Editing Rolling (section)

=== Energy ===
Since [[kinetic energy]] is entirely a function of an object mass and velocity, the above result may be used with the [[parallel axis theorem]] to obtain the kinetic energy associated with simple rolling

<math display="block">K_\text{rolling} = K_\text{translation} + K_\text{rotation}</math>

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{{show
|Derivation
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{{further|Rotation around a fixed axis}}
Let <math>r</math> be the distance between the center of mass and the point of contact; when the surface is flat, this is the radius of the object around its widest cross section. Since the center of mass has an immediate velocity as if it was ''rotating'' around the point of contact, its velocity is <math>v_\text{c.o.m.} = r\omega</math>. Due to symmetry, the object center of mass is a point in its axis. Let <math>I_\text{rotation}</math> be inertia of [[rotation|pure rotation]] around the axis of symmetry, then according to the parallel axis theorem, the [[rotational inertia]] associated with rolling is <math>I_\text{rolling} = mr^2 + I_\text{rotation}</math> (same as the rotational inertia of pure rotation around the point of contact). Using the general formula for kinetic energy of rotation, we have:

<math display="block">\begin{align}
  K_\text{rolling} &= \frac{1}{2} I_\text{rolling}\omega^2 \\
                   &= \frac{1}{2} mr^2\omega^2 + \frac{1}{2}I_\text{rotation}\omega^2 \\
                   &= \frac{1}{2} m(r\omega)^2 + \frac{1}{2}I_\text{rotation}\omega^2 \\
                   &= \frac{1}{2} mv_\text{c.o.m.}^2 + \frac{1}{2}I_\text{rotation}\omega^2 \\
                   &= K_\text{translation} + K_\text{rotation} \\
\end{align}</math>
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