Find the moment of inertia of a uniform half-disc about an axis perpendicular to the plane and passing through its centre of mass. Mass of this disc is M and radius is R.

Answer

To find the moment of inertia of a uniform half-disc about an axis perpendicular to the plane and passing through its center of mass, we can follow these steps:

### Step 1: Understand the Geometry
We have a uniform half-disc with mass \( M \) and radius \( R \). The center of mass of the half-disc is located at a distance of \( \frac{4R}{3\pi} \) from the flat edge along the axis of symmetry.

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

### Step 3: Moment of Inertia of the Half Disc
Since the half-disc is half of the full disc, its moment of inertia about the center (flat edge) is:
\[ I_{\text{half}} = \frac{1}{2} I_{\text{full}} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) = \frac{1}{4} M R^2 \]

### Step 4: Apply the Parallel Axis Theorem
To find the moment of inertia about the center of mass of the half-disc, we can use the parallel axis theorem. The theorem states:
\[ I = I_{\text{cm}} + M d^2 \]
where \( d \) is the distance from the center of mass of the half-disc to the axis of rotation.

In our case, the distance \( d \) is \( \frac{4R}{3\pi} \). Thus, we need to calculate:
\[ I = I_{\text{half}} + M \left( \frac{4R}{3\pi} \right)^2 \]

### Step 5: Calculate the Moment of Inertia
Substituting the values we have:
\[ I = \frac{1}{4} M R^2 + M \left( \frac{4R}{3\pi} \right)^2 \]
Calculating \( \left( \frac{4R}{3\pi} \right)^2 \):
\[ \left( \frac{4R}{3\pi} \right)^2 = \frac{16R^2}{9\pi^2} \]
Thus,
\[ I = \frac{1}{4} M R^2 + M \cdot \frac{16R^2}{9\pi^2} \]
Combining the terms:
\[ I = M R^2 \left( \frac{1}{4} + \frac{16}{9\pi^2} \right) \]

### Step 6: Final Expression
The final expression for the moment of inertia of the half-disc about the axis through its center of mass is:
\[ I = M R^2 \left( \frac{1}{4} + \frac{16}{9\pi^2} \right) \]