Time Evolution of a Rigid Body

Question

I've been very confused about how we get the derivation for the time evolution of a rigid body, in the resources I've seen it's given by $$\dfrac{d\vec r_{\text{rotating frame}}}{dt} = \omega \times \vec r_{\text{rotating frame}}.$$

While going through the derivation one of the parts I've gotten stuck on has to do with taking the cross product of a vector with an angle, I've understood taking cross products of vectors but taking a cross product of a vector with an angle seems a bit strange.

As of now the resource I've been using is David Tong's Lecture Notes on The Motion of Rigid Bodies. The section I'm stuck on is 3.1.1 Angular Velocities. I've been stuck on this for quite some months, but now that I made a stack exchange account I decided to ask it. Any help would be highly appreciated.

Answer 1

I think the "angle" you are referring to is $\mathbf{\omega}$?

When we first learn about angular velocity, we fix an axis and then go from there. So we only need a scalar angular velocity $\omega$.

In general, however, we want to specify an axis of rotation, too. So we allow $\omega$ to become a vector, and its direction specifies the axis of rotation. $\dot{r} = \omega r$ becomes $\mathbf{\dot{r}} = \mathbf{\omega} \times \mathbf{r}$.

Prof Tong's beginning with the angular momentum tensor $\omega_{ab}$ is explaining the natural coming about of such a quantity as the transformation of basis vectors. It is through its antisymmetry that we see that it's just the one vector we need to describe the situation.

Finally, intuition-wise, it should make sense that this is the equation. If something is rotating, then we want any point on the body not to change distance, and all orthogonal to the axis of rotation. This is satisfied by the cross product.

Hope that helps.