Question

The moment of inertia of a disc of uniform thickness of mass $M$ having internal and external radii $r$ and $R$ respectively about an axis through its centre and perpendicular to the plane of the disc is

• A. $\frac{M(R^2-r^2)}{4}$
• B. $\frac{M(R^2-r^2)}{2}$
• C. $\frac{M(R^2+r^2)}{2}$
• D. $\frac{M(R^2+r^2)}{4}$

Answer 1

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

The mass of the disc of radius $R$ (no internal radius) $=M_1=M\frac{R^2}{R^2-r^2}$
Similarly mass of the disc of radius r (no internal radius) $=M_2=M\frac{r^2}{R^2-r^2}$
$\therefore$ Moment of Inertia of Disc with radius R about an axis through its centre and perpendicular to the plane of the disc
$I_1=\frac{M}{2}\frac{R^2}{R^2-r^2}{R}^2$
& Moment of Inertia of Disc with radius r about an axis through its centre and perpendicular to the plane of the disc
$I_2=\frac{M}{2}\frac{r^2}{R^2-r^2}{r}^2$
$\therefore$ The moment of inertia of a disc of uniform thickness of mass M having internal and external radii r and R respectively about an axis through its centre and perpendicular to the plane of the disc $=I_1-I_2$
$=\frac{1}{2}(M\frac{R^2}{R^2-r^2}{R}^2-M\frac{r^2}{R^2-r^2}{r}^2)=\frac{1}{2}\left(M\frac{R^4-r^4}{R^2-r^2}\right)=\frac{1}{2}M{(R}^2+r^2)$