Question

A thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma (r)=k{r}^2$ 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. $\frac{M{R}^2}{6}$
• B. $\frac{M{R}^2}{3}$
• C. $\frac{2M{R}^2}{3}$
• D. $\frac{M{R}^2}{2}$

Answer 1

Answer: (C)

$I_{Disc}={\int }_0^R\left(dm\right)r^2\Rightarrow I_{Disc}={\int }_0^R\left(\sigma 2\pi rdr\right)r^2$
$I_{Disc}={\int }_0^R\left(k{r}^{2}2\pi rdr\right)r^2$
$I_{Disc}=2\pi k{\int }_0^R{r}^{5}drM={\int }_0^{R}2\pi rdrk{r}^2$
$I_{Dics}=2\pi k{\left(\frac{r^6}{6}\right)}_0^{R}M=2\pi k{\int }_0^R{r}^{3}dr$
$I_{Disc}=2\pi k\frac{R^6}{6}M=2\pi k\frac{r^4}{4}{|}_0^R$
$I_{Disc}=\frac{\pi k{R}^6}{3}=\left(\frac{\pi k{R}^4}{2}\right)\frac{R^{2}2}{3}M=2\pi k\frac{R^4}{4}$
$I_{Disc}=\frac{M2R^2}{3}$
$I_{Disc}=\frac{2}{3}M{R}^2$