The moment of inertia of a uniform semicircularwire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is :

Options

A. `mr^2`
B. `1/2mr^2`
C. `1/4mr^2`
D. `2/5mr^2`

Answer

To find the moment of inertia of a uniform semicircular wire of mass \( M \) and radius \( r \) about a line perpendicular to the plane of the wire through the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Visualize the semicircular wire lying in the xy-plane, with its center at the origin (0,0) and extending from \((-r, 0)\) to \((r, 0)\).

2. **Identify a Small Element**:
- Consider a small mass element \( dm \) located at an angle \( \theta \) from the center of the semicircle. The position of this element can be described in polar coordinates as:
\[ x = r \cos(\theta), \quad y = r \sin(\theta) \]
- The arc length \( ds \) for a small angle \( d\theta \) is given by:
\[ ds = r \, d\theta \]
- Therefore, the mass element can be expressed as:
\[ dm = \frac{M}{L} \, ds \]
where \( L \) is the length of the semicircular wire, which is \( \frac{1}{2} \times 2\pi r = \pi r \). Thus,
\[ dm = \frac{M}{\pi r} \cdot r \, d\theta = \frac{M}{\pi} \, d\theta \]

3. **Calculate the Moment of Inertia for the Element**:
- The moment of inertia \( dI \) of this small element \( dm \) about the axis perpendicular to the plane through the center is given by:
\[ dI = r^2 \, dm \]
- Substituting \( dm \):
\[ dI = r^2 \left( \frac{M}{\pi} \, d\theta \right) = \frac{M r^2}{\pi} \, d\theta \]

4. **Integrate Over the Semicircle**:
- The angle \( \theta \) varies from \( 0 \) to \( \pi \) for the semicircular wire. Therefore, we integrate \( dI \):
\[ I = \int_0^{\pi} dI = \int_0^{\pi} \frac{M r^2}{\pi} \, d\theta \]
- This simplifies to:
\[ I = \frac{M r^2}{\pi} \int_0^{\pi} d\theta = \frac{M r^2}{\pi} \cdot \pi = M r^2 \]

5. **Final Result**:
- The moment of inertia of the uniform semicircular wire about the specified axis is:
\[ I = M r^2 \]