Kinetic Energy

Kinetic energy represents the energy associated with the motion of an object. It is defined as the work done by a force to accelerate that object from rest to some speed $v$, in the absence of any other forces acting upon the object. Kinetic energy is a scalar and has the same units as work (i.e. Joule).

Classical mechanics

Translational kinetic energy

In classical mechanics, the translational kinetic energy of a ridid object, $K$, can be found as:

$K={\frac {1}{2}}mv^{2}$

Where

$m$ is the mass of the object

$v$ is the velocity of the object

Rotational kinetic energy

The rotational kinetic energy of a rigid object is:

$K={1 \over 2}I\omega ^{2}$

Where

$I$ is the moment of inertia of the object

$\omega$ is the angular velocity of the object

Work-Energy theorem

The change of kinetic energy is equal to the total work done on it by the resultant of all forces acting on it. For a point mass this can be expressed as:

$\Sigma W=\Delta K={\frac {1}{2}}mv_{f}^{2}-{\frac {1}{2}}mv_{I}^{2}$

Where

$v_{i}$ is the initial speed

$v_{f}$ is the final speed

Note that if the mass of an object is increased, the increase in kinetic energy increases linearly; if the velocity of an object is increased, the increase in kinetic energy increases quadratically. For example, doubling the mass of an object doubles its kinetic energy; doubling its velocity quadruples its kinetic energy.

Derivation of translational kinetic energy

The work done by a force accelerating an object from rest, which is the kinetic energy is:

$W=K=\int Fdx$

From Newton's second law, the force, $F$, is $F={\frac {dp}{dt}}$. Hence we can make the substitution and use the chain rule

$K=\int {\frac {dp}{dt}}dx=\int {\frac {dp}{dx}}{\frac {dx}{dt}}dx$

This is the same as

$K=\int vdp$

In classical mechanics, momentum is given by $p=mv$. Differentiating and substituting into the above equation results in

$K=\int _{0}^{u}mvdv$

We want to integrate between 0 and the speed of the object, $u$ as this defines kinetic energy. Performing the integration reveals that the kinetic energy is, as expected, the following:

$K={\frac {1}{2}}mv^{2}$

A similar method may be used to derive the formula for rotational kinetic energy.

Kinetic Energy in Relativity

In relativity, kinetic energy can be expressed as:

$K=(\gamma -1)m_{0}c^{2}$

where

$\gamma$ is the Lorentz factor

$m_{0}$ is the rest mass

$c$ is the speed of light

This can be shown to be equivalent to the classical equation for kinetic energy, $K={\frac {1}{2}}m_{0}v^{2}$, by performing a binomial expansion on it. Using the result:

$(1+x)^{n}=1+nx+{\frac {n(n-1)x^{2}}{2}}+...$

Expanding the Lorentz factor in this way, we see:

$K=(1+{\frac {1}{2}}{\frac {v^{2}}{c^{2}}}+{\frac {3}{8}}{\frac {v^{4}}{c^{4}}}+...-1)m_{0}c^{2}$

This simplifies to

$K={\frac {1}{2}}m_{0}v^{2}+{\frac {3}{8}}{\frac {m_{0}v^{4}}{c^{2}}}+...$

For speeds encountered everyday, which are a lot less than that of light (such that $v<<c$), all terms apart from the first are very small and can be neglected. Hence, the formula reduces to:

$K\approx {\frac {1}{2}}m_{0}v^{2}$

which is the classical formula.

Derivation

The kinetic energy is the work done accelerating a particle from rest to some speed $v$. Suppose the particle is at rest at $x_{1}$ and speed $v$ at position $x_{2}$. Hence:

$K=\int _{x_{1}}^{x_{2}}{\vec {F}}{\dot {d}}{\vec {x}}$

Since ${\vec {F}}={\vec {F}}_{\perp }+{\vec {F}}_{\parallel }$, and a perpendicular force does no work, we can ignore the perpendicular component and write:

$K=\int _{x_{1}}^{x_{2}}F_{\parallel }dx=\int _{x_{1}}^{x_{2}}\gamma ^{3}m_{0}adx$

Since $a\,dx={\frac {dv_{\parallel }}{dt}}dx={\frac {dx}{dt}}dv$, we find the integral can be rewritten as:

$K=\int _{v_{1}}^{v_{2}}{\frac {mv}{(1-{\frac {v^{2}}{c^{2}}})}}dv$

where $v_{1}$ is the initial speed and hence 0 by definition and $v_{2}$ is the final speed.

Performing the integration reveals the kinetic energy as:

$K=(\gamma -1)m_{0}c^{2}$