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 <math> v </math>, 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, <math> K </math>, can be found as:

<math> K = \frac{1}{2} m v^2</math>

Where

<math> m </math> is the mass of the object

<math> v </math> is the velocity of the object

Rotational kinetic energy

The rotational kinetic energy of a rigid object is:

<math> K = {1 \over 2} I \omega ^2 </math>

Where

<math> I </math> is the moment of inertia of the object

<math> \omega </math> 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:

<math> \Sigma W = \Delta K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{I}^{2} </math>

Where

<math> v_i </math> is the initial speed

<math> v_f </math> 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:

<math> W = K = \int F dx</math>

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

<math> K = \int \frac{dp}{dt} dx = \int \frac{dp}{dx} \frac{dx}{dt} dx </math>

This is the same as

<math> K = \int v dp </math>

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

<math> K = \int^{u}_{0} m v dv </math>

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

<math> K = \frac{1}{2} mv^2 </math>

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:

<math> K = (\gamma - 1) m_0 c^2 </math>

where

<math>\gamma</math> is the Lorentz factor

<math>m_0</math> is the rest mass

<math>c</math> is the speed of light

This can be shown to be equivalent to the classical equation for kinetic energy, <math>K=\frac{1}{2} m_0 v^2</math>, by performing a binomial expansion on it. Using the result:

<math> (1+x)^n = 1 + nx+ \frac{n(n-1)x^2}{2} + ... </math>

Expanding the Lorentz factor in this way, we see:

<math> K=(1 + \frac{1}{2} \frac{v^2}{c^2}+ \frac{3}{8} \frac{v^4}{c^4}+...-1) m_0 c^2 </math>

This simplifies to

<math> K=\frac{1}{2} m_0 v^2 + \frac{3}{8} \frac{m_0 v^4}{c^2} +... </math>

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

<math>K \approx \frac{1}{2} m_0 v^2</math>

which is the classical formula.

Derivation

The kinetic energy is the work done accelerating a particle from rest to some speed <math>v</math>. Suppose the particle is at rest at <math>x_1</math> and speed <math>v</math> at position <math>x_2</math>. Hence:

<math>K = \int^{x_2}_{x_1} \vec{F} \dot d \vec{x}</math>

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

<math>K = \int^{x_2}_{x_1} F_{\parallel} dx = \int^{x_2}_{x_1} \gamma^3 m_0 a dx</math>

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

<math>K = \int^{v_2}_{v_1} \frac{mv}{(1- \frac{v^2}{c^2})} dv</math>

where <math>v_1</math> is the initial speed and hence 0 by definition and <math>v_2</math> is the final speed.

Performing the integration reveals the kinetic energy as:

<math>K=(\gamma -1)m_0 c^2</math>