Question

A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig..The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Answer 1

$I=I_{remain}+I_{R/3}$

$\therefore I_{remian}=I-I_{R/3}$

$\therefore I_{remian}=\frac{M{R}^2}{2}-[\frac{\frac{M}{4}{\left(\frac{R}{3}\right)}^2}{2}+\frac{M}{4}{\left(\frac{R}{3}\right)}^2]$

$=\frac{M{R}^2}{2}-[\frac{{MR}^2}{72}+\frac{3{MR}^2}{36}]$

$=\frac{M{R}^2}{2}-\left[\frac{{MR}^2+2M{R}^2}{72}\right]$

$=\frac{M{R}^2}{2}-\frac{3M{R}^2}{72}$

$=\frac{36M{R}^2-3M{R}^2}{72}$

$=\frac{11}{24}M{R}^2$

Hence, the answer is $\frac{11}{24}M{R}^2.$