Question

A thin rod of mass $6m$ and length $6L$ is bent into regular hexagon. The $M.I$ of the hexagon about a normal axis to its plane and through centre of system is

• A. $m{L}^2$
• B. $3m{L}^2$
• C. $5m{L}^2$
• D. $11m{L}^2$

Answer 1

The correct option is C $5m{L}^2$

Each rod has mass m and length L and distance of their C.O.M from

center of hexagon(d) $=L\sin {60}^{\circ }=\frac{L\sqrt{3}}{2}$ (Internal angle of regular hexagon $={120}^{\circ })$

Moment of inertia of one rod about axis through its C.O.M(1) $=\frac{m{L}^2}{12}$

So its moment of inertia about axis through center of hexagon and perpendicular to the plan is:

$I_{centre}=I+m{d}^2$ (parallel axis theorem)

$=\frac{m{L}^2}{12}+\frac{3m{L}^2}{4}=\frac{5m{L}^2}{6}$

So for 6 rods the net moment of inertia this axis will be:

$6\times \frac{5m{L}^2}{6}=5m{L}^2$