Angular Momentum — Question 8

Question 8

Angular momentum reveals the signed miss distance

A particle travels along a horizontal straight path with momentum $8\hat x\,\mathrm{kg\,m/s}$. Its angular momentum about a fixed origin is $+12\hat z\,\mathrm{kg\,m^2/s}$.

1. Find the signed vertical offset of its path.

2. Find its closest approach to the origin and explain why its current distance cannot be inferred.

Question 8 - Solution

Angular momentum reveals the signed miss distance

(a) Derivation and calculation

For a rightward momentum $\vec p=8\hat x$, $$L_z=xp_y-yp_x=-8y.$$ Thus $$\boxed{y=-12/8=-1.50\ \mathrm m}.$$ The positive angular momentum therefore identifies a path below the origin.

(b) Result and interpretation

The perpendicular offset determines closest approach: $$\boxed{b=|L|/|p|=12/8=1.50\ \mathrm m}.$$ At a general location along the path, $$r=\sqrt{x^2+(1.50)^2}\ge1.50\ \mathrm m.$$ The along-path coordinate $x$ is not fixed by angular momentum, so the current distance remains undetermined.

Physical check

Closest approach occurs at $x=0$, where position and momentum are perpendicular.