Prove that the moment of inertia of a circular annulus
Prove that the moment of inertia of a circular annulus with respect to its center is twice the moment of inertia with respect to any axis through the center lying in the plane of the annulus.
Explanation of Method
The polar moment about the center and the moment about a diameter are both elementary polar-coordinate integrals over the annulus; rotational symmetry then shows the moment is the same about every diameter, and the perpendicular axis theorem $I_O=I_x+I_y$ closes the argument.
- Compute the polar moment $I_O$ about the center.
- Compute the moment $I_x$ about a diameter.
- Compare $I_O$ and $I_x$.
- Generalize to an arbitrary diameter via the perpendicular axis theorem.
Step 1 · Polar moment $I_O$
With inner radius $r_1$, outer radius $r_2$, density $\varrho=1$:
$$ I_O = \iint_D(x^2+y^2)\,dA = \int_0^{2\pi}\int_{r_1}^{r_2}r^3\,dr\,d\theta = \frac{\pi(r_2^4-r_1^4)}2. $$
Step 2 · Moment $I_x$ about a diameter
$$ I_x = \iint_Dy^2\,dA = \int_0^{2\pi}\sin^2\theta\,d\theta\cdot\int_{r_1}^{r_2}r^3\,dr = \pi\cdot\frac{r_2^4-r_1^4}4 = \frac{\pi(r_2^4-r_1^4)}4. $$
Step 3 · Comparison
$$ I_O = \frac{\pi(r_2^4-r_1^4)}2 = 2\cdot\frac{\pi(r_2^4-r_1^4)}4 = 2I_x. $$
Step 4 · Arbitrary diameter
By the same computation, $I_y=\iint_Dx^2\,dA=I_x$, so $I_O=I_x+I_y=2I_x$ (perpendicular axis theorem). For an axis at angle $\varphi$ to the $x$-axis, distance from $(x,y)$ is $|y\cos\varphi-x\sin\varphi|$, and since $\iint_Dxy\,dA=0$ by symmetry:
$$ I_\varphi = I_x\cos^2\varphi + I_y\sin^2\varphi - 2\sin\varphi\cos\varphi\iint_Dxy\,dA = I_x, $$
independent of $\varphi$. Hence $I_O=2I_\varphi$ for every diameter.
Additional Notes
- This derivation matches the source exactly.
Conclusion
For any circular annulus, the polar moment of inertia about the center equals twice the moment about any diameter: