Angular Momentum — Question 1

Question 1

Straight-line motion can carry constant angular momentum

A $2\,\mathrm{kg}$ particle moves with velocity $3\hat x\,\mathrm{m/s}$ along the line $y=4\,\mathrm m$, with $z=0$. The origin is fixed at $(0,0)$.

1. Find angular momentum about the origin at any position $x$.

2. Explain why the particle need not orbit the origin to have angular momentum.

Question 1 - Solution

Straight-line motion can carry constant angular momentum

(a) Derivation and calculation

The linear momentum is $$\vec p=m\vec v=2(3\hat x)=6\hat x\ \mathrm{kg\,m/s}.$$ At any position $\vec r=x\hat x+4\hat y$, $$\vec L=\vec r\times\vec p=(x\hat x+4\hat y)\times6\hat x =\boxed{-24\hat z\ \mathrm{kg\,m^2/s}}.$$

(b) Result and interpretation

The magnitude can also be written $$|L|=bp=4(6)=24\ \mathrm{kg\,m^2/s},$$ where $b$ is the perpendicular distance from the origin to the momentum line. The along-path coordinate cancels because $$(x\hat x)\times(6\hat x)=0.$$ Circular motion is unnecessary: a straight moving particle has angular momentum about any origin off its momentum line.

Physical check

The negative $z$ direction follows from upward lever arm crossed with rightward momentum.