Why isn't angular velocity the moment of velocity if angular momentum is moment of momentum? Angular momentum can be defined as $L$ = $\textbf{r}$ x $m\textbf{v}$.
Why is angular velocity $\omega$ then not $\textbf{r}$ x $\textbf{v}$, but instead $v = \omega \times \textbf{r}$?
 A: I don't think this is as dumb a question as everyone downvoting you seems to think.
The definitions of angular velocity ($\omega$), angular momentum ($L$), and moment of inertia ($I$) ARE defined in order to perfectly mirror Newton's laws. Angular momentum is the analogue of momentum, angular velocity is the analogue of velocity, and moment of inertia is the analogue of mass. Similarly, torque ($\tau$) is the analogue of force. Then the normal equations from linear mechanics become
$$
\begin{array}{rcl}
K=\frac{1}{2}mv^2 & \iff & K = \frac{1}{2}I\omega^2 \\
F=m\dot{v} & \iff & \tau = I\dot{\omega} \\
p=mv & \iff & L=I\omega \\
F=\dot{p} & \iff & \tau = \dot{L}
\end{array}
$$
You wanted to define $\omega$ such that $L=m\omega$. However, it turns out this simple substitution wouldn't allow ALL of Newton's equations to take the same form as before. Instead, we need to also replace the mass $m$ with the analogous quantity in angular mechanics, the moment of inertia. Thus, we ACTUALLY want to define $\omega$ such that $L=I\omega$. Once you realize that, the definition is perfectly natural. 
A: I think the distinction here is between a free vector and a line vector. 
A free vector is shared by the entire rigid body regardless of location. Examples are:


*

*angular velocity $\mathbf{\omega}$, 

*force $\mathbf{F}$

*linear momentum $\mathbf{p}$.


line vector is defined to act on a specific location (for example a point A), and along a specific direction. Examples are: 


*

*linear velocity $\mathbf{v}_A$

*pure moment $\mathbf{\tau}_A$ 

*angular momentum $\mathbf{L}_A$


The combination of a free vector and line vector defines a screw located at some point A. Examples are:


*

*Velocity Screw $v_A=(\mathbf{v}_A,\mathbf{\omega})$

*Force Screw $f_A = (\mathbf{\tau_A}, \mathbf{F})$ 

*Momentum Screw $L_A = (\mathbf{L}_A, \mathbf{p})$


Screws transform from one location to another with identical laws. For example to express the above quantities on a second point B you have


*

*Velocity Screw $v_B=(\mathbf{v}_A+\mathbf{r}\times\mathbf{\omega},\mathbf{\omega})$

*Force Screw $f_B=(\mathbf{\tau}_A+\mathbf{r}\times\mathbf{F},\mathbf{F})$

*Momentum Screw $L_B=(\mathbf{L}_A+\mathbf{r}\times\mathbf{p}, \mathbf{p})$


where $\mathbf{r}$ is the vector from B to A and ${\times}$ is the vector cross product. All of these screws are related with the following way


*

*The velocity screw defines the infinitesimal motion of a rigid body and is described by 6 unique coordinates (4 for the line of action, one for the magnitude and one for the so called pitch).

*The momentum screw is defined with a one-to-one mapping from the velocity screw using the spatial rigid body inertia. This relationship is that of a pole-polar geometrically speaking around the inertial ellipsoid located at the center of mass. $$\begin{pmatrix} \mathbf{L}_{cm} \\ \mathbf{p} \end{pmatrix} = \begin{bmatrix}0 & I_{cm} \\ m & 0 \end{bmatrix} \begin{pmatrix} \mathbf{v}_{cm} \\ \mathbf{\omega} \end{pmatrix} = \begin{pmatrix} I_{cm} \mathbf{\omega} \\ m \mathbf{v}_{cm} \end{pmatrix}$$

*Finally the force screw is the time derivative of the momentum screw $$ \begin{pmatrix} \mathbf{\tau}_{cm} \\ \mathbf{F} \end{pmatrix} = \frac{{\rm d}}{{\rm d}t} \begin{pmatrix} \mathbf{L}_{cm} \\ \mathbf{p} \end{pmatrix} = \begin{pmatrix} I_{cm} \mathbf{\alpha}+\mathbf{\omega}\times I_{cm} \mathbf{\omega}\\ m \mathbf{a}_{cm} \end{pmatrix}$$


So to answer your question, it is because $\mathbf{v}$ is a line vector and not a free vector. A free vector transforms with identity $$\mbox{(free vector)}_B=\mbox{(free vector)}_A$$, and a line vector with the cross product $$\mbox{(line vector)}_B=\mbox{(line vector)}_A+\mathbf{r}\times\mbox{(free vector)}$$
This is the origin of $\mathbf{r}\times \mathbf{p} = \mathbf{r}\times m \mathbf{v}_{cm}$ and $\mathbf{v} = \mathbf{r} \times \mathbf{\omega}$ terms.
