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 $\frac{d\theta}{dt}$, 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?
 A: Using the tangential velocity you can write $\vec{v} = \vec{\omega} \times \vec{r} $, substituting this in your expression you'll get the well-known expression for the angular momentum of a single particle: $\vec{L} = mr^2 \vec{\omega}$.
The quantity $mr^2$ is called the moment of inertia of a particle with respect to a certain axis of rotation. A generalisation can be made to a collection of particles, if they have fixed positions with respect to each other we say these particles consitute a rigid body. The general formula then becomes $\vec{L} = \bf{I} \vec{\omega}$ where $\bf{I}$ is called the inertia tensor.
Note that this has the same structure as in linear motion where $\vec{p} = m \vec{v}$ where in this case the mass $m$ takes the role of inertia.
A: Yes I convinced myself that your formula is correct. I just calculated the crossproduct, and  put in omega instead of v. so your formula gives the right answer for th absolute value of p, but i still can not see why you do not see the change of angle in the cross product.
