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{excerpt}The energy of an object's translational and/or rotational motion.{excerpt}

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h2. Motivation for Concept

One way to think of energy is the ability to do harm.  Things that are very dangerous possess considerable energy.  With that definition in mind, the concept of kinetic energy can be motivated by considering the danger posed by moving objects.  When a baseball is thrown by a child, it is not very frightening, though it can break a lamp.  When a baseball is thrown by a major league pitcher, they can cause considerable injury.  This contrast indicates that the energy of motion will depend on speed.  Similarly, although a baseball thrown at 60 mph is dangerous, a car driving 60 mph is deadly.  This contrast indicates that the energy of motion will depend upon mass.

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h2. Mathematical Definition

h4. Translational Kinetic Energy

The kinetic energy of a point particle is given by:

{latex}\begin{large}\[ K = \frac{1}{2}mv^{2}\]\end{large}{latex}

h4. Kinetic Energy of a System

Since energy is a [scalar], the kinetic energy of a system of [point particles|point particle] is the sum of the kinetic energies of the constituents:

{latex}\begin{large}\[ K^{\rm sys} = \sum_{j = 1}^{N} \frac{1}{2}m_{j}v_{j}^{2}\]\end{large}{latex}

where _N_ is the number of system constituents.

h4. Kinetic Energy of a Rigid Body

Consider a [rigid body] that can rotate and translate.  We begin by treating the rigid body as a collection of point masses that are translating with the center of mass of the body and also rotating about it with angular velocity ω.  We therefore write the velocity of each point as a sum of rotational and translational parts:

{latex}\begin{large}\[ \vec{v}_{j} = \vec{v}_{\rm cm} + \vec{\omega}\times \vec{r}_{j} \]\end{large}{latex}

where _r_~j~ is the position of the j{^}th^ particle measured from the body's axis of rotation passing through the center of mass.  

With this split, the kinetic energy of the body becomes:

{latex}\begin{large}\[ K = \sum_{j} \frac{1}{2}m_{j}(v_{\rm cm}^{2} + \vec{v}_{\rm cm}\cdot(\vec{\omega}\times \vec{r}_{j}) + \omega^{2}r_{j}^{2})\]\end{large}{latex}

The center term will equal zero, because ω and _v_~cm~ are constants, so:

{latex}\begin{large}\[ \sum_{j} \frac{1}{2}m_{j}\vec{v}_{\rm cm}\cdot(\vec{\omega}\times \vec{r}_{j}) = 
\frac{1}{2}\vec{v}_{\rm cm}\cdot\left(\vec{\omega}\times \sum_{j} m_{j}\vec{r}_{j}\right)\]\end{large}{latex}

and the sum over _m_~j~_r_~j~ is constrained to equal zero because we have assumed the center of mass is at the position _r_ = 0 in our coordinates.  With this realization, and using the definition of the [moment of inertia], we have:

{latex}\begin{large}\[ K = \frac{1}{2}m_{\rm tot}v_{\rm cm}^{2} + \frac{1}{2}I_{\rm cm}\omega^{2}\]\end{large}{latex}

This result shows that the kinetic energy of a rigid body can be broken into two parts, generally known as the translational part and the rotational part. 

h4. Rotational Kinetic Energy

The above formula suggests a definition for the kinetic energy of a rotating body:

{latex}\begin{large}\[ K^{\rm rot} = \frac{1}{2}I\omega^{2} \]\end{large}{latex}