Question

A symmetric lamina of mass $M$ consists of a square shape with a semicircular section over each of the edge of the square as shown in the figure. The side of the square is $2a.$ The moment of inertia of the lamina about an axis through its center of mass and perpendicular to the plane is $1.6M{a}^2$. The moment of Inertia about the tangent $\text{AB}$ in the plane of lamina is $\frac{n}{5}M{a}^2$. The value of $n$ .(only integer)

Answer 1

As we can see, $x$ and $y$ axis passes the COM and lies on the plane of the lamina, $z$ axis passes the COM perpendicular to the lamina.

Given, $I=I_z=1.6M{a}^2$

According to perpendicular axis theorem,

$I_z=I_x+I_y$

Due to symmetry, $I_x=I_y$

$\therefore I_z=2I_x$

$\therefore I_x=\frac{I_z}{2}=0.8M{a}^2$

Now, moment of inertia about axis $\text{AB}$ is,

Using parallel axis theorem,

$I_{AB}=I_x+M(2a{)}^2$

$=0.8M{a}^2+4M{a}^2$

$=4.8M{a}^2$

$\therefore I_{AB}=\frac{48}{10}M{a}^2=\frac{24}{5}M{a}^2$

Given, $I_{AB}=\frac{n}{5}M{a}^2\Rightarrow n=24$

Why this Question ? When moment of inertia of about any fixed axis is given, and if we need to find moment of inertia about any other axis. Try to use parallel and perpendicular axis theorem.