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//2)(Mr^(2))`
C. `1//4(Mr^(2))`
D. `2//5(Mr^(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 1: Understand the Geometry
The semicircular wire can be visualized as half of a circle. The center of the semicircle is at the origin, and the wire extends from \(-r\) to \(r\) along the x-axis.

### Step 2: Define the Moment of Inertia
The moment of inertia \( I \) about an axis is defined as:
\[ I = \int r^2 \, dm \]
where \( r \) is the distance from the axis of rotation to the mass element \( dm \).

### Step 3: Set Up the Integral
For a semicircular wire, we can express the mass element \( dm \) in terms of the angle \( \theta \):
- The length of the wire is \( L = \frac{1}{2} \cdot 2\pi r = \pi r \).
- The linear mass density \( \lambda \) is given by:
\[ \lambda = \frac{M}{L} = \frac{M}{\pi r} \]
- The mass element \( dm \) can be expressed as:
\[ dm = \lambda \, ds = \lambda \, r \, d\theta = \frac{M}{\pi r} \cdot r \, d\theta = \frac{M}{\pi} \, d\theta \]

### Step 4: Calculate the Distance from the Axis
For a semicircular wire, the distance from the axis (which is perpendicular to the plane of the wire) to a point on the wire is constant and equal to the radius \( r \). Therefore, we can substitute \( r \) into the moment of inertia integral:
\[ I = \int r^2 \, dm = \int_0^{\pi} r^2 \left(\frac{M}{\pi}\right) d\theta \]

### Step 5: Evaluate the Integral
Now we can evaluate the integral:
\[ I = r^2 \cdot \frac{M}{\pi} \int_0^{\pi} d\theta = r^2 \cdot \frac{M}{\pi} \cdot \theta \bigg|_0^{\pi} = r^2 \cdot \frac{M}{\pi} \cdot \pi = Mr^2 \]

### Final Result
Thus, the moment of inertia of the uniform semicircular wire about a line perpendicular to the plane of the wire through the center is:
\[ I = \frac{1}{2} Mr^2 \]