Angular momentum of $N$ particles

Question

I am reading Goldstein's Classical Mechanics book; I have difficulty understanding these lines. Why do the last two terms vanish? I am reading this and thinking $r'$ is a null vector, but the second term should also be zero if that is the case. Where I am going wrong? Also, If I make the center of mass as the origin, the $r'$ wouldn't become a null vector? I don't have any physical intuition. I don't understand mathematically either.

Note: I am a freshman, and English is not my first language, so please be kind.

Answer 1

Because this term is "the radius vector of the center of mass in the very coordinate system whose origin is the center of mass and is therefore a null vector". Indeed, the center-of-mass is by definition $$ \mathbf{R}=\frac{\sum_im_i\mathbf{r}_i}{\sum_im_i}, $$ whereas positions $\mathbf{r}_i'$ are defined as $\mathbf{r}_i'=\mathbf{r}_i-\mathbf{R}$, thus $$\sum_im_i\mathbf{r}_i'=0.$$

Answer 2

No, $\mathbf r_i'$ is not a null vector, but $\sum_i m_i\mathbf r_i' = 0$ is null vector, because by the very definition of center of mass. Because COM balances all unit masses along any axis which cross barycenter. For example, assuming uniform mass distribution in a body and circular shape (for building an easy intuition),- take a look at this example:

So, you can see that,

$$ m\left ( (\mathbf r_1' + \mathbf r_5') + (\mathbf r_2' + \mathbf r_6') + (\mathbf r_3' + \mathbf r_7') + (\mathbf r_4' + \mathbf r_8') + \ldots + (\mathbf r_{n-1}' + \mathbf r_n') \right) = 0 \tag 1$$

Or in short,

$$ \sum_i m_i \mathbf r_i' = 0 \tag 2$$

is a null vector.

Second term in your given equation, namely:

$$ \sum_i (\mathbf r_i' \times \mathbf p_i') \tag 3$$

(notice that there goes vector cross-product and only then - sum operation)

is an average relative angular momentum of body around it's COM, hence it's not null, because there's no cancelation of vectors like in the previous case, because each point relative angular momentum around COM "amplifies" other points momentum. Due to the fact that all points cross-product vectors are collinear and points to the same direction.

The point of 1.28 equation is that body angular momentum around any pivot point $O$ can be split into COM angular momentum about that reference point $O$ and body's angular momentum about "itself", i.e. about COM :

$$ \mathbf L_{ref} = \mathbf L_{COM \to O} + \mathbf L_{self \to COM} \tag 4$$

Answer 3

Consider the simplest example of a system of two particles masses $m_1$ and $m_2$.

If $m_1\, {\bf r}'_1 + m_2\, {\bf r}'_2 = \vec 0\Rightarrow m_1\,r'_1 = m_2\, r'_2$ then $X$ must be the centre of mass.

Thus the summation $\sum_im_i\mathbf{r}_i'$ is zero in both the last two terms.