Torque

 ( Mechanical Quantities ) 

In physics and mechanics, torque is the rotational analogue of linear force.Serway, R. A. and Jewett, J. W. Jr. (2003). Physics for Scientists and Engineers. 6th ed. Brooks Cole. . It is also referred to as the moment of force (also abbreviated to moment). The symbol for torque is typically $\backslash boldsymbol\backslash tau$, the lowercase Greek alphabet tau. When being referred to as moment of force, it is commonly denoted by . Just as a linear force is a push or a pull applied to a body, a torque can be thought of as a twist applied to an object with respect to a chosen point; for example, driving a screw uses torque to force it into an object, which is applied by the screwdriver rotating around its axis to the drives on the head.

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Historical terminology The term torque (from Latin , 'to twist') is said to have been suggested by James Thomson and appeared in print in April, 1884. Usage is attested the same year by Silvanus P. Thompson in the first edition of Dynamo-Electric Machinery. Thompson describes his usage of the term as follows:

Today, 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. Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link

In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment.Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download . This terminology can be traced back to at least 1811 in Siméon Denis Poisson's Traité de mécanique. An English translation of Poisson's work appears in 1842.

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Definition and relation to other physical quantities A force applied perpendicularly to a lever multiplied by its distance from the Lever (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 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. 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.

(2025). 9780134143323, Pearson. ISBN 9780134143323

The magnitude of torque applied to a rigid body depends on three quantities: the force applied, the lever arm vector

Paul Tipler (2025). 9780716708094, W. H. Freeman. ISBN 9780716708094

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:

$$\backslash boldsymbol\{\backslash tau\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\; \backslash implies\; \backslash tau\; =\; rF\_\{\backslash perp\}\; =\; rF\backslash sin\backslash theta$$

where

• $\backslash boldsymbol\backslash tau$ is the torque vector and $\backslash tau$ is the magnitude of the torque,
• $\backslash mathbf\{r\}$ 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,
• $\backslash mathbf\{F\}$ is the force vector, F is the magnitude of the force vector and F_⊥ is the amount of force directed perpendicularly to the position of the particle,
• $\backslash times$ denotes the cross product, which produces a vector that is perpendicular both to and to following the right-hand rule,
• $\backslash theta$ is the angle between the force vector and the lever arm vector.

The SI units for torque is the newton-metre (N⋅m). For more on the units of torque, see .

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

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

where L is the angular momentum vector and t is time. For the motion of a point particle,

$$\backslash mathbf\{L\}\; =\; I\backslash boldsymbol\{\backslash omega\},$$

where $I\; =\; mr^2$ is the moment of inertia and ω is the orbital angular velocity pseudovector. It follows that

$$\backslash boldsymbol\{\backslash tau\}\_\{\backslash mathrm\{net\}\}\; =\; I\_1\backslash dot\{\backslash omega\_1\}\backslash hat\{\backslash boldsymbol\{e\_1\}\}\; +\; I\_2\backslash dot\{\backslash omega\_2\}\backslash hat\{\backslash boldsymbol\{e\_2\}\}\; +\; I\_3\backslash dot\{\backslash omega\_3\}\backslash hat\{\backslash boldsymbol\{e\_3\}\}\; +\; I\_1\backslash omega\_1\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_1\}\}\}\{dt\}\; +\; I\_2\backslash omega\_2\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_2\}\}\}\{dt\}\; +\; I\_3\backslash omega\_3\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_3\}\}\}\{dt\}\; =\; I\backslash boldsymbol\backslash dot\{\backslash omega\}\; +\; \backslash boldsymbol\backslash omega\; \backslash times\; (I\backslash boldsymbol\backslash omega)$$

using the derivative of a Unit vector is$$\{d\backslash boldsymbol\{\backslash hat\{e\_i\}\}\; \backslash over\; dt\}\; =\; \backslash boldsymbol\backslash omega\; \backslash times\; \backslash boldsymbol\{\backslash hat\{e\_i\}\}$$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$$\backslash boldsymbol\{\backslash tau\}\; =\; I\backslash boldsymbol\{\backslash alpha\}$$where $\backslash boldsymbol\backslash alpha\; =\; \backslash dot\backslash boldsymbol\backslash omega$.

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Proof of the equivalence of definitions The definition of angular momentum for a single point particle is: $$\backslash mathbf\{L\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{p\}$$ where p is the particle's linear momentum and r is the position vector from the origin. The time-derivative of this is:

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

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 $\backslash mathbf\{F\}$ and the rate of change of position is velocity $\backslash mathbf\{v\}$,

$$\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\; +\; \backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{p\}$$

The cross product of momentum $\backslash mathbf\{p\}$ with its associated velocity $\backslash mathbf\{v\}$ is zero because velocity and momentum are parallel, so the second term vanishes. Therefore, torque on a particle is equal to the first derivative of its angular momentum with respect to time. If multiple forces are applied, according Newton's second law it follows that$$\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\_\{\backslash mathrm\{net\}\}\; =\; \backslash boldsymbol\{\backslash tau\}\_\{\backslash mathrm\{net\}\}.$$

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 integrating over the entire mass.

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Derivatives of torque In physics, rotatum is the derivative of torque with respect to time

$\backslash mathbf\; P\; =\; \backslash frac\{\backslash mathrm\; d\; \backslash boldsymbol\; \backslash tau\}\{\backslash mathrm\; d\; t\},$

where τ is torque.

This word is derived from the Latin word 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:

$\backslash mathbf\{P\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash frac\{\backslash mathrm\; d\; \backslash mathbf\{F\}\}\{\backslash mathrm\; d\; t\}\; +\; \backslash frac\{\backslash mathrm\; d\; \backslash mathbf\{r\}\}\{\backslash mathrm\; d\; t\}\; \backslash times\; \backslash mathbf\{F\}.$

Because the rate of change of force is yank $\backslash mathbf\{Y\}$ and the rate of change of position is velocity $\backslash mathbf\{v\}$, the expression can be further simplified to:

$\backslash mathbf\{P\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{Y\}\; +\; \backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{F\}.$

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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

$$W\; =\; \backslash int\_\{\backslash theta\_1\}^\{\backslash theta\_2\}\; \backslash tau\backslash \; \backslash mathrm\{d\}\backslash theta,$$

where τ is torque, and θ₁ and θ₂ represent (respectively) the initial and final of the body.

It follows from the work–energy principle that W also represents the change in the rotational kinetic energy E_r of the body, given by

$$E\_\{\backslash mathrm\{r\}\}\; =\; \backslash tfrac\{1\}\{2\}I\backslash omega^2,$$

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

Power is the work per unit time, given by

$$P\; =\; \backslash boldsymbol\{\backslash tau\}\; \backslash cdot\; \backslash boldsymbol\{\backslash omega\},$$

where P is power, τ is torque, ω is the angular velocity, and $\backslash cdot$ 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).

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

$$W\; =\; \backslash int\_\{s\_1\}^\{s\_2\}\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash mathbf\{s\}$$

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

$$\backslash mathrm\{d\}\backslash mathbf\{s\}\; =\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{r\}$$

Substitution in the above expression for work, , gives $$W\; =\; \backslash int\_\{s\_1\}^\{s\_2\}\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\; \backslash times\; \backslash mathbf\{r\}$$

The expression inside the integral is a scalar triple product $\backslash mathbf\{F\}\backslash cdot\backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{r\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$, 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

$$W\; =\; \backslash int\_\{\backslash theta\; \_1\}^\{\backslash theta\; \_2\}\; \backslash boldsymbol\{\backslash tau\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$$

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

$$W\; =\; \backslash int\_\{\backslash theta\; \_1\}^\{\backslash theta\; \_2\}\; \backslash tau\; \backslash ,\; \backslash mathrm\{d\}\backslash theta$$

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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 resultant torques due to several forces applied to about a point is equal to the sum of the contributing torques:

$$\backslash tau\; =\; \backslash mathbf\{r\}\_1\backslash times\backslash mathbf\{F\}\_1\; +\; \backslash mathbf\{r\}\_2\backslash times\backslash mathbf\{F\}\_2\; +\; \backslash ldots\; +\; \backslash mathbf\{r\}\_N\backslash times\backslash mathbf\{F\}\_N.$$

From this it follows that the torques resulting from N number of forces acting around a pivot on an object are balanced when

$$\backslash mathbf\{r\}\_1\backslash times\backslash mathbf\{F\}\_1\; +\; \backslash mathbf\{r\}\_2\backslash times\backslash mathbf\{F\}\_2\; +\; \backslash ldots\; +\; \backslash mathbf\{r\}\_N\backslash times\backslash mathbf\{F\}\_N\; =\; \backslash mathbf\{0\}.$$

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Units Torque has the dimension of force times distance, symbolically and those fundamental dimensions are the same as that for energy or mechanical work. Official SI literature indicates newton-metre, is properly denoted N⋅m, as the unit for torque; although this is dimensionally equivalent to the joule, which is not used for torque.From the official SI website , The International System of Units – 9th edition – Text in English Section 2.3.4: "For example, the quantity torque is the cross product of a position vector and a force vector. The SI unit is newton-metre. Even though torque has the same dimension as energy (SI unit joule), the joule is never used for expressing torque." 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 , which treats the radian as a base unit rather than as a dimensionless unit.

The traditional imperial units for torque are the pound foot (lbf-ft), or, for small values, the pound inch (lbf-in). In the US, torque is most commonly referred to as the foot-pound (denoted as either lb-ft or ft-lb) and the inch-pound (denoted as in-lb). Demonstration that, as in most US industrial settings, the torque ranges are given in ft-lb rather than lbf-ft.

Practitioners depend on context and the hyphen in the abbreviation to know that these refer to torque and not to energy or moment of mass (as the symbolism ft-lb would properly imply).

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Conversion to other units A conversion factor may be necessary when using different units of power or torque. For example, if rotational speed (unit: revolution per minute or second) is used in place of angular speed (unit: radian per second), we must multiply by 2 radians per revolution. In the following formulas, P is power, τ is torque, and ν (Greek letter nu) is rotational speed.

$$P\; =\; \backslash tau\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; \backslash nu$$

Showing units:

$$P\; \_\{\backslash rm\; W\}\; =\; \backslash tau\; \_\{\backslash rm\; N\; \{\backslash cdot\}\; m\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/s\}$$

Dividing by 60 seconds per minute gives us the following.

$$P\; \_\{\backslash rm\; W\}\; =\; \backslash frac\{\; \backslash tau\; \_\{\backslash rm\; N\; \{\backslash cdot\}\; m\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/min\}\; \}\; \{\backslash rm\; 60\; ~s/min\}$$

where rotational speed is in revolutions per minute (rpm, rev/min).

Some people (e.g., American automotive engineers) use horsepower (mechanical) for power, foot-pounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to:

$$P\; \_\{\backslash rm\; hp\}\; =\; \backslash frac\{\; \backslash tau\; \_\{\backslash rm\; lbf\; \{\backslash cdot\}\; ft\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/min\}\}\; \{33,000\}.$$

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.

The use of other units (e.g., BTU per hour for power) would require a different custom conversion factor.

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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:

The radius r and time t have dropped out of the equation. However, angular speed must be in radians per unit of time, 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 2 in the above derivation to give:

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

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:

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

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Special cases and other facts

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

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

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:

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

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 acting 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.

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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: and , and the torque a third equation: . That is, to solve statically determinate equilibrium problems in two-dimensions, three equations are used.

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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 the point of reference. If the net force $\backslash mathbf\{F\}$ is not zero, and $\backslash boldsymbol\{\backslash tau\}\_1$ is the torque measured from $\backslash mathbf\{r\}\_1$, then the torque measured from $\backslash mathbf\{r\}\_2$ is $$\backslash boldsymbol\{\backslash tau\}\_2\; =\; \backslash boldsymbol\{\backslash tau\}\_1\; +\; (\backslash mathbf\{r\}\_2\; -\; \backslash mathbf\{r\}\_1)\; \backslash times\; \backslash mathbf\{F\}$$

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Machine torque Torque forms part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by the angular speed of the drive shaft. Internal-combustion engines produce useful torque only over a limited range of (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. and 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.

In practice, the relationship between power and torque can be observed in : Bicycles are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a bicycle chain, and a derailleur gears if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multi-speed bicycle), all of which attached to the bicycle 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 angular speed (i.e. the number of pedal revolutions per minute times 2 π) and the torque at the Axle 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, angular speed)_input pair is converted to a (torque, angular speed)_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.

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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.

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See also

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External links

• "Horsepower and Torque" An article showing how power, torque, and gearing affect a vehicle's performance.
• Torque and Angular Momentum in Circular Motion on Project PHYSNET.
• An interactive simulation of torque
• Torque Unit Converter
• A feel for torque An order-of-magnitude interactive.

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