Question

A uniform thin bar of mass $6m$ and length $12L$ is bent to make an 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. $6m{L}^2$
• C. $\frac{12}{5}m{L}^2$
• D. $30m{L}^2$

Answer 1

The correct option is A $20m{L}^2$


Mass of one side =$\frac{1}{6}$(Total mass)$=m$
Length of one side =$\frac{1}{6}$(Total length)$=\frac{1}{6}\left(12L\right)=2L$
The desired moment of inertia $I$ about $O$ is 6 times moment of inertia due to one side $I_{oneside}$ about $O$
From Parallel axis theorem
$I=6\left[I_{oneside}\right]=6[\frac{m(2L{)}^2}{12}+m{r}^2](r=2Lsin{60}^{\circ }=L\sqrt{3})\Rightarrow 6[\frac{m{L}^2}{3}+3m{L}^2]=20m{L}^2$