Editing Moment of inertia (section)

==== Kinetic energy ====
[[File:Lever shear flywheel.jpg|thumb|This 1906 rotary shear uses the moment of inertia of two flywheels to store kinetic energy which when released is used to cut metal stock (International Library of Technology, 1906).]]
The kinetic energy of a rigid system of particles moving in the plane is given by<ref name="B-Paul"/><ref name="Uicker"/>
<math display="block">\begin{align}
  E_\text{K}
    &= \frac{1}{2} \sum_{i=1}^n m_i \mathbf{v}_i \cdot \mathbf{v}_i, \\
    &= \frac{1}{2} \sum_{i=1}^n m_i
         \left(\omega \,\Delta r_i\mathbf{\hat{t}}_i + \mathbf{V}\right) \cdot
         \left(\omega \,\Delta r_i\mathbf{\hat{t}}_i + \mathbf{V}\right), \\
    &= \frac{1}{2}\omega^2 \left(\sum_{i=1}^n m_i\, \Delta r_i^2 \mathbf{\hat{t}}_i \cdot \mathbf{\hat{t}}_i\right) +
         \omega\mathbf{V} \cdot \left(\sum_{i=1}^n m_i \,\Delta r_i\mathbf{\hat{t}}_i\right) +
         \frac{1}{2}\left(\sum_{i=1}^n m_i\right) \mathbf{V} \cdot \mathbf{V}.
\end{align}</math>

Let the reference point be the center of mass <math>\mathbf{C}</math> of the system so the second term becomes zero, and introduce the moment of inertia <math>I_\mathbf{C}</math> so the kinetic energy is given by<ref name="Beer"/>{{rp|p=1084}}
<math display="block">E_\text{K} = \frac{1}{2} I_\mathbf{C} \omega^2 + \frac{1}{2} M\mathbf{V} \cdot \mathbf{V}.</math>

The moment of inertia <math>I_\mathbf{C}</math> is the ''polar moment of inertia'' of the body.