The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 9

Question 9

Independent coaxial rotors cannot be assigned one physical angular speed

Two coaxial rotors have $I_1=0.20$, $I_2=0.30\,\mathrm{kg\,m^2}$ and signed angular speeds $+10$ and $-4\,\mathrm{rad/s}$. They are not coupled.

1. Find total axial angular momentum and total rotational energy.

2. Compare with a fictitious single rigid rotor of inertia $I_1+I_2$ carrying the same total angular momentum.

Question 9 - Solution

Independent coaxial rotors cannot be assigned one physical angular speed

(a) Derivation and calculation

Signed angular momenta add: $$\boxed{L_z=I_1\omega_1+I_2\omega_2=.20(10)+.30(-4)=.800\ \mathrm{kg\,m^2/s}}.$$ Energies add without sign cancellation: $$\boxed{K=\tfrac12(.20)(10)^2+\tfrac12(.30)(-4)^2=12.4\ \mathrm J}.$$

(b) Result and interpretation

A fictitious rigid rotor with $I_\Sigma=.50$ kg m$^2$ and the same $L_z$ would have $$\boxed{\omega_{\mathrm{common}}=L_z/I_\Sigma=.80/.50=1.60\ \mathrm{rad/s}},$$ $$\boxed{K_{\mathrm{common}}=L_z^2/(2I_\Sigma)=.640\ \mathrm J}.$$ The difference $12.4-.64=11.76$ J is associated with relative rotation. The actual uncoupled two-speed assembly is not a single rigid rotor.

Physical check

Opposite signed spin momenta partly cancel while both rotational energies remain positive.