Torque
Torque is a physical quantity that represents the rotational equivalent of linear force, measuring the tendency of a force to cause an object to rotate about a fixed axis.[1] It is calculated as the product of the force applied and the perpendicular distance from the axis of rotation to the line of action of the force, known as the lever arm or moment arm.[2] In vector form, torque \vec{\tau} is given by the cross product \vec{\tau} = \vec{r} \times \vec{F}, where \vec{r} is the position vector from the axis to the point of force application and \vec{F} is the force vector, with the magnitude \tau = r F \sin \theta (where \theta is the angle between \vec{r} and \vec{F}).[3] The direction of torque is perpendicular to both the position and force vectors, determined by the right-hand rule, and it can produce angular acceleration in rigid bodies according to Newton's second law for rotation (\vec{\tau} = I \vec{\alpha}, where I is the moment of inertia and \vec{\alpha} is angular acceleration).[1] Torque is a vector quantity, and its sign convention typically assigns positive values to counterclockwise rotations and negative to clockwise when viewed from a standard perspective.[1] In equilibrium, the net torque on an object is zero, meaning the sum of all clockwise torques equals the sum of all counterclockwise torques.[3] The SI unit of torque is the newton-meter (N·m), equivalent to a force of one newton applied at a perpendicular distance of one meter from the axis, though it is distinct from work or energy units despite sharing the same dimensions.[1][4] Torque plays a fundamental role in mechanics, appearing in applications from simple levers and pulleys to complex systems like engines, where it quantifies the twisting effectiveness of forces in generating rotational motion.[2]Definition and Formulation
Vector Definition
In physics, torque is defined as a pseudovector \vec{\tau} given by the cross product \vec{\tau} = \vec{r} \times \vec{F}, where \vec{r} is the position vector from a chosen pivot point to the point of application of the force vector \vec{F}. This formulation captures the tendency of the force to produce rotation about the pivot, distinguishing torque from linear force, which induces translational motion along the force direction. The resulting torque vector is perpendicular to the plane formed by \vec{r} and \vec{F}, with its magnitude proportional to the sine of the angle between them, emphasizing the rotational component of the force. The direction of the torque vector follows the right-hand rule for cross products: extend the fingers of the right hand along \vec{r}, curl them toward \vec{F}, and the thumb indicates the direction of \vec{\tau}, which defines the axis of rotation.[5] This convention ensures consistency in three-dimensional space, where torque acts as a pseudovector that changes sign under improper rotations like reflections. The concept of torque, historically termed the "moment of force," originated in the early 19th century with Siméon Denis Poisson's Traité de mécanique (1811), where it was formalized in analytical mechanics.[6] This development built upon ancient principles, such as Archimedes' 3rd-century BCE law of the lever, which equated the effects of forces at different distances from a fulcrum to balance rotational tendencies.[7] A practical example is the torque exerted when pushing on a door: if \vec{r} points from the hinge to the point of push and \vec{F} is applied at an angle, the torque \vec{\tau} = \vec{r} \times \vec{F} tends to rotate the door around the hinge, with maximum effect when \vec{F} is perpendicular to \vec{r}.[1]Scalar Magnitude and Direction
The scalar magnitude of the torque, denoted as |\tau|, quantifies the rotational effect of a force about a pivot point and is calculated using the formula |\tau| = r F \sin \theta, where r is the distance from the pivot to the point of force application, F is the magnitude of the force, and \theta is the angle between the position vector and the force vector.[8] This expression derives from the magnitude of the vector cross product, capturing the component of the force that contributes to rotation.[9] An equivalent formulation expresses the magnitude as |\tau| = F d, where d represents the moment arm, defined as the perpendicular distance from the pivot to the line of action of the force.[10] Geometrically, torque embodies the "twisting strength" of the force, which reaches its maximum value when \theta = 90^\circ, meaning the force acts perpendicular to the lever arm, as this alignment fully leverages the distance r for rotation.[1] In two-dimensional scenarios, the direction of torque is specified relative to the plane of motion, with the torque vector pointing perpendicular to that plane; a common sign convention assigns positive values to counterclockwise rotations and negative values to clockwise ones.[11] For instance, when using a wrench of length r = 0.25 m to turn a bolt, applying a force F = 40 N perpendicular to the handle (\theta = 90^\circ) produces a torque magnitude of |\tau| = 0.25 \times 40 \times \sin 90^\circ = 10 N·m, tending to rotate the bolt counterclockwise if viewed from above. If the same force is applied at an angle \theta = 45^\circ, the magnitude reduces to |\tau| = 0.25 \times 40 \times \sin 45^\circ \approx 7.07 N·m, illustrating how the angle diminishes the effective twisting.[12]Relations to Kinematics and Dynamics
Connection to Angular Momentum
In rotational dynamics, torque \vec{\tau} is fundamentally defined as the time rate of change of angular momentum \vec{L} of a system, expressed by the equation \vec{\tau} = \frac{d\vec{L}}{dt}.[13] This relation holds for any system of particles or rigid bodies, where the net external torque equals the total rate of change of the system's angular momentum.[14] For a rigid body rotating about a fixed axis, the angular momentum is given by \vec{L} = I \vec{\omega}, where I is the moment of inertia about that axis and \vec{\omega} is the angular velocity. Substituting this into the fundamental equation yields \vec{\tau} = I \frac{d\vec{\omega}}{dt} = I \vec{\alpha} for systems of constant mass and moment of inertia, where \vec{\alpha} is the angular acceleration.[15] This form is the rotational analog of Newton's second law for linear motion, \vec{F} = \frac{d\vec{p}}{dt}, highlighting torque's role in altering rotational motion much like force changes linear momentum.[15] When no external torque acts on a system (\vec{\tau} = 0), the angular momentum remains conserved, \vec{L} = constant, leading to applications in phenomena like the steady spin of a gyroscope or the orbital motion of planets.[13] For instance, applying torque to a wheel, such as through friction from the ground, increases its angular speed (\alpha > 0) during acceleration or decreases it during braking, directly altering its angular momentum.[16] In variable mass systems, such as rockets expelling propellant, the relationship between torque and angular momentum becomes more involved due to the changing mass distribution, requiring additional terms to account for the momentum carried away by ejected material.Derivatives and Time Evolution
In non-steady rotational systems, the time derivative of torque, denoted \dot{\tau} = \frac{d\tau}{dt}, describes the instantaneous rate of change of the applied rotational force, playing a key role in analyzing behaviors like oscillations or rapid accelerations where torque does not remain constant. This derivative is particularly relevant in transient conditions, where it influences the smoothness of motion and system stability. For instance, in oscillatory systems, \dot{\tau} captures how quickly the torque adjusts to varying inertial demands or external perturbations. The relation between \dot{\tau} and higher-order kinematics derives from the core equation of rotational dynamics, \vec{\tau} = I \vec{\alpha}, where I is the moment of inertia and \vec{\alpha} is the angular acceleration. Differentiating with respect to time yields \dot{\vec{\tau}} = \dot{I} \vec{\alpha} + I \dot{\vec{\alpha}} for cases of variable inertia, with the second term involving angular jerk \dot{\vec{\alpha}} = \frac{d\vec{\alpha}}{dt}, the rate of change of angular acceleration. When I is constant—as in most rigid-body analyses—this simplifies to \dot{\vec{\tau}} = I \dot{\vec{\alpha}}, linking the torque derivative directly to jerk and highlighting its role in quantifying rotational "jolt." This formulation extends the analogy to linear dynamics, where the force derivative relates to linear jerk, and is applied in control systems to bound motion smoothness. Applications of torque derivatives appear prominently in transient dynamics. In electric motor startups, the torque evolves rapidly from an initial high value—often 150–250% of rated torque—to accelerate the rotor, with \dot{\tau} governing the transition rate and preventing excessive mechanical stress; for induction motors, this profile typically peaks within milliseconds before decaying as speed approaches nominal values.[17] Similarly, in damped pendulums, the net torque varies temporally due to gravitational restoration and viscous damping, resulting in \dot{\tau} that modulates the exponential decay of oscillations, ensuring controlled energy dissipation.[18] A practical example is a swinging door mechanism, where applied torque must adjust dynamically to changing angular velocity; as the door accelerates from rest, initial high torque overcomes static friction and inertia, but \dot{\tau} becomes negative during deceleration phases influenced by hinges or air drag, yielding a nonlinear time profile that ensures smooth closure without slamming. In engineering design, such torque-time profiles are routinely plotted or tabulated to evaluate system performance—for instance, in automotive engines, where transient torque ramps during acceleration are optimized to balance power delivery and component durability, often showing initial surges followed by steady-state plateaus over 0.1–1 second timescales.[19]Energy and Power Aspects
Relation to Mechanical Power
In rotational systems, the instantaneous mechanical power P is the rate at which torque transfers energy to or from the system, given by the scalar product P = \tau \omega when the torque \tau and angular velocity \omega are aligned along the axis of rotation.[20] More generally, for arbitrary orientations, the power is the vector dot product P = \vec{\tau} \cdot \vec{\omega}, capturing only the component of torque parallel to the angular velocity.[21] This formulation highlights power as the instantaneous energy transfer rate driven by the torque acting at a specific angular speed.[22] The units of mechanical power align consistently with those of torque and angular velocity: power is measured in watts (W), or joules per second (J/s), derived from torque in newton-meters (N·m) multiplied by angular velocity in radians per second (rad/s), yielding \mathrm{N \cdot m \cdot rad/s = J/s} since the radian is dimensionless.[23] A practical example occurs in electric motors, where output power is computed from measured torque and rotational speed in revolutions per minute (RPM) using P = \tau \cdot \omega, with \omega converted via \omega = 2\pi \cdot \mathrm{RPM}/60; for instance, a motor delivering 10 N·m at 1500 RPM produces approximately 1570 W.[24] When angular velocity varies over time, such as during acceleration or load changes, the instantaneous power P(t) = \vec{\tau}(t) \cdot \vec{\omega}(t) fluctuates accordingly, whereas average power is the time-averaged value over an interval, providing a measure of net energy delivery.[20]Relation to Work and Energy
In rotational mechanics, the work done by a torque \vec{\tau} acting on a rigid body over an angular displacement \Delta \vec{\theta} is given by the line integral W = \int \vec{\tau} \cdot d\vec{\theta}, which is analogous to the work done by a force in linear motion, W = \int \vec{F} \cdot d\vec{x}.[20] This formulation arises because torque causes rotational displacement, transferring energy to the system. For a constant torque, the expression simplifies to W = \tau \Delta \theta, where \Delta \theta is the total angular displacement magnitude.[25] The rotational work-energy theorem states that the net work done by all torques equals the change in rotational kinetic energy of the body: \Delta K = \frac{1}{2} I (\omega_f^2 - \omega_i^2) = \int \tau \, d\theta, where I is the moment of inertia, \omega_f and \omega_i are the final and initial angular velocities, respectively.[20] This connection highlights how torque accelerates rotation, converting work into kinetic energy, much like force does in translation. To derive the integral form, consider the instantaneous power P = \frac{dW}{dt} = \tau \omega, combined with \omega = \frac{d\theta}{dt}, yielding dW = \tau \, d\theta; integrating over the displacement gives W = \int \tau \, d\theta.[26] Torque can also relate to changes in potential energy, particularly in systems involving gravitational or elastic restoring torques. For instance, in a simple pendulum, the gravitational torque \tau = -mg l \sin \theta (where m is mass, g is gravitational acceleration, l is length to the center of mass, and \theta is the angular displacement from vertical) opposes motion, and the work done against this torque increases the gravitational potential energy U = m g l (1 - \cos \theta).[27] Similarly, in a torsional spring system, applying torque to wind the spring stores elastic potential energy; the work W = \int_0^\theta \kappa \theta' \, d\theta' = \frac{1}{2} \kappa \theta^2 (with \kappa as the torsional constant) directly equals this potential increase.[20] A practical example is lifting a load using a pulley system driven by torque. Here, the applied torque \tau = r F (with r as pulley radius and F as tension) rotates the pulley through \Delta \theta, performing work W = \tau \Delta \theta = F (r \Delta \theta) = F \Delta h, where \Delta h is the linear displacement of the load; this equals the gain in gravitational potential energy m g \Delta h of the lifted mass m.[20]Units and Measurement
SI Units
In the International System of Units (SI), torque is quantified using the newton-metre (N·m), defined as the torque resulting from a force of one newton applied perpendicularly over a lever arm of one metre.[28] This unit physically interprets torque as the product of force and perpendicular distance, emphasizing its role in inducing rotation about an axis.[28] The newton-metre shares dimensional equivalence with the joule (J), the SI unit of energy, as both are kg·m²·s⁻² in base units; however, torque is distinctly not an energy measure due to its vectorial nature and dependence on the geometry of application rather than path-integrated work.[28] The Bureau International des Poids et Mesures (BIPM) maintains the coherence of the N·m through fixed definitions of the kilogram, metre, and second, ensuring traceability and precision in metrological standards. Torque measurement employs specialized instruments, including torque wrenches for applying and verifying controlled values in assembly tasks and dynamometers for assessing rotational forces in engines and machinery, with calibrations adhering to ISO 6789 for accuracy within ±4% of reading.[29] In practical engineering contexts, torque values typically span from 10^{-3} N·m in micro-mechanical devices to 10^{6} N·m in large-scale industrial systems, reflecting the unit's versatility across scales.Conversions to Other Units
Torque, as measured in the SI unit of newton-meter (N·m), can be converted to various non-SI units commonly used in engineering and industry. One prevalent imperial unit is the pound-force foot (lbf·ft), where 1 N·m = 0.737562 lbf·ft. This factor arises from the definitions of the base units: the newton (N) equals 1 kg·m/s², while the pound-force (lbf) equals 4.4482216152605 N exactly, and the foot (ft) equals 0.3048 m exactly. Thus, 1 lbf·ft = 4.4482216152605 N × 0.3048 m = 1.3558179483314004 N·m, so the inverse conversion yields 1 N·m ÷ 1.3558179483314004 ≈ 0.737562 lbf·ft.[30] Another common imperial unit for torque, particularly in smaller-scale applications like fasteners, is the inch-pound force (in·lbf), with 1 N·m = 8.85075 in·lbf. This derives similarly from the inch equaling 0.0254 m exactly, making 1 in·lbf = (4.4482216152605 N × 0.0254 m) / 12 ≈ 0.112984829 N·m, and thus 1 N·m ≈ 8.85075 in·lbf. For even finer measurements, such as in precision instruments, the ounce-force inch (ozf·in) is used, where 1 ozf·in ≈ 0.007061552 N·m, so 1 N·m ≈ 141.612 ozf·in.[30] In automotive engineering, torque specifications for engines are frequently expressed in lbf·ft in the United States or N·m internationally. This duality reflects regional standards, with conversions essential for global comparisons of vehicle performance data. Historical units like the kilogram-force meter (kgf·m), where 1 kgf·m = 9.80665 N·m and thus 1 N·m ≈ 0.101972 kgf·m, persist in some legacy engineering contexts but are avoided in modern scientific practice as they are non-SI and depend on gravitational acceleration (g ≈ 9.80665 m/s²). The SI promotes consistent use of N·m to eliminate such variability.[30]| From N·m | To lbf·ft | To in·lbf | To ozf·in | To kgf·m |
|---|---|---|---|---|
| 1 | 0.737562 | 8.85075 | 141.612 | 0.101972 |