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I came across this theorem in Morin's classical mechanics:

Theorem 7.1 The angular momentum (relative to the origin) of a body can be found by treating the body as a point mass located at the CM and finding the angular momentum of this point mass (relative to the origin), and by then adding on the angular momentum of the body, relative to the CM.

The theorem is derived by considering the coordinates of the various pieces of mass $\mathbf{r}$ as the distance to the CM $\mathbf{R}$ and the distance relative to the CM $\mathbf{r'}$ so that $\mathbf r = \mathbf R \ + \mathbf r' $. Likewise the velocity $\mathbf v = \mathbf V + \mathbf v' $ where $\mathbf V$ is the velocity of the CM and $\mathbf v'$ is the velocity relative to the CM. Therefore the angular momentum of a body rotating around the z axis can be calculated by:

$$\mathbf L = \int \mathbf{r} \ \times \mathbf{v} \ dm$$

$$= \int (\mathbf R + \mathbf r') \ \times \ (\mathbf V + \mathbf v') \ dm$$

$$ = M \mathbf R \ \times \ \mathbf V + \int \mathbf r' \ \times \ \mathbf v' dm $$

$$ = M \mathbf R \ \times V \ + (\int r'^2 w'dm)\hat{\mathbf z} $$

$$ = M \mathbf R \ \times V \ + (\mathbf I_z^{CM}w')\hat{\mathbf z} $$

In going from lines 2 to 3 the cross terms $\int (\mathbf R \ \times \ \mathbf v') $ and $\int (\mathbf r' \ \times \ \mathbf V)$ vanish. The justification in the text is that by definition of the CM $\int \mathbf r' dm = 0$, and hence $\int \mathbf v' dm = 0$. However, I'm failing to see how this implies that $\int (\mathbf R \ \times \ \mathbf v' )dm = 0$ and $\int (\mathbf r' \ \times \ \mathbf V )dm = 0$. Can someone show this explicitly?

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  • $\begingroup$ I am trying to answer this but the edition of Morin I have is different. Specify Edition? $\endgroup$
    – JohnA.
    Jan 2 at 19:22
  • $\begingroup$ Its from 2003 page VII-4 $\endgroup$
    – User13114
    Jan 2 at 19:32
  • $\begingroup$ In my edition this is chapter 8. Does the answer make sense? $\endgroup$
    – JohnA.
    Jan 2 at 19:33
  • $\begingroup$ I'm reading it right now, but it seems like it answers my question. I have to digest the index notation for the cross product though since I haven't seen it before. Is this justification given directly in your edition of the text? $\endgroup$
    – User13114
    Jan 2 at 19:36
  • $\begingroup$ The index notation is not in my book either . If you don't want to use it, I suggest you literally write out the cross product in component form and you will see that assuming $R$ is constant is enough to rewrite it as I have. $\endgroup$
    – JohnA.
    Jan 2 at 19:46

1 Answer 1

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Morin himself justifies this:

And since we can pull the constant vectors ${\bf V}$ and ${\bf R}$ out of the above integrals.

So what is happening here is $$\int {\bf R}\times {\bf v'} dm= {\bf R}\times\int {\bf v'}dm =0$$ and same for the other integral. Now, to see why this is valid, let us define $${\bf a} \equiv \int {\bf R}\times {\bf v'} dm$$ we can consider the $i^\text{th}$ component of the vector ${\bf a}$, $$a_i =\int \epsilon_{ijk}R_jv'_k dm$$ where we wrote the cross product in index notation. Now since $\epsilon_{ijk}$ is independent of $m$ (its just a number) and, as Morin says $R_j$ is too, then we take them out $$a_i =\epsilon_{ijk}R_j\int v'_k dm$$ which back in vector notation is $${\bf a}= {\bf R}\times\int {\bf v'}dm$$ as needed.

As requested, here is how one of the components would look like if we expand the index notation (call the $i=1$ component $x$ component, $i=2$ is $y$ etc...) :

$$a_x =\sum_{k = 1}^3\sum_{j =1}^3\int\epsilon_{ijk}R_j v'_k dm$$
Expand $j$ sum: $$a_x = \sum_{k = 1}^3\int\epsilon_{12k}R_y v'_k dm+\int \epsilon_{13k}R_z v'_k dm$$ where I used the fact that the $\epsilon_{ijk}=0$ anytime two indices (e.g $\epsilon_{133}=0)$ have the same value. Expand $k$ sum and take out constant $R$ components and numerical constants $\epsilon$'s: $$a_x = \epsilon_{123}R_y\int v'_z dm+\epsilon_{132}R_z\int v'_y dm$$ Use $\epsilon_{123} = 1$ and $\epsilon_{132} =-1$ so $$a_x =R_y\int v'_z dm-R_z\int v'_y dm$$ For completion, why are these vectors "constant" or mass independent as Morin says? The definition of ${\bf R}$ is the position of the center of mass and we are taking a mass integral. This mass integral can be transformed into a position integral in most cases (e.g. $dm = \lambda d^3r$) where $\int d^3r$ would be integrating over the volume of the object. The center of mass is a specific, chosen, constant vector independent of ${\bf r}$ just like say some number $a$ would be in $$\int (x-a)dx $$.

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