The Angular Momentum of a System of Particles — Question 2

Question 2

Center motion and internal motion contribute separate moments

Two $1\,\mathrm{kg}$ particles have positions $(1,0)$ and $(3,0)\,\mathrm m$ and velocities $(0,1)$ and $(0,3)\,\mathrm{m/s}$.

1. Compute total angular momentum directly.

2. Decompose it into center-of-mass orbital angular momentum and angular momentum relative to the center.

Question 2 - Solution

Center motion and internal motion contribute separate moments

(a) Derivation and calculation

Summing about the original origin, $$\boxed{L_{O,z}=1(1)(1)+1(3)(3)=10\ \mathrm{kg\,m^2/s}}.$$ The center variables are $$M=2\ \mathrm{kg},\quad\vec R_C=2\hat x\ \mathrm m,\quad\vec V_C=2\hat y\ \mathrm{m/s}.$$

(b) Result and interpretation

The orbital contribution is $$L_{{\mathrm{orb}},z}=(\vec R_C\times M\vec V_C)_z=2(4)=8.$$ Relative positions are $\mp\hat x$ m and velocities are $\mp\hat y$ m/s. Hence $$L_{C,z}=(-1)(1)(-1)+(1)(1)(1)=2,$$ $$\boxed{L_{O,z}=L_{{\mathrm{orb}},z}+L_{C,z}=8+2=10\ \mathrm{kg\,m^2/s}}.$$

Physical check

The direct particle sum and the center-plus-relative decomposition agree.