Angular Momentum — Question 3

Question 3

The origins of zero angular momentum form a line

At an instant a particle is at $(2,3,0)\,\mathrm m$ with momentum $5\hat x\,\mathrm{kg\,m/s}$. Let a possible origin have coordinates $(a,b,c)$.

1. Find angular momentum about that origin.

2. Find all origins for which it vanishes, and the nearest such origin to $(0,0,0)$.

Question 3 - Solution

The origins of zero angular momentum form a line

(a) Derivation and calculation

Relative to origin $A=(a,b,c)$, $$\vec r_{P/A}=(2-a,3-b,-c),\qquad\vec p=(5,0,0).$$ Their cross product is $$\boxed{\vec L_A=(0,-5c,-5(3-b))\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

Zero angular momentum requires $$-5c=0,\qquad-5(3-b)=0\Rightarrow c=0,\ b=3,\ a\ \text{arbitrary}.$$ These origins lie on the momentum line through the particle. Their squared distance from the original origin is $d^2=a^2+9$, minimized by $a=0$: $$\boxed{A_{\mathrm{nearest}}=(0,3,0)\ \mathrm m}.$$

Physical check

Shifting the origin along the momentum direction does not change the cross product.