Relative angular velocity and acceleration Background: (Irodov 1.55) Two bodies rotate around intersecting perpendicular axes with angular velocities $\hat\omega_1,\hat\omega_2$. Relative to one body, what is the angular-velocity and -acceleration of the other?
Irodov's answer implies that 
$$\hat \omega=\hat\omega_1-\hat \omega_2$$
$$\hat \alpha=\hat\omega_1\times\hat \omega_2.$$
I have a hard time grokking why the above are true above (for the first) vague analogies with linear velocity. Does anyone, willing to share, have an intuitive grasp on the above equations?
 A: Lets pick a coordinate system where the second body rotates about the local $Z$ axis.
The rotational kinematics if the second body are defined as
$$ E_2 = E_1 {\rm Rot}(\hat{z}, \theta)$$
where $E_i$ are the 3×3 rotation matrices, and $\hat{z}=(0,0,1)$ . If the angular velocity of the first body is $\hat{\omega}_1$ then differentiating the above expression yields the angular velocity of the second body
$$ \hat{\omega}_2 \times E_2 = \hat{\omega}_1 \times E_1 {\rm Rot}(\hat{z}, \theta) + (E_1 \hat{z} \dot\theta) \times (E_1 {\rm Rot}(\hat{z}, \theta))$$
  $$ \hat{\omega}_2 \times (E_1 {\rm Rot}(\hat{z}, \theta)) = \left( \hat{\omega}_1 + E_1 \hat{z} \dot\theta \right) \times (E_1 {\rm Rot}(\hat{z}, \theta))$$
  $$ \hat{\omega}_2 = \hat{\omega}_1 + E_1 \hat{z} \dot\theta $$
Further differentiation yields the angular acceleration kinematics
$$ \hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times ( E_1 \hat{z} \dot\theta ) $$
$$ \hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times ( \hat{\omega}_2 - \hat{\omega}_1) $$
$$ \hat{\alpha}_2 = \hat{\alpha}_1 + E_1 \hat{z} \ddot\theta + \hat{\omega}_1 \times  \hat{\omega}_2  $$
So the relative velocity and acceleration are
$$ \hat{\omega} = \hat{\omega}_2 -\hat{\omega}_1 = E_1 \hat{z} \dot\theta $$
 $$ \hat{\alpha} = \hat{\alpha}_2 -\hat{\alpha}_1 = E_1 \hat{z} \ddot\theta +   \hat{\omega}_1 \times  \hat{\omega}_2$$
If there is no relative joint acceleration ($\ddot\theta =0$) then you get the expression stated in the question.
A: ja72's answer is probably right, but I was confused by his notation, so I will give my own answer. Suppose we have two object rotating with angular velocity $\vec{\omega}_1$ and $\vec{\omega}_2$. Then the velocity of a point $\vec{r}$ of object $1$ in the lab frame is $\vec{v}_{1,lab}=\vec{\omega}_1 \times \vec{r}$. Similarly, the velocity of a point $\vec{r}$ of object $2$ in the lab frame is $\vec{v}_{2,lab}=\vec{\omega}_2 \times \vec{r}$.
Now to someone in the second object, a point $\vec{r}$ that is stationary in the lab frame will have an apparent velocity $-\vec{v}_{2,lab}=-\vec{\omega}_2 \times \vec{r}$. From this, you can see that a point $\vec{r}$ in the first object will appear to have a velocity $\vec{v}_{1,lab}-\vec{v}_{2,lab} = \vec{\omega}_1 \times \vec{r}-\vec{\omega}_2 \times \vec{r} = (\vec{\omega}_1-\vec{\omega}_2)\times \vec{r}$. Thus the first object appears to have angular velocity $\vec{\omega}_1-\vec{\omega}_2$ to an observer in the second object. 
Now since $\vec{\omega}_1$ is stationary in the lab frame, it's time derivative is $- \vec{\omega}_2 \times\vec{\omega}_1 =\vec{\omega}_1\times \vec{\omega}_2$ in the frame of the second object. Thus object one appears to have an angular acceleration of $\vec{\omega}_1\times \vec{\omega}_2$ in the frame of the second object.
