Question

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

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 is $\frac{M{l}^2}{12}$ and hence the moment of inertia of the cross is $\frac{M{l}^2}{6}$. The moment of inertia of the cross about the two bisector are equal by symmetry and according to the theorem of perpendicular axes, the moment of inertia of the cross about the bisector is $\frac{M{l}^2}{12}$