Two uniform thin rods each of mass M and length I are placed along X and Y-axis with one end of each at the origin. M.I. of the system about Z-axis is

Options

A. `3/2 ML^(2)`
B. `2/3 ML^(2)`
C. `2ML^(2)`
D. zero

Answer

To find the moment of inertia of the system about the Z-axis for two uniform thin rods, we can follow these steps:

### Step 1: Understand the Configuration
We have two uniform thin rods:
- Rod 1 is placed along the X-axis.
- Rod 2 is placed along the Y-axis.
Both rods have mass \( M \) and length \( L \), with one end of each rod at the origin (0,0).

### Step 2: Moment of Inertia of Each Rod
The moment of inertia \( I \) of a thin rod about an axis perpendicular to its length and passing through its center of mass is given by the formula:
\[ I_{cm} = \frac{1}{12} m L^2 \]
For our case:
- For Rod 1 (along the X-axis), the moment of inertia about its center of mass (which is at \( \frac{L}{2} \)) is:
\[ I_{1,cm} = \frac{1}{12} M L^2 \]
- For Rod 2 (along the Y-axis), the moment of inertia about its center of mass is also:
\[ I_{2,cm} = \frac{1}{12} M L^2 \]

### Step 3: Use the Parallel Axis Theorem
The parallel axis theorem states that if you know the moment of inertia about the center of mass, you can find the moment of inertia about any parallel axis by adding \( md^2 \), where \( d \) is the distance from the center of mass to the new axis.

For Rod 1 (X-axis):
- The distance from the center of mass to the Z-axis is \( \frac{L}{2} \).
\[ I_{1,Z} = I_{1,cm} + M \left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + M \left(\frac{L}{2}\right)^2 \]
Calculating this gives:
\[ I_{1,Z} = \frac{1}{12} M L^2 + M \frac{L^2}{4} = \frac{1}{12} M L^2 + \frac{3}{12} M L^2 = \frac{4}{12} M L^2 = \frac{1}{3} M L^2 \]

For Rod 2 (Y-axis):
- The distance from the center of mass to the Z-axis is also \( \frac{L}{2} \).
\[ I_{2,Z} = I_{2,cm} + M \left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + M \left(\frac{L}{2}\right)^2 \]
Calculating this gives:
\[ I_{2,Z} = \frac{1}{12} M L^2 + \frac{3}{12} M L^2 = \frac{4}{12} M L^2 = \frac{1}{3} M L^2 \]

### Step 4: Total Moment of Inertia of the System
The total moment of inertia \( I_{total} \) about the Z-axis is the sum of the moments of inertia of both rods:
\[ I_{total} = I_{1,Z} + I_{2,Z} = \frac{1}{3} M L^2 + \frac{1}{3} M L^2 = \frac{2}{3} M L^2 \]

### Final Answer
The moment of inertia of the system about the Z-axis is:
\[ I_{total} = \frac{2}{3} M L^2 \]