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# A circular disc of radius b has a hole of radius a at its center. if the mass per unit area of the disc varies as $\left({\sigma }_{0}/r\right)$

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Solution

## Step 1: Given data:The radius of gyration:It is the distance from the center of mass of a body at which the whole mass is concentrated.It does not change its moment of rotational inertia about an axis through the center of mass.It is an act or instance of gyrating.Something such as a coil or shell that is gyrate.A diagram to represent the radius of gyration:The formula of the radius of gyration:$I=M{K}^{2}$Step 2: To find the radius of gyration:We have,$dI=\left(dm\right){r}^{2}$Now, $dI=\left(\sigma dA\right){r}^{2}$$\left(\frac{{\sigma }_{\circ }}{r}2\mathrm{\pi rdr}\right){r}^{2}$So, $=\left({\sigma }_{\circ }2\mathrm{\pi }\right){\mathrm{r}}^{2}\mathrm{dr}$Now, $I=\int dI={\int }_{a}^{b}{\sigma }_{\circ }2{\mathrm{\pi r}}^{2}\mathrm{dr}\phantom{\rule{0ex}{0ex}}={\mathrm{\sigma }}_{\circ }2\mathrm{\pi }\left(\frac{{\mathrm{b}}^{3}-{\mathrm{a}}^{3}}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{m}=\int \mathrm{dm}=\int \mathrm{\sigma dA}\phantom{\rule{0ex}{0ex}}={\mathrm{\sigma }}_{\circ }2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{dr}\phantom{\rule{0ex}{0ex}}\mathrm{m}={\mathrm{\sigma }}_{\circ }2\mathrm{\pi }\left(\mathrm{b}-\mathrm{a}\right)$The radius of gyration:$k=\sqrt{\frac{I}{m}}=\sqrt{\frac{\left({b}^{3}-{a}^{3}\right)}{3\left(b-a\right)}}\phantom{\rule{0ex}{0ex}}k=\sqrt{\left(\frac{{a}^{3}+{b}^{3}+ab}{3}\right)}$Hence, then the radius of gyration of the disc about its axis passing through the centre is $k=\sqrt{\left(\frac{{a}^{3}+{b}^{3}+ab}{3}\right)}$.

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