Question

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

• A. $5m{I}^2$
• B. $6m{I}^2$
• C. $4m{I}^2$
• D. $m{I}^2/12$

Answer 1

Answer: (A)

$5m{I}^2$

$\begin{array}{c}\text{Clearly, length and mass of each edge}=\ell \text{and}M.\\ \text{Consider portion BC :}\\ \text{moment of inertia of BC about axis passing its}\\ \text{centre and perperdicular to its plane is}\\ \frac{m{l}^2}{12}\text{.}\end{array}$

$\begin{array}{c}\text{Thus, moment of inertia about given axis of}BC=\frac{m{l}^2}{12}+m{d}^2\\ \text{(I}{}_{\text{BC}})\text{(Parallel axis theo. )}\\ \therefore I_{BC}=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}l}{2}\right)}^2=\frac{m{l}^2}{12}+\frac{3m{l}^2}{4}=\frac{5}{6}m{l}^2\end{array}$

$\text{Hence, total moment of inertia of system}=6I_{BC}=5m{l}^2$