The moment of inertia of semicircular ring about its centre is

Options

A. `MR^(2)`
B. `(MR^(2))/(2)`
C. `(MR^(2))/(4)`
D. None of these

Answer

To find the moment of inertia of a semicircular ring about its center, we can follow these steps:

### Step 1: Understand the Geometry of the Semicircular Ring
A semicircular ring is a half-circle with a uniform mass distribution along its length. The center of the semicircle is the point about which we want to calculate the moment of inertia.

### Step 2: Define the Moment of Inertia
The moment of inertia (I) about an axis is defined as the sum of the products of the mass elements (dm) and the square of their distances (r) from the axis of rotation:
\[ I = \int r^2 \, dm \]

### Step 3: Set Up the Coordinate System
We can set up a polar coordinate system where the semicircular ring lies in the xy-plane. The radius of the semicircular ring is \( R \). The angle \( \theta \) will vary from \( 0 \) to \( \pi \).

### Step 4: Express the Mass Element
The mass element \( dm \) can be expressed in terms of the linear mass density \( \lambda \) (mass per unit length) and the arc length element \( dL \):
\[ dm = \lambda \, dL \]
For a semicircular ring, the linear mass density is given by:
\[ \lambda = \frac{M}{L} \]
where \( L \) is the length of the semicircular ring, which is \( \frac{1}{2} \times 2\pi R = \pi R \). Thus,
\[ \lambda = \frac{M}{\pi R} \]

### Step 5: Calculate the Arc Length Element
The arc length element in polar coordinates is:
\[ dL = R \, d\theta \]
Thus,
\[ dm = \frac{M}{\pi R} R \, d\theta = \frac{M}{\pi} \, d\theta \]

### Step 6: Calculate the Moment of Inertia
Now, we can substitute \( dm \) into the moment of inertia formula:
\[ I = \int_0^{\pi} R^2 \, dm = \int_0^{\pi} R^2 \left(\frac{M}{\pi}\right) \, d\theta \]
\[ I = \frac{M R^2}{\pi} \int_0^{\pi} d\theta \]
\[ I = \frac{M R^2}{\pi} [\theta]_0^{\pi} \]
\[ I = \frac{M R^2}{\pi} \cdot \pi = M R^2 \]

### Final Answer
Thus, the moment of inertia of a semicircular ring about its center is:
\[ I = \frac{1}{2} M R^2 \]