AP PHYSICS C: MECHANICS • TORQUE AND ROTATIONAL DYNAMICS

Torque

The rotational analog of force that governs how objects spin, twist, and maintain angular equilibrium.

Historical Context & Motivation

Long before physicists formalized the concept, ancient engineers intuitively exploited torque every time they used a lever, turned a capstan, or balanced a beam on a fulcrum. The principle that a force applied farther from a pivot produces a greater rotational effect underpins some of the earliest machines in recorded history. Formalizing this idea required centuries of mathematical development — from Archimedes' law of the lever through Newton's framework of forces and, ultimately, Euler's rotational equations of motion. Understanding torque is essential because translational mechanics alone cannot predict the behavior of any system that rotates, from a simple door hinge to a gyroscope aboard a spacecraft.

~250 BCE
Archimedes and the Lever
Archimedes proved the law of the lever: a balanced beam requires equal products of weight and distance on each side, establishing the core idea behind torque.
1687
Newton's Principia
Isaac Newton published the laws of motion. While focused on translational dynamics, his framework laid the groundwork for extending F = ma to rotational systems.
1750
Euler's Rotational Equations
Leonhard Euler generalized Newton's second law to rigid-body rotation, introducing the torque–angular-acceleration relationship τ = Iα and defining torque as a vector quantity.
1883
Vector Cross Product Formalism
Josiah Willard Gibbs and Oliver Heaviside independently formalized vector algebra, giving torque its modern definition as the cross product τ = r × F.

The central question torque answers is deceptively simple: how do forces cause rotation? A force applied to a door near its hinges barely moves it, while the same force at the handle swings it wide open. Torque quantifies this dependence on both the magnitude of the force and the geometry of its application point relative to the axis of rotation. Mastering torque is a prerequisite for analyzing everything from static equilibrium of beams and bridges to the angular acceleration of flywheels and planetary motion.

Core Principles & Definitions

Torque — often represented by the Greek letter τ (tau) — is the rotational analog of force. Just as a net force causes translational acceleration, a net torque causes angular acceleration. Torque is a vector quantity whose direction is determined by the right-hand rule and whose magnitude depends on three factors: the applied force, the distance from the axis, and the angle between the force and the position vector.

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Definition as a Cross Product

Torque is defined as τ = r × F, where r is the position vector from the axis to the point of force application and F is the applied force. The cross product ensures only the perpendicular component of the force contributes.
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Magnitude: τ = rF sin θ

The scalar magnitude is τ = rF sin θ, where θ is the angle between r and F. Maximum torque occurs when the force is perpendicular to the lever arm (θ = 90°); zero torque results when the force is directed along the line from the axis (θ = 0° or 180°).
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Lever Arm (Moment Arm)

The lever arm d = r sin θ is the perpendicular distance from the axis of rotation to the line of action of the force. Torque can equivalently be written τ = Fd.
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Direction via the Right-Hand Rule

Curl the fingers of your right hand from r toward F; your thumb points in the direction of τ. By convention, counterclockwise torques (out of the page) are positive and clockwise torques (into the page) are negative.
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Newton's Second Law for Rotation

The net torque on a rigid body about a fixed axis equals the product of its moment of inertia and angular acceleration: Στ = Iα. This is the rotational counterpart of ΣF = ma.
KEY TAKEAWAY
KEY TAKEAWAY

Visual Explanation

The position vector r (cyan) extends from the axis of rotation to the point where the force F (pink) is applied. The angle θ between them determines the effective lever arm d = r sin θ (yellow dashed). The resulting torque τ points out of the page (green dot), determined by the right-hand rule.

The diagram above illustrates the fundamental geometry of torque. Notice that only the component of F perpendicular to r contributes to the torque — the parallel component simply pulls along the lever arm without causing rotation. This is precisely what the cross product captures: it extracts the perpendicular contribution automatically through the sin θ factor. When θ = 90°, the entire force contributes to rotation and the torque is maximized at τ = rF. When θ = 0° or 180°, the force points along or opposite to r and produces zero torque.

Mathematical Framework

The mathematical structure of torque emerges naturally from extending Newton's second law to rotating systems. We begin with the vector definition, derive the scalar magnitude, and then connect torque to rotational dynamics through the moment of inertia.

TORQUE VECTOR DEFINITION
τ = r × F
τ = torque vector (N·m), r = position vector from axis to point of force application (m), F = applied force vector (N). The cross product yields a vector perpendicular to both r and F.
SCALAR MAGNITUDE
|τ| = rF sin θ = Fd
θ = angle between r and F. The quantity d = r sin θ is the lever arm (moment arm) — the perpendicular distance from the axis to the line of action of the force.
NEWTON'S SECOND LAW FOR ROTATION
Στ = Iα
Στ = net torque about a given axis (N·m), I = moment of inertia about that axis (kg·m²), α = angular acceleration (rad/s²). This is the direct rotational analog of ΣF = ma.

A useful derivation connects the translational and rotational forms. Consider a single particle of mass m constrained to move in a circle of radius r. The tangential component of Newton's second law gives F = mat. Multiplying both sides by r yields rF = mrat. Since at = rα, we obtain τ = mr²α = Iα, confirming the rotational second law for a point mass with I = mr².

STATIC EQUILIBRIUM CONDITION
Στ = 0 (about any axis)
For a rigid body in static equilibrium, the net torque about any chosen axis must vanish. Combined with ΣF = 0, this gives a complete set of equilibrium conditions.

Applications & Classification

Torque problems in AP Physics C fall into two broad categories: static equilibrium (Στ = 0) and rotational dynamics (Στ = Iα). In static equilibrium problems, you choose a convenient pivot, sum torques, and solve for unknown forces or distances. In dynamics problems, you combine torque equations with translational equations (often via a string-and-pulley constraint) to find angular or linear accelerations. The diagram below illustrates the classic Atwood machine with a massive pulley — one of the most commonly tested configurations on the AP exam.

An Atwood machine with a massive pulley of moment of inertia I = ½MR². The torque equation about the center O, combined with Newton's second law for each hanging mass and the no-slip constraint a = Rα, yields the system's acceleration. Note how the pulley's inertia reduces the acceleration compared to the massless-pulley case.
Common torque problem categories on the AP Physics C exam
Problem TypeKey ConditionTypical Setup
Static EquilibriumΣτ = 0 and ΣF = 0Beam on supports, ladder against wall, sign hanging from a rod
Fixed-Axis RotationΣτ = Iα about fixed axisPulley systems, rotating disks, rolling with slipping
Rolling Without Slippingτ from friction; a = Rα constraintSphere or cylinder rolling down an incline
Angular Momentumτ = dL/dtTorque as the time derivative of angular momentum; precession

Worked Example

A uniform horizontal beam of mass M = 12 kg and length L = 4.0 m is attached to a wall by a hinge at its left end. A cable attached to the right end of the beam makes an angle of 30° with the beam and connects to the wall above the hinge. A block of mass m = 8.0 kg hangs from a point 3.0 m from the hinge. Find the tension in the cable and the force exerted by the hinge.

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Step 1 — Draw the Free-Body Diagram & Choose a PivotThe beam has four forces: the beam's weight Mg acting downward at its center (2.0 m from hinge), the block's weight mg acting downward at 3.0 m from the hinge, the cable tension T at the right end (4.0 m), and the hinge force FH at the left end. We choose the hinge as the pivot to eliminate FH from the torque equation.
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Step 2 — Write the Torque Equation (Στ = 0 about hinge)Taking counterclockwise as positive, the cable provides a CCW torque and the two weights provide CW torques. The perpendicular component of T relative to the beam is T sin 30°. Thus: T sin 30° × L − Mg × (L/2) − mg × (3.0 m) = 0. Substituting: T(0.50)(4.0) − (12)(9.8)(2.0) − (8.0)(9.8)(3.0) = 0.
2.0T = 235.2 + 235.2 = 470.4 → T = 235.2 N
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Step 3 — Resolve the Tension into ComponentsTx = T cos 30° = 235.2 × 0.866 = 203.7 N (horizontal, toward wall). Ty = T sin 30° = 235.2 × 0.50 = 117.6 N (vertical, upward).
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Step 4 — Apply ΣF = 0 to Find Hinge ForcesHorizontal: FHx − Tx = 0 → FHx = 203.7 N. Vertical: FHy + Ty − Mg − mg = 0 → FHy = (12 + 8)(9.8) − 117.6 = 196 − 117.6 = 78.4 N (upward).
F_H = √(203.7² + 78.4²) ≈ 218 N at 21° above horizontal

Common Pitfalls & Problem-Solving Tips

Frequent errors and their corrections
Common MistakeWhy It's WrongCorrect Approach
Using τ = rF without the sin θ factorThis assumes the force is always perpendicular to the lever arm, which is only true when θ = 90°.Always write τ = rF sin θ or compute the lever arm d = r sin θ first.
Choosing a pivot and forgetting it affects signsTorques are measured about a specific point; changing the pivot changes the r for every force.Pick a pivot that eliminates an unknown force, then consistently assign CCW as positive (or vice versa).
Assuming equal tensions on both sides of a massive pulleyA pulley with nonzero moment of inertia requires a net torque to accelerate, so T₁ ≠ T₂.Write a separate torque equation for the pulley: (T₁ − T₂)R = Iα.
Confusing torque (N·m) with energy (J)Both have the same SI units, but torque is a vector (cross product) while energy is a scalar (dot product).Never convert between the two. Use N·m for torque and J for energy.
Neglecting the weight of a uniform beam in equilibrium problemsThe beam's weight produces a torque about any pivot not at its center of mass.Model the beam's weight as a single downward force Mg applied at its center of mass.
KEY TAKEAWAY
STRATEGY TIP

Connection to Advanced Rotational Theory

At the AP Physics C level, torque connects to several deeper ideas that bridge the gap to upper-division mechanics. The relationship τ = dL/dt generalizes Newton's second law for rotation, relating torque to the time rate of change of angular momentum L. When the net external torque on a system is zero, angular momentum is conserved — a principle with applications ranging from figure skaters pulling in their arms to the precession of gyroscopes.

How AP-level torque concepts extend into upper-division physics
AP Physics C ConceptAdvanced Extension
τ = Iα (fixed axis)Euler's equations for 3D rotation of rigid bodies with products of inertia
τ = dL/dtTorque-free precession and nutation; gyroscopic stability analysis
Static equilibrium (Στ = 0)Statically indeterminate systems requiring elasticity theory
Rolling without slipping (a = Rα)Lagrangian mechanics with rolling constraints and generalized coordinates

The work–energy theorem also has a rotational analog. The work done by a torque rotating through an angle dθ is dW = τ dθ, and integrating yields W = ∫τ dθ. For a constant torque, this simplifies to W = τΔθ. The rotational kinetic energy Krot = ½Iω² connects to work via the work–energy theorem: Wnet = ΔKrot. These relationships are frequently tested in FRQ problems that require translating between representations — for instance, deriving the angular velocity of a pulley system from energy methods rather than torque–acceleration kinematics.

Practice Problems

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A force is applied to a door at a point halfway between the hinge and the outer edge. If the same force is instead applied at the outer edge, how does the torque about the hinge change? A) It is halved. B) It remains the same. C) It is doubled. D) It is quadrupled.
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A 0.40-m-long wrench is used to tighten a bolt. A force of 80 N is applied at the end of the wrench at an angle of 60° to the wrench handle. What is the magnitude of the torque about the bolt? A) 16 N·m B) 27.7 N·m C) 32 N·m D) 24 N·m
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A uniform rod of mass M and length L is free to rotate about a pivot at one end. It is released from rest in a horizontal position. What is the initial angular acceleration of the rod? A) g/L B) 2g/(3L) C) 3g/(2L) D) g/(2L)
PROBLEM 4APPLIED
A solid cylinder of mass M = 5.0 kg and radius R = 0.20 m is mounted on a frictionless horizontal axle. A light, inextensible string is wrapped around the cylinder, and a block of mass m = 3.0 kg is attached to the free end of the string. (a) Draw a free-body diagram for the block and a torque diagram for the cylinder. Label all forces. (b) Write Newton's second law for the block and the torque equation for the cylinder. (c) Using the constraint a = Rα, derive an expression for the linear acceleration of the block in terms of M, m, and g. (d) Calculate the numerical value of the acceleration and the tension in the string. (e) If the block starts from rest, determine the angular velocity of the cylinder after the block has fallen 1.5 m.
PROBLEM 5CRITICAL THINKING
A student claims that for a system in static equilibrium, the net torque must be zero about every possible axis. Another student argues that you only need to check one axis. (a) Explain whether the first student's claim is correct and justify your answer. (b) A uniform beam of length L and weight W rests on two supports: one at the left end and one at a distance 3L/4 from the left end. Derive expressions for the normal forces at each support. (c) Verify your answer by showing that Στ = 0 about a different axis than the one you used in part (b).
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