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 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:
\[ I_{\text{full}} = \frac{1}{2} M R^2 \]

### Step 2: Relate the Moment of Inertia of the Quarter Disc
Since we are dealing with a quarter disc, we can express the moment of inertia of the quarter disc about its center of mass. The moment of inertia of the quarter disc about the same axis can be derived from the moment of inertia of the full disc:
\[ I_{\text{quarter}} = \frac{1}{4} I_{\text{full}} = \frac{1}{4} \left(\frac{1}{2} M R^2\right) = \frac{1}{8} M R^2 \]

### Step 3: Use the Parallel Axis Theorem
To find the moment of inertia about the center of mass of the quarter disc, we can use the parallel axis theorem. The parallel axis theorem states that:
\[ I = I_{\text{cm}} + Md^2 \]
where \( I_{\text{cm}} \) is the moment of inertia about the center of mass, \( M \) is the mass, and \( d \) is the distance from the center of mass to the new axis.

### Step 4: Calculate the Distance \( d \)
For a quarter disc, the center of mass is located at a distance \( d \) from the origin (the center of the full disc). The coordinates of the center of mass for a quarter disc can be derived, and it is found to be:
\[ d = \frac{4R}{3\pi} \]

### Step 5: Substitute into the Parallel Axis Theorem
Now substituting \( d \) into the parallel axis theorem:
\[ I = I_{\text{quarter}} + M \left(\frac{4R}{3\pi}\right)^2 \]
Substituting \( I_{\text{quarter}} = \frac{1}{8} M R^2 \):
\[ I = \frac{1}{8} M R^2 + M \left(\frac{4R}{3\pi}\right)^2 \]

### Step 6: Simplify the Expression
Now we simplify the expression:
\[ I = \frac{1}{8} M R^2 + M \cdot \frac{16R^2}{9\pi^2} \]
Combining the terms:
\[ I = M \left(\frac{1}{8} R^2 + \frac{16R^2}{9\pi^2}\right) \]

### Step 7: Final Result
To find the final moment of inertia, we can factor out \( MR^2 \):
\[ I = M R^2 \left(\frac{1}{8} + \frac{16}{9\pi^2}\right) \]

### Conclusion
Thus, 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 is:
\[ I = M R^2 \left(\frac{1}{8} + \frac{16}{9\pi^2}\right) \]

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