Question

Find the $\text{MOI}$ of a uniform square plate of side length $a=10\text{m}$ and mass $5\text{kg}$ about an axis passing through its centre as shown in figure. $(\theta ={30}^{\circ })$.

• A. $I_{AB}=\frac{125}{6}{\text{kg.m}}^2$
• B. $I_{AB}=\frac{125\sqrt{3}}{2}{\text{kg.m}}^2$
• C. $I_{AB}=\frac{125}{3}{\text{kg.m}}^2$
• D. $I_{AB}=\frac{250}{3}{\text{kg.m}}^2$

Answer 1

The correct option is C $I_{AB}=\frac{125}{3}{\text{kg.m}}^2$

As we know,
Moment of inertia of a square plate about an axis perpendicular to its plane passing through its geometric centre $\text{O}$ is
$I_O=\frac{m{a}^2}{6}...\left(i\right)$
Let us draw an axis $CD$ which is perpendicular to axis $AB$, in the plane of the square plate.


Axis through $\text{O}$ is $\perp$ to axis $CD\&AB$ and passing through their intersection, hence from perpendicular axis theorem,
$I_z=I_x+I_y...\left(ii\right)$
where $I_z=I_O=\frac{m{a}^2}{6}$,
$I_x\&I_y$ refers to axis $AB$ and $CD$.
$\because I_{CD}=I_{AB}$ due to symmetry of square plate
$\Rightarrow \frac{m{a}^2}{6}=I_{AB}+I_{CD}$
$\Rightarrow \frac{m{a}^2}{6}=2I_{AB}$
$\Rightarrow I_{AB}=\frac{m{a}^2}{12}$
$\therefore I_{AB}=\frac{5\times {10}^2}{12}=\frac{125}{3}{\text{kg.m}}^2$