# Centre of instantaneous rotation problem

Is there a point of Centre of Instantaneous Rotation (CIR) for every type of motion or only for cases of rolling?

• What do you think, and why? Jan 27, 2014 at 12:31
• Related post: physics.stackexchange.com/a/88597/392 Jan 27, 2014 at 20:35
• In the plane there is a point and in 3D there is a screw axis (see answer below). The point in 2D is where the 3D screw axis intersects the plane of motion. You can get to all the planar relationships from projecting the 3D problem down to a plane. Jan 28, 2014 at 15:34

For a 3D rigid body there is always an instantenous screw axis. This consists of a 3D line (with direction) and a pitch. The pitch describes how much parallel translation occurs for each rotation of the rigid body. A pure rotation has zero pitch, whereas a pure translation has an infinite pitch. ( 3D Kinematics Ref. html, University of Pennsylvania Presentation ppt, Screw Theory wiki)

## Screw Properties

1. Given a moving rigid body, a point A located at $\vec{r}_A$ at some instant has linear velocity vector at the same point $\vec{v}_A$ and angular velocity $\vec{\omega}$.
2. The screw motion axis has direction $$\vec{e} = \frac{\vec{\omega}}{|\vec{\omega}|}$$
3. The screw motion axis location closest to A is $$\vec{r}_S = \vec{r}_A + \frac{\vec{\omega}\times\vec{v}_A}{|\vec{\omega}|^2}$$
4. The screw motion pitch is $$h = \frac{\vec{\omega} \cdot \vec{v}_A}{|\vec{\omega}|^2}$$

where $\times$ is the cross product, and $\cdot$ is the dot (scalar) product.

## Proof

Image point S having a linear velocity $\vec{v}_S$ not necessarily parallel to the rotation axis $\vec{\omega}$. Working backwards (from S to A), the linear velocity of any point A on the rigid body is

$$\vec{v}_A = \vec{v}_S + \vec\omega \times ( \vec{r}_A-\vec{r}_S)$$

This is used in the screw axis position equation $|\vec{\omega}|^2 (\vec{r}_S-\vec{r}_A) = \vec{\omega} \times \vec{v}_A$ (from above) as

$$|\vec{\omega}|^2 (\vec{r}_S-\vec{r}_A) = \vec{\omega} \times \vec{v}_S - \vec{\omega} \times \vec\omega \times ( \vec{r}_S-\vec{r}_A)$$ which is expanded using the vector triple product as

$$|\vec{\omega}|^2 (\vec{r}_S-\vec{r}_A) = \vec{\omega} \times \vec{v}_S - \vec{\omega} (\vec{\omega}\cdot (\vec{r}_S-\vec{r}_A))+ |\vec{\omega}|^2 (\vec{r}_S-\vec{r}_A)$$ $$\vec{\omega} \times \vec{v}_S = \vec{\omega} (\vec{\omega}\cdot (\vec{r}_S-\vec{r}_A)) =0$$

since right hand side is always parallel to $\vec{\omega}$ and the left hand side is always perpendicular to $\vec{\omega}$. The only solution to the above is the velocity at the screw axis S to be parallel to the rotation

$$\vec{v}_S = h \vec{\omega}$$

and the velocity at A becomes

$$\vec{v}_A = h \vec{\omega} + \vec{\omega} \times (\vec{r}_A-\vec{r}_S)$$

I assume you are talking about a rigid body in motion in a plane.

Consider any two different points on the body, A and B. At any point in time, each one has a velocity vector $\vec{v_A}$ and $\vec{v_B}$ (assuming neither one is, itself, the center).

Consider the line normal to $\vec{v_A}$, call it $n_A$, and likewise $n_B$.

Where these two lines intersect is the instantaneous center. If the two lines are parallel, the motion is pure translation.

If you want to extend it to 3 dimensions, $n_A$ and $n_B$ are planes normal to $\vec{v_A}$ and $\vec{v_B}$. Where they intersect is a line, an "axle" if you like.

• I was confused with why in case of rolling motion the "instantaneous axis of rotation"(IAR) was introduced. Why couldn't we just use axis through the center of mass. Looked for (1) what IAR actually is, (2)should it always be zero (in the case of rolling motion on a surface), and (3)can there exist another such axis rather than the point of contact between the surface and the object for which the answers I were mathematical, so I couldn't get that. Could you help me with that? I didn't post the question separately cause it'd probably be closed as duplicate.
– suiz
Apr 29, 2018 at 4:47
• Well, I know what IAR is and based on your answer the third question can be answered; for an object rolling on a surface it is the only instantaneous axis of rotation, and I think for every instant it is at rest. Am I right?
– suiz
Apr 29, 2018 at 6:09
• @suiz: I'm not sure I understood your question(s), but the simplest example is a wheel rotating on a roadway. Suppose you take a photograph with very short exposure but not instantaneous. In that short exposure time, every point on the wheel will be seen to be tracing a circular path around the point of rolling contact. Sometimes these things are hard to understand if you haven't yet learned calculus, because in calculus you deal with infinitesimal quantities - quantities that are as small as possible without being zero. Apr 29, 2018 at 20:04

Instantaneous rotation axes appear just studying the motion of rigid solid bodies.

Consider a rigid solid body ${\cal B}$ moving in the three space. To study its motion, fix a point $O \in {\cal B}$ and a triple of orthonormal axes ${\bf k}_1$, ${\bf k}_2$, ${\bf k}_3$ at rest with ${\cal B}$ centred at $O$.

We can now describe the motion of ${\cal B}$ with respect to a fixed orthonormal triple of axes ${\bf e}_1$, ${\bf e}_2$, ${\bf e}_3$.

If $P\in {\cal B}$ is a particle of matter of ${\cal B}$ determined by ${\bf x}_P = \sum_{i=1}^3 x_{Pi} {\bf k}_i$, and these components do not change in time just because ${\cal B}$ is a rigid body, its position ${\bf y}_P(t)$ in the space is given by: ${\bf y}_P(t)= {\bf y}_O(t) + {\bf x}_P$ that is, in components: $$y_{Pi}(t) = y_{Oi}(t) + \sum_{j=1}^n R_{ij}(t) x_{Pj}\quad (1)$$
where ${\bf k}_j(t) = \sum_{i=1}^3 R_{ij}(t){\bf e}_i$ and $R(t) \in O(3)$ is a given rotation.

Now consider the $t$-derivative for $t=0$, when ${\bf k}\equiv {\bf e}_i$, of (1). We can fix arbitrarily the instant $t=0$ changing the origin of time so this value does not play any fundamental role and we can re-define the triple of ${\bf e}_i$ in order that ${\bf k}(0)\equiv {\bf e}_i$ is valid for $i=1,2,3$.

$$\frac{dy_{Pi}}{dt}|_{t=0} = \frac{dy_{Oi}}{dt}|_{t=0} + \sum_{j=1}^n \frac{dR_{ij}}{dt}|_{t=0} x_{Pj}\quad (2)\:.$$

This identity can be used to study the first approximation of the motion of the body ${\cal B}$ in a neighbourhood of $t=0$:

$$y_{Pi}(t) = y_{Pi}(0) + \frac{dy_{Pi}}{dt}|_{t=0} t + O(t^2)$$

so that, exploiting (2):

$$y_{Pi}(t) = y_{Pi}(0) + \frac{dy_{Oi}}{dt}|_{t=0}t + \sum_{j=1}^n \frac{dR_{ij}}{dt}|_{t=0} x_{Pj}t + O(t^2)\qquad (3)\:.$$

Using the Lie group structure of $O(3)$ (or also by direct inspection), it is possible to prove that, as $R(0)=I$, there exists a vector $\omega(0)$ such that ($^*$):

$$\frac{dR}{dt}|_{t=0} = \omega(0) \times \qquad (4)\:.$$ Finally evaluating (1) for $t=0$ we find $${\bf y}_P(0) = {\bf y}_O(0) + {\bf x}_P(0)\qquad (5)$$ where all vectors are indifferently decomposed w.r.to the basis of the ${\bf e}_i$s or that of ${\bf k}_i$s, just because they coincide for $t=0$. Inserting (4) and (5) in (3), we eventually achieve:

$${\bf y}_{P}(t) = {\bf y}_{P}(0) + {\bf v}_O(0) t + \omega(0)\times {\bf y}_p(0)t + O(t^2)\qquad (6)$$

where, obviously ${\bf v}_O(t):= \sum_i \frac{dy_{Oi}}{dt}|_{t=0} {\bf e}_i$.

For a generic instant $t_0$, defining $\Delta t = t-t_0$ we would similarly obtain:

$${\bf y}_{P}(t) = {\bf y}_{P}(t_0) + {\bf v}_O(t_0) \Delta t + \omega(t_0)\times ({\bf y}_P(t_0)- {\bf y}_O(0))\Delta t + O(\Delta t^2)\qquad (7)$$

Eq.(7) says that, in the neighbourhood of every instant ($t=t_0$ in our case), the motion of ${\cal B}$ is the superposition of a spatial translation along ${\bf v}_O(t_0)$ and a rotation around the unit vector parallel to $\omega(t)$ passing through the instantaneous centre $O(t)$. The axis is the instantaneous rotation axis by definition.

Using (7) that is valid for every choice of $O$, if the motion of not of pure translation, we can always change $O$ in order that at the interesting time ${\bf v}_O(t_0) \times \omega(t_0)=0$ so that ${\bf v}_O(t_0)$ and $\omega(t_0)$ are parallel. Notice that the new $O(t_0)$, in general, is not a point of matter of ${\cal B}$ but a geometric point in the space. In this case (7) reduces to a pure rotational motion around $O(t_0)$ plus a translation along the rotational axis (in a neighbourhood of the considered instant of time). This point $O(t_0)$ is an the instantaneous rotation center. Actually there is a whole axis with the same property: that passing for the found $O(t_0)$ directed along $\omega(t_0)$.

Footnotes.

$(^*)$ As $t \mapsto R(t)\in O(3)$ and $R(0)=I$, then $dR/dt|_{t=0}$ is an element of the Lie algebra of $O(3)$. The Lie algebra of $O(3)$ is made of all real antisymmetric $3\times 3$ matrices. If $A$ is such a matrix, it immediately arises that there is a vector $\omega_A$ such that $A{\bf u} = \omega_A \times {\bf u}$ for all vectors ${\bf u}$.

• You mentioned using the Lie group structure of $O(3)$.Could you please elaborate further on that subtle point?.Because that statement was really the crux of your proof. Jan 27, 2014 at 16:12
• The Lie algebra of $O(n)$ is made of all antisymmetric $n\times n$ matrices (this can be easily proved). For $n=3$ an antisymmetric matrix $A$ is always of the form $\omega_A \times$ for some vector $\omega_A$. Jan 27, 2014 at 16:16
• As $t \mapsto R(t)\in O(3)$ and $R(0)=I$, $dR/dt|_{t=0}$ is an element of the Lie algebra of $O(3)$. Jan 27, 2014 at 16:17

The fact you are stating is quite general in fact and even extends in a related form to 3 dimensions also.

It is known as Chasles's rotation theorem: Any general displacement of a rigid body can be represented by a translation plus a rotation.

In the case of motion of a body in a plane,the axis intersects the given plane in a point which we can call the instantaneous centre of rotation.Even in the case if doesn't intersect,we say the centre of rotation is at infinity.

So,yes any motion of a body in a plane has an instantaneous axis of rotation.