Question

A rectangular lamina $ABCD$ of mass $m=3\text{kg}$ and sides $AB=2\text{m}$ and $BC=4\text{m}$ is shown in the figure. Find its moment of inertia about an axis passing through point $A$ and perpendicular to the plane of lamina.

• A. $5{\text{kg.m}}^2$
• B. $20{\text{kg.m}}^2$
• C. $15{\text{kg.m}}^2$
• D. $10{\text{kg.m}}^2$

Answer 1

The correct option is B $20{\text{kg.m}}^2$

Given:-
$a=2\text{m}$ and $b=4\text{m}$
$I_z$ is the moment of inertia of lamina about an axis passing through its centre of mass and $\perp$ to its plane.


Moment of inertia about $x-$axis $\left(I_x\right)=\frac{m{a}^2}{12}$
Simillarly,
Moment of inertia about $y-$ axis $\left(I_y\right)=\frac{m{b}^2}{12}$
Applying perpendicular axis theorem,
$I_z=I_x+I_y$

$\Rightarrow I_z=\frac{m{a}^2}{12}+\frac{m{b}^2}{12}$

$I_z=\frac{m}{12}(a^2+b^2)$

$I_z=\frac{3\times (2^2+4^2)}{12}$

$\therefore I_z=5{\text{kg.m}}^2$


Let $I_A$ be the moment of inertia about axis passing through $A$ and $\perp$ to the plane of lamina, hence $I_A$ and $I_z$ be moment of inertia about $\parallel$ axes .

$\Rightarrow$Using parallel axis theorem
$I_A=I_Z+m{d}^2$
Where $d=\frac{\sqrt{a^2+b^2}}{2}$

$I_A=5+m\left(\frac{a^2+b^2}{4}\right)$
$=5+3\times \left(\frac{2^2+4^2}{4}\right)$
$I_A=5+\left(\frac{3\times 20}{4}\right)$
$\therefore I_A=20{\text{kg.m}}^2$