Question

A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O

Answer 1

$I_0=\frac{m{r}^2}{2}$ $I_{removes}=\frac{1}{2}\frac{m}{16}\frac{r^2}{16+m}\frac{a{r}^2}{16(m+md)}$

$=\frac{m{r}^2+18m{r}^2}{512}$

$=\frac{19m{r}^2}{512}=1$

$I_{remaining}=m{r}^2=\frac{19}{512}.m{r}^2$

$=\frac{237}{512}m{r}^2$