The Angular Momentum of a Rigid Body Rotating About a Fixed Axis — Question 10

Question 10

Reflected inertia is an energy coefficient rather than total angular momentum

Two gears on fixed parallel axes mesh externally with $R_2=2R_1$. Their inertias are $I_1=.10$, $I_2=.40\,\mathrm{kg\,m^2}$. Gear 1 turns at $+6\,\mathrm{rad/s}$. Their centers are stationary.

1. Find their total axial angular momentum and kinetic energy.

2. Find the inertia referred to gear 1 using energy; explain why multiplying it by $\omega_1$ is not the physical total $L_z$.

Question 10 - Solution

Reflected inertia is an energy coefficient rather than total angular momentum

(a) Derivation and calculation

External meshing imposes opposite rim velocities: $$R_1\omega_1+R_2\omega_2=0\Rightarrow\omega_2=-6/2=-3\ \mathrm{rad/s}.$$ Thus $$\boxed{L_z=.10(6)+.40(-3)=-.600\ \mathrm{kg\,m^2/s}},$$ $$\boxed{K=\tfrac12(.10)(6)^2+\tfrac12(.40)(-3)^2=3.60\ \mathrm J}.$$

(b) Result and interpretation

Referring kinetic energy to gear 1 gives $$K=\tfrac12\left[I_1+I_2(R_1/R_2)^2\right]\omega_1^2,$$ $$\boxed{I_{\mathrm{eff}}=.10+.40(1/2)^2=.200\ \mathrm{kg\,m^2}}.$$ Its product $I_{\mathrm{eff}}\omega_1=1.20$ differs from the physical $L_z=-.60$. Energy weights squared speed ratios; physical angular momentum uses signed first powers.

Physical check

The stationary centers contribute no orbital angular momentum, so the physical total is the sum of the two signed spins.