Is there a point of Centre of Instantaneous Rotation (CIR) for every type of motion or only for cases of rolling?
[Physics] Centre of instantaneous rotation problem
newtonian-mechanicsrigid-body-dynamicsrotational-dynamics
Related Solutions
I think the orientation of the IAOR is mainly due to mathematical reasons. With the rotation equation: $\vec{v_A}=\vec{\omega}\times\vec{r_{CA}}$ I.e. for a pure rotation about the z-axis, there must not be a velocity along the z-axis, and the z-component of $\vec{v_A}$ is only zero, if $\vec{\omega}$ is along the z-axis.
The ICOR has by definition zero velocity at this moment, with respect to a certain frame of reference. Take for axample a rolling cylinder (wheel). If you take the non-moving floor as reference frame, the point of the wheel touching the floor must be the ICOR, because if that point was moving, slipping would occur. But if you take the moving wheel itself as the reference frame, the center of the wheel becomes the reference frame, as from it's perspective the wheel is only rotating and the ground is moving with the wheel, so slipping doesn't occur.
Consider this graphic:
The red x marks the respective frame of reference.
In case 2 of the image the frame of reference is at the center of the wheel. From it's perspective it is simply rotating around the center, thus the given velocity distribution (in blue).
But from the frame of reference of a person standing on the floor, e.g. the point of contact (case 1 and 2) the whole wheel is also moving to the right, consisting of an overall translational velocity to the right (case 3). So the velocity distributions of case 2 and 3 (literally) add up to case 1. Now the ICOR is simply the point where there is no velocity at this instant, which turns out to be the point of contact. This is true for the point on the wheel where it touches the ground, as well as for the point on the ground.
Best Answer
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
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
$$ \vec{v}_A = h \vec{\omega} + \vec{\omega} \times (\vec{r}_A-\vec{r}_S) $$