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 horizontal diameter, we can follow these steps:

### Step 1: Understand the Moment of Inertia about the Diameter
The moment of inertia of a circular loop (ring) about an axis passing through its center and perpendicular to its plane (which is the diameter in this case) is given by the formula:
\[ I_{\text{cm}} = \frac{1}{2} m R^2 \]
where \( m \) is the mass of the loop and \( R \) is its radius.

### Step 2: Apply the Parallel Axis Theorem
To find the moment of inertia about an axis that is parallel to the diameter and at a distance of \( \frac{R}{2} \) from it, we use the Parallel Axis Theorem. This theorem states:
\[ I = I_{\text{cm}} + m d^2 \]
where \( d \) is the distance between the two axes.

### Step 3: Calculate the Distance \( d \)
In our case, the distance \( d \) is \( \frac{R}{2} \).

### Step 4: Substitute the Values
Now, substituting the values into the Parallel Axis Theorem:
\[ I_{xx'} = I_{\text{cm}} + m \left(\frac{R}{2}\right)^2 \]
Substituting \( I_{\text{cm}} = \frac{1}{2} m R^2 \):
\[ I_{xx'} = \frac{1}{2} m R^2 + m \left(\frac{R^2}{4}\right) \]

### Step 5: Simplify the Expression
Now, simplify the expression:
\[ I_{xx'} = \frac{1}{2} m R^2 + \frac{1}{4} m R^2 \]
To combine these, we need a common denominator:
\[ I_{xx'} = \frac{2}{4} m R^2 + \frac{1}{4} m R^2 = \frac{3}{4} m R^2 \]

### Step 6: Conclusion
Thus, the moment of inertia of the circular loop at a distance of \( \frac{R}{2} \) from the diameter is:
\[ I_{xx'} = \frac{3}{4} m R^2 \]

### Final Answer
The required moment of inertia is \( \frac{3}{4} m R^2 \).

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