Moment of inertia of a circular wire of mass M and radius R about its diameter is

Options

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

Explanation

To find the moment of inertia of a circular wire of mass \( M \) and radius \( R \) about its diameter, we can follow these steps:

### Step 1: Understand the Geometry
We have a circular wire that can be visualized as a thin ring. The wire has a mass \( M \) and a radius \( R \). We need to calculate the moment of inertia about an axis that passes through the diameter of the ring.

### Step 2: Use the Perpendicular Axis Theorem
The perpendicular axis theorem states that for a planar body, the moment of inertia about an axis perpendicular to the plane (let's call it the \( z \)-axis) is equal to the sum of the moments of inertia about two perpendicular axes (the \( x \)-axis and \( y \)-axis) lying in the plane of the body. Mathematically, this can be expressed as:
\[ I_z = I_x + I_y \]

### Step 3: Moment of Inertia About the Center Axis
For a thin ring, the moment of inertia about an axis perpendicular to the plane of the ring (the \( z \)-axis) and passing through its center is given by:
\[ I_z = M R^2 \]

### Step 4: Symmetry of the Ring
Due to the symmetry of the ring, the moments of inertia about the two axes in the plane (the \( x \)-axis and \( y \)-axis) are equal:
\[ I_x = I_y \]
Let’s denote the moment of inertia about the \( x \)-axis (or \( y \)-axis) as \( I \). Therefore, we can rewrite the equation from the perpendicular axis theorem as:
\[ I_z = I + I \]
or
\[ I_z = 2I \]

### Step 5: Substitute and Solve
Now, substituting \( I_z \) into the equation:
\[ M R^2 = 2I \]
To find \( I \), we rearrange the equation:
\[ I = \frac{M R^2}{2} \]

### Conclusion
Thus, the moment of inertia of a circular wire of mass \( M \) and radius \( R \) about its diameter is:
\[ I = \frac{M R^2}{2} \]