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Torque
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== Definition and relation to other physical quantities == [[File:Torque, position, and force.svg|thumb|right|A particle is located at position '''r''' relative to its axis of rotation. When a force '''F''' is applied to the particle, only the perpendicular component '''F'''<sub>⊥</sub> produces a torque. This torque {{math|1='''''τ''''' = '''r''' × '''F'''}} has magnitude {{math|1=''τ'' = {{abs|'''r'''}} {{abs|'''F'''<sub>⊥</sub>}} = {{abs|'''r'''}} {{abs|'''F'''}} sin ''θ''}} and is directed outward from the page.]] A force applied perpendicularly to a lever multiplied by its distance from the [[Lever|lever's fulcrum]] (the length of the [[lever arm]]) is its torque. Therefore, torque is defined as the product of the magnitude of the perpendicular component of the force and the distance of the [[line of action]] of a force from the point around which it is being determined. In three dimensions, the torque is a [[pseudovector]]; for [[point particles]], it is given by the [[cross product]] of the [[Euclidean vector|displacement vector]] and the force vector. The direction of the torque can be determined by using the [[right hand grip rule]]: if the fingers of the right hand are curled from the direction of the lever arm to the direction of the force, then the thumb points in the direction of the torque.<ref>{{cite web|url=http://hyperphysics.phy-astr.gsu.edu/hbase/tord.html|title=Right Hand Rule for Torque|access-date=2007-09-08|archive-date=2007-08-19|archive-url=https://web.archive.org/web/20070819141440/http://hyperphysics.phy-astr.gsu.edu/HBASE/tord.html|url-status=live}}</ref> It follows that the ''torque vector'' is perpendicular to both the ''position'' and ''force'' vectors and defines the plane in which the two vectors lie. The resulting ''torque vector'' direction is determined by the right-hand rule. Therefore any force directed parallel to the particle's position vector does not produce a torque.<ref name="halliday_184-85">{{cite book |last1=Halliday |first1=David |title=Fundamentals of Physics |last2=Resnick |first2=Robert |publisher=John Wiley & Sons |year=1970 |pages=184–85}}</ref><ref>{{Cite book |last1=Knight |first1=Randall |title=College Physics: A Strategic Approach |last2=Jones |first2=Brian |last3=Field |first3=Stuart |publisher=Pearson |year=2016 |isbn=9780134143323 |edition=3rd technology update |location=Boston |pages=199 |oclc=922464227}}</ref> The magnitude of torque applied to a [[rigid body]] depends on three quantities: the force applied, the ''lever arm vector''<ref>{{cite book |last=Tipler |first=Paul |title=Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics |publisher=W. H. Freeman |year=2004 |isbn=0-7167-0809-4 |edition=5th}}</ref> connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols: <math display="block">\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \implies \tau = rF_{\perp} = rF\sin\theta</math> where * <math>\boldsymbol\tau</math> is the torque vector and <math>\tau</math> is the magnitude of the torque, * <math> \mathbf{r} </math> is the [[position vector]] (a vector from the point about which the torque is being measured to the point where the force is applied), and ''r'' is the magnitude of the position vector, * <math> \mathbf{F} </math> is the force vector, ''F'' is the magnitude of the force vector and ''F''<sub>⊥</sub> is the amount of force directed perpendicularly to the position of the particle, * <math> \times </math> denotes the [[cross product]], which produces a vector that is [[perpendicular]] both to {{math|'''r'''}} and to {{math|'''F'''}} following the [[right-hand rule]], * <math> \theta</math> is the angle between the force vector and the lever arm vector. The [[SI units|SI unit]] for torque is the [[newton-metre]] (N⋅m). For more on the units of torque, see {{slink||Units}}. === 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. === Derivatives of torque === In [[physics]], '''rotatum''' is the derivative of [[torque]] with respect to [[time]]<ref>{{Cite journal |title=Survey of Human–Robot Collaboration in Industrial Settings: Awareness, Intelligence, and Compliance |date=2021 |doi=10.1109/TSMC.2020.3041231 |url=https://ieeexplore.ieee.org/abstract/document/9302892/keywords#keywords |last1=Kumar |first1=Shitij |last2=Savur |first2=Celal |last3=Sahin |first3=Ferat |journal=IEEE Transactions on Systems, Man, and Cybernetics: Systems |volume=51 |pages=280–297 |url-access=subscription }}</ref><blockquote><math>\mathbf P = \frac{\mathrm d \boldsymbol \tau}{\mathrm d t},</math></blockquote>where '''τ''' is torque. This word is derived from the [[Latin]] word {{lang|la|rotātus}} meaning 'to rotate'. The term ''rotatum'' is not universally recognized but is commonly used. There is not a universally accepted lexicon to indicate the successive derivatives of rotatum, even if sometimes various proposals have been made. Using the cross product definition of torque, an alternative expression for rotatum is: <blockquote><math>\mathbf{P} = \mathbf{r} \times \frac{\mathrm d \mathbf{F}}{\mathrm d t} + \frac{\mathrm d \mathbf{r}}{\mathrm d t} \times \mathbf{F}.</math></blockquote> Because the rate of change of force is yank <math display="inline">\mathbf{Y}</math> and the rate of change of position is velocity <math display="inline">\mathbf{v}</math>, the expression can be further simplified to: <blockquote><math>\mathbf{P} = \mathbf{r} \times \mathbf{Y} + \mathbf{v} \times \mathbf{F}.</math></blockquote> === 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>
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