Editing Torque (section)

=== Relationship with power and energy ===
The law of [[conservation of energy]] can also be used to understand torque. If a [[force]] is allowed to act through a distance, it is doing [[mechanical work]]. Similarly, if torque is allowed to act through an angular displacement, it is doing work. Mathematically, for rotation about a fixed axis through the [[center of mass]], the work ''W'' can be expressed as

<math qid=Q104145165 display="block"> W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,</math>

where ''τ'' is torque, and ''θ''<sub>1</sub> and ''θ''<sub>2</sub> represent (respectively) the initial and final [[angular position]]s of the body.<ref name="kleppner_267-68">{{cite book|last1=Kleppner |first1=Daniel |last2=Kolenkow |first2=Robert|title=An Introduction to Mechanics |url=https://archive.org/details/introductiontome00dani |url-access=registration|publisher=McGraw-Hill |year=1973|pages=[https://archive.org/details/introductiontome00dani/page/267 267–268]|isbn=9780070350489 }}</ref>

It follows from the [[work–energy principle]] that ''W'' also represents the change in the [[Rotational energy|rotational kinetic energy]] ''E''<sub>r</sub> of the body, given by

<math qid=Q104145205 display="block">E_{\mathrm{r}} = \tfrac{1}{2}I\omega^2,</math>

where ''I'' is the [[moment of inertia]] of the body and ''ω'' is its [[angular speed]].<ref name="kleppner_267-68" />

[[Power (physics)|Power]] is the work per unit [[time]], given by

<math qid=Q104145185 display="block">P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},</math>

where ''P'' is power, '''''τ''''' is torque, '''''ω''''' is the [[angular velocity]], and <math>\cdot </math> represents the [[scalar product]].

Algebraically, the equation may be rearranged to compute torque for a given angular speed and power output. The power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).

==== Proof ====
The work done by a variable force acting over a finite linear displacement <math>s</math> is given by integrating the force with respect to an elemental linear displacement <math>\mathrm{d}\mathbf{s}</math>

<math display="block">W = \int_{s_1}^{s_2} \mathbf{F} \cdot \mathrm{d}\mathbf{s}</math>

However, the infinitesimal linear displacement <math>\mathrm{d}\mathbf{s}</math> is related to a corresponding angular displacement <math>\mathrm{d}\boldsymbol{\theta}</math> and the radius vector <math>\mathbf{r}</math> as 

<math display="block">\mathrm{d}\mathbf{s} = \mathrm{d}\boldsymbol{\theta}\times\mathbf{r}</math>

Substitution in the above expression for work, , gives
<math display="block">W = \int_{s_1}^{s_2} \mathbf{F} \cdot \mathrm{d}\boldsymbol{\theta} \times \mathbf{r}</math>

The expression inside the integral is a [[scalar triple product]] <math>\mathbf{F}\cdot\mathrm{d}\boldsymbol{\theta}\times\mathbf{r} = \mathbf{r} \times \mathbf{F} \cdot \mathrm{d}\boldsymbol{\theta}</math>, but as per the definition of torque, and since the parameter of integration has been changed from linear displacement to angular displacement, the equation becomes

<math display="block">W = \int_{\theta _1}^{\theta _2} \boldsymbol{\tau} \cdot \mathrm{d}\boldsymbol{\theta}</math>

If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., <math>\boldsymbol{\tau}\cdot \mathrm{d}\boldsymbol{\theta} = \left|\boldsymbol{\tau}\right| \left| \mathrm{d}\boldsymbol{\theta}\right|\cos 0 = \tau \, \mathrm{d}\theta</math> giving

<math display="block">W = \int_{\theta _1}^{\theta _2} \tau \, \mathrm{d}\theta</math>