Definitions

# torque

torque, in physics, that which tends to change the rate of rotation of a body; also called the moment of force. The torque produced by rotating parts of an electric motor or internal-combustion engine is often used as a measure of its ability to do useful work. The magnitude of the torque acting on a body is equal to the product of the force acting on the body and the distance from its point of application to the axis around which the body is free to rotate. Only the component of the force lying in the plane of rotation and perpendicular to the radius from the axis of rotation to the point of application contributes to the torque. This radius is called the moment arm, or lever arm. The net torque acting on a body is always equal to the product of the body's moment of inertia about its axis of rotation and its observed angular acceleration. If a body undergoes no angular acceleration, there is no net torque acting on it. Units of torque are units of force multiplied by units of distance, e.g., newton-meters, dyne-centimeters, and foot-pounds (or pound-feet).
or moment

In physics, the tendency of a force to rotate the body to which it is applied. Torque is always specified with regard to the axis of rotation. It is equal to the magnitude of the component of the force lying in the plane perpendicular to the axis of rotation, multiplied by the shortest distance between the axis and the direction of the force component. Torque is the force that affects rotational motion; the greater the torque, the greater the change in this motion.

A torque (τ) in physics, also called a moment (of force), is a pseudo-vector that measures the tendency of a force to rotate an object about some axis (center). The magnitude of a torque is defined as the product of a force and the length of the lever arm (radius). Just as a force is a push or a pull, a torque can be thought of as a twist.

The SI unit for torque is the newton meter (N m). In Imperial and U.S. customary units, it is measured in foot pounds (ft·lbf) (also known as 'pound feet') and for smaller measurement of torque: inch pounds (in·lbf) or even inch ounces (in·ozf) . The symbol for torque is τ, the Greek letter tau. .

## History

The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.

## Explanation

The force applied 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 meters from the fulcrum, for example, exerts the same torque as one newton applied six meters from the fulcrum. This assumes the force is in a direction at right angles to the straight lever. The direction of the torque can be determined by using the right hand grip rule: curl the fingers of your right hand the direction of rotation and stick your thumb out so it is aligned with the axis of rotation. Your thumb points in the direction of the torque vector.

Mathematically, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:

$mathbf\left\{tau\right\} = mathbf\left\{r\right\} times mathbf\left\{F\right\}$

where

r is the particle's position vector relative to the fulcrum
F is the force acting on the particle.

The torque on a body determines the rate of change of its angular momentum,

$mathbf\left\{tau\right\}=frac\left\{mathrm\left\{d\right\}mathbf\left\{L\right\}\right\}\left\{mathrm\left\{d\right\}t\right\}$

where

L is the angular momentum vector
t stands for time.

As can be seen from either of these relationships, torque is a vector, which points along the axis of the rotation it would tend to cause.

### Proof of the equivalence of definitions

The definition of angular momentum for a single particle is:

$mathbf\left\{L\right\} = mathbf\left\{r\right\} times mathbf\left\{p\right\}$

where "×" indicates the vector cross product. The time-derivative of this is:

$frac\left\{dmathbf\left\{L\right\}\right\}\left\{dt\right\} = mathbf\left\{r\right\} times frac\left\{dmathbf\left\{p\right\}\right\}\left\{dt\right\} + frac\left\{dmathbf\left\{r\right\}\right\}\left\{dt\right\} times mathbf\left\{p\right\}$

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that:

$frac\left\{dmathbf\left\{L\right\}\right\}\left\{dt\right\} = mathbf\left\{r\right\} times m frac\left\{dmathbf\left\{v\right\}\right\}\left\{dt\right\} + mathbf\left\{v\right\} times mmathbf\left\{v\right\}$

But the cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma, (Newton's 2nd law) we obtain:

$frac\left\{dmathbf\left\{L\right\}\right\}\left\{dt\right\} = mathbf\left\{r\right\} times mathbf\left\{F\right\}$

And by definition, torque τ = r×F.

Note that there is a hidden assumption that mass is constant — this is quite valid in non-relativistic mechanics. Also, total (summed) forces and torques have been used — it perhaps would have been more rigorous to write:

$frac\left\{dmathbf\left\{L\right\}\right\}\left\{dt\right\}$ $= mathbf\left\{tau\right\}_\left\{tot\right\}$
$= sum_\left\{i\right\} mathbf\left\{tau\right\}_i$

## Units

Torque has dimensions of force times distance and the SI unit of torque is the "newton meter" (N m). Even though the order of "newton" and "meter" are mathematically interchangeable, the BIPM (Bureau International des Poids et Mesures) specifies that the order should be N m not m N. N·m is also acceptable.

The joule, which is the SI unit for energy or work, is also defined as 1 N m, but this unit is not used for torque. Since energy can be thought of as the result of "force times distance", energy is always a scalar whereas torque is "force cross distance" and so is a (pseudo) vector-valued quantity. The dimensional equivalence of these units, of course, is not simply a coincidence: a torque of 1 N m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,

$E= tau theta$

where

E is the energy
τ is torque
θ is the angle moved, in radians.

Other non-SI units of torque include "pound-force-feet" or "foot-pounds-force" or "inch-pounds-force" or "ounce-force-inches" or "meter-kilograms-force" or "kilogrammeter" (kgm).

## Extended units in relation with rotation angles

As a consequence of the previous equation, if you introduce the radian (rad) as part of the dimensional units in the SI units system, the torque could be measured using "newton meters per radian" (N m/rad), or "joules per radian" (J/rad), while the energy needed and spent to perform the rotation would be measured simply in "newton meters" or "joules".

In the strict SI system, angles are not given any dimensional unit, because they do not designate physical quantities, despite the fact that they are measurable indirectly simply by dividing two distances (the arc length and the radius): one way to conciliate the two systems would be to say that arc lengths are not measures of distances (given they are not measured over a straight line, and a full circle rotation returns to the same position, i.e. a null distance). So arc lengths should be measured in "radian meter" (rad·m), differently from straight segment lengths in "meters" (m). In such extended SI system, the perimeter of a circle whose radius is one meter, will be two pi rad·m, and not just two pi meters.

If you apply this measure to a rotating wheel in contact with a plane surface, the center of the wheel will move across a distance measured in meters with the same value, only if the contact is efficient and the wheel does not slide on it: this does not happen in practice, unless the surface of contact is constrained and is then not perfectly plane (and can resist to the horizontal linear forces applied to the irregularities of the pseudo-plane surface of movement and to the surface of the pseudo-circular rotating wheel); but then the system generates friction that loses some energy spent by the engine: this lost energy does not change the measurement of the torque or the total energy spent in the system but the effective distance that has been made by the center of the wheel.

The difference between the efficient energy spent by the engine and the energy produced in the linear movement is lost in friction and sliding, and this explains why, when applying the same non-null torque constantly to the wheel, so that the wheel moves at a constant speed according to the surface in contact, there may be no acceleration of the center of the wheel: in that case, the energy spent will be directly proportional to the distance made by the center of the wheel, and equal to the energy lost in the system by friction and sliding.

For this reason, when measuring the effective power produced by a rotating engine and the energy spent in the system to generate a movement, you will often need to take into account the angle of rotation, and then, adding the radian in the unit system is necessary as well as making a difference between the measurement of arcs (in radian meter) and the measurement of straight segment distances (in meters), as a way to effectively compute the efficiency of the mobile system and the capacity of a motor engine to convert between rotational power (in radian watt) and linear power (in watts): in a friction-free ideal system, the two measurements would have equal value, but this does not happen in practice, each conversion losing energy in friction (it's easier to limit all losses of energy caused by sliding, by introducing mechanical constraints of forms on the surfaces of contacts).

Depending on works, the extended units including radians as a fundamental dimension may or may not be used.

## Special cases and other facts

### Moment arm formula

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

$|tau| = \left(textrm\left\{moment arm\right\}\right) cdot textrm\left\{force\right\}$

The construction of the "moment arm" is shown in the figure below, 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:

$|tau| = \left(textrm\left\{distance to center\right\}\right) cdot textrm\left\{force\right\}$

For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.

### Force at an angle

If a force of magnitude F is at an angle θ from the displacement arm of length r (and within the plane perpendicular to the rotation axis), then from the definition of cross product, the magnitude of the torque arising is:

$tau=rF sintheta$

### 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, we use three equations.

### Torque as a function of time

Torque is the time-derivative of angular momentum, just as force is the time derivative of linear momentum:

$boldsymbol\left\{tau\right\} =\left\{mathrm\left\{d\right\}mathbf\left\{L\right\} over mathrm\left\{d\right\}t\right\} ,!$

where

L is angular momentum.

Angular momentum on a rigid body can be written in terms of its moment of inertia $boldsymbol I ,!$ and its angular velocity $boldsymbol\left\{omega\right\}$:

$mathbf\left\{L\right\}=I,boldsymbol\left\{omega\right\} ,!$

so if $boldsymbol I ,!$ is constant,

$boldsymbol\left\{tau\right\}=I\left\{mathrm\left\{d\right\}boldsymbol\left\{omega\right\} over mathrm\left\{d\right\}t\right\}=Iboldsymbol\left\{alpha\right\} ,!$

where α is angular acceleration, a quantity usually measured in radians per second squared.

## Machine torque

Torque is 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). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve usually occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.

Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is typically a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics.

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). Therefore, these types of engines usually have quite different types of drivetrains from internal combustion engines.

Torque is also the easiest way to explain mechanical advantage in just about every simple machine.

## 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. Power is the work per unit time. However, time and rotational distance are related by the angular speed where each revolution results in the circumference of the circle being travelled by the force that is generating the torque. The power injected by the applied torque may be calculated as:

$mbox\left\{Power\right\}=mbox\left\{torque\right\} cdot mbox\left\{angular speed\right\} ,$

On the right hand side, this is a scalar product of two vectors, giving a scalar on the left hand side of the equation. Mathematically, the equation may be rearranged to compute torque for a given 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 power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.

Consistent units must be used. For metric SI units power is watts, torque is newton meters and angular speed is radians per second (not rpm and not revolutions per second).

Also, the unit newton meter 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

For different units of power, torque, or angular speed, a conversion factor must be inserted into the equation. Also, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), a conversion factor of $2 pi$ must be added because there are $2 pi$ radians in a revolution:

$mbox\left\{Power\right\} = mbox\left\{torque\right\} times 2 pi times mbox\left\{rotational speed\right\} ,$,

where rotational speed is in revolutions per unit time.

Useful formula in SI units:

$mbox\left\{Power \left(kW\right)\right\} = frac\left\{ mbox\left\{torque \left(N\right\}cdotmbox\left\{m\right)\right\} times 2 pi times mbox\left\{rotational speed \left(rpm\right)\right\}\right\} \left\{60000\right\}$

where 60,000 comes from 60 seconds per minute times 1000 watts per kilowatt.

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

$mbox\left\{Power \left(hp\right)\right\} = frac\left\{ mbox\left\{torque\left(lbf\right\}cdotmbox\left\{ft\right)\right\} times 2 pi times mbox\left\{rotational speed \left(rpm\right)\right\} \right\}\left\{33000\right\}.$

The constant below in, ft·lbf./min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.

Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.

### Derivation

For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius x angular speed. By definition, linear distance=linear speed x time=radius x angular speed x time.

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

$mbox\left\{power\right\} = frac\left\{mbox\left\{force\right\} times mbox\left\{linear distance\right\}\right\}\left\{mbox\left\{time\right\}\right\}=frac\left\{left\left(frac\left\{mbox\left\{torque\right\}\right\}\left\{r\right\}right\right) times \left(r times mbox\left\{angular speed\right\} times t\right)\right\} \left\{t\right\} = mbox\left\{torque\right\} times mbox\left\{angular speed\right\}$

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 $2 pi$ in the above derivation to give:

$mbox\left\{power\right\}=mbox\left\{torque\right\} times 2 pi times mbox\left\{rotational speed\right\} ,$

If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft·lbf/min per horsepower:

$mbox\left\{power\right\} = mbox\left\{torque \right\} times 2 pi times mbox\left\{ rotational speed\right\} cdot frac\left\{mbox\left\{ft\right\}cdotmbox\left\{lbf\right\}\right\}\left\{mbox\left\{min\right\}\right\} times frac\left\{mbox\left\{horsepower\right\}\right\}\left\{33000 cdot frac\left\{mbox\left\{ft \right\}cdotmbox\left\{ lbf\right\}\right\}\left\{mbox\left\{min\right\}\right\} \right\} approx frac \left\{mbox\left\{torque\right\} times mbox\left\{RPM\right\}\right\}\left\{5252\right\}$

because $5252.113122... = frac \left\{33,000\right\} \left\{2 pi\right\} ,$.