Two identical concentric rings each of mass m and radius R are placed perpendicularly. What is the moment of inertia of the system about the axis of one of the rings ?

Options

A. `3MR^(2)`
B. `(3)/(2)MR^(2)`
C. `(1)/(4)MR^(2)`
D. `2MR^(2)`

Explanation

To find the moment of inertia of the system consisting of two identical concentric rings placed perpendicularly about the axis of one of the rings, we can follow these steps:

### Step 1: Understand the Configuration
We have two identical rings, each with mass \( m \) and radius \( R \). One ring lies in the xy-plane, and the other lies in the xz-plane, making them perpendicular to each other.

### Step 2: Moment of Inertia of One Ring
The moment of inertia \( I \) of a single ring about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I = mR^2 \]
For the first ring (lying in the xy-plane), the moment of inertia about its own axis is:
\[ I_1 = mR^2 \]

### Step 3: Moment of Inertia of the Second Ring
For the second ring (lying in the xz-plane), we need to find the moment of inertia about the same axis as the first ring. Since the second ring is also a ring with mass \( m \) and radius \( R \), we can use the perpendicular axis theorem.

According to the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the plane of the ring is the sum of the moments of inertia about two perpendicular axes in the plane of the ring. Thus, the moment of inertia about the axis of the first ring is:
\[ I_2 = \frac{1}{2} mR^2 + \frac{1}{2} mR^2 = mR^2 \]

### Step 4: Total Moment of Inertia
Now, we can find the total moment of inertia of the system about the axis of one of the rings. Since the moment of inertia of each ring about the same axis is \( mR^2 \), we can add them together:
\[ I_{\text{total}} = I_1 + I_2 = mR^2 + mR^2 = 2mR^2 \]

### Step 5: Conclusion
Thus, the moment of inertia of the system about the axis of one of the rings is:
\[ I_{\text{total}} = 2mR^2 \]