Expected Moment (part 3)

The disk is parameterised in polar coordinates $r$ and $\alpha$ . Consider a mass element on the disk as follows:

$dM = \frac{M}{\pi R^2}r \ dr \ d\alpha$

The position of a point on the disk is:

$\vec{r}_p = \left(r\cos{\alpha}, r\sin{\alpha},0\right)$

The axis makes an angle of $\theta$ to the x-axis. The unit vector along which the axis is directed is:

$\hat{n} = \cos{\theta} \hat{i} + \sin{\theta} \hat{k}$

The projection of the position vector $\vec{r}_p$ along the axis is:

$\left(\vec{r}_p \cdot \hat{n}\right) \hat{n}$

The vector perpendicular to the axis to the point on the disk is therefore:

$\vec{r}_c = \vec{r}_p - \left(\vec{r}_p \cdot \hat{n}\right) \hat{n}$

The elementary moment of inertia of the disk about this axis is, therefore:

$dI = dM \lvert \vec{r}_c \rvert^2$

Substituting and simplifying gives:

$dI = \frac{M}{\pi R^2} r^3 \left(\cos^2{\alpha}\sin^2{\theta} + \sin^2{\alpha}\right) \ dr \ d\alpha$

Integrating gives:

$I = \frac{M}{\pi R^2} \int_{0}^{R} \int_{0}^{2 \pi} r^3 \left(\cos^2{\alpha}\sin^2{\theta} + \sin^2{\alpha}\right) \ dr \ d\alpha$

$I = \frac{MR^2}{4}\left(1 + \sin^2{\theta}\right)$

Having computed the moment of inertia, the next step is to find the average value in the specified range of $\theta$ given that $\theta$ is uniformly distributed. The expression for that is:

$I_E = \frac{MR^2}{2 \pi} \int_{0}^{\pi/2} \left(1 + \sin^2{\theta}\right) d\theta$

$I_E = \frac{3}{8} MR^2$

Therefore:

$\boxed{\alpha = \frac{3}{8}=0.375}$