Question

A square plate is of mass $M$ and length of the edge is $2a$. Its Moment of inertia about its centre of mass axis, perpendicular to its plane is equal to $l_0$. Four identical disks are cut from the plane. Find the Moment of inertia of leftover square about the same axis.

Answer 1

The moment of inertia of the square plate about an axis passing through its centre of mass and perpendicular to its plane.

$I_0=\frac{M{\left(2a\right)}^2}{6}=\frac{4}{6}M{a}^2{I}_0=\frac{2}{3}M{a}^2$

Let $d$ is the density of the plate then $M=d.4a^2$

$d=\frac{M}{4a^2}$

Mass of each disc,

$d=d.\pi {\left(\frac{a}{2}\right)}^2=\frac{\pi a^2}{4}d=\frac{\pi a^2}{4}.\frac{M}{4a^2}=\frac{\pi }{16}M$

and radius $r=\frac{a}{2}$

Moment of inertia of the disc about the axis passing through $o$ and perpendicular to the plane of the plate.

$=MI$ of disc about axis through $O$ and perpendicular to the palne of plate.

$+m{\left(\frac{a}{\sqrt{2}}\right)}^2$

$+m{\left(\frac{a}{\sqrt{2}}\right)}^2=\frac{1}{2}m{\left(\frac{a}{2}\right)}^2+m\frac{a^2}{2}=\frac{5m{a}^2}{8}$

So for all disc

$=4\times \frac{5m{a}^2}{8}=\frac{5\pi }{32}M{a}^2$

Hence net $MI$ after cut the disc.

$=\frac{2}{3}M{a}^2-\frac{5\pi }{32}M{a}^2=(\frac{2}{3}-\frac{5\pi }{32})M{a}^2$