3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A mechanic pushes on a wrench, exerting a torque of 40 N·m on a rusted bolt, but the bolt does not turn. How much work does the torque exerted by the wrench do on the bolt?
Answer and reasoning
A40 J, as work done by a torque equals the torque, 40 N·m. A student who reads the torque as an amount of work, because both can be written with N·m, picks this. Torque and work are different quantities: work also needs an angular displacement, W = τΔθ, and here Δθ = 0.
BA positive amount, as the wrench exerts a large torque on the bolt. A student who equates exerting a torque with doing work picks this. Effort is not work: the torque transfers energy only while the bolt turns through an angle, and the bolt does not turn.
CNone, as the bolt turns through no angle as the torque acts.Correct A torque transfers energy only if it is exerted over an angular displacement: W = τΔθ = (40 N·m)(0 rad) = 0. However hard the mechanic pushes, no energy is transferred to the bolt while it does not turn.
DIt depends on how long the mechanic keeps pushing on the wrench. A student who treats work like impulse, which depends on how long a force acts, picks this. Work depends on the angle turned, W = τΔθ, which is zero here however long the push lasts.
A light rope wrapped around a large drum of radius 0.80 m is pulled with a constant tension of 1.5 N, tangent to the drum. Friction in the drum's axle exerts an opposing torque, so the drum turns at constant angular velocity. How much work does the rope's force do on the drum while the drum turns through 1.0 revolution?
Answer and reasoning
A1.2 J A student who uses 1.0 revolution as if it were 1.0 rad gets (1.2 N·m)(1.0) = 1.2 J. W = τΔθ needs the angle in radians: 1.0 rev = 2π rad ≈ 6.28 rad.
B7.5 JCorrect τ = rF = (0.80 m)(1.5 N) = 1.2 N·m. Δθ = 1.0 rev × 2π rad/rev = 6.28 rad. W = τΔθ = (1.2 N·m)(6.28 rad) ≈ 7.5 J. (Check: the rope moves 2π(0.80 m) = 5.0 m, and (1.5 N)(5.0 m) ≈ 7.5 J.) The friction torque does about −7.5 J, so the net work, and the change in kinetic energy, is zero.
C9.4 J A student who uses the tension, 1.5 N, as the torque gets (1.5)(6.28) ≈ 9.4 J. The torque is the force times the radius at which it is exerted: τ = (0.80 m)(1.5 N) = 1.2 N·m.
D0.0 J A student who reasons that constant angular velocity means no work is done picks this. Constant angular velocity means the NET work is zero: the rope's torque does +7.5 J while the friction torque does −7.5 J. The rope's force on its own does positive work.
Working τ = rF = 0.80 × 1.5 = 1.2 N·m. Δθ = 1.0 × 2π = 6.283 rad. W = 1.2 × 6.283 = 7.54 ≈ 7.5 J. Errors: Δθ = 1.0 → 1.2 J; τ = F = 1.5 → 9.42 ≈ 9.4 J; net work → 0.0 J.
The graph shows the torque τ exerted on a wheel by a motor as a function of the wheel's angular position θ. How much work does the motor's torque do on the wheel as the wheel turns from θ = 0 to θ = 4.0 rad?
Answer and reasoning
A8.0 J A student who multiplies the largest torque by the whole angular displacement, (2.0 N·m)(4.0 rad), picks this. The torque falls to zero after 2.0 rad, so the area from 2.0 rad to 4.0 rad is a triangle, not a rectangle.
B2.0 J A student who reads the torque value off the graph, 2.0 N·m, and gives it as the work picks this. The height of the graph is the torque at one position; the work is the area under the graph over the whole 4.0 rad.
C1.0 J A student who uses the slope of the sloping section, (2.0 N·m)/(2.0 rad) = 1.0, as the work picks this. The slope shows how fast the torque changes with angle; the work is the area under the graph.
D6.0 JCorrect The work is the area under the τ–θ graph: a rectangle (2.0 N·m)(2.0 rad) = 4.0 J from 0 to 2.0 rad, plus a triangle (1/2)(2.0 N·m)(2.0 rad) = 2.0 J from 2.0 rad to 4.0 rad. Total 6.0 J.
Working Area = (2.0)(2.0) + (1/2)(2.0)(4.0 − 2.0) = 4.0 + 2.0 = 6.0 J. Errors: largest torque × 4.0 rad → 8.0 J; torque value 2.0 → 2.0 J; slope magnitude 2.0/2.0 → 1.0 J.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.2.A.1 Torque Fix
Torque
The rotational effect of a force about an axis: τ = rF sin θ = rF⊥, where r is the distance from the axis to the point where the force is exerted. Unit N·m. A force directed along the line to the axis exerts no torque.
Angular displacement
The angle, in radians, through which a rigid system turns about an axis: Δθ = θ − θ0. One revolution is 2π rad.
Energy transfer by a torque
A torque transfers energy into or out of a rigid system only while the system turns through an angle. If the torque is in the same sense as the rotation, it does positive work and energy goes in; if it opposes the rotation, it does negative work and energy comes out; if the system does not turn, the torque does no work however large it is.
Net work on a rotating rigid system
The sum of the work done by all the torques exerted on a rigid system turning about a fixed axis equals the change in its rotational kinetic energy. One torque can do positive work while another does an equal amount of negative work, so that the kinetic energy stays constant.
Students often think Whenever a force or torque is exerted on an object, it does positive work on the object and transfers energy into it, as effort does. In fact No. A torque does work only if the object turns through an angle while the torque is exerted, and the work is positive only if the torque is in the same sense as the rotation. A torque on an object that does not turn does no work; a torque that opposes the rotation does negative work.
Students often think No work is done on an object whose center of mass does not move, so torques on a wheel turning on a fixed axle do no work. In fact Yes. A torque does work W = τΔθ whenever the wheel turns through an angle; the points where the forces are exerted move even though the center of mass does not.
6.2.A.2 Work done by a torque Fix
Work done by a torque
For a constant torque, W = τΔθ, the torque multiplied by the angular displacement during the interval in which the torque is exerted. Δθ must be in radians; the result is in joules. It is the work done by that one torque, not by the net torque.
Tangential force and work
For a force F exerted tangent to a wheel of radius r, τΔθ = (rF)Δθ = F(rΔθ): the work equals the force multiplied by the distance moved by its point of application along the rim, as W = Fd gives in Unit 3.
Newton-meter and joule
Torque and work can both be reduced to the base units kg·m²/s², but they are different quantities. Torque is written in N·m and is not an energy; work is written in J and requires an angular displacement.
Students often think Work and torque are the same quantity, because both are measured in newton-meters, so the work done by a torque equals the value of the torque. In fact Not necessarily. Torque is not work. W = τΔθ, so the work depends on the angle turned as well: 40 N·m through 0 rad is 0 J, and through 2.0 rad is 80 J.
Students often think The work done by a torque behaves like an impulse: it depends on how long the torque is exerted, and τΔθ equals Iω, the rotational version of mv. In fact On the angle. W = τΔθ: a torque exerted for any length of time on an object that does not turn does no work. The work equals the change in kinetic energy, (1/2)Iω² − (1/2)Iω0².
6.2.A.3 Area under a torque–angular position graph Fix
Area under a torque–angular position graph
The work done by a torque as a rigid system turns from θ1 to θ2 equals the area between the graph of τ against θ and the θ axis over that interval. For a constant torque the area is the rectangle τΔθ; for a varying torque it is the area of the shape, counted as negative where the torque is negative.
Students often think W = τΔθ can be used for a torque that changes during the rotation by multiplying its largest value by the whole angular displacement. In fact No. W = τΔθ applies directly only while the torque is constant. For a varying torque the work is the area under the τ–θ graph, found by splitting it into rectangles and triangles; using the largest torque over the whole interval gives too much.
Students often think The work done by a torque can be read from the height of the torque–angular position graph, so the larger the torque reached, the more work is done. In fact No. The value read from the graph is the torque at one angular position, in N·m. The work done over an interval is the area under the graph over that interval, in J.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
The diagram shows a wheel turning counterclockwise on a fixed axle while a brake pad presses on its rim. At point P the pad exerts two forces on the wheel: a normal force FN, directed toward the axle, and a friction force Ff, tangent to the rim. How does the torque exerted on the wheel by the brake pad affect the wheel's energy while the wheel turns?
Answer and reasoning
AIt transfers energy into the wheel, as it is exerted while the wheel turns. A student who thinks any torque exerted during a rotation puts energy in picks this. The sign of the work depends on the sense of the torque: Ff's torque is clockwise while the wheel turns counterclockwise, so it does negative work.
BIt transfers no energy, as the wheel's center of mass does not move. A student who thinks work needs the center of mass to move picks this. The wheel turns through an angle, and a torque exerted over an angular displacement does work, W = τΔθ, even though the axle stays put.
CIt transfers no energy, as the friction force is perpendicular to the radius. A student who applies 'perpendicular forces do no work' to the radius picks this. A force does no work if it is perpendicular to the DISPLACEMENT of its point of application; the rim at P moves tangentially, parallel to Ff, so Ff does work (negative work, as it opposes the motion).
DIt transfers energy out of the wheel, as it opposes the rotation.Correct FN points along the line to the axle, so it exerts no torque. Ff is tangent to the rim and points down at P, where the counterclockwise-turning rim is moving up; its torque is clockwise, opposite to the rotation. A torque exerted against the angular displacement does negative work, so energy is transferred out of the wheel, which slows down.
A string is wrapped around the rim of a flywheel of radius R and rotational inertia I that can turn on a fixed axle with negligible friction. Starting from rest, the string is pulled with a constant force of magnitude F, tangent to the rim, while the flywheel turns through an angle Δθ. Which expression gives the flywheel's final angular speed?
Answer and reasoning
A√(2FRΔθ/I)Correct The string's torque is τ = RF, and it does work W = τΔθ = FRΔθ. With no friction this is the only torque that does work, so it equals the gain in rotational kinetic energy: FRΔθ = (1/2)Iω². Solving, ω = √(2FRΔθ/I).
B√(2FR/I) A student who takes the work done by the string's torque to be the torque itself, FR, because both can be written in newton-meters, picks this. Work also needs the angle turned through: W = τΔθ = FRΔθ, so the final angular speed depends on Δθ as well.
C2FRΔθ/I A student who writes the kinetic energy as (1/2)Iω, without the square, picks this. Rotational kinetic energy is (1/2)Iω², so solving for ω needs a square root.
DFRΔθ/I A student who treats τΔθ like an impulse and sets it equal to Iω, the rotational version of mv, picks this. τΔθ is work and equals the gain in kinetic energy, (1/2)Iω², so solving for ω needs the ½ and a square root.
Working τ = FR. W = τΔθ = FRΔθ = ΔK = (1/2)Iω² − 0 → ω = √(2FRΔθ/I). Errors: work taken as the torque itself, W = FR → √(2FR/I); K = (1/2)Iω → 2FRΔθ/I; τΔθ = Iω (impulse analogy) → FRΔθ/I.
Motor 1 and motor 2 each turn a separate wheel. The graph shows the torque that each motor exerts on its wheel, τ1 and τ2, as a function of the wheel's angular position θ, from θ = 0 to θ = 4.0 rad. Which statement correctly compares the work done by the two torques over this interval?
Answer and reasoning
ATorque τ1 does more work, as more area lies under its graph.Correct Work is the area under each τ–θ graph. τ1: (3.0 N·m)(4.0 rad) = 12 J. τ2: (1/2)(5.0 N·m)(4.0 rad) = 10 J. So τ1 does more work over the interval, although τ2 ends at the larger torque.
BTorque τ2 does more work, as it reaches the greater torque. A student who compares the largest values on the graph picks this. τ2 reaches 5.0 N·m only at the end of the interval; for most of the rotation it is smaller than τ1, and the area under its graph, 10 J, is less than τ1's 12 J.
CTorque τ2 does more work, as its graph has the steeper slope. A student who takes the slope of the graph as a measure of the work picks this. The slope shows how quickly the torque changes with angle; the work is the area under the graph, which is greater for τ1.
DThe two torques do equal work, as their graphs cross. A student who takes the crossing point to mean the torques are equivalent picks this. The graphs cross where the two torques are equal at one angular position, 2.4 rad; the work over the whole interval is the area under each graph, 12 J for τ1 and 10 J for τ2.
Working W1 = (3.0)(4.0) = 12 J. W2 = (1/2)(5.0)(4.0) = 10 J. W1 > W2. The graphs cross where 5.0θ/4.0 = 3.0, θ = 2.4 rad.
A rope wrapped around a drum that turns on a fixed axle is pulled with a constant force F, tangent to the drum, until a length ℓ of rope has unwound. The drum is then replaced by one of the same mass and shape but twice the radius, and the rope is again pulled with force F until the same length ℓ has unwound. The work done on the new drum by the rope's force equals the work done on the original drum multiplied by which factor?
Answer and reasoning
A0.50 A student who takes the torque to be the force, unchanged at F, but halves the angle gets 1 × 1/2 = 0.50. The same force at twice the radius exerts twice the torque, which makes up for the smaller angle.
B2.00 A student who doubles the torque but keeps the angle the same gets 2 × 1 = 2.00. The same length of rope unwinding from a drum of twice the radius turns it through only half the angle, Δθ = ℓ/r.
C1.00Correct At twice the radius the torque doubles, τ = (2R)F. Unwinding the same length ℓ turns the larger drum through half the angle, Δθ = ℓ/(2R). W = τΔθ = (2RF)(ℓ/2R) = Fℓ, the same as before: factor 1.00.
D4.00 A student who finds the angle as Δθ = ℓ·r doubles the angle as well as the torque and gets 2 × 2 = 4.00. The angle is Δθ = ℓ/r, so the same length of rope turns a drum of twice the radius through half the angle, which exactly offsets the doubled torque.
An electric motor keeps a flywheel turning at constant angular velocity. The motor exerts a torque on the flywheel in the sense of its rotation, and friction in the axle exerts an opposing torque of equal magnitude. A student claims: “The flywheel's kinetic energy does not change, so the motor's torque does no work on the flywheel.” Which reasoning correctly evaluates the student's claim?
Answer and reasoning
AIt is right: a torque does work only if it changes the flywheel's angular speed. A student who applies 'no change in kinetic energy, no work' to one torque picks this. The work–energy theorem is about the net work from all the torques; the motor's torque on its own does positive work as the flywheel turns.
BIt is wrong: the motor's torque does positive work; friction's does negative work.Correct The motor's torque is exerted in the sense of the rotation as the flywheel turns through an angle, so it does positive work, τΔθ, transferring energy in. The friction torque does an equal amount of negative work, transferring energy out. The NET work is zero, which is why the kinetic energy stays constant.
CIt is right: the flywheel's center of mass stays at rest, so no work is done. A student who thinks work requires the center of mass to move picks this. The flywheel turns through an angle, and a torque exerted over an angular displacement does work, W = τΔθ, even though the axle does not move.
DIt is wrong: both torques transfer energy into the flywheel, as both are exerted on it. A student who thinks every torque exerted puts energy in picks this. The friction torque opposes the rotation, so it does negative work and transfers energy out of the flywheel; if both transferred energy in, the flywheel would speed up.
A flywheel of rotational inertia I starts from rest and turns on a fixed axle with negligible friction. A motor exerts a torque τ on the flywheel. The graph shows τ as a function of the flywheel's angular position θ, from θ = 0 to θ = θ₀. Which expression gives the flywheel's angular speed at θ = θ₀?
Answer and reasoning
A√(2τ₀θ₀/I) A student who multiplies the largest torque, τ₀, by the whole angular displacement, θ₀, takes W = τ₀θ₀ and picks this. The torque rises from zero, so the area under the graph is a triangle, (1/2)τ₀θ₀, half of that rectangle.
Bτ₀θ₀/I A student who finds the work correctly, (1/2)τ₀θ₀, but sets it equal to (1/2)Iω, without the square on ω, picks this. Rotational kinetic energy is (1/2)Iω², so a square root is needed to find ω.
C√(τ₀θ₀/I)Correct The work done by the torque is the area under the graph, a triangle of area (1/2)τ₀θ₀. With negligible friction this work is the flywheel's gain in kinetic energy from rest: (1/2)Iω² = (1/2)τ₀θ₀, so ω = √(τ₀θ₀/I).
D√(2τ₀/(Iθ₀)) A student who takes the slope of the graph, τ₀/θ₀, as the work and sets (1/2)Iω² = τ₀/θ₀ picks this. The slope shows how fast the torque changes with angle; the work is the area under the graph.
Working The work done by the motor's torque is the area under the τ–θ graph, a triangle: W = (1/2)τ₀θ₀. Friction is negligible, so this is the only torque doing work, and it equals the gain in rotational kinetic energy from rest: (1/2)Iω² = (1/2)τ₀θ₀, so ω = √(τ₀θ₀/I). Errors: largest torque × whole angle, W = τ₀θ₀ → √(2τ₀θ₀/I); correct work but W = (1/2)Iω without the square → ω = τ₀θ₀/I; slope τ₀/θ₀ taken as the work → √(2τ₀/(Iθ₀)).
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
