= Solution
Assume <Newton's second law>, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summing
$$
m_i\ddot r_i=F_i+\sum_{j\ne i}F_{ij}
$$
cancels internal pairs and gives $dP/dt=F:=\sum_iF_i$. Taking moments about fixed $a$ cancels the internal <torques> pairwise and gives
$$
\frac{dL}{dt}=G:=\sum_i(r_i-a)\times F_i.
$$
The <angular momentum about the centre of mass> result follows similarly: differentiating $\sum_i(r_i-R)\times m_i(\dot r_i-\dot R)$ introduces no extra term because both total relative position weighted by mass and total relative <momentum> vanish. Thus its <derivative> is the external <torque> about $R$.
Finally, if every mass is $m$ and $F_i=-k\dot r_i$, then about a fixed point
$$
G=-k\sum_i(r_i-a)\times\dot r_i=-\frac{k}{m}L.
$$
Therefore
$$
\boxed{L(t)=L(0)e^{-kt/m}.}
$$
Back to article page
