The Angular Momentum of a System of Particles — Question 1

Question 1

Zero total momentum does not imply zero total angular momentum

Two $1\,\mathrm{kg}$ particles are at $(0,1,0)$ and $(0,-1,0)\,\mathrm m$. Their velocities are respectively $(2,0,0)$ and $(-2,0,0)\,\mathrm{m/s}$.

1. Find total linear and angular momentum about the origin.

2. Would translating the origin change total angular momentum at this instant?

Question 1 - Solution

Zero total momentum does not imply zero total angular momentum

(a) Derivation and calculation

The two momenta are $\vec p_1=2\hat x$ and $\vec p_2=-2\hat x$ kg m/s, so $$\boxed{\vec P=\vec p_1+\vec p_2=\vec0}.$$ Their angular momenta add: $$\vec L_1=\hat y\times2\hat x=-2\hat z,\qquad \vec L_2=(-\hat y)\times(-2\hat x)=-2\hat z,$$ $$\boxed{\vec L_O=-4\hat z\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

For a translated origin $A$ at $\vec a$, $$\vec L_A=\sum(\vec r_i-\vec a)\times\vec p_i =\vec L_O-\vec a\times\vec P.$$ Since $\vec P=0$, $$\boxed{\vec L_A=\vec L_O}\quad\text{for every }\vec a.$$ This comparison keeps the same velocity frame.

Physical check

Equal opposite momenta cancel their resultant but can add angular momentum through opposite lever arms.