Ring (Annular Cylinder)
Mass
We can build the ring using values of mass that we calculated in the previous post. Recall that $M_{\text{disc}}(R, L) = \rho \cdot \pi R^2 L$.
\[\begin{align} M_{\text{ring}} &= M_{\text{disc}}(R_2, L) - M_{\text{disc}}(R_1, L) \\[10pt] &= \rho \cdot \pi R_2^2 L - \rho \cdot \pi R_1^2 L \\[10pt] &= \rho \cdot \pi (R_2^2 - R_1^2) L \end{align}\]As an exercise, evaluate the mass of the ring using an integral and see if you get the same result.
Moment of Inertia About Central Axis
Now, it turns out we can do exactly the same thing as the moment of inertia. We have to be careful though. We cannot conflate the masses. Thus, instead of using $I_{\text{disc}}(R, L) = \frac{1}{2} M_{\text{disc}} R^2$ we have to use $I_{\text{disc}}(R, L) = \rho \cdot \frac{1}{2} \pi R^4 L$, since we are assuming both have a uniform mass density.
\[\begin{align} I_{\text{ring}} &= I_{\text{disc}}(R_2, L) - I_{\text{disc}}(R_1, L) \\[10pt] &= \rho \cdot \frac{1}{2} \pi R_2^4 L - \rho \cdot \frac{1}{2} \pi R_1^4 L\\[10pt] &= \rho \cdot \frac{1}{2} \pi (R_2^4 - R_1^4) L \\[10pt] &= \frac{1}{2} M (R_2^2 + R_1^2) \\[10pt] \end{align}\]Again, as an exercise you can evaluate this from scratch using an integral and see if you can get the same result.
Moment of Inertia About Central Diameter
We do likewise here, using $I_{\text{disc}}(R, L) = \frac{1}{4} M_{\text{disc}} R^2 + \frac{1}{12} M_{\text{disc}} L^2 = \rho \cdot \left ( \frac{1}{4} \pi R^4 L + \frac{1}{12} \pi R^2 L^3 \right )$.
\[\begin{align} I_{\text{ring}} &= I_{\text{disc}}(R_2, L) - I_{\text{disc}}(R_1, L) \\[10pt] &= \rho \cdot \left ( \frac{1}{4} \pi R_2^4 L + \frac{1}{12} \pi R_2^2 L^3 \right ) - \rho \cdot \left ( \frac{1}{4} \pi R_1^4 L + \frac{1}{12} \pi R_1^2 L^3 \right )\\[10pt] &= \rho \cdot \left ( \frac{1}{4} \pi (R_2^4 - R_1^4) L + \frac{1}{12} \pi (R_2^2 - R_1^2) L^3 \right )\\[10pt] &= \rho \cdot \pi (R_2^2 - R_1^2) L \left ( \frac{1}{4} (R_2^2 + R_1^2) + \frac{1}{12} L^2 \right )\\[10pt] &= \frac{1}{4} M (R_2^2 + R_1^2) + \frac{1}{12} M L^2 \end{align}\]Evaluating this integral from scratch becomes very tedious. However, using the linearity method it’s not too bad.