Derive the expression of Kinetic energy in rotation.

Explanation

(i) Let us consider a rigid body rotating with angular velocity `omega` about an axis as shown in Figure.
(ii) Every particle of the body will have the same angular velocity `omega` and different tangential velocities v based on its positions from the axis of rotation.

(iii) Let us choose a particle of mass `m_(i)` situated at distance `r_(i)` from the axis of rotation. If 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_(t)=1/2m_(i)v_(i)^(2)`
(iv) Writing the expression with the angular velocity,
`KE_(t)=1/2m_(i)(r_(i)omega)^(2)=1/2 (m_(i)r_(i)^(2))omega^(2)`
(v) For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
`KE=1/2(summ_(i)r_(i)^(2))`
(vi) where, the term `summ_(i)r_(i)^(2)` is the moment of inertia I of the whole body. `I=summ_(i)r_(i)^(2)`
(vii) Hence, the expression for KE of the rigid body in rotational motion is,
`KE=1/2Iomega^(2)`
This is analogous to the expression for kinetic energy in translational motion.
`KE=1/2Mv^(2)`
(viii) Relational between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity `omega`.
The angular momentum of rigid body is,
`L=Iomega`
(ix) The rotational kinetic energy of the rigid body is, `KE=1/2Iomega^(2)`
(x) By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,
`KE=1/2(I^(2)omega^(2))/I=1/2((Iomega^(2)))/I`
`KE=L^(2)/(2I)`