Summarized Problem: A system consists of $$N$$ masses $$m_{\alpha}$$ at positions $$\textbf{r}_{\alpha}$$ relative to fixed origin $$O$$. Let $$\textbf{r}'_{\alpha}$$ denote the position of $$m_{\alpha}$$ relative to the center of mass (CM). Prove that $$\sum m_{\alpha} \textbf{r}'_{\alpha} = 0$$. Explain why this relation is nearly obvious. Prove that the rate of change of the angular momentum about the CM is equal to the total external torque about the CM. (This result is surprising since the CM may be accelerating, so that it is not necessarily a fixed point in any inertial frame.)

Solution: The first part of this problem is pretty straight forward. If $$\textbf{r}'_{\alpha}$$ is the position of $$m_{\alpha}$$ relative to the CM, then it must be equal to $$\textbf{r}_{\alpha} - \textbf{R}$$, where $$\textbf{R}$$ denotes the CM. This makes sense since $$\textbf{r}'_{\alpha}$$ would then simply be the vector pointing from $$\textbf{R}$$ to $$\textbf{r}_{\alpha}$$. Therefore, $\sum m_{\alpha} \textbf{r}'_{\alpha} = \sum m_{\alpha} (\textbf{r}_{\alpha} - \textbf{R}) = \sum m_{\alpha} \textbf{r}_{\alpha} - \sum m_{\alpha} \textbf{R}$ Remembering that the CM is defined as $$\textbf{R} = \frac{1}{M} \sum m_{\alpha} \textbf{r}_{\alpha}$$ where $$M$$ is the total mass of the system, $\sum m_{\alpha} \textbf{r}_{\alpha} - \sum m_{\alpha} \textbf{R} = \sum m_{\alpha} \textbf{r}_{\alpha} - \sum m_{\alpha} \frac{1}{M} \sum m_{\alpha} \textbf{r}_{\alpha}$ $= \sum m_{\alpha} \textbf{r}_{\alpha} - \sum m_{\alpha} \frac{1}{\sum m_{\alpha}} \sum m_{\alpha} \textbf{r}_{\alpha} = \sum m_{\alpha} \textbf{r}_{\alpha} - \sum m_{\alpha} \textbf{r}_{\alpha} = 0$ This result makes sense because you'd expect the average weighted distance from the CM, which is the average weighted distance from the origin, to be zero.

For the next part of the question, we need to prove that the torque about the CM is equal to the total external torque about the CM. Expressed as an equation, we are trying to prove that $\dot{\textbf{L}} = \boldsymbol{\Gamma}^{\textrm{ext}}$ when evaluated about the CM, where $$\dot{\textbf{L}}$$ is the derivative of total angular momentum about the center of mass with respect to time, and $$\boldsymbol{\Gamma}^{\textrm{ext}}$$ is the total external torque about the CM.

Before trying to prove this, let's think about why it might make sense intuitively. The problem is suggesting that, barring external torque, angular momentum about the CM is conserved, even if the CM is accelerating. One way this could happen is if every particle in the system is accelerated equally. In this case, every particle's angular velocity relative to the CM remains constant, and angular momentum is thus conserved. One (albeit imperfect) example of this scenario could be molecules of air accelerating to the back of a car as it starts driving.

A slightly less trivial example of an accelerating CM is when only one particle in the system is being accelerated centrally and away from the CM (any non-central components of this acceleration would be be considered external torque). In this case, angular momentum is conserved because the further away from the CM the particle gets (that is, the more $$r'$$ increases), the lower its angular velocity $$\omega$$ about the CM becomes, and since angular momentum (about the CM) is given by $$\ell = mr'^2 \omega$$, angular momentum is conserved.

This example naturally extends to a scenario where multiple particles are being acted upon by central forces. Now, onto the proof.

We can define $$\dot{\textbf{L}}$$ (about the CM) as follows: $\dot{\textbf{L}} = \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha}$ Separating the internal and external forces, $\dot{\textbf{L}} = \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \left(\sum\limits_{\beta \neq \alpha} \textbf{F}_{\alpha \beta} + \textbf{F}_{\alpha}^{\textrm{ext}}\right)$ $\dot{\textbf{L}} = \sum\limits_{\alpha} \sum\limits_{\beta \neq \alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha \beta} + \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha}^{\textrm{ext}}$ $\dot{\textbf{L}} = \sum\limits_{\alpha} \sum\limits_{\beta > \alpha} (\textbf{r}'_{\alpha} \times \textbf{F}_{\alpha \beta} + \textbf{r}'_{\beta} \times \textbf{F}_{\beta \alpha}) + \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha}^{\textrm{ext}}$ Now, assuming, by Newton's Third Law, that $$\textbf{F}_{\alpha \beta} = -\textbf{F}_{\beta \alpha}$$, $\dot{\textbf{L}} = \sum\limits_{\alpha} \sum\limits_{\beta > \alpha} (\textbf{r}'_{\alpha} - \textbf{r}'_{\beta}) \times \textbf{F}_{\alpha \beta} + \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha}^{\textrm{ext}}$ Since $$\textbf{r}'_{\alpha} - \textbf{r}'_{\beta} = \textbf{r}_{\alpha} - \textbf{R} - \textbf{r}_{\beta} + \textbf{R} = \textbf{r}_{\alpha} - \textbf{r}_{\beta}$$, and since both $$\textbf{r}_{\alpha} - \textbf{r}_{\beta}$$ and $$\textbf{F}_{\alpha \beta}$$ point along the line that connects particles $$\alpha$$ and $$\beta,$$ their cross product is zero, leaving us with $\dot{\textbf{L}} = \sum\limits_{\alpha} \textbf{r}'_{\alpha} \times \textbf{F}_{\alpha}^{\textrm{ext}} = \boldsymbol{\Gamma}^{\textrm{ext}}$