Question

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

Answer 1

The Correct Option is C

$I_{Disc } =\int^{R}_{0} \left(dm\right)r^{2} \Rightarrow I_{Disc } =\int^{R}_{0} \left(\sigma2\pi rdr\right)r^{2} $
$ I_{Disc } = \int^{R}_{0} \left(kr^{2} 2\pi rdr\right)r^{2} $
$ I_{Disc } =2\pi k \int^{R}_{0} r^{5} dr M = \int^{R}_{0} 2\pi rdr kr^{2} $
$ I_{Dics } = 2\pi k\left(\frac{r^{6}}{6}\right)^{R}_{0} M = 2\pi k \int^{R}_{0} r^{3}dr $
$ I_{Disc } = 2\pi k \frac{R^{6}}{6} M = 2\pi k \frac{r^{4}}{4} \bigg|^{R}_{0} $
$I_{Disc } =\frac{\pi kR^{6}}{3} = \left(\frac{\pi kR^{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} MR^{2} $