Rolling without slipping means by definition that the contacting points have the same velocity. For a stationary ground the velocity is zero, so the contact point on the body must also have zero velocity.

We begin by observing that the sign conventions in Figure #rko-ff mean that $\vec\omega = -\omega\,\hat{e}_b$. Now rolling without slipping means the contact point $A$ must instantaneously have zero velocity, so using #rkg-er gives: $$\begin{aligned} \vec{v}_C &= \vec{v}_A + \vec{\omega} \times \vec{r}_{AC} \\ &= (-\omega \,\hat{e}_b) \times r \,\hat{e}_n \\ &= r \omega \,\hat{e}_t. \end{aligned}$$ Because the surface is flat, the tangential basis vector $\hat{e}_t$ is constant, and the radius $r$ is also constant. Differentiating the velocity expression thus results in: $$\begin{aligned} \vec{v}_C &= r \omega \,\hat{e}_t \\ \dot{\vec{v}}_C &= r \dot\omega \,\hat{e}_t \\ \vec{a}_C &= r \alpha \,\hat{e}_t. \end{aligned}$$

The contact point $P$ and center $C$ are offset by the constant vector $r\,\hat{e}_n$, so $\dot{s} = v_P = v_C = r\omega$, from #rko-ef. Because $r$ is constant, differentiating the velocity expression gives $\ddot{s} = r \dot\omega = r \alpha$, while integrating with zero initial displacement gives $s = r \theta$.

By definition of non-slip rolling contact, the point of contact $P$ has zero velocity. The acceleration can be computed from the center $C$ with #rkg-e2: $$\begin{aligned} \vec{a}_P &= \vec{a}_C + \vec{\alpha} \times \vec{r}_{CP} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP}) \\ &= \alpha r \,\hat{e}_t + (-\alpha\,\hat{e}_b) \times (-r\,\hat{e}_n) + (-\omega\,\hat{e}_b) \times \Big((-\omega\,\hat{e}_b) \times (-r\,\hat{e}_n)\Big) \\ &= \alpha r \,\hat{e}_t - \alpha r \,\hat{e}_t + \omega^2 r \,\hat{e}_n \\ &= \omega^2 \,\vec{r}_{PC}. \end{aligned}$$

These formulas are all choices of sign conventions for $\omega$ and $\alpha$, and definitions of $R$. Figure #rko-fc shows the appropriate geometry. Note that $\omega$ and $\alpha$ are defined with positive values corresponding to motion in the tangential direction.

The two configurations shown in Figure #rko-eg must be considered individually. We will do the derivation for the case of rolling inside (the left diagram), as the rolling-outside case is very similar, but with different signs.

We begin with the point of contact $P$, which has zero velocity and so it is the instantaneous center of the body. Using #rkg-er with the expressions #rko-eg thus gives the velocity expression: $$\begin{aligned} \vec{v}_C &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PC} \\ &= (-\omega\,\hat{e}_b) \times r\,\hat{e}_n \\ &= r \omega \,\hat{e}_t. \end{aligned}$$ Differentiating this velocity and using #rkt-ed gives the acceleration: $$\begin{aligned} \vec{a}_C &= r \dot{\omega} \,\hat{e}_t + r \omega \,\dot{\hat{e}}_t \\ &= r \alpha \,\hat{e}_t + r \omega \frac{\dot{s}}{\rho} \,\hat{e}_n. \end{aligned}$$ To replace $\dot{s}$ in this acceleration expression, we first consider the instantaneous center $M$, which moves along the surface with speed $\dot{s}$ and tangential acceleration $\ddot{s}$. Thus from #rkt-ev: $$\begin{aligned} \vec{v}_M &= \dot{s} \,\hat{e}_t. \end{aligned}$$ We can now compute the velocity of $C$ from $M$, as follows. Note that we cannot use equation #rkg-er because $M$ and $C$ are not fixed to the same body ($M$ is not fixed to the body). Thus: $$\begin{aligned} \vec{r}_C &= \vec{r}_M + r\,\hat{e}_n \\ \vec{v}_C &= \vec{v}_M + r\,\dot{\hat{e}}_n \\ &= \dot{s}\,\hat{e}_t - r \frac{\dot{s}}{\rho} \,\dot{\hat{e}}_t \\ &= \frac{(\rho - r)}{\rho} \dot{s} \,\hat{e}_t \\ &= \frac{R}{\rho} \dot{s} \,\hat{e}_t. \end{aligned}$$ Comparing this expression for $\vec{v}_C$ to the one above, we see that: $$\begin{aligned} r \omega &= \frac{R}{\rho} \dot{s} \\ \dot{s} &= \frac{\rho r}{R} \omega. \end{aligned}$$ This expression for $\dot{s}$ can be substituted back into our earlier equation for $\vec{a}_C$, giving: $$\begin{aligned} \vec{a}_C &= r \alpha \,\hat{e}_t + r \omega \frac{1}{\rho} \left( \frac{\rho r}{R} \omega \right) \,\hat{e}_n \\ &= r \alpha \,\hat{e}_t + \frac{(r \omega)^2}{R} \,\hat{e}_n. \end{aligned}$$ The case of rolling on the outside of the curved surface (the right diagram in Figure #rko-fc) is similar to the above but with the different sign conventions from #rko-eg. The final expressions in #rko-ec apply to rolling on both the inside and outside.

See the derivation for #rko-ec, where the relationship between $\omega$ and $\dot{s}$ is obtained. If the surface is exactly circular then by definition $\rho$ and $R$ are constant, so differentiating the angular velocity expression immediately gives the angular acceleration equation.

The derivations for rolling on the inside or outside of a surface are different. Here we show the rolling inside case, as the outside case is very similar.

By definition the contact point $P$ has zero velocity. The acceleration can be computed from the center $C$: $$\begin{aligned} \vec{a}_P &= \vec{a}_C + \vec{\alpha} \times \vec{r}_{CP} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP}) \\ &= r \alpha \,\hat{e}_t + \frac{(r\omega)^2}{R} \,\hat{e}_n + (-\alpha \,\hat{e}_b) \times (-r\,\hat{e}_n) + (-\omega \,\hat{e}_b) \times \Big( (-\omega \,\hat{e}_b) \times (-r\,\hat{e}_n) \Big) \\ &= r \alpha \,\hat{e}_t + \frac{(r\omega)^2}{R} \,\hat{e}_n - r \alpha \,\hat{e}_t + r \omega^2 \,\hat{e}_n \\ &= \frac{(r + R)}{R} r \omega^2 \,\hat{e}_n \\ &= \frac{\rho}{R} \omega^2 \,\vec{r}_{PC}. \end{aligned}$$