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kinetic energy

 Definition/Summary Kinetic energy (KE) is energy associated with movement as a whole (linear or rotational or both). Kinetic energy of a body therefore does not include thermal energy (associated with internal movement, of its molecules), nor any other form of potential energy (such as chemical or nuclear energy, or energy associated with its own deformation or with its position in a gravitational or other field). Since total energy is always conserved, a gain (or loss) in kinetic energy is always accompanied by an equal loss (or gain) in other forms of energy.

 Equations Linear motion: $$KE\ =\ \frac{1}{2}\,mv^2\ =\ \frac{p^2}{2m}$$ Rotational motion: $$KE\ =\ \frac{1}{2}\,I_{c.o.r.}\omega^2\ =\ \frac{1}{2}\,I_{c.o.m.}\omega^2\ +\ \frac{1}{2}\,m\mathbf{v}_{c.o.m.}^2$$ (c.o.r. = centre of rotation, c.o.m. = centre of mass) Conservation of energy: $$KE\ +\ PE\ =\ constant$$ Linear motion (relativistic): $$KE\ =\ mc^2\,\sqrt{1\ +\ \frac{p^2}{m^2c^2}}\ -mc^2=\ \frac{mc^2}{\sqrt{1\ -\ v^2/c^2}}-mc^2\ \approx\ \frac{1}{2}\,mv^2\;,$$ where the final approximate equality holds for $v<  Scientists  Recent forum threads on kinetic energy  Breakdown Physics > Classical Mechanics >> Newtonian Dynamics  See Also elastic collisioninelastic collisioncoefficient of restitutpotential energywork done  Images  Extended explanation When a (non-relativistic) particle of mass [itex]m$ moves with velocity $\vec v$ (of magnitude $v$) the particle's kinetic energy $KE$ is given by $$KE=\frac{1}{2}mv^2\;.\qquad (1)$$ Because the relationship between the momentum $p$ and velocity is $p=mv$ equation (1) can also be written as $$KE=\frac{p^2}{2m}\;.$$ For a collection of particles (labelled by index $i$) the total kinetic energy is given by $$KE=\sum_i KE_{i} = \sum_{i}\frac{1}{2}m_{(i)}v_{(i)}^2\;,$$ where $m_{(i)}$ is the mass of the ith particle and v_{(i)} is the magnitude of the ith particle's velocity. For the case of a continuous distribution of particles $$KE=\frac{1}{2}\int d^3 r \rho(\mathbf{r}){|\mathbf{v}(\mathbf{r})|}^2\;.$$ For the case of a rigid body $\mathbf{v}(\mathbf{r})= \mathbf{\omega}\times\mathbf{r}$ for constant $\mathbf{\omega}$, and the above equation reduces to $$KE=\frac{1}{2}\omega^i\omega^j I_{ij}\;,$$ where $$I^{ij}=\int d^3 r \rho(\mathbf{r})\left(r^2\delta^{ij}-r^i r^j\right).$$ In many cases (e.g., cubic, spherical) the system is symmetric enough that $I^{ij}=I\delta^{ij}$, in which case the above equation for kinetic energy reduces to $$\frac{1}{2}I\omega^2$$

Commentary

 sanka @ 08:09 AM Jan31-12 Are you saying that the kinetic energy associated with movement as a whole is the kinetic energy of the continuum? This would be the velocity of the continuum. My understanding was that each individual molecule in the continuum has a specific kinetic energy associated with it and the thermal energy (which is given by Temperature) is a measure of the average or mean knietic energy of said molecules. Thus thermal energy is a continuum concept.

 tiny-tim @ 06:59 AM Nov16-10 Added centre-of-mass version to rotational KE equation.

 martdon @ 08:02 PM Nov9-10 what computer program can be used to calculate kinetic energy?

 bilal_khan @ 03:42 AM Oct25-10 why do we take 2 times the distance that a gas molecules covers in a vessel between two faces in the derivation of kinetic gas equation?

 khalilzade @ 04:01 PM Sep11-10 what's relation between negativ kinetic energy and tunneling effect?

 tiny-tim @ 05:42 PM Nov12-08 Amended to a more general definition, and included comparison of KE with other forms of energy.

 olgranpappy @ 03:38 PM Nov11-08 added derivation of the expression given for rotation KE

 olgranpappy @ 03:21 PM Nov11-08 fixed mistake in relativistic KE... now it has the correct NR limit.

 DrGreg @ 11:55 AM Oct16-08 The relativistic equation quoted isn't quite right: mc2 needs to be subtracted, as that isn't considered part of the kinetic energy. In other words, KE ought to zero for zero velocity.

 tiny-tim @ 03:16 AM Oct12-08 Added ½Iω² mc² and conservation of energy.