Centripetal acceleration of centre of mass towards instantaneous axis of rotation (IAOR)

Question

When an object purely rolls on a horizontal surface, the centre-of-mass moves in purely circular motion, with respect to the instantaneous axis of rotation (IAOR).

So does the centre of mass have an acceleration of ${\omega}^2 R$ towards the IAOR?

Like in this case, should the normal contact force be:

$$F = mg - \frac{mv^2}{R} \ ?$$

If no, why not? (The body is purely rolling)

I found this question here: Centripetal acceleration of Centre of mass of rolling body

In the answer given by Bill, I don't understand why $R$ has to be the curvature of the floor. Shouldn't it be the radius of the circle, followed by the centre-of-mass of the body which is $R$ in this case?

Answer 1

At a particular instant, every point of a rolling body has the same velocity as an identical body that is rotating around the point of contact. But the points have different accelerations, and therefore different forces.

It might help to think of a body rolling along on a surface in space without gravity. The COM travels in a straight line at uniform velocity. Individual points travel in circles around the COM. The forces that deflect them from straight paths are internal forces.

Suppose you add an external force. You could make the rolling body travel along a circular surface. For each point, the external force is added to the internal forces. The centripetal acceleration is added to the acceleration from internal forces.