Can angular momentum directly be defined in terms of angular velocity?

I don’t like it being defined as $vec{r} times vec{mv}$ as the angular nature is not obvious in that definition.

Suppose there’s a single particle moving around. We choose an arbitrary origin. We define the angular momentum at time $t$ as $m|vec{r(t)}|^2$ times its angular velocity. Angular velocity at time $t$ is defined as the vector perpendicular to both $vec{v(t)}$ and $vec{r(t)}$ (according to some conventional rule), and having the magnitude $frac{dtheta}{dt}$, where $theta (t)$ is the angular position of the particle at time $t$ in the plane of $vec{r(t)}$ and $vec{v(t)}$, with respect to the chosen origin.

So this defines it for a single particle. For a system of particles, we just sum up the angular momenta. The formula $vec{r}times vec{mv}$ is arrived at as a means of calculating it. Is this definition equivalent to $vec{r}times vec{mv}$? Can either of these definitions be used for any general problem?