Question

A rod of length $2.4\text{m}$ and mass $6\text{kg}$ is bent to form a regular hexagon. Find the moment of inertia of the hexagon about the axis perpendicular to the plane of hexagon passing through its centre.

• A. $0.3{\text{kg-m}}^2$
• B. $0.5{\text{kg-m}}^2$
• C. $0.8{\text{kg-m}}^2$
• D. $0.6{\text{kg-m}}^2$

Answer 1

The correct option is C $0.8{\text{kg-m}}^2$


Since, regular hexagon have equal sides.
So, for $\text{AB},l=\frac{2.4}{6}=0.4\text{m}$
and mass of $\text{AB},m=\frac{6}{6}=1\text{kg}$

On using parallel axis theorem for $\text{AB}$,

$I=I_0+m{d}^2$

where, $d$ is the perpendicular distance between centre and side of hexagon.

From above diagram,
$d=l\cos {30}^{\circ }$

$\therefore I=\frac{m{l}^2}{12}+m(l\cos {30}^{\circ }{)}^2$

$\Rightarrow I=\frac{1\times {0.4}^2}{12}+1\times {(0.4\times \frac{\sqrt{3}}{2})}^2$

$\Rightarrow I=0.13{\text{kg-m}}^2$

So, for the whole system, moment of inertia will be,

$I_0=6I$

$\therefore I_0=6\times 0.13=0.78\approx 0.8{\text{kg-m}}^2$

Hence, option $\left(c\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - A regular hexagon is a geometric figure}\\ \text{having all six sides of equal length. Also, lines drawn from}\\ \text{the centre to the ends of one of the side from an equilateral triangle}\end{array}$