Question

Two uniform identical rods each of mass $M$ and length $l$ are joined to form a cross as shown in figure. Find the moment of inertia of the cross about a bisector shown by dotted lines in the figure.

Answer 1

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) $I^{\prime }=M{l}^2/12$. Hence the moments of inertia of the cross $I_z=2I_z^{\prime }=M{l}^2/6$. The moments of inertia of the cross about the bisector are equal by symmetry $I_X=I_Y$ according to the theorem of perpendicular axis, $I_Z=I_X+I_Y$ the moment of inertia of the cross about the bisector is $M{l}^2/12$.