# Radius of curvature and Instantaneous Axis of Rotation

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.