Conceptual Doubts regarding centripetal acceleration of rolling objects Centripetal acceleration of a point with velocity $v$ and moving on a path of radius of curvature $r$ is given by $v^2/r$. If $v= rw$ where $w$ is angular velocity  of a body , then centripetal acceleration is given by $w^2r$.
So, my doubts are as follows:
Doubt 1: in pure rolling motion as velocity of lowermost point is zero so it’s centripetal acceleration should be equal to zero but it is traversing a path of radius of curvature r about the centre of object so it seems contradictory to me . Also, I was taught that acceleration of lowermost point is $w^2r$ and hence it has an acceleration towards the centre of the object which also confuses me.
Doubt 2 : Also, if the lowermost point has a centripetal acceleration and we solve any problem from the frame of reference of the lowermost point in a purely rolling object, then is a pseudo force $mw^2r$ applied on the centre of mass towards the lowermost point ? Or does the centre of mass have a centripetal acceleration towards the lowermost point that means a centrifugal force away from it?
Any help would be greatly appreciated!
 A: Suppose we have a nice rolling wheel, with its center of mass (CoM) at its geometric center. The wheel is rolling on flat ground (so CoM moving at constant velocity).
For doubt 1: there are two types of motion going on. The wheel is rolling, and also translating.
The speed given by $v=\omega r$ is for rolling, and is measured in the center of mass frame (i.e. frame that is moving with the CoM). If you want the velocity with respect to an inertial frame (e.g. you standing still, observing the motion), you also have to account for the velocity of the center of mass.
The acceleration vector in the CoM frame will be centripetal acceleration, $\omega^2 r$. However, if $\alpha\neq0$, within the CoM frame, you will also have tangential acceleration, $r\alpha$ (such as rolling down an incline). If you want the acceleration with respect to an inertial frame, you'd have to account for the acceleration of the CoM.
For doubt 2: depends which frame you're considering.
If you're in an inertial frame (e.g. you, observing a rolling wheel) then there is no pseudo force.
If your frame is moving with a point (i.e. the frame follows a point on the rim), then there will be an outwards pseudo force.
If you're considering a frame that tracks the point of contact on the ground (I think this is the one you're talking about), then there may or may not be a pseudo force.

*

*Case 1: the wheel is rolling on flat ground and CoM has constant velocity. In this case, there are no pseudo forces, since the frame of reference is inertial. There will be a net centripetal force of $\omega^2 r$ applied to the CoM, directed towards the point on the ground.

*Case 2: the wheel is rolling down an incline. In this case, you're in an accelerating frame of reference (since $\bf \vec a_{\rm CoM}\neq 0$, so there will be a pseudo force (it will actually be a force that opposes static friction such that $\sum \bf \vec F= 0$).

Hope this clears your doubts.
