Question

Four identical rods each of mass $M$ are joined to form as square frame, the moment of inertia of the system about one of the diagonals is

• A. $\frac{13M{l}^2}{3}$
• B. $\frac{2M{l}^2}{3}$
• C. $\frac{M{l}^2}{6}$
• D. $\frac{13M{l}^2}{6}$

Answer 1

Answer: (B)

$\frac{2M{l}^2}{3}$

Consider the frame made using four rods as shown.
We have the moment of inertia of rod AB about the center of mass as $\frac{M{l}^2}{12}+M(\frac{l}{2}{)}^2=\frac{M{l}^2}{3}$
Thus we get the moment of inertia of all the four rods as $\frac{4}{3}M{l}^2$
Now using perpendicular axis theorem we see the the moment of inertia about the two diagonals (which are perpendicular to each other) is given as $2I=\frac{4}{3}M{l}^2$
Thus we get $I=\frac{2}{3}M{l}^2$