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I know that Angular Momentum is defined as the cross product of Linear Momentum and Position Vector $$\vec{L}=\vec{r}\times\vec{p}$$ However, in the book Essential Astrophysics, Angular Momentum is written as $$\vec{L}=\vec{r}\times m\vec{v}$$ Therefore, my question is, for the sake of calculations, can you treat the above Formulas as the same?

For example, if $\vec{r}=2 m$, $\vec{v}=4 m/s$, and $m=5 kg$ could you do the computation: $$L=(2 m)×(20 kgm/s)=(2 m)(4 m/s)(5 kg)=40 kgm^2/s$$ I suspect the answer is no because the cross product has a more complex definition, but the book seems too indicate otherwise.

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    $\begingroup$ The second formula should be $m(\vec{r} \times \vec{v})$. $\endgroup$ – Mitchell Oct 15 '17 at 16:04
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    $\begingroup$ Yes, the two formulae are same. $|\vec{L}|=|\vec{r}|.(m|\vec{v}|)\sin{\theta}=|\vec{r}||\vec{P}|\sin{\theta}$. $\endgroup$ – Mitchell Oct 15 '17 at 16:12
  • $\begingroup$ Oh, I see. As long as $\sin{\theta}=1$, right? $\endgroup$ – K Ferreira Oct 15 '17 at 16:15
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    $\begingroup$ No, the formulae are same for all theta because the angle between $\vec{r}$ and $\vec{v}$ is equal to the angle between $\vec{r}$ and $\vec{P}$, as $\hat{P}$=$\hat{v}$. $\endgroup$ – Mitchell Oct 15 '17 at 16:17
  • $\begingroup$ Consider mentioning author, title, page, eq #, etc for reference. $\endgroup$ – Qmechanic Oct 15 '17 at 16:42
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Yes, both formulas are the same in classical mechanics and for a single particle or equivalent, which is the case. That's because

$\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times(m\vec{v})=m(\vec{r}\times \vec{v})$

since the cross product is linear in both components (you can extract scalars outside and introduce them inside one of the factors).

As for your example, it is wrong because you have forgotten the vectors! Let the hat ^ denote unit vector in that direction, i.e. $\hat{r}$ is the unit vector in the radial direction and $\hat{v}$ is the unit vector in the velocity direction.

Then you can write

$\vec{L}=5 kg \cdot (2m\cdot \hat{r})\times (4 m/s \ \ \hat{v})$

and then extract scalars outside:

$\vec{L}=5 kg\cdot 2m \cdot 4 m/s \cdot (\hat{r}\times \hat{v})$

So yes, you can do the multiplication equally, but without forgetting the vector. Of course that cross product Is also a unit vector perpendicular to both $r$ and $v$ at the same time, provided that $\vec{r}\perp\vec{v}$ (else the result is $\sin \theta$ times the vector. Let's call it $\hat{n}$. You have

$\vec{L}=(5 kg\cdot 2m \cdot 4 \frac{m}{s}) \cdot \sin \theta \ \hat{n}\\$

and

$\\ |\vec{L}|=5 kg\cdot 2m \cdot 4 \frac{m}{s}\cdot \sin \theta \ $

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    $\begingroup$ $\hat{r}\times\hat{v}$ is not a unit vector! It has magnitude $\sin \theta$, so the last section of your answer is totally wrong, and it should intuitively feel wrong to you as well. $\endgroup$ – CDCM Oct 15 '17 at 23:00
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    $\begingroup$ It seems I'm having bad days. For some reason I was thinking about circular motion exclusively. Edited. $\endgroup$ – FGSUZ Oct 16 '17 at 13:11
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    $\begingroup$ no worries it happens. Switched my downvote to an upvote :) $\endgroup$ – CDCM Oct 16 '17 at 21:04
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Both formulae are correct.

Whenever you face a difficulty in formulae, your first resort should be to dimensional analysis. The dimension of $L$ is $[M L^2 T^{-1}]$.

The dimension of R.H.S. of the second formula is: $[L]×[M]×[L T^{-1}] = [M L^2 T^{-1}]$, which is the dimensions of L.H.S. So, the second formula is correct.

By vector notation, the second formula is actually $\vec{L} = m(\vec{r} × \vec{v})$. This is derived from the first formula by simply taking mass out from the cross product as mass is a scalar quantity.

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  • $\begingroup$ In the book, it's written exactly as $L=M×R×V$, but I assumed that the author used × as meaning normal multiplication because you can't use the cross product with a scalar quantity like mass. $\endgroup$ – K Ferreira Oct 15 '17 at 16:24
  • $\begingroup$ @KFerreira Yes. It seems your book hasn't used the vector notation, and simply wants to denote the magnitude. $\endgroup$ – Wrichik Basu Oct 15 '17 at 16:38
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Just as @WrichikBasu stated in his answer, the correct formula for angular momentum is $$ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m \vec{v}) = m ( \vec{r} \times \vec{v} )$$

The above is valid for a system of particles each located $\vec{r}_i$ from the origin, with the total angular momentum about the origin

$$ \vec{L}_{\rm origin} = \sum_i m_i (\vec{r}_i \times \vec{v}_i ) $$

After doing a lot of math the above is evaluated as

$$ \vec{L}_{\rm origin} = {\rm I}_C \omega + \vec{r}_C \times \vec{p} = {\rm I}_C \omega + \vec{r}_C \times m\vec{v}_C $$ where $\vec{r}_C$ is the position of the center of mass, $\vec{v}_C$ the velocity of the center of mass and ${\rm I}_C$ the mass moment of inertia about the center of mass.

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