# 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.