A thin, uniform rod is bent into a square of side length a.

If the total mass is M, find the moment of inertia about an axis

through the center and perpendicular to the plane of the square.

(Hint:Use the parallel axis theorem.)

The Mass of the rod, M

Length of the side of the square a

Since the rod is bent into a square, therefore the total moment of inertia of the rod must be the sum of moments of inertia of each side of the square.

To find the total moment of inertia formula used are given below

$I_{cm}=\frac{1}{12}M{L}^2$

Where M is mass and L is the length

According to this theorem moment of inertia about any axis I

can be given as

$I=I_{cm}+m{d}^2$

Where$I_{cm}$is a moment of inertia about the center of mass, m is mass d is the distance between two axes.

The moment of inertia of the rod is

$I_{rod}=4I_{side}$

The moment of inertia of each side about an axis through the center and perpendicular to the plane of the square can be given by using the parallel axis theorem.

$I=I_{cm}+m{d}^2$

Where$I_{cm}$ is a moment of inertia about the center of mass, m is mass d is the distance between two axes.

Since the rod is bent in the shape of a square therefore the mass of each side of the square is $M/4$

The moment of inertia of each side about its center of mass is

$I_{cm}=\frac{1}{12}\left(\frac{M}{4}\right)a^2$

The distance between two axes $d=a/2$

Therefore, the moment of inertia of the side about the parallel axis is

$\begin{array}{c}{I}_{side}=\frac{1}{12}\left(\frac{M}{4}\right)a^2+\frac{M}{4}{\left(\frac{a}{2}\right)}^2\\ =\frac{1}{12}M{a}^2\end{array}$

Therefore, the moment of inertia of the rod

$\begin{array}{c}{I}_{rod}=4I_{side}\\ =4\left(\frac{1}{12}M{a}^2\right)\\ =\frac{1}{3}M{a}^2\end{array}$

Hence the moment of inertia of the rod is $\frac{1}{3}M{a}^2$