Instantaneous Centre of Rotation Let's say a body is undergoing both rotational and translational motion.
I know that ICR of the body as a whole will be the point about which the body is doing pure rotation, so basically will be the point with zero velocity, and that it will lie where perpendiculars to the velocities of each point intersect. I heard about a concept of ICR of each point, defined as "The point about which the given point rotates". Can you please explain this concept, of individual ICRs? Also, how can we calculate its location, and is it same as the centre of curvature for that point? 
Please take a look at this - 
As you can see, the ICR of the yellow point lies on the yellow broken line, and its distance is given by 4R and not 2R. Similarly of the green one is double that of its distance from the common ICR. The point where they both intersect is the ICR of the body as a whole, but then the individual points have different ICRs too. Please explain.
 A: Well, you've answered yourself a little bit. The instantaneous center of rotation is the point about which "the whole body" is performing pure rotational motion, so the ICR of each individual point of that body will be the same as the ICR for the entire body. To find this ICR, I think this image from wikipedia says it all:
"Pole-object-A1-A2" by VanBuren - Own work. Licensed under CC BY-SA 3.0 via Commons.
You take two points in the body, and find their velocity vectors. Bisect each velocity vector, and the intersection point of the bisections is the ICR. Therefore, to find the ICR, you need at least two points in the body, but once you've found it, you can speak of "the ICR of each point in the body".
The center of curvature is the point about which a point is moving in a circle. Since we find the ICR by using perpendicular vectors to the velocity vectors, the velocity vectors are tangents to a circle of radius equal to the distance to the ICR. Therefore, the ICR is the instantaneous center of curvature.
A: In the 2D case each point (let's call it A) on the rigid body has two velocity components $(v_x,v_y)$ and the body itself rotates by $\omega$. The location of the ICR relative to A is $$icr = \begin{pmatrix} -\frac{v_y}{\omega} & \frac{v_x}{\omega} \end{pmatrix}$$
To prove this check that $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times \vec{r} = 0$ or with planar vectors $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} =
\begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} + \begin{pmatrix}0\\0\\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} $$
$$ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v_x - \omega \,y \\ v_y + \omega \,x \\ 0 \end{pmatrix} $$
$$ \begin{matrix} x=-\frac{v_y}{\omega} \\ y=\frac{v_x}{\omega} \end{matrix} $$
Also, since the body is rotating about the ICR the point A prescribes a circle (instantaneously) and thus yes the ICR is the center of curvature of all points on the rigid body.

In the 3D case a rigid body will rotate about an axis and it might have a parallel translation to that axis as well. Given a point A (on the body) with velocity $\vec{v}_A = (v_x,v_y,v_z)$ and rotational velocity of the body $\vec{\omega} = (\omega_x,\omega_y,\omega_z)$ the location if the ICR is given by
$$\vec{r}_{icr} = \frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} $$
In addition, the direction of rotation is $\vec{e} = \frac{\vec{\omega}}{\|\vec{\omega}\|}$ and the vector of any parallel translational motion is $$\vec{v}_\parallel = \left( \frac{\vec{\omega} \cdot \vec{v}_A}{\|\vec{\omega}\|^2}\right) \vec{\omega}$$
The scaling factor in the parenthesis above is called the screw pitch because the motion of rotation with parallel translation is a screw motion.
To prove this, look at the well known transformation law $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times  \vec{r}_{icr} =0 $ and cross each side with  $\vec{\omega}$
$$  \begin{array}{rcl}0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}_{icr}\right)\\0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\left(\vec{\omega}\cdot\vec{r}_{icr}\right)-\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)\\\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)&=&\vec{\omega}\times\vec{v}_{A}\\\vec{r}_{icr}&=&\frac{\vec{\omega}\times\vec{v}_{A}}{\vec{\omega}\cdot\vec{\omega}}=\frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} \end{array} $$
In step 3 above it is assumed that $\vec{\omega}\cdot \vec{r}_{icr} = 0$ which turns out to be true given that $\vec{r}_{icr}$ is perpendicular to the rotation axis.
