Two uniform identicla rods each of mass M and length l are joined to form a cross as shown in figure. Find the momet of inertia of the cross about a bisector as shown doted in the figure

Explanation

Consider the line perpendicular to the plane of the figure through the centre of the cross. The moment of inertia of each rod about this line (say about `z`- axis) `I'=Ml^(2)//12.` Hence the moments of inertia of the cross `I_(z)=2I_()z^(')=Ml^(2)//6`. The moments of inertia of the cross about the two bisector are equla by symmetry `I_(X)=I_(Y)` according to the theorem of perpnedicular axis `I_(z)=I_(X_(2))+I_(Y)` the moment of inertia of the cross about the bisector is `Ml^(2)//12`.