Centre of instantaneous rotation problem

Question

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

Answer 1

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

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

Answer 2

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.

Answer 3

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.

Answer 4

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 $\omega(t_0)$.

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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 ${\bf u}$.