Consider a thin uniform square sheet made of a rigid material. If its side is $$a$$, mass m and moment of inertia $$I$$ about one of its diagonals, then:

We are given a thin uniform square sheet with side length $$a$$, mass $$m$$, and we need to find the moment of inertia $$I$$ about one of its diagonals. The options are provided, and we will derive the moment of inertia step by step.

First, recall that for a thin square sheet, we can use the perpendicular axis theorem. This theorem states that for a planar body lying in the $$xy$$-plane, the moment of inertia about the $$z$$-axis (perpendicular to the plane) is the sum of the moments of inertia about the $$x$$-axis and $$y$$-axis (both in the plane and perpendicular to each other):

$$ I_z = I_x + I_y $$

For a square sheet of side $$a$$ and mass $$m$$, the moment of inertia about an axis through the center and parallel to one side (say, the $$x$$-axis) is the same as that of a rectangular lamina. The formula for a rectangular lamina about an axis through its center and parallel to one side is $$\frac{m}{12} \times (\text{length of the perpendicular side})^2$$. Since the square has equal sides, both axes yield the same moment of inertia:

$$ I_x = \frac{m a^2}{12} $$

$$ I_y = \frac{m a^2}{12} $$

Here, $$I_x$$ is the moment of inertia about the $$x$$-axis (where the distance is measured along the $$y$$-direction), and $$I_y$$ is about the $$y$$-axis (distance along the $$x$$-direction). Adding these using the perpendicular axis theorem:

$$ I_z = I_x + I_y = \frac{m a^2}{12} + \frac{m a^2}{12} = \frac{2m a^2}{12} = \frac{m a^2}{6} $$

So, $$ I_z = \frac{m a^2}{6} $$.

Now, we need the moment of inertia about a diagonal. Consider the diagonal along the line $$y = x$$ in the coordinate system where the square is centered at the origin, with sides parallel to the axes. The vertices are at $$\left(\frac{a}{2}, \frac{a}{2}\right)$$, $$\left(\frac{a}{2}, -\frac{a}{2}\right)$$, $$\left(-\frac{a}{2}, \frac{a}{2}\right)$$, and $$\left(-\frac{a}{2}, -\frac{a}{2}\right)$$.

The moment of inertia about an axis is given by $$ I = \int r^2 dm $$, where $$r$$ is the perpendicular distance from a point to the axis. For the axis $$y - x = 0$$, the distance $$r$$ from a point $$(x, y)$$ is:

$$ r = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{2}} $$

Thus,

$$ I_d = \int r^2 dm = \int \left( \frac{|x - y|}{\sqrt{2}} \right)^2 dm = \int \frac{(x - y)^2}{2} dm $$

Due to the symmetry of the square, $$|x - y|^2 = (x - y)^2$$, so we can write:

$$ I_d = \frac{1}{2} \int (x - y)^2 dm $$

Expanding the square:

$$ (x - y)^2 = x^2 - 2xy + y^2 $$

So,

$$ I_d = \frac{1}{2} \int (x^2 - 2xy + y^2) dm = \frac{1}{2} \left[ \int x^2 dm - 2 \int xy dm + \int y^2 dm \right] $$

Now, evaluate each integral:

• $$\int x^2 dm$$ is the moment of inertia about the $$y$$-axis, $$I_y = \frac{m a^2}{12}$$.
• $$\int y^2 dm$$ is the moment of inertia about the $$x$$-axis, $$I_x = \frac{m a^2}{12}$$.
• $$\int xy dm$$ is the product of inertia $$I_{xy}$$. Due to the symmetry of the square about both axes, for every point $$(x, y)$$, there are symmetric points $$(x, -y)$$, $$(-x, y)$$, and $$(-x, -y)$$, making the integral of $$xy$$ over the entire mass zero. Thus, $$I_{xy} = 0$$.

Substituting these values:

$$ I_d = \frac{1}{2} \left[ \frac{m a^2}{12} - 2 \cdot 0 + \frac{m a^2}{12} \right] = \frac{1}{2} \left[ \frac{m a^2}{12} + \frac{m a^2}{12} \right] = \frac{1}{2} \left[ \frac{2m a^2}{12} \right] = \frac{1}{2} \cdot \frac{m a^2}{6} = \frac{m a^2}{12} $$

Therefore, the moment of inertia about the diagonal is $$ I = \frac{m a^2}{12} $$.

Comparing with the options:

• A. $$ I = \frac{m a^2}{24} $$
• B. $$ \frac{m a^2}{24} < I < \frac{m a^2}{12} $$
• C. $$ I > \frac{m a^2}{12} $$
• D. $$ I = \frac{m a^2}{12} $$

Hence, the correct answer is Option D.