Torque, moment, or moment of force is rotational force.[1] Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object. In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The symbol for torque is typically τ displaystyle tau , the lowercase Greek letter tau. When it is called moment of force, it is commonly denoted by M. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector[2] connecting the origin to the point of force application, and the angle between the force and lever arm vectors. In symbols: τ = r × F displaystyle boldsymbol tau =mathbf r times mathbf F ,! τ = ‖ r ‖ ‖ F ‖ sin θ displaystyle tau =mathbf r ,mathbf F sin theta ,! where τ displaystyle boldsymbol tau is the torque vector and τ displaystyle tau is the magnitude of the torque, r is the position vector (a vector from the origin of the coordinate system defined to the point where the force is applied) F is the force vector, × denotes the cross product,which is defined as magnitudes of the respective vectors times sin θ. θ is the angle between the force vector and the lever arm vector. The
Contents 1 Defining terminology 2 History 3 Definition and relation to angular momentum 3.1 Proof of the equivalence of definitions 4 Units
5
5.1 Moment arm formula 5.2 Static equilibrium 5.3 Net force versus torque 6 Machine torque 7 Relationship between torque, power, and energy 7.1 Proof 7.2 Conversion to other units 7.3 Derivation 8 Principle of moments
9
Defining terminology[edit]
See also: Couple (mechanics)
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⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sin θ and is directed outward from the page. A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. 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.[6] More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product: τ = r × F , displaystyle boldsymbol tau =mathbf r times mathbf F , where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by τ = r F sin θ , displaystyle tau =rFsin theta ,! where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively, τ = r F ⊥ , displaystyle tau =rF_ perp , where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[7][8] It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. The torque vector points along the axis of the rotation that the force vector (starting from rest) would initiate. The resulting torque vector direction is determined by the right-hand rule.[7] The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum, τ = d L d t displaystyle boldsymbol tau = frac mathrm d mathbf L mathrm d t where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum: τ 1 + ⋯ + τ n = τ n e t = d L d t . displaystyle boldsymbol tau _ 1 +cdots + boldsymbol tau _ n = boldsymbol tau _ mathrm net = frac mathrm d mathbf L mathrm d t . For the motion of a point particle, L = I ω , displaystyle mathbf L =I boldsymbol omega , where I is the moment of inertia and ω is the angular velocity. It follows that τ n e t = d L d t = d ( I ω ) d t = I d ω d t + d I d t ω = I α + d ( m r 2 ) d t ω = I α + 2 r p
ω , displaystyle boldsymbol tau _ mathrm net = frac mathrm d mathbf L mathrm d t = frac mathrm d (I boldsymbol omega ) mathrm d t =I frac mathrm d boldsymbol omega mathrm d t + frac mathrm d I mathrm d t boldsymbol omega =I boldsymbol alpha + frac mathrm d (mr^ 2 ) mathrm d t boldsymbol omega =I boldsymbol alpha +2rp_ boldsymbol omega , where α is the angular acceleration of the particle, and p is the
radial component of its linear momentum. This equation is the
rotational analogue of
L = r × p displaystyle mathbf L =mathbf r times boldsymbol p where "×" indicates the vector cross product, p is the particle's linear momentum, and r is the displacement vector from the origin (the origin is assumed to be a fixed location anywhere in space). The time-derivative of this is: d L d t = r × d p d t + d r d t × p . displaystyle frac mathrm d mathbf L mathrm d t =mathbf r times frac mathrm d boldsymbol p mathrm d t + frac mathrm d mathbf r mathrm d t times boldsymbol p . This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force F = d p d t displaystyle mathbf F = frac mathrm d boldsymbol p mathrm d t (whether or not mass is constant) and the definition of velocity d r d t = v displaystyle frac mathrm d mathbf r mathrm d t =mathbf v d L d t = r × F + v × p . displaystyle frac mathrm d mathbf L mathrm d t =mathbf r times mathbf F +mathbf v times boldsymbol p . The cross product of momentum p displaystyle boldsymbol p with its associated velocity v displaystyle mathbf v is zero because velocity and momentum are parallel, so the second term vanishes. By definition, torque τ = r × F. Therefore, torque on a particle is equal to the first derivative of its angular momentum with respect to time. If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that d L d t = r × F n e t = τ n e t . displaystyle frac mathrm d mathbf L mathrm d t =mathbf r times mathbf F _ mathrm net = boldsymbol tau _ mathrm net . This is a general proof.
Units[edit]
E = τ θ displaystyle E=tau theta where E is the energy, τ is magnitude of the torque, and θ is the
angle moved (in radians). This equation motivates the alternate unit
name joules per radian.[9]
In Imperial units, "pound-force-feet" (lbf⋅ft), "foot-pounds-force",
"inch-pounds-force", "ounce-force-inches" (ozf⋅in)[citation needed]
are used, and other non-
Moment arm diagram A very useful special case, often given as the definition of torque in fields other than physics, is as follows: τ = ( moment arm ) ( force ) . displaystyle tau =( text moment arm )( text force ). The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in three-dimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force: τ = ( distance to centre ) ( force ) . displaystyle tau =( text distance to centre )( text force ). For example, if a person places a force of 10 N at the terminal end of a wrench that is 0.5 m long (or a force of 10 N exactly 0.5 m from the twist point of a wrench of any length), the torque will be 5 N.m – assuming that the person moves the wrench by applying force in the plane of movement and perpendicular to the wrench. The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess. Static equilibrium[edit] For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in two-dimensions, three equations are used. Net force versus torque[edit] When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a current-carrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force F displaystyle mathbf F is not zero, and τ 1 displaystyle boldsymbol tau _ 1 is the torque measured from r 1 displaystyle mathbf r _ 1 , then the torque measured from r 2 displaystyle mathbf r _ 2 is … τ 2 = τ 1 + ( r 1 − r 2 ) × F displaystyle boldsymbol tau _ 2 = boldsymbol tau _ 1 +(mathbf r _ 1 -mathbf r _ 2 )times mathbf F Machine torque[edit]
W = ∫ θ 1 θ 2 τ d θ , displaystyle W=int _ theta _ 1 ^ theta _ 2 tau mathrm d theta , where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[12] Proof[edit] The work done by a variable force acting over a finite linear displacement s displaystyle s is given by integrating the force with respect to an elemental linear displacement d s → displaystyle mathrm d vec s W = ∫ s 1 s 2 F → ⋅ d s → displaystyle W=int _ s_ 1 ^ s_ 2 vec F cdot mathrm d vec s However, the infinitesimal linear displacement d s → displaystyle mathrm d vec s is related to a corresponding angular displacement d θ → displaystyle mathrm d vec theta and the radius vector r → displaystyle vec r as d s → = d θ → × r → displaystyle mathrm d vec s =mathrm d vec theta times vec r Substitution in the above expression for work gives W = ∫ s 1 s 2 F → ⋅ d θ → × r → displaystyle W=int _ s_ 1 ^ s_ 2 vec F cdot mathrm d vec theta times vec r The expression F → ⋅ d θ → × r → displaystyle vec F cdot mathrm d vec theta times vec r is a scalar triple product given by [ F → d θ → r → ] displaystyle left[ vec F ,mathrm d vec theta , vec r right] . An alternate expression for the same scalar triple product is [ F → d θ → r → ] = r → × F → ⋅ d θ → displaystyle left[ vec F ,mathrm d vec theta , vec r right]= vec r times vec F cdot mathrm d vec theta But as per the definition of torque, τ → = r → × F → displaystyle vec tau = vec r times vec F Corresponding substitution in the expression of work gives, W = ∫ s 1 s 2 τ → ⋅ d θ → displaystyle W=int _ s_ 1 ^ s_ 2 vec tau cdot mathrm d vec theta Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving W = ∫ θ 1 θ 2 τ → ⋅ d θ → displaystyle W=int _ theta _ 1 ^ theta _ 2 vec tau cdot mathrm d vec theta If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., τ → ⋅ d θ → =
τ →
d θ →
cos 0 = τ d θ displaystyle vec tau cdot mathrm d vec theta =left vec tau rightleft,mathrm d vec theta rightcos 0=tau ,mathrm d theta giving W = ∫ θ 1 θ 2 τ d θ displaystyle W=int _ theta _ 1 ^ theta _ 2 tau ,mathrm d theta It follows from the work-energy theorem that W also represents the change in the rotational kinetic energy Er of the body, given by E r = 1 2 I ω 2 , displaystyle E_ mathrm r = tfrac 1 2 Iomega ^ 2 , where I is the moment of inertia of the body and ω is its angular speed.[12] Power is the work per unit time, given by P = τ ⋅ ω , displaystyle P= boldsymbol tau cdot boldsymbol omega , where P is power, τ is torque, ω is the angular velocity, and ⋅ represents the scalar product. Algebraically, the equation may be rearranged to compute torque for a given angular speed and power output. Note that 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). In practice, this relationship can be observed in bicycles: Bicycles are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a circular chain, and a derailleur mechanism if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multi-speed bicycle), all of which attached to the frame. A cyclist, the person who rides the bicycle, provides the input power by turning pedals, thereby cranking the front sprocket (commonly referred to as chainring). The input power provided by the cyclist is equal to the product of cadence (i.e. the number of pedal revolutions per minute) and the torque on spindle of the bicycle's crankset. The bicycle's drivetrain transmits the input power to the road wheel, which in turn conveys the received power to the road as the output power of the bicycle. Depending on the gear ratio of the bicycle, a (torque, rpm)input pair is converted to a (torque, rpm)output pair. By using a larger rear gear, or by switching to a lower gear in multi-speed bicycles, angular speed of the road wheels is decreased while the torque is increased, product of which (i.e. power) does not change. Consistent units must be used. For metric SI units, power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second). Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar. Conversion to other units[edit] A conversion factor may be necessary when using different units of power or torque. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution. In the following formulas, P is power, τ is torque, and ν (Greek letter nu) is rotational speed. P = τ ⋅ 2 π ⋅ ν displaystyle P=tau cdot 2pi cdot nu Showing units: P ( W ) = τ ( N ⋅ m ) ⋅ 2 π ( r a d / r e v ) ⋅ ν ( r e v / s e c ) displaystyle P( rm W )=tau rm (Ncdot m) cdot 2pi rm (rad/rev) cdot nu rm (rev/sec) Dividing by 60 seconds per minute gives us the following. P ( W ) = τ ( N ⋅ m ) ⋅ 2 π ( r a d / r e v ) ⋅ ν ( r p m ) 60 displaystyle P( rm W )= frac tau rm (Ncdot m) cdot 2pi rm (rad/rev) cdot nu rm (rpm) 60 where rotational speed is in revolutions per minute (rpm). Some people (e.g., American automotive engineers) use horsepower (imperial mechanical) for power, foot-pounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to: P ( h p ) = τ ( l b f ⋅ f t ) ⋅ 2 π ( r a d / r e v ) ⋅ ν ( r p m ) 33 , 000 . displaystyle P( rm hp )= frac tau rm (lbfcdot ft) cdot 2pi rm (rad/rev) cdot nu ( rm rpm ) 33,000 . The constant below (in foot pounds per minute) changes with the
definition of the horsepower; for example, using metric horsepower, it
becomes approximately 32,550.
Use of other units (e.g.,
power = force ⋅ linear distance time = ( torque r ) ⋅ ( r ⋅ angular speed ⋅ t ) t = torque ⋅ angular speed . displaystyle begin aligned text power &= frac text force cdot text linear distance text time \[6pt]&= frac left( dfrac text torque r right)cdot (rcdot text angular speed cdot t) t \[6pt]&= text torque cdot text angular speed .end aligned The radius r and time t have dropped out of the equation. However, angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give: power = torque ⋅ 2 π ⋅ rotational speed . displaystyle text power = text torque cdot 2pi cdot text rotational speed ., If torque is in newton metres and rotational speed in revolutions per second, the above equation gives power in newton metres per second or watts. If Imperial units are used, and if torque is in pounds-force feet and rotational speed in revolutions per minute, the above equation gives power in foot pounds-force per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower: power = torque ⋅ 2 π ⋅ rotational speed ⋅ ft ⋅ lbf min ⋅ horsepower 33 , 000 ⋅ ft ⋅ lbf min ≈ torque ⋅ RPM 5 , 252 displaystyle begin aligned text power &= text torque cdot 2pi cdot text rotational speed cdot frac text ft cdot text lbf text min cdot frac text horsepower 33,000cdot frac text ft cdot text lbf text min \[6pt]&approx frac text torque cdot text RPM 5,252 end aligned because 5252.113122 ≈ 33 , 000 2 π . displaystyle 5252.113122approx frac 33,000 2pi ., Principle of moments[edit]
The Principle of Moments, also known as
( r × F 1 ) + ( r × F 2 ) + ⋯ = r × ( F 1 + F 2 + ⋯ ) . displaystyle (mathbf r times mathbf F _ 1 )+(mathbf r times mathbf F _ 2 )+cdots =mathbf r times (mathbf F _ 1 +mathbf F _ 2 +cdots ).
Moment
Conversion of units
References[edit] ^ Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. ISBN 0-534-40842-7. ^ Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4. ^ a b Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link ^ a b c Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download ^ Thomson, James; Larmor, Joseph (1912). Collected Papers in Physics and Engineering. University Press. p. civ. , at Google books ^ "Right Hand Rule for Torque". Retrieved 2007-09-08. ^ a b Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons, Inc. pp. 184–85. ^ Knight, Randall; Jones, Brian; Field, Stuart (2016). College Physics: A Strategic Approach. Jones, Brian, 1960-, Field, Stuart, 1958- (Third edition, technology update ed.). Boston: Pearson. p. 199. ISBN 9780134143323. OCLC 922464227. ^ a b c From the official SI website: "...For example, the quantity torque may be thought of as the cross product of force and distance, suggesting the unit newton metre, or it may be thought of as energy per angle, suggesting the unit joule per radian." ^ "SI brochure Ed. 8, Section 5.1". Bureau International des Poids et Mesures. 2006. Retrieved 2007-04-01. ^ See, for example: "CNC Cookbook: Dictionary: N-Code to PWM". Retrieved 2008-12-17. ^ a b Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics. McGraw-Hill. pp. 267–68. External links[edit] Look up torque in Wiktionary, the free dictionary. Wikimedia Commons has media related to Torque. "
v t e
Linear/translational quantities Angular/rotational quantities Dimensions 1 L L2 Dimensions 1 1 1 T time: t s absement: A m s T time: t s 1 distance: d, position: r, s, x, displacement m area: A m2 1 angle: θ, angular displacement: θ rad solid angle: Ω rad2, sr T−1 frequency: f s−1, Hz speed: v, velocity: v m s−1 kinematic viscosity: ν, specific angular momentum: h m2 s−1 T−1 frequency: f s−1, Hz angular speed: ω, angular velocity: ω rad s−1 T−2 acceleration: a m s−2 T−2 angular acceleration: α rad s−2 T−3 jerk: j m s−3 T−3 angular jerk: ζ rad s−3 M mass: m kg ML2 moment of inertia: I kg m2 MT−1 momentum: p, impulse: J kg m s−1, N s action: 𝒮, actergy: ℵ kg m2 s−1, J s ML2T−1 angular momentum: L, angular impulse: ΔL kg m2 s−1 action: 𝒮, actergy: ℵ kg m2 s−1, J s MT−2 force: F, weight: Fg kg m s−2, N energy: E, work: W kg m2 s−2, J ML2T−2 torque: τ, moment: M kg m2 s−2, N m energy: E, work: W kg m2 s−2, J MT−3 yank: Y kg m s−3, N s−1 power: P kg m2 s−3, W ML2T−3 rotatum: P kg m2 s−3, N m s−1 power: P kg m2 s |