Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.

Explanation

Let us consider a rigid body rotating with angular velocity `omega` about an axis as shown in figure. 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 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)= omega` . The kinetic energy `KE_(i)` of the particle is

`KE=(1)/(2)m_(i)v_(i)^(2)`
Writing the expression with the angular velocity,
`KE=(1)/(2)m_(i)(r_(i) omega)^(2)=(1)/(2)(m_(i) r_(1)^(2)) omega^(2)`
For the kinetic energy of the whole body, which is made up of large number of such parti the equation is written with summation 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 inertial of the who.e body. `sum m_(i) r_(i)^(2)` Hence, the expression for KE of the rigid body in rotational motion is,
`KE=(1)/(2)I omega^(2)`
his is analogous to the expression for kinetic energy in translational motion
`KE=(1)/(2) Mv^(2)`
Relation 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 a rigid body is, `L=I omega`
The rotational kinetic energy of the rigid body is, `K.E=(1)/(2)I omega^(2)`
By multiplying the numerator and denominator of the above equation with I, we relation between Land KE as,
`KE=(1)/(2)(I^(2) omega^(2))/(I)=(1)/(2)((I omega)^(2))/(I)`
`KE=(L^(2))/(2I)`