Question

A thin uniform disc of mass $M$ and radius $R$ has concentric hole of radius $r$. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane.

• A. $\frac{1}{2R}M(R^3+r^3)$
• B. $\frac{1}{2}M(R^2-r^2)$
• C. $\frac{1}{3}M(R^2+r^2)$
• D. $\frac{1}{2}M(R^2+r^2)$

Answer 1

The correct option is D $\frac{1}{2}M(R^2+r^2)$

Mass per unit area of the disc is
$m=\frac{M}{\pi (R^2-r^2)}$
Mass of the disc if it was complete (i.e without hole) is
$M_1=m\times \pi R^2=\frac{M}{\pi (R^2-r^2)}\times \pi R^2$
$=\frac{M{R}^2}{R^2-r^2}$
Mass of the removed portion is


$M_2=m\times \pi r^2=\frac{M{r}^2}{R^2-r^2}$
Since the two portions are concentric, the moment of inertia of the given disc about the given axis is
$I=\frac{1}{2}{M}_1{R}^2-\frac{1}{2}{M}_2{r}^2$

$=\frac{1}{2}[\frac{M{R}^4}{R^2-r^2}-\frac{M{r}^4}{R^2-r^2}]$

$=\frac{M}{2}\left[\frac{R^4-r^4}{R^2-r^2}\right]=\frac{1}{2}M(R^2+r^2)$