Moment of Inertia can be tough

Note: This is not a complete solution because it assumes that the general equation given is correct. However, the solution process is still interesting, so here it is:

Step 1: Examine the $(\theta = 0)$ case to determine the "p" parameter:

When $(\theta = 0)$ , the general expression reduces to:

$I = \frac{M a^2}{p}$

It is easy to integrate and determine that the moment in this case is $\frac{M a^2}{3}$ , implying that $\boxed{p = 3}$ .

Step 2: Examine boundary cases to determine the value of the "r" parameter.

By inspection, the moments for $\theta = 0$ and $\theta = \frac{\pi}{2}$ must be identical. This implies that the "r" parameter must be a multiple of 2 (i.e. 2,4,6 etc.). However, we also know that 4 doesn't work, because the moments for $\theta = 0$ and $\theta = \frac{\pi}{4}$ must be different. Similar arguments can be made to disqualify all higher multiples of 2. The "r" parameter is therefore equal to 2. $\boxed{r = 2}$ .

Step 3: Examine the $(\theta = \frac{\pi}{4})$ case to determine the "q" parameter:

When $(\theta = \frac{\pi}{4})$ , the general expression reduces to:

$I = Ma^2 (\frac{1}{3} + \frac{1}{q})$

It is relatively easy to integrate and determine that the moment in this case is $\frac{7}{12} M a^2$ , implying that $\boxed{q = 4}$ .

Step 4: Use numerical integration to confirm these numbers for some less ideal angles (like $\frac{\pi}{6}$ and $\frac{\pi}{7}$ ) . These checks pass, indicating that the parameters are correct.