A uniform thin bar


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 _____________ $\times 10^{-1} \mathrm{~kg} \mathrm{~m}^{2}$.


$61=2.4          \ell=0.4 \mathrm{~m}$

$\sin 60^{\circ}=\frac{r}{\ell}$

$r=1 \sin 60^{\circ}=\frac{\ell \sqrt{3}}{2}$

MOI, $\quad \mathrm{I}=\left[\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{mr}^{2}\right] 6$

$=\left[\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{m}\left(\frac{\ell \sqrt{3}}{2}\right)^{2}\right] 6$

$=5 \mathrm{ml}]^{2}$

$=5 \times 1 \times 0.16$


$\mathrm{I}=8 \times 10^{-1} \mathrm{~kg} \mathrm{~m}^{2}$

Ans. 8

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now