Question

# A circular disc of radius b has a hole of radius a at its centre. If the mass per unit area of the disc varies as $(\dfrac{{{\sigma }_{0}}}{r})$, then the radius of gyration of the disc about its axis passing through the centre is:

Hint: For a small radius dr, we have to consider a small distribution of mass dm. We need to formulate for small change in moment of inertia and integrate it along smaller radius (a) and bigger radius (b). After obtaining a moment of inertia of the disc I, we can find the radius of gyration.

Formula used:
Radius of gyration, $k=\sqrt{\dfrac{I}{m}}$

Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia which is equal to the moment of inertia of the bodyâ€™s actual distribution of mass. This is conditional to the total mass of the body being concentrated.

Let us consider the moment of inertia of the disc to be I
dI is a partial derivative of moment of inertia.

\begin{align} & dI=(dm){{r}^{2}} \\ & =(\sigma dA){{r}^{2}} \\ & =(\dfrac{{{\sigma }_{0}}}{r}2\pi rdr){{r}^{2}} \\ & =({{\sigma }_{0}}2\pi ){{r}^{2}}dr \\ & I=\int{dI=\int_{a}^{b}{{{\sigma }_{0}}}}2\pi {{r}^{2}}dr \\ & ={{\sigma }_{0}}2\pi (\dfrac{{{b}^{3}}-{{a}^{3}}}{3}) \\ & m=\int{dm}=\int{\sigma dA} \\ & ={{\sigma }_{0}}2\pi \int_{a}^{b}{dr} \\ & m={{\sigma }_{0}}2\pi (b-a) \\ \end{align}
Radius of gyration of the disc is
\begin{align} & k=\sqrt{\dfrac{I}{m}}=\sqrt{\dfrac{({{b}^{3}}-{{a}^{3}})}{3(b-a)}} \\ & =\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})} \\ \end{align}
Hence the radius of gyration of the disc is $\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})}$