{r}_2$ respectively, so that $\vec{v}_1$ and $\vec{v}_2$ are not parallel, then $$\vec{\omega}\, =\, \pm \, |\vec{\omega}| \, \frac{(\vec{v}_1 - \vec{v}_Q) \times (\vec{v}_2 - \vec{v}_Q)} {\big|(\vec{v}_1 - \vec{v}_Q) \times (\vec{v}_2 - \vec{v}_Q)\big|}$$ This is because $$\vec{v}_j \, = \, \vec{v}_Q \, + \, \vec{\omega} \times (\vec{r}_j - \vec{r}_Q)$$ for both points $j=1,2$ and therefore, $\vec{\omega}$ should be perpendicular to both $\vec{v}_1 - \vec{v}_Q$ and $\vec{v}_2 - \vec{v}_Q$, a direction determined by the cross-product of the two latter vectors. Let us denote, for abbreviation, $\vec{v}_j - \vec{v}_Q = \vec{v}_{Qj}$ and $\vec{r}_j - \vec{r}_Q = \vec{r}_{Qj}$ for $j=1,2$. Since $$\vec{v}_{Q1} \, = \, \vec{\omega} \times \vec{r}_{Q1}$$ you plug the expression for the angular velocity $$\vec{v}_{Q1} \, = \, \pm\, |\vec{\omega}| \, \frac{\vec{v}_{Q1} \times \vec{v}_{Q2}}{|\vec{v}_{Q1} \times \vec{v}_{Q2}|} \times \vec{r}_{Q1}$$ take the norm $$|\vec{v}_{Q1}| \, = \, |\vec{\omega}| \, \left|\frac{\vec{v}_{Q1} \times \vec{v}_{Q2}}{|\vec{v}_{Q1} \times \vec{v}_{Q2}|} \times \vec{r}_{Q1}\right| $$ and you can manipulate the expression if you want $$|\vec{v}_{Q1}| \, = \, |\vec{\omega}| \, \frac{\sqrt{|\vec{r}_{Q1}|^2|\vec{v}_{Q1} \times \vec{v}_{Q2}|^2 \, - \, \big(\, (\vec{v}_{Q1} \times \vec{v}_{Q2})\cdot \vec{r}_{Q1} \,\big)^2 \, }}{|\vec{v}_{Q1} \times \vec{v}_{Q2}|} $$ and get $$|\vec{\omega}| \, = \, \frac{|\vec{v}_{Q1} \times \vec{v}_{Q2}| \, |\vec{v}_1| }{\sqrt{|\vec{r}_{Q1}|^2|\vec{v}_{Q1} \times \vec{v}_{Q2}|^2 \, - \, \big(\, (\vec{v}_{Q1} \times \vec{v}_{Q2})\cdot \vec{r}_{Q1} \,\big)^2 \, }}$$

$$\vec{\omega}\, =\, \pm \, \left( \, \frac{ |\vec{v}_{Q1}| }{\sqrt{|\vec{r}_{Q1}|^2|\vec{v}_{Q1} \times \vec{v}_{Q2}|^2 \, - \, \big(\, (\vec{v}_{Q1} \times \vec{v}_{Q2})\cdot \vec{r}_{Q1} \,\big)^2 \, }}\right) \, \vec{v}_{Q1} \times \vec{v}_{Q2}$$ And the sign of $\vec{\omega}$ can be determined so that $\vec{\omega} \cdot (\vec{r}_{Q1} \times \vec{v}_{Q1}) > 0$.

A word of caution with regards to the instantaneous axis of motion at a given moment of time, discussed in the other answers. I would just like to point out that one needs to be careful with the instantaneous axis of motion. In some sense, it is not a physical axis, i.e. it is not an axis attached to the rigid body. In general, it changes its position in both the inertial and the body-fixed coordinate systems and as such, it travels from point to point on the body as time passes. In both the inertial frame and the body fixed frame the instantaneous axis traverses ruled surfaces.

End of edit.

  • 1
    $\begingroup$ This is an excellent answer. I did not ask about the dynamics, only the kinematics though. The dynamics part is clean and concise. I do still want to know how to find U(t) if the path of each point r(R,t) is given? In other words, given a velocity field of the whole body, how to find its rotation path in terms of the rotation matrix U(t)? $\endgroup$ – user138668 Jul 26 '20 at 2:08
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    $\begingroup$ @user138668 I added some extra paragraphs, hopefully they can help you. I can only write general stuff as I am not sure what the precise formulation of the probelm you are trying to solve is. $\endgroup$ – Futurologist Jul 27 '20 at 2:29
  • $\begingroup$ Thanks a lot. Your answer definitely helped me a lot. I was mainly trying to understand the motion and its mathematical representation, where your formulation is precisely it. The other answer below helped me with this instantaneous rotation axis puzzle. The two answers put together completes the picture for me. $\endgroup$ – user138668 Jul 27 '20 at 2:46

The decomposition is not unique for a general linear and rotational motion.

Any point A riding on the rigid body will have linear motion $\boldsymbol{v}_A$ and shared rotational motion $\boldsymbol{\omega}$. But any other point B will also have linear motion $\boldsymbol{v}_B = \boldsymbol{v}_A + \boldsymbol{\omega}\times ( \boldsymbol{r}_B - \boldsymbol{r}_A)$ and the same rotational motion. Here $\boldsymbol{r}_A$ is the instantenous location of A and $\boldsymbol{r}_B$ the instantenous location of B.

But the more interesting thing is to look at the entire motion of the rigid body as a velocity field.

There is a unique axis in space that has the property that the velocity vectors anywhere long this axis are parallel to the rotation vector. A special case to this is when the velocity vectors on this line are zero. This is why this axis is commonly referred to as the instant axis of rotation (IAR).

This axis has direction parallel to $\omega$, But is located at a specific place in space.

Consider such a point C located at $\boldsymbol{r}_C$ where velocities are parallel to the rotation $\boldsymbol{\omega}$. So $$ \boldsymbol{v}_C = h \, \boldsymbol{\omega} \tag{1}$$ where $h$ is any scalar value. The quantity $h$ is called the pitch and it represents the linear distance traveled by the body per one radian rotation.

Now the velocity of any other point, such as A, can be uniquely calculated from the motion at the IAR

$$ \boldsymbol{v}_A = \boldsymbol{v}_C + \boldsymbol{\omega} \times ( \boldsymbol{r}_A - \boldsymbol{r}_C) = h\; \boldsymbol{\omega} - (\boldsymbol{r}_A - \boldsymbol{r}_C) \times \boldsymbol{\omega} \tag{2}$$

The above defines all three components of $\boldsymbol{v}_A$, from the geometry (the pitch $h$ and the locations) and the rotation $\boldsymbol{\omega}$.

More commonly we start with the known motion of a point, like A, and need to find the location and properties of the IAR. This is done with

$$ \begin{aligned} h & = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_A}{\|\boldsymbol{\omega}\|^2} \\ \boldsymbol{r}_C - \boldsymbol{r}_A & = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_A} {\|\boldsymbol{\omega}\|^2} \end{aligned} \tag{3} $$

In addition, to $\text{(direction)} = \boldsymbol{\omega} / \| \boldsymbol{\omega} \|$.

Proof comes by using (3) into (2) and remembering the vector triple product $a\times(b \times c) = b(a\cdot c) - c(a \cdot b)$.

$$ \begin{aligned}\boldsymbol{v}_{A} & =\frac{\boldsymbol{\omega}\cdot\boldsymbol{v}_{A}}{\|\boldsymbol{\omega}\|^{2}}\,\boldsymbol{\omega}+\frac{\boldsymbol{\omega}\times\boldsymbol{v}_{A}}{\|\boldsymbol{\omega}\|^{2}}\times\boldsymbol{\omega}\\ & =\frac{\boldsymbol{\omega}\left(\boldsymbol{\omega}\cdot\boldsymbol{v}_{A}\right)-\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times\boldsymbol{v}_{A}\right)}{\|\boldsymbol{\omega}\|^{2}}\\ & =\frac{\boldsymbol{\omega}\left(\boldsymbol{\omega}\cdot\boldsymbol{v}_{A}\right)-\boldsymbol{\omega}\left(\boldsymbol{\omega}\cdot\boldsymbol{v}_{A}\right)+\boldsymbol{v}_{A}\left(\boldsymbol{\omega}\cdot\boldsymbol{\omega}\right)}{\|\boldsymbol{\omega}\|^{2}}\\ & =\frac{\boldsymbol{v}_{A}\|\boldsymbol{\omega}\|^{2}}{\|\boldsymbol{\omega}\|^{2}}=\boldsymbol{v}_{A}\;\checkmark \end{aligned} \tag{4} $$

So in summary, take the velocity of an arbitrary point A and the rotational vector and you can find the IAR location and pitch with (3). Or we can take the known IAR location and pitch and find the velocity of an arbitrary point with (2). The two situations are equivalent to each other.

You can use this to categorize the motion of a rigid body into three categories:

$$\begin{array}{r|lll} & \text{pure rotation} & \text{screw motion} & \text{pure translation}\\ \hline \text{pitch} & h=0 & h\neq0 & h=\infty\\ \text{rotation} & \boldsymbol{\omega}\neq0 & \boldsymbol{\omega}\neq0 & \boldsymbol{\omega}=0\\ \text{velocity field} & \boldsymbol{v}(\boldsymbol{r})=-\boldsymbol{r}\times\boldsymbol{\omega} & \boldsymbol{v}(\boldsymbol{r})=h\,\boldsymbol{\omega}-\boldsymbol{r}\times\boldsymbol{\omega} & \boldsymbol{v}(\boldsymbol{r})=\text{(const)} \end{array}$$

PS. The center of mass of a body isn't special in terms of the kinematics of motion. It is only special because it simplifies the equations of motion.

PS2. Something similar can happen with decomposing momentum and decomposing forces into special axes in space. We can the first one the axis of percussion, and the second one the line of action of a force.

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    $\begingroup$ Thanks for the good insight. I suppose the angular velocity $\omega$ depends on the choice of the pivot. So the IAR should be the very specific collection of pivot points $\{C: v_C \| \omega_C\}$. What you say is this IAR is a unique straight line. How to prove its existence and uniqueness? $\endgroup$ – user138668 Jul 26 '20 at 21:47
  • $\begingroup$ From (2) you can see that if you choose the arbitrary point A being on this axis the result is parallel velocity and if not on the axis, general velocity. So there could not be another axis somewhere else not-coincident to IAR that would also produce parallel velocities. This is because the non parallel components are composed with the cross product $\omega \times r$ which is non-zero for non-parallel vectors. $\endgroup$ – John Alexiou Jul 27 '20 at 1:52
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    $\begingroup$ "the angular velocity ω depends on the choice of the pivot" - that is not a correct statement unless I misunderstood it. Angular velocity is shared among all points on a rigid body (in fact the entire rotating frame for that manner), and it is the core property of the motion vector field.Knowing the vector $\boldsymbol{\omega}$ only gives us the magnitude (speed) of rotation, and the direction of rotation. You need at least the velocity vector at some other known location to establish the complete picture of motion of a rigid body. $\endgroup$ – John Alexiou Jul 27 '20 at 1:59
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    $\begingroup$ You are right about $\omega$ being common to all choice of "pivot". I have figured out how to "solve" the following equation to obtain the IAR axis: $$\omega\times v_C=\omega\times v_A+\omega\times(\omega\times(r_C-r_A))=0$$ This equation satisfied by $r_C$ represents a straight line, namely the IAR. Thanks! $\endgroup$ – user138668 Jul 27 '20 at 2:39

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