Screw Axis (ISA) and its usefulness the instantaneous screw axis is a useful concept in kinematics. For example, in planar motion, the instantaneous displacement is a pure rotation if we consider the center of instantaneous rotation (CIR) which is the 2D equivalent of the ISA with the axis parallel to the plane.
 In many kinematic problems involving linkages, link chain problems (multiple bodies connected to each other and constrained), the CIR is a very used concept to find the velocity and angular velocities of the various parts...Why is the CIR so useful? I am missing the point...
Also, I was thinking about a different concept, i.e. the axis whose point has the same acceleration and the acceleration vectors are parallel to the axis and have the smallest magnitude... Does this concept have any use? 
Thanks!
 A: The concept if ISA brings forth the geometry of motion. The geometry is important because you can establish relationships between concepts. The whole idea of screws as presented by Stalwell Ball and Chasles is that physical concepts of rigid body motion (and later momentum and forces) has the geometry of a line in space together with a scalar magnitude and pitch (parallel motion).
The general instantaneous motion of a 3D body is fully described by the ISA. Thus the geometry of kinematics is encoded into the ISA.
Kinematics ≡ ISA

$$ \boldsymbol{\hat{v}} = \pmatrix{ \boldsymbol{v} \\ \boldsymbol{\omega} } =  \omega \pmatrix{ \boldsymbol{r} \times \boldsymbol{z} + h \boldsymbol{z} \\ \boldsymbol{z} } $$
$\boldsymbol{v}$ velocity, $\boldsymbol{\omega}$ angular velocity, $\boldsymbol{z}$ screw axis direction, $h$ screw pitch, $\boldsymbol{r}$ location of screw.
There is also another screw axis in space that is important in dynamics. This is the instant axis of percussion (IAP) and it encodes the geometry of momentum.
Dynamics ≡ IAP

$$ \boldsymbol{\hat{p}} = \pmatrix{\boldsymbol{p} \\ \boldsymbol{L} } = p \pmatrix{ \boldsymbol{z} \\ \boldsymbol{r} \times \boldsymbol{z} + h \boldsymbol{z} } $$
$\boldsymbol{p}$ momentum, $\boldsymbol{L}$ angular momentum, $\boldsymbol{z}$ screw axis direction, $h$ screw pitch, $\boldsymbol{r}$ location of screw.
The relationship between these two axes is quite important in dynamics. A rigid body has a 1:1 map between all the ISA and all the IAP through the 6×6 mass matrix $\mathbf{\hat{I}}$
$$ \boldsymbol{\hat{p}} = \mathbf{\hat{I}} \boldsymbol{\hat{v}} $$
and its time derivative (the equations of motion in screw theory)
$$ \boldsymbol{\hat{f}} = \mathbf{\hat{I}} \boldsymbol{\hat{a}}  + \boldsymbol{\hat{v}} \times  \mathbf{\hat{I}} \boldsymbol{\hat{v}} $$
So, for example, a constraint force that does no work should obey the algebraic relationship $$ \boldsymbol{\hat{f}} \cdot \boldsymbol{\hat{v}} = 0$$ 
but the above has a geometric interpretation, that of a reciprocal screw. Think of it as a general perpendicularity. In 2D a force screw is a line on the plane, and the motion is a rotation about a point. All the forces that do no work, are the lines passing through that point. 
Also in 2D if the ISA of a body is a distance $c$ from the center of mass, and the body has radius of gyration  $\kappa$, then the IAP line goes through a point $\kappa^2/c$ on the other side of the center of mass.

Think about this. For any rigid body, if you know where the ISA is, then you know where the percussion axis is, which means that applying an impulse along the IAP can instantaneously stop the body from moving.

Link to a primer on screws answer in [Physics.SE] for completeness.


Back to the more specific question of why ISA is useful, well in 2D see this section I authored in Wikipedia about relative centers of rotations and how they can be used to describe the gearing (relative rotational motion) between two contacting bodies.
Since the relative motion of two bodies is also a screw, in 2D the relative motion of two bodies is a point that is colinear to the two ISA of the two bodies.

Consider two bodies in contact, where to contact point is only allowed to slide. Where the contact normal line intersects the line connecting the two ISA centers, is where the relative ISA is. If the distances from the relative center to the two ISA points are $r_1$ and $r_2$ respectively, then the relative gearing between the bodies is $$ \gamma = \frac{\omega_1}{\omega_2} = \frac{r_2}{r_1} $$
