Calculating the Rotational Inertia — Question 8

Question 8

Parallel-axis measurements locate an unknown center

A body of mass $2\,\mathrm{kg}$ has inertia $I(x)=I_c+2(x-c)^2$ about a family of parallel axes at coordinate $x$ in meters. Measurements give $I(0)=0.50$ and $I(1)=1.30\,\mathrm{kg\,m^2}$.

1. Infer center coordinate $c$ and central inertia $I_c$.

2. Predict $I(0.60)$ and explain the equal-distance symmetry.

Question 8 - Solution

Parallel-axis measurements locate an unknown center

(a) Derivation and calculation

Subtract the two measurements to eliminate $I_c$: $$1.30-0.50=2[(1-c)^2-c^2]=2-4c.$$ $$\boxed{c=0.30\ \mathrm m},\qquad \boxed{I_c=0.50-2(0.30)^2=0.32\ \mathrm{kg\,m^2}}.$$

(b) Result and interpretation

At the requested coordinate, $$\boxed{I(0.60)=0.32+2(0.60-0.30)^2=0.50\ \mathrm{kg\,m^2}}.$$ More generally, $$I(c+d)=I_c+2d^2=I(c-d).$$ Equal offsets on opposite sides of the center have equal inertia.

Physical check

The reconstructed law also gives $I(1)=0.32+2(0.70)^2=1.30$ kg m$^2$.