A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as $$\frac{\sigma_0}{r}$$, then the radius of gyration of the disc about its axis passing through the center is

$$\text{Consider an elemental concentric ring of radius } r \text{ and thickness } dr \text{ within the limits } r \in [a, b]:$$

$$dA = 2\pi r \, dr \implies dm = \sigma(r) \, dA = \left(\frac{\sigma_0}{r}\right) (2\pi r \, dr) = 2\pi \sigma_0 \, dr$$

$$\text{Total Mass, } M = \int_{a}^{b} dm = 2\pi \sigma_0 \int_{a}^{b} dr = 2\pi \sigma_0 (b - a)$$

$$\text{Moment of Inertia, } I = \int_{a}^{b} r^2 \, dm = 2\pi \sigma_0 \int_{a}^{b} r^2 \, dr = 2\pi \sigma_0 \left[ \frac{r^3}{3} \right]_{a}^{b} = \frac{2\pi \sigma_0}{3} (b^3 - a^3)$$

$$\text{Using } I = M k^2:$$

$$k^2 = \frac{I}{M} = \frac{\frac{2\pi \sigma_0}{3} (b^3 - a^3)}{2\pi \sigma_0 (b - a)} = \frac{b^3 - a^3}{3(b - a)} = \frac{(b - a)(a^2 + b^2 + ab)}{3(b - a)} = \frac{a^2 + b^2 + ab}{3}$$

$$k = \sqrt{\frac{a^2 + b^2 + ab}{3}}$$