Moment of Inertia of Annular Quadrant I am measuring the moments of inertia for various numbers of annular quadrants placed on a torsional oscillator. I know $\displaystyle{I=\frac{1}{2}M(R^2+r^2)}$ for a whole annulus. If I want the moment of inertia of only an annular quadrant, would I divide the formula above by four?
 A: It sounds like you are doing the following TeachSpin experiment.
If I understand correctly, brass quadrants are placed symmetrically on a rotation table, at the same distance from the axis as for the annulus.
When the axis is the same, the distribution of mass about the axis is the same for the quadrant as for the whole annulus. Because of this the same formula can be used, but using the mass of the quadrant instead of the whole annulus. 
The quadrant could be stretched out to 360 degrees, every part of it remaining at the same distance from the axis, without changing the moment of inertia. This is known as the Stretch Rule. After doing that, you would be left with an annulus with the same dimensions (and volume) but $\frac14$ the mass (and density) of the whole annulus.
A: *

*The MMOI about the geometric center is still $$I_{center} = \frac{m}{2} (r_1^2+r_2^2) $$


This is a result of the relationship ${\rm d}m = \rho z r \,{\rm d}r {\rm d}\theta$ and ${\rm I} = r^2 {\rm d}m$
$$ \begin{aligned} m & = \rho z \int \limits_{r_1}^{r_2}  \int \limits_{0}^\Theta r\, {\rm d}\theta {\rm d}r  & I_{center} & = \rho z \int \limits_{r_1}^{r_2}  \int \limits_{0}^\Theta r^3\, {\rm d}\theta {\rm d}r \end{aligned} $$


*

*The MMOI about the center of mass is found from the center of mass distance $c$ and the parallel axis theorem $I_{cm} = I_{center} - m c^2$


$$ \begin{aligned} c &= \frac{4 (r_1^2+r_1 r_2 + r_2^2) \sin \left(\tfrac{\Theta}{2}\right)}{3 \Theta (r_1+r_2) } \\ \\
I_{cm} & = \frac{m}{2} (r_1^2+r_2^2) - \frac{8 m (r_1^2+r_1 r_2 + r_2^2)^2 (1-\cos\Theta)}{9 \Theta^2 (r_1 +r_2)^2 }
\end{aligned} $$
