The M.I. of a thin ring of mass M and radius R about an axis through the diameter in its plane will be:

Options

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

Answer

To find the moment of inertia (M.I.) of a thin ring of mass \( M \) and radius \( R \) about an axis through the diameter in its plane, we can use the perpendicular axis theorem. Here’s a step-by-step solution:

### Step 1: Understand the Geometry of the Ring
- A thin ring is a circular object with all its mass concentrated at a distance \( R \) from its center. The axis we are considering is through the diameter of the ring and lies in the plane of the ring.

### Step 2: Identify the Relevant Axes
- Let’s denote:
- \( I_z \): Moment of inertia about an axis perpendicular to the plane of the ring (through the center).
- \( I_x \): Moment of inertia about an axis along the diameter of the ring (in the plane).
- \( I_y \): Moment of inertia about another axis along the diameter of the ring (in the plane).

### Step 3: Apply the Perpendicular Axis Theorem
- The perpendicular axis theorem states that for a planar object:
\[ I_z = I_x + I_y \]
- Since the ring is symmetrical, \( I_x = I_y \). Therefore, we can write:
\[ I_z = 2I_x \]

### Step 4: Calculate \( I_z \)
- The moment of inertia of a thin ring about an axis perpendicular to its plane (the z-axis) is given by:
\[ I_z = MR^2 \]

### Step 5: Substitute \( I_z \) into the Equation
- From the perpendicular axis theorem, we have:
\[ MR^2 = 2I_x \]
- Solving for \( I_x \):
\[ I_x = \frac{MR^2}{2} \]

### Conclusion
- The moment of inertia of the thin ring about an axis through the diameter in its plane is:
\[ I_x = \frac{1}{2} MR^2 \]

### Final Answer
- The correct answer is \( \frac{1}{2} MR^2 \).

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