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 a bisector in the plane of rods as shown by dotted line in the figure.

• A. $\frac{M{L}^2}{12}$
• B. $\frac{M{L}^2}{4}$
• C. $\frac{M{L}^2}{6}$
• D. $\frac{M{L}^2}{3}$

Answer 1

The correct option is A $\frac{M{L}^2}{12}$

Given, length of the rod is $L$ and mass of the rod is $M$

As we know that, $I_z$ of a rod is given as $\frac{M{L}^2}{12}$
So, $\text{MOI}$ of the two rod shown will be $(2\times I_z)$, which is $\frac{M{L}^2}{6}$
Now, from the perpendicular axis theorem we have
$I_z=I_x+I_y$
and due to symmetry, $I_x=I_y$
$\Rightarrow I_z=2I_x$ or $I_x=I_y=\frac{I_z}{2}$
Hence, $\text{MOI}$ of the cross about the bisector will be
$\frac{\left(\frac{M{L}^2}{6}\right)}{2}=\frac{M{L}^2}{12}$