Question

A uniform thin bar of mass $6kg$and length $2.4m$ is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the hexagon is ______$\times {10}^{-1}kg{m}^2$.

Answer 1

Step 1. Given data:

Mass of bar, $M=6kg$

Length of the bar, $l=2.4m$

Step 2. Finding a moment of inertia of hexagon:

Since the rod that is bent to form regular hexagon is uniform,

Side of hexagon, $a=\frac{l}{6}=\frac{2.4}{6}m$

$a=0.4m$

Mass of one side of hexagon is, $m=\frac{M}{6}=\frac{6}{6}kg$

$m=1kg$

Moment of inertia about side $AB$ about point $P$ is, $I_p=\frac{m{a}^2}{12}$

Using parallel axis theorem,

Moment of inertia about side $AB$ about point $O$ is, $I_o=\frac{m{a}^2}{12}+m{\left(a\cos {30}^o\right)}^2$ $\left[\because a=\frac{l}{6}\right]$

$I_o=\frac{m{a}^2}{12}+\frac{m{\left(a\sqrt{3}\right)}^2}{4}$

$I_o=\frac{1\times 0.4^2}{12}+\frac{1{\left(0.4\times \sqrt{3}\right)}^2}{4}$

$I_o=\frac{0.16}{12}+\frac{1.44}{12}=\frac{1.6}{12}$

Moment of inertia of hexagon about the centre is, $I_{hex}=6\times I_o=6\times \frac{1.6}{12}$

$I_{hex}=0.8=8\times {10}^{-1}kg{m}^2$

Hence, the answer is 8.