Question

A thin rod of mass $6m$ and length $6L$ is bent into a regular hexagon. The M.I. of the hexagon about a normal axis to its plane and through the centre of the system is
(A) $mL^2$
(B) $3mL^2$
(C) $5mL^2$
(D) $11mL^2$

Answer 1

Hint
The moment of inertia of one side of the thin rod of a hexagon can be calculated by the parallel axes theorem which then be multiplied by 6 for the moment of inertia of the whole hexagon.

Complete step by step solution
Given, the mass of the thin rod is 6m. Thus, the mass of each side of the hexagon is m. The length of the thin rod is 6L. Thus, the length of each side of the hexagon is L.
The internal angles of a regular hexagon are $120^\circ $.
The center of mass of each thin rod from the center of the hexagon is given by,
${\rm{d}} = {\rm{L}}\sin 60^\circ = \dfrac{{\sqrt 3 {\rm{L}}}}{2}$
We know that the moment of inertia of one rod about an axis through the center of mass of the rod is given by,
${\rm{I}} = \dfrac{{{\rm{m}}{{\rm{L}}^2}}}{{12}}$
According to parallel axis theorem of the moment of inertia, the moment of inertia about an axis through the centre of hexagon and perpendicular to the plane will be,
${{\rm{I}}_{\rm{c}}} = {\rm{I}} + {\rm{m}}{{\rm{d}}^2}$
$ = \dfrac{{{\rm{m}}{{\rm{L}}^2}}}{{12}} + {\rm{m}}{\left( {\dfrac{{\sqrt 3 {\rm{L}}}}{2}} \right)^2}$
$ = \dfrac{{{\rm{m}}{{\rm{L}}^2}}}{{12}} + \dfrac{{3{\rm{m}}{{\rm{L}}^2}}}{4}$
$ = \dfrac{{10}}{{12}}{\rm{m}}{{\rm{L}}^2} = \dfrac{5}{6}{\rm{m}}{{\rm{L}}^2}$
Therefore, the net moment of inertia of 6 rods $ = 6 \times \dfrac{{5{\rm{m}}{{\rm{L}}^2}}}{6} = 5{\rm{m}}{{\rm{L}}^2}$
Therefore, (C) $5{\rm{m}}{{\rm{L}}^2}$ is the required solution.

Note The parallel axes theorem stated that the moment of inertia along any axes drawn parallel to the axis passing through the centre of mass is equal to the sum of moment of inertia along the centre of the mass axis and the product of mass and the square of the distance between two axes. While solving the question, we should remember the moment of inertia of a thin rod.