# Torque

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Torque
Relationship between force F, torque τ, linear momentum p, and angular momentum L in a system which has rotation constrained to only one plane (forces and moments due to gravity and friction not considered).
Common symbols
${\displaystyle \tau }$, M
SI unitN⋅m
Other units
pound-force-feet, lbf⋅inch, ozf⋅in
In SI base unitskg⋅m2⋅s−2
DimensionM L2T−2

Torque, moment, moment of force, rotational force or "turning effect" is the rotational equivalent of linear force.[1] The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation. The symbol for torque is typically ${\displaystyle {\boldsymbol {\tau }}}$, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.

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 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:

${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \,\!}$
${\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!}$

where

${\displaystyle {\boldsymbol {\tau }}}$ is the torque vector and ${\displaystyle \tau }$ 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 produces a vector that is perpendicular to both r and F following the right-hand rule,
${\displaystyle \theta }$ is the angle between the force vector and the lever arm vector.

The SI unit for torque is N⋅m. For more on the units of torque, see Units.

## Defining terminology

James Thomson, the brother of Lord Kelvin, introduced the term torque into English scientific literature in 1884.[3] However, torque is referred to using different vocabulary depending on geographical location and field of study. This article follows the definition used in US physics in its usage of the word torque.[4] In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment.[5] In US physics[4] and UK physics terminology these terms are interchangeable, unlike in US mechanical engineering, where the term torque is used for the closely related "resultant moment of a couple".[5]

Torque is defined mathematically as the rate of change of angular momentum of an object. The definition of torque states that one or both of the angular velocity or the moment of inertia of an object are changing. Moment is the general term used for the tendency of one or more applied forces to rotate an object about an axis, but not necessarily to change the angular momentum of the object (the concept which is called torque in physics).[5] For example, a rotational force applied to a shaft causing acceleration, such as a drill bit accelerating from rest, results in a moment called a torque. By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the angular momentum of the beam is not changing, this bending moment is not called a torque. Similarly with any force couple on an object that has no change to its angular momentum, such moment is also not called a torque.

## Definition and relation to angular momentum

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 perpendicularly 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 point particle (which has the position r in some reference frame) can be defined as the cross product:

${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} ,}$

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

${\displaystyle \tau =rF\sin \theta ,\!}$

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,

${\displaystyle \tau =rF_{\perp },}$

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. Conversely, the torque vector defines the plane in which the position and force vectors lie. The resulting torque vector direction is determined by the right-hand rule.[7]

The net torque on a body determines the rate of change of the body's angular momentum,

${\displaystyle {\boldsymbol {\tau }}={\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}}$

where L is the angular momentum vector and t is time.

For the motion of a point particle,

${\displaystyle \mathbf {L} =I{\boldsymbol {\omega }},}$

where I is the moment of inertia and ω is the orbital angular velocity pseudovector. It follows that

${\displaystyle {\boldsymbol {\tau }}_{\mathrm {net} }={\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}={\frac {\mathrm {d} (I{\boldsymbol {\omega }})}{\mathrm {d} t}}=I{\frac {\mathrm {d} {\boldsymbol {\omega }}}{\mathrm {d} t}}+{\frac {\mathrm {d} I}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }}+{\frac {\mathrm {d} (mr^{2})}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }}+2rp_{||}{\boldsymbol {\omega }},}$

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 Newton's Second Law for point particles, and is valid for any type of trajectory. Note that although force and acceleration are always parallel and directly proportional, the torque τ need not be parallel or directly proportional to the angular acceleration α. This arises from the fact that although mass is always conserved, the moment of inertia in general is not.

### Proof of the equivalence of definitions

The definition of angular momentum for a single point particle is:

${\displaystyle \mathbf {L} =\mathbf {r} \times {\boldsymbol {p}}}$

where p is the particle's linear momentum and r is the position vector from the origin. The time-derivative of this is:

${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}+{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times {\boldsymbol {p}}.}$

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force ${\displaystyle \mathbf {F} ={\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}}$ (whether or not mass is constant) and the definition of velocity ${\displaystyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {v} }$

${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} +\mathbf {v} \times {\boldsymbol {p}}.}$

The cross product of momentum ${\displaystyle {\boldsymbol {p}}}$ with its associated velocity ${\displaystyle \mathbf {v} }$ 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

${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} _{\mathrm {net} }={\boldsymbol {\tau }}_{\mathrm {net} }.}$

This is a general proof for point particles.

The proof 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 integrating over the entire mass.

## Units

Torque has the dimension of force times distance, symbolically L2MT−2. Official SI literature suggests using the unit newton metre (N⋅m), or, joule per radian (J/rad).[9] The unit newton metre is properly denoted N⋅m rather than m·N to avoid confusion with mN, millinewtons.[10]

The SI unit for energy or work is the joule. The presence of "joule" in the unit name joules per radian for torque is not a coincidence: a torque of 1 N⋅m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,

${\displaystyle E=\tau \theta \ }$

where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This explains the physical significance of the name joule per radian.[9]

The traditional Imperial units for torque are pound feet (lb-ft) or for small values inch pound (in-lb).

## Special cases and other facts

### Moment arm formula

Moment arm diagram

A very useful special case, often given as the definition of torque in fields other than physics, is as follows:

${\displaystyle \tau =({\text{moment arm}})({\text{force}}).}$

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:

${\displaystyle \tau =({\text{distance to centre}})({\text{force}}).}$

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

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

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 ${\displaystyle \mathbf {F} }$ is not zero, and ${\displaystyle {\boldsymbol {\tau }}_{1}}$ is the torque measured from ${\displaystyle \mathbf {r} _{1}}$, then the torque measured from ${\displaystyle \mathbf {r} _{2}}$ is … ${\displaystyle {\boldsymbol {\tau }}_{2}={\boldsymbol {\tau }}_{1}+(\mathbf {r} _{1}-\mathbf {r} _{2})\times \mathbf {F} }$

## Machine torque

Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis shows the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in newton metres) that the engine is capable of providing at that speed.

Torque forms part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internal-combustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). One can measure the varying torque output over that range with a dynamometer, and show it as a torque curve.

Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam-engines and electric motors can start heavy loads from zero rpm without a clutch.

## Relationship between torque, power, and energy

If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass, the work W can be expressed as

${\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \ \mathrm {d} \theta ,}$

where τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[11]

### Proof

The work done by a variable force acting over a finite linear displacement ${\displaystyle s}$ is given by integrating the force with respect to an elemental linear displacement ${\displaystyle \mathrm {d} {\vec {s}}}$

${\displaystyle W=\int _{s_{1}}^{s_{2}}{\vec {F}}\cdot \mathrm {d} {\vec {s}}}$

However, the infinitesimal linear displacement ${\displaystyle \mathrm {d} {\vec {s}}}$ is related to a corresponding angular displacement ${\displaystyle \mathrm {d} {\vec {\theta }}}$ and the radius vector ${\displaystyle {\vec {r}}}$ as

${\displaystyle \mathrm {d} {\vec {s}}=\mathrm {d} {\vec {\theta }}\times {\vec {r}}}$

Substitution in the above expression for work gives

${\displaystyle W=\int _{s_{1}}^{s_{2}}{\vec {F}}\cdot \mathrm {d} {\vec {\theta }}\times {\vec {r}}}$

The expression ${\displaystyle {\vec {F}}\cdot \mathrm {d} {\vec {\theta }}\times {\vec {r}}}$ is a scalar triple product given by ${\displaystyle \left[{\vec {F}}\,\mathrm {d} {\vec {\theta }}\,{\vec {r}}\right]}$. An alternate expression for the same scalar triple product is

${\displaystyle \left[{\vec {F}}\,\mathrm {d} {\vec {\theta }}\,{\vec {r}}\right]={\vec {r}}\times {\vec {F}}\cdot \mathrm {d} {\vec {\theta }}}$

But as per the definition of torque,

${\displaystyle {\vec {\tau }}={\vec {r}}\times {\vec {F}}}$

Corresponding substitution in the expression of work gives,

${\displaystyle W=\int _{s_{1}}^{s_{2}}{\vec {\tau }}\cdot \mathrm {d} {\vec {\theta }}}$

Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving

${\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}{\vec {\tau }}\cdot \mathrm {d} {\vec {\theta }}}$

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

${\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \,\mathrm {d} \theta }$

It follows from the work-energy theorem that W also represents the change in the rotational kinetic energy Er of the body, given by

${\displaystyle E_{\mathrm {r} }={\tfrac {1}{2}}I\omega ^{2},}$

where I is the moment of inertia of the body and ω is its angular speed.[11]

Power is the work per unit time, given by

${\displaystyle P={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}$

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. This means that the dimensional equivalence of the newton metre and the joule may be applied in the former, but not in the latter case. This problem is addressed in orientational analysis which treats radians as a base unit rather than a dimensionless unit.[12]

### Conversion to other units

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.

${\displaystyle P=\tau \cdot 2\pi \cdot \nu }$

Showing units:

${\displaystyle P({\rm {W}})=\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}$

Dividing by 60 seconds per minute gives us the following.

${\displaystyle P({\rm {W}})={\frac {\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}{60}}}$

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:

${\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{33,000}}.}$

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., BTU per hour for power) would require a different custom conversion factor.

### Derivation

For a rotating object, the linear distance covered at the circumference of rotation is the product of the radius with the angle covered. That is: linear distance = radius × angular distance. And by definition, linear distance = linear speed × time = radius × angular speed × time.

By the definition of torque: torque = radius × force. We can rearrange this to determine force = torque ÷ radius. These two values can be substituted into the definition of power:

{\displaystyle {\begin{aligned}{\text{power}}&={\frac {{\text{force}}\cdot {\text{linear distance}}}{\text{time}}}\\[6pt]&={\frac {\left({\dfrac {\text{torque}}{r}}\right)\cdot (r\cdot {\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:

${\displaystyle {\text{power}}={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}.\,}$

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:

{\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text{RPM}}}{5,252}}\end{aligned}}}

because ${\displaystyle 5252.113122\approx {\frac {33,000}{2\pi }}.\,}$

## Principle of moments

The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:

${\displaystyle (\mathbf {r} \times \mathbf {F} _{1})+(\mathbf {r} \times \mathbf {F} _{2})+\cdots =\mathbf {r} \times (\mathbf {F} _{1}+\mathbf {F} _{2}+\cdots ).}$

## Torque multiplier

Torque can be multiplied via three methods: by locating the fulcrum such that the length of a lever is increased; by using a longer lever; or by the use of a speed reducing gearset or gear box. Such a mechanism multiplies torque, as rotation rate is reduced.

## References

1. ^ Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. ISBN 0-534-40842-7.
2. ^ Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.
3. ^ Thomson, James; Larmor, Joseph (1912). Collected Papers in Physics and Engineering. University Press. p. civ., at Google books
4. ^ a b Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link
5. ^ a b c Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download.
6. ^ "Right Hand Rule for Torque". Retrieved 2007-09-08.
7. ^ a b Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons, Inc. pp. 184–85.
8. ^ 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.
9. ^ a b 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."
10. ^ "SI brochure Ed. 8, Section 5.1". Bureau International des Poids et Mesures. 2014. Retrieved 2018-07-06.
11. ^ a b Kleppner, Daniel; Kolenkow, Robert (1973). An Introduction to Mechanics. McGraw-Hill. pp. 267–68.
12. ^ Page, Chester H. (1979), "Rebuttal to de Boer's "Group properties of quantities and units"", American Journal of Physics, 47 (9): 820, doi:10.1119/1.11704