The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition

equation=12550

Applying Newton's second law, this expression can be rewritten as

$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$

Using the definition of angular velocity

equation=3679

we get

$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$

The difference in angular velocities is

$\Delta\omega=\omega_2-\omega_1$

On the other hand, angular velocity itself can be approximated with the average angular velocity

$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$

Using both expressions, we obtain the equation

$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$

Thus, the change in energy is given by

$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$

This allows us to define kinetic energy as

equation

The work variance ($\Delta W$) required for an object to change from the initial Angular Speed ($\omega_0$) to the angular Speed ($\omega$) is obtained by applying a the torque ($T$) that produces an angular displacement the difference of Angles ($\Delta\theta$), according to:

equation=12550

Applying Newton's second law for rotation, in terms of the moment of inertia for axis that does not pass through the CM ($I$) and the mean Angular Acceleration ($\bar{\alpha}$):

equation=3253

this expression can be rewritten as:

$\Delta W = I \alpha \Delta\theta$

or, using the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$):

equation=3234

we get:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta$

Using the definition of the mean angular velocity ($\bar{\omega}$) and the time elapsed ($\Delta t$):

equation=3679

results in:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta = I\omega \Delta\omega$

where the difference in Angular Speeds ($\Delta\omega$) is expressed as:

equation=3681

On the other hand, the angular velocity can be approximated by the average angular velocity:

$\bar{\omega}=\displaystyle\frac{\omega_1 + \oemga_2}{2}$

By combining both expressions, we obtain the equation:

$\Delta W = I \omega \Delta\omega = I(\omega_2 - \omega_1) \displaystyle\frac{(\omega_1 + \omega_2)}{2} = \displaystyle\frac{I}{2}(\omega_2^2 - \omega_1^2)$

Therefore, the change in energy is expressed as:

$\Delta W = \displaystyle\frac{I}{2}\omega_2^2 - \displaystyle\frac{I}{2}\omega_1^2$

This allows us to define the rotational kinetic energy as:

equation

When an object rolls, its angular velocity is related to its translational velocity by

equation=3233

resulting in the rotational kinetic energy

equation=3255

which becomes

$K_r=\displaystyle\frac{1}{2}I \omega^2=\displaystyle\frac{1}{2} I \displaystyle\frac{v^2}{r^2}=\displaystyle\frac{1}{2}\left(\displaystyle\frac{I}{r^2}\right)v^2$

Thus, when combining the translational kinetic energy

equation=3244

the kinetic energy of a rotating body is calculated by the sum

equation=3686

meaning,

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The total Kinetic Energy ($K$) corresponds to the sum of the translational Kinetic Energy ($K_t$) and the kinetic energy of rotation ($K_r$):

equation=3686

Since the translational Kinetic Energy ($K_t$) is expressed as a function of the inertial Mass ($m_i$) and the speed ($v$):

equation=3244

and the kinetic energy of rotation ($K_r$), as a function of the moment of inertia for axis that does not pass through the CM ($I$) and the angular Speed ($\omega$), is defined as:

equation=3255

the final expression is:

equation