Representation Of Linear Velocity as Cross Product Why is linear velocity represented as cross product of angular velocity of the particle and its position vector? Why not vice versa?
(Consider rigid body rotation)
 A: OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes:

Let the particle rotate about the axis OO' ...
Within time interval $dt$ let its motion be represented by the vector $d\varphi$ whose direction is along axis obeying the right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ.
Now, if the elementary displacement of particle at a be specified by radius vector $r$,
From the diagram, it is easy to see that, for infinitesimal rotation, $dr= d\varphi\times r \tag{1}$
By definition, $ω = dφ/dt$
Thus taking the elementary time interval as $dt$, all given equations surely hold!
Thus we can divide both sides of the equation $(1)$ by $dt$ which is the corresponding time interval!
So we get $dr/dt = dφ/dt \times r$   of course $r$ value won't change WRT the particle and axis, so $r$/dt  is essentially $r$!
So result is, $$\boxed{v = ω \times r}$$
A: $\omega$ is a axial vector, not an ordinary or polar vector, and its sense depends on the handedness of the coordinate system.  For a right handed coordinate system v=ω×r.  See the textbook   Symon Mechanics.
