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

Options

A. `20 mL^2`
B. `30 mL^2`
C. `((12)/(5)) mL^2`
D. `6 mL^2`

Answer

A

Explanation

Length of each side of hexagon = 2L and mass of each side = m . Let O be the centre of mass of hexagon . Perpendicular distance of O from each side ,
r = L tan `60^(@) = L sqrt3`
The desired moment of inertia of hexagon about O is
`I = 6[I_("one side") ] = 6 [ (m (2L)^(2))/(12) + mr^(2)]`
`= 6 [(mL^2)/(3) + m(Lsqrt3)^(2)] = 20 mL^(2)`