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

The energy of a particle in relativity is:

$E=\gamma m_{0}c^{2}$

where $\gamma$ is the Lorentz factor, $m_{0}$ is the rest mass and c is the speed of light.

Since this includes the mass energy of the particle, we must subtract a factor of $m_{0}c^{2}$ to get the energy due to the particle's motion, the kinetic energy, as:

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