Editing Kinetic energy (section)

==Kinetic energy for non-relativistic velocity==
Treatments of kinetic energy depend upon the relative velocity of objects compared to the fixed [[speed of light]]. Speeds experienced directly by humans are '''non-relativisitic'''; higher speeds require the [[theory of relativity]]. 

===Kinetic energy of rigid bodies===
In [[classical mechanics]], the kinetic energy of a ''point object'' (an object so small that its mass can be assumed to exist at one point), or a non-rotating [[rigid body]] depends on the [[mass]] of the body as well as its [[speed]]. The kinetic energy is equal to half the [[Multiplication|product]] of the mass and the square of the speed. In formula form:

<math display="block">E_\text{k} = \frac{1}{2} mv^2</math>

where <math>m</math> is the mass and <math>v</math> is the speed (magnitude of the velocity) of the body. In [[SI]] units, mass is measured in [[kilogram]]s, speed in [[metres per second]], and the resulting kinetic energy is in [[joule]]s.

For example, one would calculate the kinetic energy of an 80&nbsp;kg mass (about 180&nbsp;lbs) traveling at 18&nbsp;metres per second (about 40&nbsp;mph, or 65&nbsp;km/h) as

<math display="block">E_\text{k} = \frac{1}{2} \cdot 80 \,\text{kg} \cdot \left(18 \,\text{m/s}\right)^2 = 12,960 \,\text{J} = 12.96 \,\text{kJ}</math>

When a person throws a ball, the person does [[work (physics)|work]] on it to give it speed as it leaves the hand. The moving ball can then hit something and push it, doing work on what it hits. The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: '''net force × displacement = kinetic energy''', i.e.,

<math display="block">Fs = \frac{1}{2} mv^2</math>

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force. As a consequence of this quadrupling, it takes four times the work to double the speed.

The kinetic energy of an object is related to its [[momentum]] by the equation:
<math display="block">E_\text{k} = \frac{p^2}{2m}</math>

where:
*<math>p</math> is momentum
*<math>m</math> is mass of the body

For the ''translational kinetic energy,'' that is the kinetic energy associated with [[rectilinear motion]], of a [[rigid body]] with constant [[mass]] <math>m</math>, whose [[center of mass]] is moving in a straight line with speed <math>v</math>, as seen above is equal to

<math display="block"> E_\text{t} = \frac{1}{2} mv^2 </math>

where:
*<math>m</math> is the mass of the body
*<math>v</math> is the speed of the [[center of mass]] of the body.

The kinetic energy of any entity depends on the reference frame in which it is measured. However, the total energy of an isolated system, i.e. one in which energy can neither enter nor leave, does not change over time in the reference frame in which it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called the [[Oberth effect]]. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames would however disagree on the value of this conserved energy.

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is the [[center of momentum]] frame, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to the [[invariant mass]] of the system as a whole.

====Derivation====
=====Without vector calculus=====
The work W done by a force ''F'' on an object over a distance ''s'' parallel to ''F'' equals 

<math display="block">W = F \cdot s.</math>

Using [[Newton's second law]] 

<math display="block">F = m a</math> 

with ''m'' the mass and ''a'' the [[Acceleration#Uniform_acceleration|acceleration]] of the object and 

<math display="block">s = \frac{a t^2}{2}</math>

the distance traveled by the accelerated object in time ''t'', we find with <math>v = a t</math> for the velocity ''v'' of the object

<math display="block">W = m a \frac{a t^2}{2} = \frac{m (at)^2}{2} = \frac{m v^2}{2}.</math>

=====With vector calculus=====
The work done in accelerating a particle with mass ''m'' during the infinitesimal time interval ''dt'' is given by the dot product of ''force'' '''F''' and the infinitesimal ''displacement'' ''d'''''x'''

<math display="block">\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})\,,</math>

where we have assumed the relationship '''p'''&nbsp;=&nbsp;''m''&nbsp;'''v''' and the validity of [[Newton's second law]]. (However, also see the special relativistic derivation [[#Relativistic kinetic energy of rigid bodies|below]].)

Applying the [[product rule]] we see that:

<math display="block">d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) =  2(\mathbf{v} \cdot d\mathbf{v}).</math>

Therefore (assuming constant mass so that ''dm'' = 0), we have

<math display="block">\mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2  = d \left(\frac{m v^2}{2}\right).</math>

Since this is a [[total differential]] (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

<math display="block">E_\text{k} = \int_{v_1}^{v_2}\mathbf{p}d\mathbf{v} =  \int_{v_1}^{v_2}m\mathbf{v}d\mathbf{v} = {mv^2\over 2}\bigg\vert_{v_1}^{v_2} = {1\over 2}m(v_2^2-v_1^2).</math>

This equation states that the kinetic energy (''E''<sub>k</sub>) is equal to the [[integral]] of the [[dot product]] of the [[momentum]] ('''p''') of a body and the [[infinitesimal]] change of the [[velocity]] ('''v''') of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

===Rotating bodies===
If a rigid body Q is rotating about any line through the center of mass then it has [[rotational energy|''rotational kinetic energy'']] (<math>E_\text{r}\,</math>) which is simply the sum of the kinetic energies of its moving parts, and is thus given by:

<math display="block">E_\text{r} = \int_Q \frac{v^2 dm}{2} = \int_Q \frac{(r \omega)^2 dm}{2} = \frac{\omega^2}{2} \int_Q {r^2}dm = \frac{\omega^2}{2} I = \frac{1}{2} I \omega^2</math>

where:
* ω is the body's [[angular velocity]]
* ''r'' is the distance of any mass ''dm'' from that line
* <math>I</math> is the body's [[moment of inertia]], equal to <math display="inline">\int_Q {r^2} dm</math>.

(In this equation the moment of [[inertia]] must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape).

===Kinetic energy of systems===
A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in the [[Solar System]] the planets and planetoids are orbiting the Sun. In a tank of gas, the molecules are moving in all directions. The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains.

A macroscopic body that is stationary (i.e. a reference frame has been chosen to correspond to the body's [[center of momentum]]) may have various kinds of [[internal energy]] at the molecular or atomic level, which may be regarded as kinetic energy, due to molecular translation, rotation, and vibration, electron translation and spin, and nuclear spin. These all contribute to the body's mass, as provided by the special theory of relativity. When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only. However, all internal energies of all types contribute to a body's mass, inertia, and total energy.

===Fluid dynamics===
In [[fluid dynamics]], the kinetic energy per unit volume at each point in an incompressible fluid flow field is called the [[dynamic pressure]] at that point.<ref>A. M. Kuethe and J. D. Schetzer (1959). ''Foundations of Aerodynamics'', 2nd edition, p.53. John Wiley & Sons {{ISBN|0-471-50952-3}}</ref>

<math display="block">E_\text{k} = \frac{1}{2} mv^2</math>

Dividing by V, the unit of volume:

<math display="block">\begin{align}
  \frac{E_\text{k}}{V} &= \frac{1}{2} \frac{m}{V}v^2 \\
                     q &= \frac{1}{2} \rho v^2
\end{align}</math>

where <math>q</math> is the dynamic pressure, and ρ is the density of the incompressible fluid.

===Frame of reference===

The speed, and thus the kinetic energy of a single object is frame-dependent (relative): it can take any non-negative value, by choosing a suitable [[inertial frame of reference]]. For example, a bullet passing an observer has kinetic energy in the reference frame of this observer. The same bullet is stationary to an observer moving with the same velocity as the bullet, and so has zero kinetic energy.<ref>{{cite book
|title=Introduction to the theory of relativity
|url=https://archive.org/details/introductiontoth0000sear_c2m9
|url-access=registration
|first1=Francis Weston
|last1=Sears
|first2=Robert W.
|last2=Brehme
|publisher=Addison-Wesley
|year=1968
|page=[https://archive.org/details/introductiontoth0000sear_c2m9/page/127 127]
}}, [https://books.google.com/books?id=cpzvAAAAMAAJ&q=%22in+its+own+rest+frame%22 Snippet view of page 127] {{Webarchive|url=https://web.archive.org/web/20200804041331/https://books.google.com/books?id=cpzvAAAAMAAJ&dq=%22in+its+own+rest+frame%22+%22kinetic+energy%22&q=%22in+its+own+rest+frame%22 |date=2020-08-04 }}</ref> By contrast, the total kinetic energy of a system of objects cannot be reduced to zero by a suitable choice of the inertial reference frame, unless all the objects have the same velocity. In any other case, the total kinetic energy has a non-zero minimum, as no inertial reference frame can be chosen in which all the objects are stationary. This minimum kinetic energy contributes to the system's [[invariant mass]], which is independent of the reference frame.

The total kinetic energy of a system depends on the [[inertial frame of reference]]: it is the sum of the total kinetic energy in a [[center of momentum frame]] and the kinetic energy the total mass would have if it were concentrated in the [[center of mass]].

This may be simply shown: let <math>\textstyle\mathbf{V}</math> be the relative velocity of the center of mass frame ''i'' in the frame ''k''. Since

<math display="block">
  v^2 = \left(v_i + V\right)^2
      = \left(\mathbf{v}_i + \mathbf{V}\right) \cdot \left(\mathbf{v}_i + \mathbf{V}\right)
      = \mathbf{v}_i \cdot \mathbf{v}_i + 2 \mathbf{v}_i \cdot \mathbf{V} + \mathbf{V} \cdot \mathbf{V}
      = v_i^2 + 2 \mathbf{v}_i \cdot \mathbf{V} + V^2,
</math>

Then,

<math display="block">  E_\text{k} = \int \frac{v^2}{2} dm = \int \frac{v_i^2}{2} dm + \mathbf{V} \cdot \int \mathbf{v}_i dm + \frac{V^2}{2} \int dm.</math>

However, let <math display="inline"> \int \frac{v_i^2}{2} dm = E_i </math> the kinetic energy in the center of mass frame, <math display="inline"> \int \mathbf{v}_i dm </math> would be simply the total momentum that is by definition zero in the center of mass frame, and let the total mass: <math display="inline"> \int dm = M </math>. Substituting, we get:<ref>[http://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1/node64.html Physics notes – Kinetic energy in the CM frame] {{webarchive|url=https://web.archive.org/web/20070611231147/http://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1/node64.html |date=2007-06-11 }}. [[Duke University|Duke]].edu. Accessed 2007-11-24.</ref>

<math display="block">E_\text{k} = E_i + \frac{M V^2}{2}.</math>

Thus the kinetic energy of a system is lowest to center of momentum reference frames, i.e., frames of reference in which the center of mass is stationary (either the [[center of mass frame]] or any other [[center of momentum frame]]). In any different frame of reference, there is additional kinetic energy corresponding to the total mass moving at the speed of the center of mass. The kinetic energy of the system in the [[center of momentum frame]] is a quantity that is invariant (all observers see it to be the same).

===Rotation in systems===
It sometimes is convenient to split the total kinetic energy of a body into the sum of the body's center-of-mass translational kinetic energy and the energy of rotation around the center of mass ([[rotational energy]]):

<math display="block">E_\text{k} = E_\text{t} + E_\text{r} </math>

where:
*''E''<sub>k</sub> is the total kinetic energy
*''E''<sub>t</sub> is the translational kinetic energy
*''E''<sub>r</sub> is the ''rotational energy'' or ''angular kinetic energy'' in the rest frame

Thus the kinetic energy of a tennis ball in flight is the kinetic energy due to its rotation, plus the kinetic energy due to its translation.