The moment of inertia of a uniform semicircular disc of mass $$M$$ and radius $$r$$ about a line perpendicular to the plane of the disc through the centre is

Let a full circular disc of radius $$r$$ and uniform surface-mass density $$\sigma$$ be imagined. For a complete disc, the standard moment of inertia about an axis perpendicular to the plane and passing through the centre (the $$z$$-axis) is

$$I_{\text{full}} = \frac{1}{2}\,M_{\text{full}}\,r^{2} \qquad -(1)$$

where $$M_{\text{full}}$$ is the mass of the full disc.

1. Relate the masses of the full disc and the given semicircular disc. The area of the full disc is $$\pi r^{2}$$, while that of a semicircle is $$\tfrac{1}{2}\pi r^{2}$$. Because the surface density is uniform, the mass is proportional to the area:

$$M_{\text{semicircle}} = \frac{1}{2}\,M_{\text{full}} \qquad -(2)$$

Given in the problem, the mass of the semicircular disc is $$M$$, so from $$-(2)$$

$$M_{\text{full}} = 2M \qquad -(3)$$

2. Insert $$M_{\text{full}}$$ from $$-(3)$$ into the formula $$-(1)$$:

$$I_{\text{full}} = \frac{1}{2}\,(2M)\,r^{2} = M r^{2} \qquad -(4)$$

3. Because the semicircular disc occupies exactly half the area of the full disc and has the same surface density, its moment of inertia about the same axis is exactly half of $$I_{\text{full}}$$:

$$I_{\text{semicircle}} = \frac{1}{2}\,I_{\text{full}} = \frac{1}{2}\,M r^{2}$$

Thus, the moment of inertia of the uniform semicircular disc about a line perpendicular to its plane through the centre is $$\tfrac{1}{2} M r^{2}$$.

Option D which is: $$\frac{1}{2}\,Mr^{2}$$