Question

A square lamina of mass $m=4\text{kg}$ having sides of length $a=2\text{m}$. Find the moment of inertia of the lamina along one of its side.

• A. $\frac{4}{3}{\text{kg-m}}^2$
• B. $\frac{8}{3}{\text{kg-m}}^2$
• C. $\frac{16}{3}{\text{kg-m}}^2$
• D. $\frac{2}{3}{\text{kg-m}}^2$

Answer 1

The correct option is C $\frac{16}{3}{\text{kg-m}}^2$


As we know,
$I_{com}=\frac{m{a}^2}{12}$
Applying parallel axis theorem,
$I=I_{com}+m{d}^2$ $[\because \text{Here}d=\frac{a}{2}]$
$I=\frac{m{a}^2}{12}+m{\left(\frac{a}{2}\right)}^2$
$=\frac{m{a}^2}{12}+\frac{m{a}^2}{4}$
$I=\frac{m{a}^2}{3}=\frac{4\times 2^2}{3}{\text{kg-m}}^2=\frac{16}{3}{\text{kg-m}}^2$

So, moment of inertia of square lamina along one of its side is $\frac{16}{3}{\text{kg-m}}^2$.