Question

a uniform thin bar of mass 6m and length 12l bent to make regular hexagon. its moment of inertia about an axis passing throw the center of mass and perpendicular to the plane

Answer 1

Final Answer : $20 \:mL^2$

Steps:
1) Regular Hexagon of side :
a = 2l
Mass of each bar = m

Moment of Inertia about centre of mass of each bar ( Perpendicular to plane of Hexagon) ,
$I_{com} = \frac{m(2L)^2}{12} =\frac{mL^2}{3}$

2) Then, Moment of inertia of each bar about centre of Regular Hexagon.
By Parallel Axis Theorem,
$I = I_ {com} + md^2 \\ \\ => \frac{mL^2}{3} + m[2L sin(60\degree)]^2 \\ \\ => \frac{mL^2}{3} + 3mL^2$

3) We have such 6 identical bars which are symmetrical from Centre of Regular Hexagon.
Then,
$I_{req. } = 6I \\ \\ = 2mL^2 + 18mL^2 \\ \\ = 20mL^2$