Moment of inertia of a uniform quarter disc of radius `R` and mass `M` about an axis through its centre of mass and perpendicular to its plane is `:`

Options

A. `(MR^(2))/(2)-M((4R)/(3pi))^(2)`
B. `(MR^(2))/(2)-M(sqrt(2)(4R)/(3pi))^(2)`
C. `(MR^(2))/(2)+M((4R)/(3pi))^(2)`
D. `(MR^(2))/(2)+M(sqrt(2)(4R)/(3pi))^(2)`

Explanation

To find the moment of inertia of a uniform quarter disc of radius \( R \) and mass \( M \) about an axis through its center of mass and perpendicular to its plane, we can follow these steps:

### Step 1: Understand the Geometry
We have a quarter disc, which is one-fourth of a full disc. The radius of the quarter disc is \( R \) and its mass is \( M \). The center of mass of the quarter disc lies along the line that bisects the right angle of the quarter disc.

### Step 2: Moment of Inertia of a Full Disc
The moment of inertia \( I \) of a full disc about an axis through its center and perpendicular to its plane is given by the formula:
\[ I = \frac{1}{2} M R^2 \]

### Step 3: Use the Parallel Axis Theorem
To find the moment of inertia of the quarter disc about its center of mass, we can use the parallel axis theorem. The moment of inertia of the quarter disc can be expressed as:
\[ I_{quarter} = I_{full} + M d^2 \]
where \( d \) is the distance from the center of the full disc to the center of the quarter disc.

### Step 4: Calculate the Distance \( d \)
The center of mass of a quarter disc is located at a distance \( \frac{4R}{3\pi} \) from the vertex along both axes (x and y). Therefore, the distance \( d \) from the center of the full disc to the center of the quarter disc can be calculated using the Pythagorean theorem:
\[ d = \sqrt{\left(\frac{4R}{3\pi}\right)^2 + \left(\frac{4R}{3\pi}\right)^2} = \sqrt{2 \left(\frac{4R}{3\pi}\right)^2} = \frac{4R}{3\pi} \sqrt{2} \]

### Step 5: Substitute Values into the Moment of Inertia Equation
Now we can substitute \( I_{full} \) and \( d \) into the equation:
\[ I_{quarter} = \frac{1}{2} M R^2 + M \left(\frac{4R}{3\pi} \sqrt{2}\right)^2 \]
Calculating \( d^2 \):
\[ d^2 = \left(\frac{4R}{3\pi} \sqrt{2}\right)^2 = \frac{32R^2}{9\pi^2} \]
Thus,
\[ I_{quarter} = \frac{1}{2} M R^2 + M \cdot \frac{32R^2}{9\pi^2} \]

### Step 6: Combine Terms
Now we can combine the terms:
\[ I_{quarter} = \frac{1}{2} M R^2 + \frac{32M R^2}{9\pi^2} \]
To combine these, we can find a common denominator:
\[ I_{quarter} = \frac{9\pi^2}{18\pi^2} M R^2 + \frac{64M R^2}{18\pi^2} = \frac{(9\pi^2 + 64)M R^2}{18\pi^2} \]

### Final Result
Thus, the moment of inertia of the uniform quarter disc about an axis through its center of mass and perpendicular to its plane is:
\[ I_{quarter} = \frac{(9\pi^2 + 64)M R^2}{18\pi^2} \]