Find the expression of KE of the rigid body in rotational motion.

Explanation

Let us consider a rigid body rotating with angualar velocity `omega` about an axis as shown in the figure. Every particle of the bdoy will have the same angular velocity `omega` and different tangential velocities v based on its positions from the axis of rotation.
Let us choose a particle of mass `m_(i)` situated at distance `r_(i)` from the axis of rotation. It has a tangential velocity`v_(i)` given by the relation, `v_(i)=r_(i) omega`. The kinetic energy `KE_(i)` of the particle is
`KE_(i)=1/2 m_(i)v_(i)^(2)`
Since `v_(i)=r_(i)omega`
`KE_(i)=1/2m_(i)(r_(i)omega)^(2)=1/2m_(i)(r_(i)^(2))omega^(2)`
For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with simmation as
`KE=1/2(sum m_(i)r_(i)^(2))omega^(2)`
where the term `sum m_(i)r_(i)^(2)` is the moment of inertia I of the whole body. `I=sum m_(i)r_(i)^(2)`
In rotational motion kinetic energy is
`KE=1/2 I omega^(2)`