Show that angular momentum can be divided into center of mass and internal coordinate

Question

This might belong in physics but I want to be sure I am approaching the math right.

Given: show angular momentum can be divided into separate parts of the center of mass and internal coordinates.

So I start with vector $\mathbf R$ as the coordinate for the center of mass and $\mathbf r_i$ for the $i^{th}$ particle, and say that $\mathbf r_i'$ is the coordinate of the $i^{th}$ particle in terms of the internal coordinates, and further $\mathbf r_i = \mathbf R + \mathbf r_i'$

Angular momentum $\mathbf L$ is $$\mathbf L = \sum_i \mathbf r_i \times \mathbf p = \sum_i \mathbf r_i \times (m_i \mathbf{\dot r_i})$$ because p=mv and v is the derivative of the position vector.

OK, so I know that $\mathbf r_i = \mathbf R + \mathbf r_i'$ which means I can plug that into the summation. I get $$\mathbf L = \sum_i (\mathbf R + \mathbf r_i') \times (m_i \mathbf{\dot r_i})$$ further I know that if I take a derivative w/r/t time of my original expression $\mathbf r_i = \mathbf R + \mathbf r_i'$ I should get $\mathbf {\dot r_i} = \mathbf {\dot R} + \mathbf {\dot r_i'}$ and I can plug that in to get $$\mathbf L = \sum_i (\mathbf R + \mathbf r_i') \times (m_i \mathbf {\dot R} + m_i \mathbf {\dot r_i'})$$

and it's the last step where I kept getting answers that didn't make sense to me, unless there's some stupid-simple identity I am missing. I feel that two terms should go to zero and that should be that. But I wasn't sure how that kind of vector multiple distributes. I have my parts there -- I can see them (assuming I did everything else right). But I feel like the very last step is stopping me.

Thanks in advance for the help.

Answer 1

By definition of centre of mass, you have $\sum_i R m_i = \sum_i m_i r_i = \sum_i m_i (R + r_i')$, and so $\sum_i m_i r_i' = 0$. It follows that $\sum_i m_i \dot{r}_i' = 0$ also.

Expanding \begin{eqnarray} L &=& \sum_i (R + r_i') \times (m_i \dot{R}+ m_i \dot{r}_i')\\ &=& \sum_i R \times m_i \dot{R} + \sum_i r_i' \times m_i \dot{R} + \sum_i R \times m_i \dot{r}_i' + \sum_i r_i' \times m_i \dot{r}_i' \\ &=& R \times (\sum_i m_i) \dot{R} + (\sum_i m_i r_i') \times \dot{R} + R \times (\sum_i m_i \dot{r}_i') + \sum_i r_i' \times m_i \dot{r}_i' \\ &=& R \times (\sum_i m_i) \dot{R} + 0 \times \dot{R} + R \times 0 + \sum_i r_i' \times m_i \dot{r}_i' \\ &=& R \times (\sum_i m_i) \dot{R} + \sum_i r_i' \times m_i \dot{r}_i' \\ &=& L_{\text{com}} + L_{\text{internal}} \end{eqnarray}

Answer 2

Hint: To eliminate the cross terms, use the definition of centre of mass, viz. $$\begin{align} \sum_i m_i \mathbf {r_i'} = 0 && \sum_i m_i \mathbf {\dot r_i'} = 0 \end{align}$$