Calculating the Rotational Inertia — Question 5

Question 5

An offset hole requires subtracting its shifted inertia

A uniform disk originally has mass $M$ and radius $R$. A circular hole of radius $R/2$ is removed with its center at $(R/2,0)$. Let $O$ denote the original disk center. All axes below are perpendicular to the plate.

1. Find the remaining inertia about $O$.

2. Find the remaining center of mass and inertia about that center.

Question 5 - Solution

An offset hole requires subtracting its shifted inertia

(a) Derivation and calculation

Uniform density makes the removed mass proportional to area: $$m_h=M\frac{\pi(R/2)^2}{\pi R^2}=\frac M4.$$ Its inertia about $O$ includes both its own central inertia and its offset: $$I_{h,O}=\frac12\frac M4\left(\frac R2\right)^2 +\frac M4\left(\frac R2\right)^2=\frac{3MR^2}{32}.$$ $$\boxed{I_O=\frac12MR^2-I_{h,O}=\frac{13MR^2}{32}}.$$

(b) Result and interpretation

Treat the removed disk as negative mass for the first moment: $$M_r=\frac{3M}{4},\qquad \boxed{x_c=\frac{0-(M/4)(R/2)}{3M/4}=-\frac R6},\quad y_c=0.$$ $$\boxed{I_c=I_O-M_rx_c^2 =\frac{13MR^2}{32}-\frac{3M}{4}\frac{R^2}{36} =\frac{37MR^2}{96}}.$$

Physical check

The remaining center shifts away from the hole, and translating to it reduces the inertia.