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>

Kinetic Energy in Relativity

The energy of a particle in relativity is:

<math> E = \gamma m_0 c^2 </math>

where <math>\gamma</math> is the Lorentz factor, <math>m_0</math> 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 <math> m_0 c^2 </math> to get the energy due to the particle's motion, the kinetic energy, as:

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