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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 $\sqrt{(\dfrac{{{a}^{2}}+{{b}^{2}}+ab}{3})}$

The dynamics and kinematics of rotation around a fixed axis of a rigid body are algebraically much easier than those for free rotation of a rigid body. They are completely analogous to those of linear motion along a single fixed direction, which is not true for free rotation of a rigid body.

A rigid body is an object of limited reach in which all the distances between the component particles doesnâ€™t change. Perfect rigid bodies do not exist. External forces can deform any solid. So we define a rigid body as an object that can be deformed under large forces.