A circular disc of radius b has a hole of rad | vedclass

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(B) Consider a thin elemental ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2\pi r dr$. The mass of this element is $dm = \sig

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

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(B) Consider a thin elemental ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2\pi r dr$. The mass of this element is $dm = \sigma(r) dA = \left(\frac{\sigma_0}{r}\right) (2\pi r dr) = 2\pi \sigma_0 dr$.The total mass $M$ of the disc is $M = \int_a^b dm = \int_a^b 2\pi \sigma_0 dr = 2\pi \sigma_0 (b - a)$.The moment of inertia $I$ about the central axis is $I = \int_a^b r^2 dm = \int_a^b r^2 (2\pi \sigma_0 dr) = 2\pi \sigma_0 \int_a^b r^2 dr = 2\pi \sigma_0 \left(\frac{b^3 - a^3}{3}\right)$.The radius of gyration $k$ is defined by $I = Mk^2$,so $k^2 = \frac{I}{M}$.$k^2 = \frac{2\pi \sigma_0 (b^3 - a^3) / 3}{2\pi \sigma_0 (b - a)} = \frac{b^3 - a^3}{3(b - a)}$.Using the identity $b^3 - a^3 = (b - a)(b^2 + a^2 + ab)$,we get $k^2 = \frac{(b - a)(b^2 + a^2 + ab)}{3(b - a)} = \frac{a^2 + b^2 + ab}{3}$.Therefore,$k = \sqrt{\frac{a^2 + b^2 + ab}{3}}$.

$A$ 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 $\sigma(r) = \frac{\sigma_0}{r}$,then the radius of gyration of the disc about its axis passing through the centre is

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