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 as shown dotted in the figure.

• A. $\frac{M{L}^2}{6}$
• B. $\frac{M{L}^2}{24}$
• C. $\frac{M{L}^2}{12}$
• D. $\frac{M{L}^2}{3}$

Answer 1

The correct option is C

$\frac{M{L}^2}{12}$

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 bisectors are equal by symmetry

and according to the theorem of perpendicular axes, the moment of inertia of the cross about

the bisector is $\frac{M{l}^2}{12}$.