Torque

In US mechanical engineering, torque is defined mathematically as the rate of change of angular momentum of an object (in physics it is called "net torque"). 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 forces 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).[7] 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.

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

$${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$$

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

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

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force ${\textstyle \mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}}$ (whether or not mass is constant) and the definition of velocity ${\textstyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {v} }$

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

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.

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 F_net = ma, and it follows that

$${\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} _{\mathrm {net} }={\boldsymbol {\tau }}_{\mathrm {net} }.}$$

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.

Moment arm diagram

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

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

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 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:

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

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 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.

The torque caused by the two opposing forces F_g and −F_g causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.

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: ∑H = 0 and ∑V = 0, and the torque a third equation: ∑τ = 0. That is, to solve statically determinate equilibrium problems in two-dimensions, three equations are used.

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

$${\displaystyle {\boldsymbol {\tau }}_{2}={\boldsymbol {\tau }}_{1}+(\mathbf {r} _{1}-\mathbf {r} _{2})\times \mathbf {F} }$$

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

${\displaystyle W=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} \mathbf {s} }$

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

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

Substitution in the above expression for work gives

${\displaystyle W=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }$

The expression ${\displaystyle \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }$ is a scalar triple product given by ${\displaystyle \left[\mathbf {F} \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]}$. An alternate expression for the same scalar triple product is

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

But as per the definition of torque,

${\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} }$

Corresponding substitution in the expression of work gives,

${\displaystyle W=\int _{s_{1}}^{s_{2}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}$

Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving

${\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}$

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

${\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \,\mathrm {d} \theta }$

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

where I is the moment of inertia of the body and ω is its angular speed.[15]

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 scalar product.

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: 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 torque 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 (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 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.

Consistent units must be used. For metric SI units, power is watts, torque 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 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 orientational analysis which treats radians as a base unit rather than a dimensionless unit.[16]

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

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

Showing units:

${\displaystyle P({\rm {W}})=\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}$

Dividing by 60 seconds per minute gives us the following.

${\displaystyle P({\rm {W}})={\frac {\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}{60}}}$

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 torque and rpm for rotational speed. This results in the formula changing to:

${\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{33,000}}.}$

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.

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:

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

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:

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

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:

${\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text{RPM}}}{5,252}}\end{aligned}}}$

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