Rotational force

τ

{\displaystyle \tau}

In physics and mechanics, **rotational force** or **torque** is the rotational equivalent of linear force. It is also referred to as the **moment**, **moment of force**, **rotational force** or **turning effect**, depending on the field of study. 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 rotational force can be thought of as a twist to an object around a specific axis. Another definition of rotational force is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for rotational force is typically

τ

{\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 rotational force 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 rotational force of a rigid body depends on three quantities: the force applied, the *lever arm vector* connecting the point about which the rotational force is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:

# τ

r

×

F

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

# τ
‖

r

‖

‖

F

‖
sin
⁡
θ

where

τ

{\displaystyle \boldsymbol{\tau}}

 is the rotational force vector and

τ

{\displaystyle \tau}

 is the magnitude of the rotational force,

r

{\displaystyle \mathbf{r}}

 is the position vector (a vector from the point about which the rotational force is being measured to the point where the force is applied),

F

{\displaystyle \mathbf F }

 is the force vector,

×

{\displaystyle \times}

 denotes the cross product, which produces a vector that is perpendicular to both r and F following the right-hand rule,

θ

{\displaystyle \theta}

 is the angle between the force vector and the lever arm vector.
The SI unit for rotational force is the Newton-metre (N⋅m). For more on the units of rotational force, see Units.

## Defining terminology

See also: Couple (mechanics)
James Thomson, the brother of Lord Kelvin, introduced the term *rotational force* into English scientific literature in 1884. However, rotational force 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 *rotational force*. In the UK and in US mechanical engineering, rotational force is referred to as *moment of force*, usually shortened to *moment*. These terms are interchangeable in US physics and UK physics terminology, unlike in US mechanical engineering, where the term *rotational force* is used for the closely related "resultant moment of a couple".

### Rotational force and moment in the US mechanical engineering terminology

In US mechanical engineering, *rotational force* is defined mathematically as the rate of change of rotational momentum of an object (in physics it is called "net rotational force"). The definition of rotational force states that one or both of the rotational velocity or the rotational mass 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 rotational momentum of the object (the concept which is called *rotational force* in physics).
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 *rotational force*. By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the rotational momentum of the beam is not changing, this bending moment is not called a *rotational force*. Similarly with any force couple on an object that has no change to its rotational momentum, such moment is also not called a *rotational force*.

## Definition and relation to rotational momentum

A force applied perpendicularly to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its rotational force. A force of three newtons applied two metres from the fulcrum, for example, exerts the same rotational force as a force of one newton applied six metres from the fulcrum. The direction of the rotational force 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 rotational force.

More generally, the rotational force on a point particle (which has the position **r** in some reference frame) can be defined as the cross product:

# τ

r

×

F

,

{\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 rotational force is given by

# τ
r
F
sin
⁡
θ
,

{\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,

# τ
r

F

⊥

,

{\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 rotational force.

It follows from the properties of the cross product that the *rotational force vector* is perpendicular to both the *position* and *force* vectors. Conversely, the *rotational force vector* defines the plane in which the *position* and *force* vectors lie. The resulting *rotational force vector* direction is determined by the right-hand rule.

The net rotational force on a body determines the rate of change of the body's rotational momentum,

# τ

d

L

d

t

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

where **L** is the rotational momentum vector and *t* is time.

For the motion of a point particle,

L

=
I
ω
,

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

where *I* is the rotational mass and **ω** is the orbital rotational velocity pseudovector. It follows that

τ

n
e
t

=

d

L

d

t

=

d

(
I
ω
)

d

t

=
I

d

ω

d

t

d

I

d

t

# ω
I
α

d

(
m

r
2

)

d

t

# ω
I
α
2
r

p

ω
,

{\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 rotational 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 rotational force **τ** need not be parallel or directly proportional to the rotational acceleration **α**. This arises from the fact that although mass is always conserved, the rotational mass in general is not.

### Proof of the equivalence of definitions

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

L

=

r

×
p

{\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:

d

L

d

t

=

r

×

d

p

d

t

d

r

d

t

×
p
.

{\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

F

=

d

p

d

t

{\displaystyle \mathbf{F}=\frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t}}

 (whether or not mass is constant) and the definition of velocity

d

r

d

t

=

v

{\displaystyle \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \mathbf{v}}

d

L

d

t

=

r

×

F

v

×
p
.

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

The cross product of momentum

p

{\displaystyle \boldsymbol{p}}

 with its associated velocity

v

{\displaystyle \mathbf{v} }

 is zero because velocity and momentum are parallel, so the second term vanishes.

By definition, rotational force **τ** = **r** × **F**. Therefore, rotational force on a particle is *equal* to the
first derivative of its rotational momentum with respect to time.

If multiple forces are applied, Newton's second law instead reads **F**net = *m***a**, and it follows that

d

L

d

t

=

r

×

F

n
e
t

=

τ

n
e
t

.

{\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

Rotational force has the dimension of force times distance, symbolically L2MT−2. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit *newton metre* (N⋅m) and never the joule. The unit *newton metre* is properly denoted N⋅m.

The traditional Imperial and U.S. customary units for rotational force are the pound foot (lbf-ft), or for small values the pound inch (lbf-in). Confusingly, in US practice rotational force 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). Practitioners depend on context and the hyphen in the abbreviation to know that these refer to rotational force and not to energy or moment of mass (as the symbolism ft-lb would properly imply).

## Special cases and other facts

### Moment arm formula

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

# τ
(
moment arm
)
(
force
)
.

{\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 rotational force 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 rotational force will be a maximum for the given force. The equation for the magnitude of a rotational force, arising from a perpendicular force:

# τ
(
distance to centre
)
(
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 rotational force 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.

### 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 rotational forces (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: Σ*H* = 0 and Σ*V* = 0, and the rotational force a third equation: Σ*τ* = 0. That is, to solve statically determinate equilibrium problems in two-dimensions, three equations are used.

### Net force versus rotational force

When the net force on the system is zero, the rotational force measured from any point in space is the same. For example, the rotational force on a current-carrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force

F

{\displaystyle \mathbf F }

 is not zero, and

τ
1

{\displaystyle \boldsymbol{\tau}\_1}

 is the rotational force measured from

r

1

{\displaystyle \mathbf{r}\_1}

, then the rotational force measured from

r

2

{\displaystyle \mathbf{r}\_2}

 is …

τ
2

=

τ
1

(

r

1

−

r

2

)
×

F

{\displaystyle \boldsymbol{\tau}\_2 = \boldsymbol{\tau}\_1 + (\mathbf{r}\_1 - \mathbf{r}\_2) \times \mathbf{F}}

## Machine rotational force

Rotational force forms part of the basic specification of an engine: the power output of an engine is expressed as its rotational force multiplied by its rotational speed of the axis. Internal-combustion engines produce useful rotational force 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 rotational force output over that range with a dynamometer, and show it as a rotational force curve.

Steam engines and electric motors tend to produce maximum rotational force close to zero rpm, with the rotational force 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 rotational force, power, and energy

If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if rotational force 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

# W

∫

θ
1

θ
2

τ

d

θ
,

{\displaystyle W = \int\_{\theta\_1}^{\theta\_2} \tau\ \mathrm{d}\theta,}

where *τ* is rotational force, and *θ*1 and *θ*2 represent (respectively) the initial and final angular positions of the body.

### Proof

The work done by a variable force acting over a finite linear displacement

s

{\displaystyle s}

 is given by integrating the force with respect to an elemental linear displacement

d

s
→

{\displaystyle \mathrm{d}\vec{s}}

# W

∫

s
1

s
2

F
→

⋅

d

s
→

{\displaystyle W = \int\_{s\_1}^{s\_2} \vec{F}\cdot\mathrm{d}\vec{s}}

However, the infinitesimal linear displacement

d

s
→

{\displaystyle \mathrm{d}\vec{s}}

 is related to a corresponding angular displacement

d

θ
→

{\displaystyle \mathrm{d}\vec{\theta}}

 and the radius vector

r
→

{\displaystyle \vec{r}}

 as

d

s
→

=

d

θ
→

×

r
→

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

Substitution in the above expression for work gives

# W

∫

s
1

s
2

F
→

⋅

d

θ
→

×

r
→

{\displaystyle W = \int\_{s\_1}^{s\_2} \vec{F} \cdot \mathrm{d}\vec{\theta} \times \vec{r}}

The expression

F
→

⋅

d

θ
→

×

r
→

{\displaystyle \vec{F}\cdot\mathrm{d}\vec{\theta}\times\vec{r}}

 is a scalar triple product given by

[

F
→

d

θ
→

r
→

]

{\displaystyle \left[\vec{F}\,\mathrm{d}\vec{\theta}\,\vec{r}\right]}

. An alternate expression for the same scalar triple product is

[

F
→

d

θ
→

r
→

]

=

r
→

×

F
→

⋅

d

θ
→

{\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 rotational force,

τ
→

=

r
→

×

F
→

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

Corresponding substitution in the expression of work gives,

# W

∫

s
1

s
2

τ
→

⋅

d

θ
→

{\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

# W

∫

θ
1

θ
2

τ
→

⋅

d

θ
→

{\displaystyle W = \int\_{\theta \_1}^{\theta \_2} \vec{\tau}\cdot \mathrm{d}\vec{\theta}}

If the rotational force and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e.,

τ
→

⋅

d

θ
→

=

τ
→

d

θ
→

cos
⁡
# 0
τ

d

θ

 giving

# W

∫

θ
1

θ
2

τ

d

θ

{\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 *E*r of the body, given by

E

r

=

1
2

I

ω
2

,

{\displaystyle E\_{\mathrm{r}} = \tfrac{1}{2}I\omega^2,}

where *I* is the rotational mass of the body and *ω* is its angular speed.

Power is the work per unit time, given by

# P
τ
⋅
ω
,

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

where *P* is power, *τ* is rotational force, *ω* is the rotational velocity, and

⋅

{\displaystyle \cdot}

 represents the scalar product.

Algebraically, the equation may be rearranged to compute rotational force for a given angular speed and power output. Note that the power injected by the rotational force depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the rotational force 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 rotational force 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 (rotational force, rpm)input pair is converted to a (rotational force, 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 rotational force is increased, product of which (i.e. power) does not change.

Consistent units must be used. For metric SI units, power is watts, rotational force 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 rotational force, 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.

### Conversion to other units

A conversion factor may be necessary when using different units of power or rotational force. 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 rotational force, and *ν* (Greek letter nu) is rotational speed.

# P
τ
⋅
2
π
⋅
ν

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

Showing units:

P
(

W

# )
τ

(
N
⋅
m
)

⋅
2
π

(
r
a
d

/

r
e
v
)

⋅
ν

(
r
e
v

/

s
e
c
)

{\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.

P
(

W

# )

τ

(
N
⋅
m
)

⋅
2
π

(
r
a
d

/

r
e
v
)

⋅
ν

(
r
p
m
)

60

{\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 (mechanical) for power, foot-pounds (lbf⋅ft) for rotational force and rpm for rotational speed. This results in the formula changing to:

P
(

h
p

# )

τ

(
l
b
f
⋅
f
t
)

⋅
2
π

(
r
a
d

/

r
e
v
)

⋅
ν
(

r
p
m

)

33
,
000

.

{\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.

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

power

=

force
⋅
linear distance

time

=

(

rotational force
r

)

⋅
(
r
⋅
angular speed
⋅
t
)

t

=
rotational force
⋅
angular speed
.

{\displaystyle
\begin{align}
\text{power} & = \frac{\text{force} \cdot \text{linear distance}}{\text{time}} \\[6pt]
& = \frac{\left(\dfrac{\text{rotational force}} r \right) \cdot (r \cdot \text{angular speed} \cdot t)} t \\[6pt]
& = \text{rotational force} \cdot \text{angular speed}.
\end{align}
}

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:

# power
rotational force
⋅
2
π
⋅
rotational speed
.

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

If rotational force 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 rotational force 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:

power

=
rotational force
⋅
2
π
⋅
rotational speed
⋅

ft
⋅
lbf

min

⋅

horsepower

33
,
000
⋅

ft
⋅
lbf

min

≈

rotational force
⋅
RPM

5
,
252

{\displaystyle
\begin{align}
\text{power} & = \text{rotational force} \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{rotational force} \cdot \text{RPM}}{5,252}
\end{align}
}

because

5252.113122
≈

33
,
000

2
π

.

{\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 rotational forces due to several forces applied to *a single* point is equal to the rotational force due to the sum (resultant) of the forces. Mathematically, this follows from:

(

r

×

F

1

)
(

r

×

F

2

)
# ⋯

r

×
(

F

1

F

2

⋯
)
.

{\displaystyle (\mathbf{r}\times\mathbf{F}\_1) + (\mathbf{r}\times\mathbf{F}\_2) + \cdots = \mathbf{r}\times(\mathbf{F}\_1+\mathbf{F}\_2 + \cdots). }

From this it follows that if a pivoted beam of zero mass is balanced with two opposed forces then:

(

r

×

F

1

# )
(

r

×

F

2

)
.

{\displaystyle (\mathbf{r}\times\mathbf{F}\_1) = (\mathbf{r}\times\mathbf{F}\_2) . }

## Rotational force multiplier

Main articles: Rotational force multiplier
Rotational force 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 rotational force, as rotation rate is reduced.

## See also

* Moment
* Conversion of units
* Friction rotational force
* Mechanical equilibrium
* Rigid body dynamics
* Statics
* Rotational force converter
* Rotational force limiter
* Rotational force screwdriver
* Rotational force tester
* Rotational force wrench
* Torsion (mechanics)