Question

A uniform thin bar of mass $6m$ and length $12L$ is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

• A. $20 m L^2$
• B. $30 m L^2$
• C. $\left( \frac{12}{5}\right) m L^2$
• D. $6 m L^2$

Answer 1

The Correct Option is A

Length of each side of hexagon = \(2L\) and mass of each side = \(m\). 

Let \(O\) be the centre of mass of hexagon. Therefore, perpendicular distance of \(O\) from each side, \(r = L \tan 60? = L \sqrt{3}\) . 
The desired moment of inertia of hexagon about \(O\) is 
\(I = 6\left[I_{one side}\right] =6\left[\frac{m\left(2L\right)^{2}}{12}+mr^{2}\right] = 6\left[\frac{mL^{2}}{3} +m\left(L\sqrt{3}\right)^{2}\right] =20mL^2\)
So, the correct options is (A) : \(20mL^2\)

Answer 2

As per the formula,
\(I_1=\frac{m(2L)^2}{12}\)
\(=\frac{mL^3}{3}\)
Now,
\(I_1'=\frac{mL^3}{3}+m(\sqrt3L)^2\)
\(=\frac{10}{3}mL^3\)
So, the total moment of system about axis is 6I :
\(=6\times\frac{10}{3}mL^3\)
\(=20mL^3\)
So, the correct option is (A) : \(20mL^3\)