Angular Momentum — Question 2

Question 2

Spatial angular momentum has a direction beyond the orbital sketch

A particle has $\vec r=(1,2,3)\,\mathrm m$ and momentum $\vec p=(4,-2,1)\,\mathrm{kg\,m/s}$ relative to a fixed origin.

1. Compute angular momentum and its magnitude.

2. Check its orthogonality to both $\vec r$ and $\vec p$.

Question 2 - Solution

Spatial angular momentum has a direction beyond the orbital sketch

(a) Derivation and calculation

Expand the cross product: $$L_x=2(1)-3(-2)=8,\qquad L_y=3(4)-1(1)=11,$$ $$L_z=1(-2)-2(4)=-10.$$ Therefore $$\boxed{\vec L=(8,11,-10)\ \mathrm{kg\,m^2/s}},\qquad \boxed{|L|=\sqrt{8^2+11^2+(-10)^2}=\sqrt{285}}\ \mathrm{kg\,m^2/s}.$$

(b) Result and interpretation

Check orthogonality to the two input vectors: $$\vec L\cdot\vec r=8(1)+11(2)-10(3)=\boxed0,$$ $$\vec L\cdot\vec p=8(4)+11(-2)-10(1)=\boxed0.$$ The magnitude alone cannot specify the angular-momentum direction; all three signed components are needed.

Physical check

The squared norm is 285, independently consistent with $r^2p^2-(\vec r\cdot\vec p)^2=14(21)-3^2$.