Consider the system shown. Find the moment of inertia about the diagonal shown.

The correct answer is 4 kg.m².

Analysis of the System from the Image:
Based on the provided diagram, we can observe the following details:

• The system consists of four point masses, each of mass $m=1\mathrm{kg}$, placed at the four corners of a square.
• The sides of the square have length $a=2\mathrm{m}$ (as indicated by the horizontal length label "$2\mathrm{m}$" at the top and the vertical height label "$2\mathrm{m}$" on the left).
• The axis of rotation is represented by the solid diagonal line passing through the bottom-left corner and the top-right corner.

Step-by-Step Derivation:

1. Identify the position of each mass relative to the diagonal axis:
Let us label the four corners of the square:

• Mass 1: Located at the bottom-left corner (on the diagonal axis).
• Mass 2: Located at the top-right corner (on the diagonal axis).
• Mass 3: Located at the top-left corner.
• Mass 4: Located at the bottom-right corner.

Since Mass 1 and Mass 2 lie directly on the diagonal axis of rotation, their perpendicular distances from the axis are zero:
$r_1=0$
$r_2=0$

2. Determine the perpendicular distance for Mass 3 and Mass 4:
In a square of side length $a$, the diagonals are perpendicular and bisect each other. The perpendicular distance $d$ from the remaining two corners (top-left and bottom-right) to the diagonal axis is equal to half the length of the other diagonal of the square.

The total length of the diagonal of a square with side $a=2\mathrm{m}$ is:
$L=a\sqrt{2}=2\sqrt{2}\mathrm{m}$

Therefore, the perpendicular distance $d$ of Mass 3 and Mass 4 from the diagonal axis is:
$d=r_3=r_4=\frac{L}{2}=\frac{2\sqrt{2}}{2}=\sqrt{2}\mathrm{m}$

3. Calculate the Total Moment of Inertia ($I$):
The moment of inertia of a system of discrete particles is given by the formula:
$I=\sum m_i{r}_i^2$
Substituting the values:
$I=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2+m_4{r}_4^2$
$I=\left(1\mathrm{kg}\right)·0^2+\left(1\mathrm{kg}\right)·0^2+\left(1\mathrm{kg}\right)·{\left(\sqrt{2}\mathrm{m}\right)}^2+\left(1\mathrm{kg}\right)·{\left(\sqrt{2}\mathrm{m}\right)}^2$
$I=0+0+1·2+1·2$
$I=2+2=4\mathrm{kg}·{\mathrm{m}}^2$

Thus, the moment of inertia of the system about the diagonal is 4 kg.m².