# What is kinetic energy

1. Jul 24, 2014

### Greg Bernhardt

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 $v<<c$

Extended explanation

When a (non-relativistic) particle of mass $m$ 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
$$\frac{1}{2}I\omega^2$$

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