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

Answer 1

Answer: (A)

$20 m L^2$

Answer 2

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
$I = 6\left[I_{one side}\right] =6\left[\frac{m\left(2L\right)^{2}}{12}+mr^{2}\right] = 6\left[\frac{mL^{2}}{3} +m\left(L\sqrt{3}\right)^{2}\right] =20mL^2$
So, the correct options is (A) : $20mL^2$