Calculating the Rotational Inertia — Question 7

Question 7

A compound object needs every part referred to the same axis

A uniform rod of mass $2\,\mathrm{kg}$ and length $1\,\mathrm m$ extends from a pivot to the center of a uniform disk of mass $3\,\mathrm{kg}$ and radius $0.20\,\mathrm m$. The rigid assembly lies in a plane; the pivot axis is perpendicular to it.

1. Find its total inertia about the pivot.

2. What fraction of the disk contribution is missed by modeling the disk as a point at its center?

Question 7 - Solution

A compound object needs every part referred to the same axis

(a) Derivation and calculation

Refer both parts to the pivot axis: $$I_{\mathrm{rod}}=\frac13mL^2=\frac23\ \mathrm{kg\,m^2}.$$ $$I_{\mathrm{disk}}=\frac12MR^2+ML^2 =\tfrac12(3)(0.20)^2+3(1)^2=3.06\ \mathrm{kg\,m^2}.$$ $$\boxed{I_{\mathrm{total}}=\frac23+3.06=\frac{559}{150}=3.7267\ \mathrm{kg\,m^2}}.$$

(b) Result and interpretation

Replacing the disk by a central point preserves $ML^2$ but omits $MR^2/2$: $$\boxed{f_{\mathrm{missed}}=\frac{MR^2/2}{MR^2/2+ML^2} =\frac{0.06}{3.06}=\frac1{51}=1.96\%}.$$ This fraction refers to the disk contribution, not the total assembly inertia.

Physical check

The finite-size correction vanishes as $R/L\to0$.