Two identical semicircular discs of mass 'm' each and radius 'R' are placed in the XY (horizontal) plane and the YZ (vertical) plane, respectively. They are so placed that they have their common diameter along the y-axis. Then, the moment of inertia `I_(n)` of the system about the appropriate axis is given by (`I_(n)` refers to moment of inertia about axis n-where n is X, Y, Z)

Options

A. `l_(x)=(1)/(2)mR^(2)`
B. `l_(Y)=(1)/(2)mR^(2)`
C. `l_(Z)=(3)/(4)mR^(2)`
D. `l_(X)=l_(Y)=l_(Z)`

Answer

To find the moment of inertia of the system consisting of two identical semicircular discs, we will calculate the moment of inertia about the X, Y, and Z axes step by step.

### Step 1: Understanding the Configuration
We have two semicircular discs:
- The first semicircular disc is in the XY plane (horizontal).
- The second semicircular disc is in the YZ plane (vertical).
Both discs have a common diameter along the Y-axis.

### Step 2: Moment of Inertia about the X-axis (I_x)
1. **For the first disc (in the XY plane)**:
- The moment of inertia of a semicircular disc about an axis perpendicular to its plane and passing through its center is given by:
\[ I_{x1} = \frac{mR^2}{2} \]

2. **For the second disc (in the YZ plane)**:
- The moment of inertia about the X-axis can be found using the perpendicular axis theorem. The moment of inertia about the Z-axis (perpendicular to the plane) is:
\[ I_{z2} = \frac{mR^2}{2} \]
- Using the perpendicular axis theorem:
\[ I_{z2} = I_{y2} + I_{x2} \Rightarrow I_{x2} + I_{y2} = \frac{mR^2}{2} \]
- Since \(I_{x2} = I_{y2}\) (both axes are along the diameter of the disc):
\[ 2I_{x2} = \frac{mR^2}{2} \Rightarrow I_{x2} = \frac{mR^2}{4} \]

3. **Total moment of inertia about the X-axis**:
\[ I_x = I_{x1} + I_{x2} = \frac{mR^2}{2} + \frac{mR^2}{4} = \frac{3mR^2}{4} \]

### Step 3: Moment of Inertia about the Y-axis (I_y)
1. **For the first disc (in the XY plane)**:
- The Y-axis is in the plane of the first disc, so:
\[ I_{y1} = \frac{mR^2}{4} \]

2. **For the second disc (in the YZ plane)**:
- The Y-axis is also in the plane of the second disc, so:
\[ I_{y2} = \frac{mR^2}{4} \]

3. **Total moment of inertia about the Y-axis**:
\[ I_y = I_{y1} + I_{y2} = \frac{mR^2}{4} + \frac{mR^2}{4} = \frac{mR^2}{2} \]

### Step 4: Moment of Inertia about the Z-axis (I_z)
1. **For the first disc (in the XY plane)**:
- The Z-axis is in the plane of the first disc:
\[ I_{z1} = \frac{mR^2}{4} \]

2. **For the second disc (in the YZ plane)**:
- The Z-axis is perpendicular to the plane of the second disc:
\[ I_{z2} = \frac{mR^2}{2} \]

3. **Total moment of inertia about the Z-axis**:
\[ I_z = I_{z1} + I_{z2} = \frac{mR^2}{4} + \frac{mR^2}{2} = \frac{mR^2}{4} + \frac{2mR^2}{4} = \frac{3mR^2}{4} \]

### Summary of Results
- Moment of inertia about the X-axis:
\[ I_x = \frac{3mR^2}{4} \]
- Moment of inertia about the Y-axis:
\[ I_y = \frac{mR^2}{2} \]
- Moment of inertia about the Z-axis:
\[ I_z = \frac{3mR^2}{4} \]