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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?

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  • $\begingroup$ So your definition, for one particle, would be $\vec{p_{ang}}=m{|\vec r(t)|}^2\vec{\omega}$? $\endgroup$
    – Deschele Schilder
    Aug 23, 2020 at 4:21
  • $\begingroup$ @descheleschilder yes $\endgroup$
    – Ryder Rude
    Aug 23, 2020 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? $\endgroup$
    – Deschele Schilder
    Aug 23, 2020 at 5:09
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    $\begingroup$ 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, 2020 at 11:09
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    $\begingroup$ 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, 2020 at 15:44

2 Answers 2

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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.

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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.

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  • $\begingroup$ I'm not sure this is appropriate as an answer, since it is more a question than an answer. $\endgroup$
    – NDewolf
    Aug 23, 2020 at 16:49

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