Question

A thin disc of mass M and radius R has mass per unit area $\sigma \left(r\right)=k{r}^2$, where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is

• A. $\frac{M{R}^2}{2}$
• B. $\frac{M{R}^2}{6}$
• C. $\frac{M{R}^2}{3}$
• D. $\frac{2M{R}^2}{3}$

Answer 1

The correct option is D $\frac{2M{R}^2}{3}$

Given, surface mass density
$\sigma =k{r}^2$
So, mass of the disc can be calculated by considering small element of area $2r\pi dr$ on it and then integrating it for complete disc, i.e,

$dm=\sigma dA=\sigma \times 2\pi rdr$
$\int dm=M={\int }_0^R\left(k{r}^2\right)2\pi rdr$
$\Rightarrow M=2\pi k\frac{R^4}{4}=\frac{1}{2}\pi k{R}^4$

Moment of inertia about the axis of the disc,
$I=\int dI=\int dm{r}^2=\int \sigma dA{r}^2$
$={\int }_0^{R}k{r}^2\left(2\pi rdr\right)r^2$
$\Rightarrow I=2\pi k{\int }_0^R{r}^{5}dr=\frac{2\pi k{R}^6}{6}=\frac{\pi k{R}^6}{3}....\left(ii\right)$
From Eqs. (i) and (ii), we get
$I=\frac{2}{3}M{R}^2$