Question

A thin disc of mass M and radius R has 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 ti its plane is :

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

Answer 1

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

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