# 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?

• So your definition, for one particle, would be $\vec{p_{ang}}=m{|\vec r(t)|}^2\vec{\omega}$? Commented Aug 23, 2020 at 4:21
• @descheleschilder yes Commented Aug 23, 2020 at 4:30
• I can't see why not. Why do you think it wouldn't apply to a collection of particles? Or is this exactly what you ask? Commented Aug 23, 2020 at 5:09
• for me it seems you just try to explain the crossproduct, where r and v are perpendicular ? try your approach for a planet moving on an ellipse. Commented Aug 23, 2020 at 11:09
• your definition of tangential velocity is new. The common definition is: tangential to the curve, the point is traveling on, and only in circles the tangential velocity is p always perpendicular to the position vector. Commented Aug 23, 2020 at 15:44

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.