its momentarily tangent speed is 0 (since we are considering pure rotation), hence the normal acceleration is also 0
This is an incorrect argument. A velocity of zero in no way implies an acceleration of zero. Also, the tangential speed and the centripetal acceleration are only related by that simple formula in the frame where the axle is at rest.
To calculate the acceleration you must take $\frac{d^2}{dt^2}x$. There is no shortcut by calculating only $\frac{d}{dt}x$, even for rolling.
For the point $\phi$ on the edge of the wheel the position is $$\left( r \cos(\omega t+\phi)+r\omega t,r \sin(\omega t +\phi) \right)$$
The acceleration and velocity can easily be calculated, but it is clear that the $r\omega t$ term has a non-zero first derivative but a zero second derivative