Radius of curvature and Instantaneous Axis of Rotation For a rigid body undergoing planar motion I can associate to it the concept of Instantaneous Axis of rotation (IAOR), the point from which I can see the body undergoing pure rotation. Consider any point on the rigid body. 
Here is my concern. Why is the radius of curvature of the point's trajectory and the it's distance from the IAOR not the same? Take for example a rolling disc on a smooth horizontal surface. 
The radius of curvature of the top most point is $4r$ but the distance of that point from the axis of rotation is $2r$ (Here the IAOR is the bottom most point of the disc). What I think is that it's distance of the point from IAOR should be the same as the radius of curvature. Where am I wrong? 
I calculated the radius of curvature of the topmost point. It's velocity w.r.t. ground is $2v$, so according to the equation of kinematics centripetal acceleration of a particle whose trajectory is known beforehand would be 
$$R_c=\frac{(2V_0)^{2}}{a_c}.$$
The centripetal acceleration would remain the same $=\omega^{2}r$ or $\frac{V_0 ^2}{R}$. Plugging in the values we get the radius of curvature $=4R$ ( where $R$ is the radius of the rolling body).
 A: Let us start by setting some parameters.
The body is rotating with angular speed $\omega$ and its centre of mass is moving translationally with velocity $\omega r$.
Centripetal acceleration of the uppermost point about centre will be given by, 
$$a_c=\omega^2r \tag 1$$
For a purely rolling body, point of contact will be the instantaneous axis of rotation. The angular speed of the rigid body will still be $\omega$ about this axis. (You can prove it by dividing the velocity of uppermost point with respect to the IAOR which will be $2v$ and its the distance which will be $2r$, and that will leave you with $\omega$.)
About this axis, centripetal acceleration of the upper most point will be given by, $A_c=\omega^{2} (2r)$
$$\therefore A_c=2a_c\tag 2$$
The centripetal acceleration of the uppermost point about IAOR can also be given by, $A_c=\frac{v^2}{R'}$, where $R'$ is the radius of curvature.
$$R'=\frac{v^2}{A_c}$$
For a purely rolling body, velocity of the uppermost point is $2\omega r$ and zero of the POC.
From equations $(1)$ and $(2)$,
$$R'=\frac{4\omega^2r^2}{2\omega^2 r}=2r$$
I thing the mistake you were doing is assuming the centripetal acceleration of the uppermost point about the IAOR to be equal to $\omega^2 r$, but actually it is $2\omega^2 r$.
Now this kinda makes sense doesn't it? Let's try doing a reverse calculation and try to find $A_c$ about the IAOR.
\begin{align}
A_c&=\frac{v^2}{2r} \\
&=\frac{4\omega^2r^2}{2r} \\
&=2\omega^2r
\end{align}
And from equation $(1)$, $A_c=2a_c$, which completely makes sense.
A: The radius of curvature equals the distance to the instantaneous axis of rotation (IAOR). This is true due to the definition of the IAOR: At a given instant there is an axis perpendicular to the plane of motion such that any point of the body is in pure rotation, i.e. has circular trajectory about that axis. If a point is in circular motion about the IAOR, then the radius of curvature equals the distance to that axis as it is shown in the figure bellow:

A: My answer is very simple. In this answer, I will not show any rigorous calculations, just the concept.
In the first case, you should understand that the IAOR is also known as the point of ZERO VELOCITY(in the ground frame).
HOWEVER- Please note that IAOR might not be the point of zero acceleration
IAOR is that point about which it can be assumed that the body performs pure rolling only at that instant. It has made finding velocity easier if you go by vectors.
However you should be aware of the fact that radius of curvature is a mathematical concept.
Radius of curvature is actually the radius of the largest circle that can touch the curve at a single point.It is related to the trajectory of the point
CONCLUSION - We can clearly see that they are two different parameters one being completely physical while the other mathematical (however can be computed by physical means as Mitchell has done)
I stumbled upon this problem myself at high school and got over it after some struggle and brainwracking
You might try this problem out to understand the concept better.

Hope this helps 
