Question

The uniform disc shown in the figure has a moment of inertia of $0.6{\text{kg-m}}^2$ around the axis that passes through $O$ and is perpendicular to the plane. If a segment is cut out from the disc as shown, what is the moment of inertia of the remaining disc in ${\text{kg-m}}^2$ ?

Answer 1

Required moment of inertia is
$I=\frac{1}{2}M{r}^2-\frac{1}{2}.m{r}^2=\frac{5}{6}.\frac{1}{2}M{r}^2$
($m=M/6$), where $m$ is the mass of the removed segment.
$=\frac{5}{6}\times 0.6{\text{kg-m}}^2=0.5{\text{kg-m}}^2$