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

Physics Asked on July 9, 2021

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?

2 Answers

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.

Correct answer by JulianDeV on July 9, 2021

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.

Answered by trula on July 9, 2021

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