Question

A rectangular lamina of mass $2\text{kg}$ is having side $a=6\text{m}$ and $b=3\text{m}$. What is the moment of inertia of the rectangular lamina about an axis passing through the side $3\text{m}$ as shown in figure below?

• A. $6{\text{kg-m}}^2$
• B. $15{\text{kg-m}}^2$
• C. $24{\text{kg-m}}^2$
• D. $14.8{\text{kg-m}}^2$

Answer 1

The correct option is C $24{\text{kg-m}}^2$

Let given lamina as lamina 1.

Assume another lamina identical to lamina 1 is attached to form a system as shown in figure.



Mass of system, $M=(m+m)\text{kg}=2m\text{kg}$
We know,
Moment of inertia of system about $x{x}^{\prime };I_{net}=\frac{M{l}^2}{12}{I}_{net}=\frac{2m\times (12{)}^2}{12}=24m{\text{kg-m}}^2=(24\times 2){\text{kg-m}}^2{I}_{net}=48{\text{kg-m}}^2..\left(i\right)$

We know,
$I_{net}=I_1+I_2$
As both lamina are identical and having same axis of rotation.
$\therefore I_1=I_2\Rightarrow I_{net}=2I_1\Rightarrow I_1=\frac{I_{net}}{2}{I}_1=\frac{48}{2}{\text{kg-m}}^2=24{\text{kg-m}}^2$

Alternate:
MOI of the lamina about the given axis is given by
$I=\frac{m{a}^2}{3}=\frac{2\times 6^2}{3}=24{\text{kg-m}}^2$