The moment of inertia of a circular loop of radius R, at a distance of `R//2` around a rotating axis parallel to horizontal diameter of loop is

Options

A. `MR^(2)`
B. `(1)/(2)MR^(2)`
C. `2MR^(2)`
D. `(3)/(4)MR^(2)`

Explanation

To find the moment of inertia of a circular loop of radius \( R \) at a distance of \( \frac{R}{2} \) from the center around a rotating axis parallel to the horizontal diameter of the loop, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of the Loop:**
The moment of inertia \( I \) of a circular loop (ring) of radius \( R \) and mass \( m \) about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{center}} = mR^2 \]

2. **Use the Parallel Axis Theorem:**
Since we need to find the moment of inertia about an axis that is parallel to the horizontal diameter and at a distance of \( \frac{R}{2} \) from the center, we will use the parallel axis theorem. The parallel axis theorem states:
\[ I = I_{\text{center}} + md^2 \]
where \( d \) is the distance from the center of mass to the new axis. Here, \( d = \frac{R}{2} \).

3. **Calculate the Moment of Inertia Using the Parallel Axis Theorem:**
Substitute \( I_{\text{center}} \) and \( d \) into the equation:
\[ I = mR^2 + m\left(\frac{R}{2}\right)^2 \]
Simplifying the second term:
\[ I = mR^2 + m\left(\frac{R^2}{4}\right) \]
\[ I = mR^2 + \frac{mR^2}{4} \]
\[ I = \frac{4mR^2}{4} + \frac{mR^2}{4} = \frac{5mR^2}{4} \]

4. **Final Result:**
Thus, the moment of inertia of the circular loop at a distance of \( \frac{R}{2} \) from the center around the specified axis is:
\[ I = \frac{5mR^2}{4} \]