1
$\begingroup$

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?

$\endgroup$
11
  • $\begingroup$ So your definition, for one particle, would be $\vec{p_{ang}}=m{|\vec r(t)|}^2\vec{\omega}$? $\endgroup$ Aug 23 '20 at 4:21
  • $\begingroup$ @descheleschilder yes $\endgroup$
    – Ryder Rude
    Aug 23 '20 at 4:30
  • $\begingroup$ 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? $\endgroup$ Aug 23 '20 at 5:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – trula
    Aug 23 '20 at 11:09
  • 1
    $\begingroup$ 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. $\endgroup$
    – trula
    Aug 23 '20 at 15:44
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I'm not sure this is appropriate as an answer, since it is more a question than an answer. $\endgroup$
    – NDewolf
    Aug 23 '20 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.