Question

A uniform thin bar of mass $6kg$ and length 2.4 meter 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 of hexagon is_____ $\times 10^{-1}kg{m}^2$

Answer 1

Answer: 8

$m=$ mass of one side of hexagon
$=1kg$
$61=2.4\ell =0.4m$
$\sin 60^{\circ }=\frac{r}{\ell }$
$r=1\sin 60^{\circ }=\frac{\ell \sqrt{3}}{2}$
MOI, $I=\left[\frac{m{\ell }^2}{12}+m{r}^2\right]6$
$=\left[\frac{m{\ell }^2}{12}+m{\left(\frac{\ell \sqrt{3}}{2}\right)}^2\right]6$
$=5m{l}^2$
$=5\times 1\times 0.16$
$=0.8$
$I=8\times 10^{-1}kg{m}^2$