A wheel of radius $R$ spins about its center with a centripetal acceleration of $v^2/R$. I get that the acceleration at all points on the rim of the wheel point towards the center of the wheel. But, what happens when the acceleration of the wheel is slowing down? Let's say the wheel is spinning clockwise.
So, at the top, in the case where the wheel is not spinning, there is only one acceleration vector (pointing down towards the center of the wheel). When the wheel starts to slow down, I would guess that the centripetal acceleration remains at $v^2/R$ but now there's another component - the acceleration vector that points to the left, tangent to the circle, and perpendicular to the centripetal acceleration. So... ←↓
Adding these up gives me the new acceleration, right? How does this relate to the new period of the wheel though? Since the wheel is slowing down, I cannot just use $T = 2 \times \pi \times R / v$... Or can I?