Question

A rectangular lamina of mass $20\text{kg}$ having sides $6\text{m}$ and $3\text{m}$ is placed on the horizontal surface. Find moment of inertia of the lamina about the axis passing through the center and perpendicular to its plane.

• A. $75{\text{kg m}}^2$
• B. $45{\text{kg m}}^2$
• C. $15{\text{kg m}}^2$
• D. $55{\text{kg m}}^2$

Answer 1

The correct option is A $75{\text{kg m}}^2$

The rectangular lamina at horizontal surface having origin as a center is shown in the figure.

Given, sides of lamina is $a=6\text{m}$ and $b=3\text{m}$.
Mass of lamina is $M=20\text{kg}$.
$\text{MOI}$ of rectangular lamina about the $x-$axis is given by $I_x$.
$\Rightarrow I_x=\frac{M{b}^2}{12}=\frac{\left(20\right)(3{)}^2}{12}=15{\text{kg m}}^2$
Similarly, $\text{MOI}$ of rectangular lamina about the $y-$ axis is given by $I_y$.
$\Rightarrow I_y=\frac{M{a}^2}{12}=\frac{\left(20\right)(6{)}^2}{12}=60{\text{kg m}}^2$
From the perpendicular axis theorem, the moment of inertia of the lamina about the axis passing through the center and perpendicular to its plane is $I_z$, which is given by
$I_z=I_x+I_y$
$\Rightarrow I_z=15+60=75{\text{kg m}}^2$