Question

Two rods of equal mass $M$ and length $L$ lie along the X-axis And Y-Axis with their centres at the origin. What is the moment of inertia of the rods about the line $\mathrm{X}=\mathrm{Y}$?

Answer 1

Step 1: Drawing of the orientation of the rods

Step 2: Moment of inertia for the rods

The moment of inertia of a rod of about an axis passing through the center and perpendicular to its length can be expressed as follows:

$\mathrm{I}=\frac{{\mathrm{ML}}^2}{12}$

Here,

$\mathrm{I}$ is the moment of inertia of the rod.

$M$ is the mass of the rod.

$L$ is the length of the rod.

Step 3: Total moment of inertia due to two rods

Since there are two rods present, the moment of inertia will be doubled.

${\mathrm{I}}_{\mathrm{total}}=2\mathrm{I}$

Here,

${\mathrm{I}}_{\mathrm{total}}$ is the total moment of inertia due to the two rods.

Step 4: Moment of inertia at $X=Y$

The line $\mathrm{X}=\mathrm{Y}$ makes an angle of ${45}^{\circ }$.

The moment of inertia of the rods about an axis inclined at an angle $\theta$ to the original axis can be expressed as follows:

$\mathrm{I}\text{'}={\mathrm{I}}_{\mathrm{total}}{\sin }^2\mathrm{\theta }$

Here, $\mathrm{I}\text{'}$is the moment of inertia of the rods about an axis inclined at an angle $\theta$.

Calculation of the moment of inertia of the rods about the line $X=Y$:

$$\begin{gathered} \mathrm{I}\text{'}=2\mathrm{I}\times {\sin }^2\left({45}^{\circ }\right) \\ \mathrm{I}\text{'}=2\mathrm{I}\times \frac{1}{2} \\ \mathrm{I}\text{'}=\frac{{\mathrm{ML}}^2}{12} \end{gathered}$$

Hence, the moment of inertia of the rods about the line $\mathrm{X}=\mathrm{Y}$ is $\frac{{\mathrm{ML}}^2}{12}$.