Question

A uniform thin bar of mass $6m$ and length $12L$ is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

• A. $20m{L}^2$
• B. $30m{L}^2$
• C. $\left(\frac{12}{5}\right)m{L}^2$
• D. $6m{L}^2$

Answer 1

Answer: (A)

Length of each side of hexagon = $2L$ and mass of each side = $m$.

Let $O$ be the centre of mass of hexagon. Therefore, perpendicular distance of $O$ from each side, $r=L\tan 60°=L\sqrt{3}$ .
The desired moment of inertia of hexagon about $O$ is
$I=6\left[I_{oneside}\right]=6\left[\frac{m{\left(2L\right)}^2}{12}+m{r}^2\right]=6\left[\frac{m{L}^2}{3}+m{\left(L\sqrt{3}\right)}^2\right]=20m{L}^2$