Question

Consider a disc of radius $R$ from which a small circular disc of radius $r$ has been cut with its centre $C_2$ at a distance $a$ from the centre of the full disc. Calculate the moment of inertia of the holed disc about an axis perpendicular to the plane of the disc and passing through the centre $C_2$ of the circular hole. The mass of the circular disc without hole is $M$.

Answer 1

We have to find out the moment of inertia of held about axis 'XOK' which is passing through the centre of hole and perpendicular to its plane.

The Moment of inertia of the disk about an axis passing through its centre and perpendicular to its plane is $=\frac{1}{12}M{R}^2$

So, the moment of inertia of the disk about axis XO'X' is

$I=\frac{1}{2}M{R}^2+M{a}^2$ (applying parallel axis theorem )

Let the mass distributed uniformly over the surface of disk.So mass per unit area$=\frac{M}{\pi R^2}$

Hence,the mass of hole which has been cut is

$⟹\frac{M}{\pi R^2}\pi r^2=\frac{M{r}^2}{R^2}$