Rolling without Slipping - axes of rotation and centripital acceleration

[lightlightsup]

Therefore, if someone were to ask what the magnitude of centripetal acceleration is at the top of the wheel at a given instant (relative to the ground):
$v_{cm} = v_{translational, center-of-mass/wheel}$
$ω = ω_{point-of-contact}$
$v_{top} = 2(v_{cm}) = 2(rω)$
$a_{c(top)} = \frac{v_{top}^2}{R} = \frac{(2v_{cm})^2}{2r} = \frac{2(v_{cm})^2}{r} = \frac{2(ωr)^2}{r} = 2ω^2r$
I think this is correct.

But, what is the $ω$ and $α$ about the center of the wheel?
Also, what is the $a_{c(top)}$ relative to the center of wheel? Would that even make sense?See this image:

 

[A.T.]

Therefore, if someone were to ask what the magnitude of centripetal acceleration is at the top of the wheel at a given instant (relative to the ground):
$v_{cm} = v_{translational, center-of-mass/wheel}$
$ω = ω_{point-of-contact}$
$v_{top} = 2(v_{cm}) = 2(rω)$
$a_{c(top)} = \frac{v_{top}^2}{R} = \frac{(2v_{cm})^2}{2r} = \frac{2(v_{cm})^2}{r} = \frac{2(ωr)^2}{r} = 2ω^2r$
I think this is correct.

For a wheel rolling at constant speed, the accelerations at the rim must be same in the ground frame and in the wheel center frame: $ω^2r$. You cannot derive the acceleration in the ground frame like you did, by using a moving center for the centripetal acceleration.

See also:
https://www.raeng.org.uk/publications/other/20-wheels-bssc