DaveC426913 said:
Note the "B" in the edumakashun level for this thread.
This is perfectly B-level.The proof is not that difficult:
To describe a rigid body, take a model that it consists of a discrete set of point particles connected by massless rigid rods. Let's describe the rigid body with respect to an inertial refrence frame. To describe the body's particles take an arbitrary point fixed in the body. The position vector of this body-fixed origin is denoted with ##\vec{R}##, and any point of the body is described by
$$\vec{x}_i=\vec{R}+\vec{r}_i.$$
Since the ##\vec{r}_i## are vectors connecting the body-fixed origin with another body-fixed point, the rigidity of the body means that ##r_i=|\vec{r}_i|=\text{const}##. It's pretty obvious without calculation that thus the ##\vec{r}_i## can only rotate. To see this formally, we write
$$r_i^2=\vec{r}_i^2=(\vec{x}_i-\vec{R})^2.$$
Since this is constant, taking the derivative with respect to time yields
$$\dot{\vec{r}}_i \cdot \vec{r}_i=0,$$
which means that the velocity of the body-fixed point relative to the body-fixed reference point is always perpendicular to the position vector ##\vec{r}_i## relative to the body-fixed position. That means that there must be a vector ##\vec{\omega}## such that
$$\dot{\vec{r}}_i=\vec{\omega} \times \vec{r}_i.$$
This implies that the velocity of this point in the body relative to the inertial frame,
$$\dot{\vec{x}}_i=\dot{\vec{R}}+\dot{\vec{r}_i}=\dot{\vec{R}} + \vec{\omega} \times \vec{r}_i.$$
The change of the relative vector in an infinitesimal step of time is
$$\mathrm{d} \vec{r}_i=\mathrm{d} t \vec{\omega} \times \vec{r}_i,$$
and this describes an infinitesimal rotation around a momentary rotation axis in direction of ##\vec{\omega}## by an angle ##|\vec{\omega}|\mathrm{d} t##. Thus ##\vec{\omega}## is the angular velocity, describing the rotation of the body around the body-fixed origin, ##\vec{R}##.
Finally we show that this relative angular velocity is independent of the choice of this body-fixed origin. So let ##\vec{R}'## be another choice. Then
$$\vec{x}_i=\vec{R}+\vec{r}_i=\vec{R}'+\vec{r}_i'.$$
Now as before we have
$$\dot{\vec{r}}'=\vec{\omega}' \times \vec{r}_i',$$
and we have to show that ##\vec{\omega}'=\vec{\omega}##.
To see this we note that
$$\vec{R}'-\vec{R}=\vec{r}=\vec{r}_i-\vec{r}_i'$$
is also a body-fixed vector, because it connects two body fixed points. Thus there must be a vector ##\vec{\omega}''## such that
$$\dot{\vec{R}}'-\dot{\vec{R}}=\vec{\omega}'' \times (\vec{R}'-\vec{R}) = \vec{\omega}'' \times (\vec{r}_i-\vec{r}_i')=\vec{\omega} \times \vec{r}_i-\vec{\omega}' \times \vec{r}_i,$$
and this can hold true for all points ##\vec{x}_i## making up the body, if
$$\vec{\omega}''=\vec{\omega}=\vec{\omega}'.$$
So finally we also have
$$\dot{\vec{r}}_i'=\vec{\omega} \times \vec{r}_i,$$
and that was to be shown.
