Rigid body motion decomposition

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 $atimes(b times c) = b(acdot c) - c(a cdot b)$.

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

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:

$$begin{array}{r|lll} & text{pure rotation} & text{screw motion} & text{pure translation}\ hline text{pitch} & h=0 & hneq0 & h=infty\ text{rotation} & boldsymbol{omega}neq0 & boldsymbol{omega}neq0 & boldsymbol{omega}=0\ text{velocity field} & boldsymbol{v}(boldsymbol{r})=-boldsymbol{r}timesboldsymbol{omega} & boldsymbol{v}(boldsymbol{r})=h,boldsymbol{omega}-boldsymbol{r}timesboldsymbol{omega} & boldsymbol{v}(boldsymbol{r})=text{(const)} end{array}$$

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.

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 $$vec{r}(t) = begin{bmatrix} x(t)\y(t)\z(t)end{bmatrix} , ,, vec{r}_Q(t) = begin{bmatrix} x_Q(t)\y_Q(t)\z_Q(t)end{bmatrix} , , text{ and } ,, vec{R} = begin{bmatrix} X\Y\Zend{bmatrix}$$ 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 $$frac{dvec{r}}{dt} = frac{dvec{r}_Q}{dt} , + , frac{d U}{dt}, vec{R} = frac{dvec{r}_Q}{dt} , + , U big(vec{Omega} times vec{R}big)$$ 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 $$U^{-1}frac{d U}{dt} vec{R} = U^Tfrac{d U}{dt} vec{R} = vec{Omega} times vec{R}$$

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 $vec{f}big(vec{r}, ,tbig)$, 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^2vec{r}_Q}{dt^2} , + , m, Uleft(frac{dvec{Omega}}{dt} times vec{R}_Gright) , &+ , m , U left( vec{Omega} times Big( , vec{Omega} times vec{R}_G , Big)right) , = , vec{f}_abig(vec{r}_Q, U, tbig) , + , sum_j, U vec{F}_j\ J_Qfrac{dvec{Omega}}{dt} , + , m, vec{R}_G times left( U^Tfrac{d^2vec{r}_Q}{dt^2}right) , &+ , vec{Omega} times J_Qvec{Omega} , + , m, vec{Omega} times Big(, vec{R}_G times left( U^Tfrac{dvec{r}_Q}{dt}right),Big)\ &= , vec{T}_fbig(vec{r}_Q, U, tbig) + sum_j, vec{R}_j times vec{F}_j \ & frac{dU}{dt} , = , Ubig(vec{Omega} times cdotbig) end{align*} where $$vec{f}_abig(vec{r}_Q, U, tbig) , = , int_{text{Body}},vec{f}big(vec{r}_Q + Uvec{R}, , tbig), dR$$ $$vec{T}_fbig(vec{r}_Q, U, tbig) , = , int_{text{Body}},Big{,vec{R} times Big[ , U^Tvec{f}big(vec{r}_Q + Uvec{R}, , tbig) Big] ,Big}, dR$$ 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^2vec{r}_G}{dt^2} , = , vec{f}_abig(vec{r}_G, U, tbig) , + , sum_j, U vec{F}_j\ & J_Gfrac{dvec{Omega}}{dt} , + , vec{Omega} times J_Gvec{Omega} , = , vec{T}_fbig(vec{r}_G, U, tbig) , + ,sum_j, vec{R}_j times vec{F}_j \ & frac{dU}{dt} , = , Ubig(vec{Omega} times cdotbig) end{align*}

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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 $$vec{R}, = , U(t)^Tbig(vec{r}(t) - vec{r}_Q(t)big)$$ Then $$frac{dvec{r}}{dt} , = , frac{dvec{r}_Q}{dt} , + , frac{dU}{dt} , vec{R} , = , frac{dvec{r}_Q}{dt} , + , left(frac{dU}{dt} ,U^Tright) big(vec{r}(t) - vec{r}_Q(t)big)$$ 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.