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 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 1: Understand the Geometry
We have a semicircular wire with radius \( r \) and mass \( M \). The total length of the semicircular wire is given by:
\[ L = \pi r \]

### Step 2: Determine the Elemental Mass
The mass per unit length of the wire can be calculated as:
\[ \text{Mass per unit length} = \frac{M}{L} = \frac{M}{\pi r} \]
For an elemental length \( dl \), the elemental mass \( dm \) can be expressed as:
\[ dm = \frac{M}{\pi r} \cdot dl \]

### Step 3: Relate Elemental Length to Angular Displacement
The elemental length \( dl \) can be expressed in terms of the angular displacement \( d\theta \):
\[ dl = r \, d\theta \]
Substituting this into the expression for \( dm \):
\[ dm = \frac{M}{\pi r} \cdot (r \, d\theta) = \frac{M}{\pi} \, d\theta \]

### Step 4: Set Up the Moment of Inertia Integral
The moment of inertia \( I \) about the axis perpendicular to the plane through the center is given by:
\[ I = \int r^2 \, dm \]
Here, the distance \( r \) from the axis to the elemental mass \( dm \) is constant for all elements along the semicircle, so we can factor it out:
\[ I = r^2 \int dm \]
Substituting \( dm \):
\[ I = r^2 \int_0^{\pi} \frac{M}{\pi} \, d\theta \]

### Step 5: Evaluate the Integral
Now we evaluate the integral:
\[ I = r^2 \cdot \frac{M}{\pi} \cdot \int_0^{\pi} d\theta \]
The integral \( \int_0^{\pi} d\theta = \pi \):
\[ I = r^2 \cdot \frac{M}{\pi} \cdot \pi = M r^2 \]

### Final Result
Thus, the moment of inertia of the uniform semicircular wire about the specified axis is:
\[ I = \frac{M r^2}{2} \]