Problem 146
Find the moment of inertia of a circle of radius R
Find the moment of inertia of a circle of radius $R$ with respect to one of its diameters.
Explanation of Method
Evaluate $I_x=\iint y^2\,dA$ in polar coordinates over the disk, separating the radial and angular integrals.
Step 1: Set up in polar coordinates
\[I_x=\int_0^{2\pi}\int_0^R(\rho\sin\theta)^2\rho\,d\rho\,d\theta=\left(\int_0^{2\pi}\sin^2\theta\,d\theta\right)\left(\int_0^R\rho^3\,d\rho\right)\]
Step 2: Evaluate each factor
\[\int_0^{2\pi}\sin^2\theta\,d\theta=\pi,\qquad \int_0^R\rho^3\,d\rho=\frac{R^4}4\]
Step 3: Combine
\[I_x=\pi\cdot\frac{R^4}4=\frac{\pi R^4}4\]
\[\boxed{I_x=\frac{\pi R^4}4}\]
By symmetry, $I_y=\dfrac{\pi R^4}4$ as well, and with $A=\pi R^2$ this is $\dfrac{AR^2}4$; the radius of gyration is $k=\dfrac R2$.
Result
Final answer:
\[\boxed{I = \dfrac{\pi R^4}{4} = \dfrac{AR^2}{4}}\]