Angular Acceleration about stationary point in pure rolling

Question

I was reading David Morin's "Introduction to classical mechanics ", Chapter Angular Momentum Part I, where I found this paragraph. " Invariably, we will calculate angular momentum and torque around either the CM or a fixed point (or a point that moves with constant velocity, but this doesn’t come up often). These are the “safe” origins, in the sense that Eq. (8.46) holds. As long as you always use one of these safe origins, you can simply apply Eq. (8.46) and not worry much about its derivation."

Here Eq. (8.46) is the equation $ \tau = I\alpha $ where $ \tau$ represents torque about the axis, I is the moment of inertia about the axis and $ \alpha $ is the angular acceleration. I want to know why is the angular acceleration same about the Centre of mass and the stationary point( Which is the point of contact in pure rolling ). I cannot get the intuition behind this concept. I tried searching in "Introduction to Mechanics " By Kleppner and Kolenkow but could not find there to. If you guys could just help me to understand why is angular acceleration same in COM frame and the stationary point frame, I would be very grateful. There is another question on this topic but it is quite unclear.

Answer 1

Angular acceleration is the same with respect to any two parallel axes crossing the rigid body, not only those that cross it through the points you mention. Its easier to prove first that angular velocities are the same for every time $t$, and then by differentiation you get that angular accelerations are also the same.

Consider a body rotating with respect to point $O$ with angular velocity $\vec{\omega}$, and two points $A$ and $B$. Point $B$ performs a roto-translation with respect to $A$ so $\vec{v}_{BA} = \vec{\omega}^* \times \vec{r}_{BA}$. We need to show that $\vec{\omega}^* = \vec{\omega}$

We can write $\vec{v}_A = \vec{\omega} \times \vec{r}_{A0}$ and $\vec{v}_B = \vec{\omega} \times \vec{r}_{B0}$, and subtracting the first equation from the second one $\vec{v}_{BA} = \vec{\omega} \times \vec{r}_{BA}$, so we have $\vec{\omega}^* \times \vec{r}_{BA} = \vec{\omega} \times \vec{r}_{BA}$. From here and the fact that $\vec{\omega}^*$ and $\vec{\omega}$ are parallel (because we are considering parallel axes) it follows that $\omega^* = \omega \implies \vec{\omega}^* = \vec{\omega}$. This derivation holds for every time $t$ so you can differentiate and get $\frac{d\vec{\omega}^*}{dt} = \frac{d\vec{\omega}}{dt}$.