Question

A uniform disc of radius $R$ ($3m$) has a round disc of radius $R/3$ cut from it as shown in the figure. The mass of the remaining (shaded) portion of the disc equals $M$ ($2kg$). The moment of inertia of such a disc relative to the axis passing through the geometrical centre of the original disc and perpendicular to the plane of the disc is $xkg{m}^2$. Then the value of $x$ is

Answer 1

Let the mass per unit area of the material of disc be $\sigma$ . Now the empty space can be considered as having density $-\sigma$.

Now $I_0=I_{\sigma }+I_{-\sigma }$
$I_{\sigma }=\left(\sigma \pi R^2\right)\frac{R^2}{2}$
$I_{-\sigma }=-\left(\sigma \pi \right(\frac{R}{3}{)}^2\left)\frac{(\frac{R}{3}{)}^2}{2}\right)-\left(\sigma \pi \right(\frac{R}{3}{)}^2{\left(\frac{2R}{3}\right)}^2)$
$=\frac{-\sigma \pi R^4}{18}$

$\therefore I_0=\frac{4\sigma \pi R^4}{9}$
Now, substituting $\sigma =\frac{M}{\pi r^2}$
$I_0=\frac{4M{R}^2}{9}$

Substituting $M=2kg,R=3m$,
$I_0=8kg{m}^2$

So $x=8$