Question:
A uniform thin bar of mass $6 \mathrm{~kg}$ 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 _____________ $\times 10^{-1} \mathrm{~kg} \mathrm{~m}^{2}$.
Solution:
$61=2.4 \ell=0.4 \mathrm{~m}$
$\sin 60^{\circ}=\frac{r}{\ell}$
$r=1 \sin 60^{\circ}=\frac{\ell \sqrt{3}}{2}$
MOI, $\quad \mathrm{I}=\left[\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{mr}^{2}\right] 6$
$=\left[\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{m}\left(\frac{\ell \sqrt{3}}{2}\right)^{2}\right] 6$
$=5 \mathrm{ml}]^{2}$
$=5 \times 1 \times 0.16$
$=0.8$
$\mathrm{I}=8 \times 10^{-1} \mathrm{~kg} \mathrm{~m}^{2}$
Ans. 8