Question

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

Answer 1

MOI of $AB$ about $P$,
$I_{{\text{AB}}_P}=\frac{\left(\frac{M}{6}\right)(\frac{l}{6}{)}^2}{12}$

MOI of AB about O,
$I_{{AB}_O}=\frac{\left(\frac{M}{6}\right)(\frac{l}{6}{)}^2}{12}$+$\left(\frac{M}{6}\right)(\frac{L}{6}\times \frac{\sqrt{3}}{2}{)}^2$

$I_{{hexagon}_0}=6I_{{AB}_0}$

On putting the value of $M=6\text{kg}$ and $L=24\text{m}$ in the above equation, we get,

$I_{{hexagon}_0}$= $0.8{\text{kg m}}^2$ = $8\times {10}^{-1}{\text{kg m}}^2$.