Torque, moment, moment of force or "turning effect" is the rotational equivalent of linear force.Serway, R. A. and Jewett, Jr. J.W. (2003). Physics for Scientists and Engineers. 6th Ed. Brooks Cole. . The concept originated with the studies by Archimedes of the usage of . 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 $\backslash boldsymbol\backslash tau$, the lowercase Greek alphabet 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 (Euclidean vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector
connecting the origin to the point of force application, and the angle between the force and lever arm vectors. In symbols:
The SI units for torque is N⋅m. For more on the units of torque, see Units.
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 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). Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download. 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.
More generally, the torque on a point particle (which has the position r in some reference frame) can be defined as the cross product:
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 righthand rule.
The net torque on a body determines the rate of change of the body's angular momentum,
For the motion of a point particle,
where p is the particle's linear momentum and r is the position vector from the origin. The timederivative of this is:
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force $\backslash mathbf\{F\}=\backslash frac\{\backslash mathrm\{d\}\backslash boldsymbol\{p\}\}\{\backslash mathrm\{d\}t\}$ (whether or not mass is constant) and the definition of velocity $\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{r\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{v\}$
The cross product of momentum $\backslash boldsymbol\{p\}$ with its associated velocity $\backslash mathbf\{v\}$ 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 , and it follows that
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.
The SI unit for energy or mechanical 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,
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in ). This explains the physical significance of the name joule per radian.
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 threedimensional 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:
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.
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 steamengines and electric motors can start heavy loads from zero rpm without a clutch.
However, the infinitesimal linear displacement $\backslash mathrm\{d\}\backslash vec\{s\}$ is related to a corresponding angular displacement $\backslash mathrm\{d\}\backslash vec\{\backslash theta\}$ and the radius vector $\backslash vec\{r\}$ as
Substitution in the above expression for work gives
The expression $\backslash vec\{F\}\backslash cdot\backslash mathrm\{d\}\backslash vec\{\backslash theta\}\backslash times\backslash vec\{r\}$ is a scalar triple product given by $\backslash left\backslash vec\{F\}\backslash ,\backslash mathrm\{d\}\backslash vec\{\backslash theta\}\backslash ,\backslash vec\{r\}\backslash right$. An alternate expression for the same scalar triple product is
But as per the definition of torque,
Corresponding substitution in the expression of work gives,
Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving
If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., $\backslash vec\{\backslash tau\}\backslash cdot\; \backslash mathrm\{d\}\backslash vec\{\backslash theta\}\; =\; \backslash left\backslash vec\{\backslash tau\}\backslash right\backslash left\; \backslash ,\; \backslash mathrm\{d\}\backslash vec\{\backslash theta\}\backslash right\backslash cos\; 0\; =\; \backslash tau\; \backslash ,\; \backslash mathrm\{d\}\backslash theta$ giving
It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy E_{r} of the body, given by
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
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 are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a circular bicycle chain, and a derailleur gears if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multispeed 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 cadence (i.e. the number of pedal revolutions per minute) and the torque on 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, 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 multispeed 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 , torque is and angular speed is 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 which treats radians as a base unit rather than a dimensionless unit.
Showing units:
Dividing by 60 seconds per minute gives us the following.
where rotational speed is in revolutions per minute (rpm).
Some people (e.g., American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to:
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.
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, 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:
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 poundsforce feet and rotational speed in revolutions per minute, the above equation gives power in foot poundsforce per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:
because $5252.113122\; \backslash approx\; \backslash frac\; \{33,000\}\; \{2\; \backslash pi\}.\; \backslash ,$

