Torque

Torque
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 unit	N⋅m
Other units	pound-force-feet, lbf⋅inch, ozf⋅in
In SI base units	kg⋅m2⋅s−2
Dimension	${\displaystyle {\mathsf {M}}{\mathsf {L}}^{2}{\mathsf {T}}^{-2}}$

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In

It is also referred to as the moment of force (also abbreviated to moment). The symbol for torque is typically ${\displaystyle {\boldsymbol {\tau }}}$, the lowercase from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum.

History

The concept originated with the studies by Archimedes of the usage of levers, which is reflected in his famous quote: "Give me a lever and a place to stand and I will move the Earth". The term torque (from Latin torquēre, 'to twist') is said to have been suggested by James Thomson and appeared in print in April, 1884.^[2][3][4] Usage is attested the same year by Silvanus P. Thompson in the first edition of Dynamo-Electric Machinery.^[4] Thompson motivates the term as follows:^[3]

Just as the Newtonian definition of force is that which produces or tends to produce motion (along a line), so torque may be defined as that which produces or tends to produce torsion (around an axis). It is better to use a term which treats this action as a single definite entity than to use terms like "couple" and "moment", which suggest more complex ideas. The single notion of a twist applied to turn a shaft is better than the more complex notion of applying a linear force (or a pair of forces) with a certain leverage.

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.^[5]

In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment.^[6] This terminology can be traced back to at least 1811 in Siméon Denis Poisson's Traité de mécanique.^[7] An English translation of Poisson's work appears in 1842.

Definition and relation to other physical quantities

A force applied perpendicularly to a lever multiplied by its distance from the

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:

$${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \implies \tau =rF_{\perp }=rF\sin \theta }$$

where

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

The

.

Relationship with the angular momentum

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

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

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

using the derivative of a vector is

This equation is the rotational analogue of 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 bewhere ${\displaystyle {\boldsymbol {\alpha }}={\dot {\boldsymbol {\omega }}}}$.

Proof of the equivalence of definitions

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

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

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

The cross product of momentum ${\displaystyle \mathbf {p} }$ with its associated velocity ${\displaystyle \mathbf {v} }$ is zero because velocity and momentum are parallel, so the second term vanishes. Therefore, torque on a particle is equal to the

it follows that

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

over the entire mass.

Derivatives of torque

In physics, rotatum is the derivative of torque with respect to time^[12]

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

where τ is torque.

This word is derived from the Latin word rotātus meaning 'to rotate', but the term rotatum is not universally recognized but is commonly used. There isn't an universally accepted lexicon to indicate the successive derivatives of rotatum, even if sometimes various proposals have been made.

Relationship with power and energy

The law of

, the work W can be expressed as

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

It follows from the

E_r of the body, given by

where I is the

Power is the work per unit time, given by

where P is power, τ is torque, ω is the angular velocity, and ${\displaystyle \cdot }$ represents the

.

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).

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 ${\displaystyle \mathrm {d} \mathbf {s} }$

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

Substitution in the above expression for work, , gives

The expression inside the integral is a

${\displaystyle \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} =\mathbf {r} \times \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}}$, 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

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

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:

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

Units

Torque has the

The traditional imperial and U.S. customary 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).^[17][18] 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).

Conversion to other units

A conversion factor may be necessary when using different units of power or torque. For example, if

) is rotational speed.

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

Showing units:

$${\displaystyle P_{\rm {W}}=\tau _{\rm {N{\cdot }m}}\cdot 2\pi _{\rm {rad/rev}}\cdot \nu _{\rm {rev/s}}}$$

Dividing by 60 seconds per minute gives us the following.

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

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:

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.,

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:

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

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:

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:

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.

precess

.

Static equilibrium

For an object to be in

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 the 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

Machine torque

rotational 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

.

In practice, the relationship between power and torque can be observed in

of the road wheels is decreased while the torque is increased, product of which (i.e. power) does not change.

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

. Such a mechanism multiplies torque, as rotation rate is reduced.

See also

References

External links

Look up torque in Wiktionary, the free dictionary.

Wikimedia Commons has media related to Torque.

• "Horsepower and Torque" Archived 2007-03-28 at the Wayback Machine 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 Archived 2021-05-08 at the Wayback Machine An order-of-magnitude interactive.