Let us consider a rigid body consisting of n particles of masses $m_1, m_2, m_3, …, m_n$ which lie at distance $r_1, r_2, r_3, …, r_n$ respectively from the axis of rotation $XY$ as shown in figure.

The moment of inertia $(I)$ about the axis of rotation is, \[I=m_1r_1^2+m_2r_2^2+m_3r_3^2+…+m_nr_n^2\] \[I=\sum_{i=1}^n m_ir_i^2\] Thus, moment of inertia of a body about a given axis of rotation is the sum of the product of the masses of the various particles and squares of their perpendicular distances from the axis of rotation.

Relation of Moment of Inertia with Torque

We know, torque is the product of force and its perpendicular distance from the axis of rotation. \[τ=Fr\] where, $F =$ force and $r =$ perpendicular distance from the axis of rotation. \[τ=mar\;\;\;[∵F=ma]\] \[τ=m\;αr\;r\;\;\; [∵a=αr]\] where $α =$ angular acceleration \[τ=(mr^2)α\] \[τ=Iα\]

Radius of Gyration

Radius of gyration of a body is the distance from the axis of rotation at which its whole mass can be assumed concentrated. It is denoted by $K$. At the radius of gyration, the moment of inertia is same as the actual distribution of the mass of the body into small particles.

For a body of mass $M$, we can write, \[I=MK²\] also, \[I=m_1r_1^2+m_2r_2^2+m_3r_3^2+…+m_nr_n^2\]

If all the particles have same mass $m$, \[I=m(r_1^2+r_2^2+r_3^2+…+r_n^2)\] \[I=\frac{m×n(r_1²+r_2²+r_3²+…+r_n²)}{n}\] Here, $m×n=M$ (Total mass of the body) \[∴I=\frac{M(r_1^2+r_2^2+r_3^2+…+r_n^2)}{n}\]

From above equations, \[MK^2=\frac{M(r_1^2+r_2^2+r_3^2+…+r_n^2)}{n}\] \[K=\sqrt{\frac{r_1^2+r_2^2+r_3^2+…+r_n^2}{n}}\]

$\therefore K=$ root mean square distance of the particles from the axis of rotation.