Moment of Inertia of a Rectangular Plate

Results

For a uniform rectangular plate of sides $a$ (along $x$) and $b$ (along $y$):

$I_x = \frac{Mb^2}{12} \quad \text{(about x-axis through centre)}$

$I_y = \frac{Ma^2}{12} \quad \text{(about y-axis through centre)}$

$I_z = \frac{M(a^2+b^2)}{12} \quad \text{(about z-axis through centre, perpendicular to plate)}$

Derivation of $I_x$

$I_x = \frac{Mb^2}{12}$ follows from treating the plate as a collection of horizontal rods of length $a$, stacked along $y$ from $-b/2$ to $b/2$.

Each rod at position $y$ has mass $dm = \frac{M}{b} \, dy$ and contributes $y^2 \, dm$ to $I_x$ (its MI about the $x$-axis is from its distance $y$ from the $x$-axis, since the rod runs parallel to the $x$-axis and is thin):

$I_x = \int_{-b/2}^{b/2} y^2 \cdot \frac{M}{b} \, dy = \frac{M}{b} \cdot \frac{b^3}{12} = \frac{Mb^2}{12}$

Derivation of $I_z$

By the perpendicular axis theorem:

$I_z = I_x + I_y = \frac{Mb^2}{12} + \frac{Ma^2}{12} = \frac{M(a^2+b^2)}{12}$

Special case: square plate ($a = b$)

$I_x = I_y = \frac{Ma^2}{12}, \quad I_z = \frac{Ma^2}{6}$

About an edge

Parallel axis theorem, $d = b/2$ (distance from centre to edge parallel to $x$):

$I_{edge} = \frac{Mb^2}{12} + M\left(\frac{b}{2}\right)^2 = \frac{Mb^2}{12} + \frac{Mb^2}{4} = \frac{Mb^2}{3}$

Same as a rod of length $b$ about its end — makes sense, since the plate rotates about one edge exactly as each strip (a rod) rotates about its end.