Find the polar moment of inertia of an annular ring

Find the polar moment of inertia of an annular ring with inner radius $R_1$ and outer radius $R_2$ ($0\le R_1 \lt R_2$) with respect to the center.

Step 1: Set up and evaluate

\[I_O=2\pi\int_{R_1}^{R_2}\rho^3\,d\rho = 2\pi\cdot\frac{R_2^4-R_1^4}4=\frac{\pi(R_2^4-R_1^4)}2\]

\[\boxed{I_O=\frac\pi2(R_2^2-R_1^2)(R_2^2+R_1^2)}\]

Step 2: In terms of area

With $A=\pi(R_2^2-R_1^2)$: \[I_O=\frac{A(R_2^2+R_1^2)}2\]

Step 3: Special cases

Solid disk ($R_1=0$): reduces to $\dfrac{\pi R_2^4}2$, matching Problem 149. Thin ring ($R_2-R_1=t\ll R$): $I_O\approx AR^2$, as expected when nearly all area sits at radius $R$.

Step 4: Radius of gyration

\[k_O=\sqrt{\frac{R_2^2+R_1^2}2}\]