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 $$t = 3.5 \ s$$.

$$\frac{\mathrm{d}\vec\omega_0}{\mathrm{d}t} = \vec\omega' \times \vec\omega_0$$

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.

• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Commented Jun 10, 2022 at 17:07
• It'll be better if you insert an image diagram for this question to get a much better clarification. Commented Jun 11, 2022 at 0:19

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 $$\vec \omega' \times \vec \omega_0$$.