Angular Momentum — Question 10

Question 10

The swept-area rate measures angular momentum per unit mass

A particle of mass $0.50\,\mathrm{kg}$ moves in a plane. Its radius vector sweeps signed area at constant rate $0.80\,\mathrm{m^2/s}$ counterclockwise about a fixed point.

1. Derive the relation between area rate and angular momentum.

2. Find angular momentum and the time required to sweep $2.0\,\mathrm{m^2}$ of signed area.

Question 10 - Solution

The swept-area rate measures angular momentum per unit mass

(a) Derivation and calculation

Over an infinitesimal displacement $d\vec r=\vec v\,dt$, the signed swept triangle has area $$dA=\tfrac12(\vec r\times d\vec r)_z.$$ Hence $$\frac{dA}{dt}=\tfrac12(\vec r\times\vec v)_z =\frac{L_z}{2m},\qquad\boxed{L_z=2m\dot A}.$$

(b) Result and interpretation

The counterclockwise area rate is positive: $$\boxed{L_z=2(.50)(.80)=+.800\ \mathrm{kg\,m^2/s}}.$$ At constant rate, $$\boxed{\Delta t=\frac{\Delta A}{\dot A}=\frac{2.0}{.80}=2.50\ \mathrm s}.$$ Since $\dot A=r^2\dot\theta/2$, constant area rate permits varying angular speed when radius changes.

Physical check

The units $m\dot A$ are kg m$^2$/s, exactly those of angular momentum.