Is the tangential/linear acceleration in rotational dynamics equal to the COM acceleration only in a pure rolling case?

Question

So after I learnt rotational dynamics I understood that the linear velocity is equal to the angular velocity times the radius and if we derive with respect to time, we get the linear/tangential acceleration:

$ a_{t} = \alpha R $

However, after reading about the pure rolling/slipping/skidding cases, I'm not sure I understand if the previous equation is valid for all three cases.

My logic is that if the ball spins faster (or slower) then it moves, i.e. not pure rolling:

$ v_{cm} \neq \omega R $

Therefore:

$ a_{cm} \neq \alpha R $

And

$ a_{t} \neq a_{cm} $

Answer 1

You are not completely right. $v\neq \omega r$ is possible even in the case of pure rolling. Consider a case where a sphere is rolling over a rough plank, which itself is moving forward with some speed $v_p$. In the case of pure rolling, the equation would be: $$v_{com}-\omega r=v_p$$ Which means that $v\neq \omega r$. By extension of this same concept it is possible that $a\neq \alpha r$ even in a pure rolling case, when the surface on which the object is rolling is itself accelerating. The main concept behind pure rolling is that the point of contact should be at relative rest with the surface on which it rolls. While it is rare to encounter cases where the surface itself would accelerate, it is very much a possibility.