Angular Acceleration of Spinning Body About a Rotating Axis How would you find the angular acceleration of a body spinning about an axis that is itself rotating? Specifically, how would you find the angular acceleration in  question 1.58 of Irodov's physics book.

A solid body rotates with a constant angular velocity $\omega_0 = 0.50$ rad/s about a horizontal axis $AB$. At the moment $t = 0$, the axis $AB$ starts turning about the vertical with a constant angular acceleration $\alpha = 0.10 \ rad \ s^{-2}$. Find the angular velocity and angular acceleration of the body after $t = 3.5 \ s$.

The answer key gives:
$$\frac{\mathrm{d}\vec\omega_0}{\mathrm{d}t} = \vec\omega' \times \vec\omega_0$$
where $\vec\omega' = \alpha t$, is the angular velocity of the axis $AB$, but I have no idea why this is true. Any help would be appreciated.
 A: Let's first start with a picture.  The object (blue ball) is rotating around the AB axis with angular velocity $\vec\omega_0$.

Next, the $AB$ axis is rotating around the vertical with an angular velocity $\vec \omega '$. We are told that $\omega'$ is constantly being accelerated by an angular acceleration $\alpha$, and from our kinematic equations (recall $\displaystyle \alpha = \dfrac{d\omega'}{dt} \implies \omega' = \int\alpha \, \mathrm dt = \alpha t $) we find that $\vec \omega ' =
\alpha t \hat z$.
We need to also find out what $\vec \omega_0$ is -- and let's work in cylindricalcoordinates here -- since the $AB$ axis is always rotating in the $xy$ plane, we can write $\vec\omega_0 = \omega_0 \hat r$ where $\hat r$ is the radial unit vector pointing along $\vec\omega_0$.
The angular acceleration of the body is
\begin{align}
\dfrac{d\vec\omega_0}{dt} &= \dfrac d{dt} \left( \omega_0\hat r \right)\\
&= \omega_0\dfrac{d\hat r}{dt}
\end{align}
Using the cartesian basis, one can show that $\dfrac{d\hat r}{dt}=\omega' \hat \varphi$ to get $$\dfrac{d\vec\omega_0}{dt} = \omega_0 \omega'\hat\varphi.$$
Next, recall in cylindrical coordinates that $\hat\varphi = \hat z \times \hat r$ so we can substitute that in and get $$\dfrac{d\vec\omega_0}{dt} = \omega' \hat z\times \omega_0\hat r$$ which is exactly $\vec \omega' \times \vec \omega_0$.
