Question

A uniform rod of mass $6M$ and length $6I$ is bent to make an equilateral hexagon. Its $M.I.$ about an axis passing through the centre of mass and perpendicular to the place of hexagon is:

• A. $22m{I}^2$
• B. $6m{I}^2$
• C. $4m{I}^2$
• D. $m{I}^2/12$

Answer 1

The correct option is A $22m{I}^2$

We have,

$mass=6m,length=6L$

Since each side of the hexagon has side 2L and mass m. The mole of each side of the hexagon about an axis perpendicular to the plane of the hexagon through the center of the side is:

$I_1=\frac{M(2L{)}^2}{6}=\frac{2M{L}^2}{3}$

The distance between the center of the hexagon and the center of the rod is $|sqrt{3L}. By using parallel

axis theorem the mole about the axis is:

$I^{\prime }=\frac{2M{L}^2}{3}+m(\sqrt{3L}{)}^2=\frac{11M{L}^2}{3}$

So, the Total mole of the system about the axis is:

$I=6I^{\prime }=6\times \frac{11M{L}^2}{3}$

$=22M{L}^2$