Question

Four thin uniform rods, each of length $L$ and mass $M$ are joined to form a square. The moment of inertia of square about an axis along one of its diagonal is

• A. $\frac{M{L}^2}{6}$
• B. $\frac{4M{L}^2}{3}$
• C. $\frac{3M{L}^2}{4}$
• D. $\frac{2M{L}^2}{3}$

Answer 1

The correct option is D $\frac{2M{L}^2}{3}$

The given condition can be shown as


From Perpendicular axis theorem,
$I_z=I_x+I_y$
Since, due to symmetry $I_x=I_y$
$\Rightarrow I_z=2I_x$ or, $I_x=I_y=\frac{I_z}{2}$
Now, for a single rod (say $BC$),
$I_{\text{COM}}=\frac{M{L}^2}{12}$
Also, from parallel axis theorem we have
$I_z=I_{\text{COM}}+M{d}^2$
$\Rightarrow I_z=\frac{M{L}^2}{12}+M\times {\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{3}$
Similarly, for four rods
$I_z=4\times \frac{M{L}^2}{3}=\frac{4M{L}^2}{3}$
Hence, the moment of inertia of square about an axis along its one diagonal is
$I_x=I_y=\frac{I_z}{2}=\frac{\left(\frac{4M{L}^2}{3}\right)}{2}=\frac{2M{L}^2}{3}$