Winged Disk Moment

Consider any point on the disk with x-coordinate greater than 0.5. Let such a point be $\vec{r}=(x,y)$ . Transformed coordinates as per the guidelines can be computed as such:

$\vec{r} = \left[\begin{matrix} x \\ y \end{matrix} \right]$ $\vec{p}= \frac{1}{2}\left[\begin{matrix} 1 \\ \sqrt{3} \end{matrix} \right]$ $R = \left[\begin{matrix} \cos{\pi/4} & -\sin{\pi/4} \\ \sin{\pi/4} & \cos{\pi/4} \end{matrix} \right]$

$\left[\begin{matrix} x_n \\ y_n \end{matrix} \right] = \vec{p} + R\left(\vec{r}-\vec{p}\right)$

Therefore:

$x_n = \frac{1}{2} + \frac{1}{\sqrt{2}}\left(x-\frac{1}{2} - y + \frac{\sqrt{3}}{2}\right)$

The moment of inertia of the region of transformed coordinates about the Y-axis can be computed as such:

$I_t = \int_{0.5}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x_n^2 \ dy \ dx$

The moment of inertial contribution due to the transformed region left of the line $x = -0.5$ is equal to $I_t$ due to symmetry. For the rest of the disk, the moment of inertia about the Y-axis is:

$I_d = \int_{-0.5}^{0.5} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2 \ dy \ dx$

All integrals are outsourced to Wolfram-Alpha. The total moment of inertia of this winged disk is, therefore:

$I = I_d + 2I_t$ $\boxed{I \approx 2.205546}$