Question 76 of 280

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 inertia about an axis going through its centre of mass and perpendicular to its plane is:

A$\dfrac{M{R}^{2}}{3}$B$\dfrac{M{R}^{2}}{2}$C$\dfrac{M{R}^{2}}{6}$D$\dfrac{2M{R}^{2}}{3}$

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Answer: (D) $\dfrac{2M{R}^{2}}{3}$

$dm=\sigma \cdot dA=\sigma 2\pi r dr$
$=2\pi k {r}^{3} dr$
$M=\int dm=2\pi k\int _{0}^{R}{r}^{3}dr$
$=2\pi k\left(\dfrac{{R}^{4}}{4}\right)=\dfrac{\pi k{R}^{4}}{2}$
$dI=dm {r}^{2}=2\pi k{r}^{5} dr$
$I=2\pi k\int _{0}^{R}{r}^{5}dr=2\pi k\dfrac{{R}^{6}}{6}=\dfrac{2\pi k{R}^{6}}{3\times 2}$
since, $\text{M}=\dfrac{{\pi kR}^{4}}{2}$
so,
$I=\dfrac{2M{R}^{2}}{3}$