Editing Torque (section)

=== Relationship with the angular momentum ===
The net torque on a body determines the rate of change of the body's [[angular momentum]],

<math display="block">\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}</math>

where '''L''' is the angular momentum vector and ''t'' is time. For the motion of a point particle,

<math display="block">\mathbf{L} = I\boldsymbol{\omega},</math>

where <math display="inline">I = mr^2   </math> is the [[moment of inertia]] and '''ω''' is the orbital [[angular velocity]] pseudovector. It follows that

<math display="block">\boldsymbol{\tau}_{\mathrm{net}} = I_1\dot{\omega_1}\hat{\boldsymbol{e_1}} + I_2\dot{\omega_2}\hat{\boldsymbol{e_2}} + I_3\dot{\omega_3}\hat{\boldsymbol{e_3}} + I_1\omega_1\frac{d\hat{\boldsymbol{e_1}}}{dt} + I_2\omega_2\frac{d\hat{\boldsymbol{e_2}}}{dt} + I_3\omega_3\frac{d\hat{\boldsymbol{e_3}}}{dt}
= I\boldsymbol\dot{\omega} + \boldsymbol\omega \times (I\boldsymbol\omega)</math>

using the derivative of a [[Unit vector|vector]] is<math display="block">{d\boldsymbol{\hat{e_i}} \over dt} = \boldsymbol\omega \times \boldsymbol{\hat{e_i}}</math>This equation is the rotational analogue of [[Newton's second law]] for point particles, and is valid for any type of trajectory. In some simple cases like a rotating disc, where only the moment of inertia on rotating axis is, the rotational Newton's second law can be<math display="block">\boldsymbol{\tau} = I\boldsymbol{\alpha}  </math>where <math>\boldsymbol\alpha = \dot\boldsymbol\omega </math>. 

==== Proof of the equivalence of definitions ====
The definition of angular momentum for a single point particle is:
<math display="block">\mathbf{L} = \mathbf{r} \times \mathbf{p}</math>
where '''p''' is the particle's [[linear momentum]] and '''r''' is the position vector from the origin. The time-derivative of this is:

<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} + \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \times \mathbf{p}.</math>

This result can easily be proven by splitting the vectors into components and applying the [[product rule]]. But because the rate of change of linear momentum is force <math display="inline">\mathbf{F}</math> and the rate of change of position is velocity <math display="inline">\mathbf{v}</math>,

<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \mathbf{F} + \mathbf{v} \times \mathbf{p} </math>

The cross product of momentum <math>\mathbf{p}</math> with its associated velocity <math>\mathbf{v}</math> is zero because velocity and momentum are parallel, so the second term vanishes. Therefore, torque on a particle is ''equal'' to the [[Derivative#Notation for differentiation|first derivative]] of its angular momentum with respect to time. If multiple forces are applied, according [[Newton's second law]] it follows that<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} =  \boldsymbol{\tau}_{\mathrm{net}}.</math>

This is a general proof for point particles, but it can be generalized to a system of point particles by applying the above proof to each of the point particles and then summing over all the point particles. Similarly, the proof can be generalized to a continuous mass by applying the above proof to each point within the mass, and then [[Integral calculus|integrating]] over the entire mass.