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 constant torque of 40 N·m on a rusted bolt for 5.0 s, but the bolt does not turn. Which statement about the energy the torque transfers to the bolt, with its reasoning, is correct?
Answer and reasoning
ASome energy, as the torque was exerted on the bolt for a time interval of 5.0 s. A student who links the energy a torque transfers to how long it acts picks this. Work depends on the angular displacement, not the time: with no rotation the work is exactly zero, however long the torque acts.
BNone, as the rust exerts an equal and opposite torque that balances it. A student who confuses the work done by one torque with the net work picks this. A balanced torque does work whenever the object turns, as a motor's torque does on a wheel turning steadily against friction; here it does none only because the bolt does not turn.
CNone, as the bolt turned through no angle while the torque was exerted on it.Correct The work done by a torque is W = ∫τ dθ. The bolt did not rotate, so Δθ = 0 and the torque, however large and however long it is exerted, did no work on the bolt.
DNone, as the bolt's angular speed did not change during the 5.0 s. A student who thinks energy is transferred only when speed changes picks this. A motor turning a wheel at constant speed against friction transfers energy to it; what makes the work zero here is the zero angular displacement.
Working W = ∫τ dθ = (40 N·m)(0 rad) = 0 J: no angular displacement, so no work, independent of the 5.0 s duration.
A rope wound around a winch drum of radius 0.20 m is pulled with a constant force of magnitude 50 N, tangent to the drum, while the drum turns through 3.0 revolutions. How much work does the torque exerted by the rope do on the drum?
Answer and reasoning
A1.9 × 10² JCorrect The torque is τ = rF = (0.20 m)(50 N) = 10 N·m, and the angular displacement is 3.0 rev × 2π rad/rev = 6.0π rad ≈ 18.8 rad. W = τΔθ = (10 N·m)(18.8 rad) ≈ 1.9 × 10² J. As a check, this equals the force times the length of rope pulled off, (50 N)(2π × 0.20 m × 3.0).
B3.0 × 10¹ J A student who uses the number of revolutions as the angular displacement picks this: (10 N·m)(3.0) = 30 J. W = τΔθ needs Δθ in radians, and 3.0 rev is 6.0π rad.
C9.4 × 10² J A student who leaves the lever arm out of the torque picks this: (50 N)(6.0π rad) ≈ 940 J. The torque of the rope is rF = 10 N·m, not 50 N·m.
D1.0 × 10¹ J A student who takes the torque itself, 10 N·m, as the energy transferred picks this. Torque and work share the unit N·m, but the work is the torque multiplied by the angular displacement, 6.0π rad.
Working τ = rF = (0.20 m)(50 N) = 10 N·m. Δθ = 3.0 × 2π = 18.85 rad. W = τΔθ = 188 J ≈ 1.9 × 10² 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 this torque do on the wheel as the wheel turns from θ = 0 to θ = 6.0 rad?
Answer and reasoning
A36 J A student who treats the torque as constant at its greatest value picks this: (6.0 N·m)(6.0 rad) = 36 J. The torque is less than 6.0 N·m everywhere except at θ = 2.0 rad, so the area under the graph is smaller.
B24 J A student who multiplies the torque at the final position by the whole angular displacement picks this: (4.0 N·m)(6.0 rad) = 24 J. The torque varies, so the work is the whole area under the graph, not one value times the width.
C12 J A student who averages the first and last torques, (0 + 4.0)/2 = 2.0 N·m, picks this: (2.0 N·m)(6.0 rad) = 12 J. The torque does not change linearly over the interval, so the mean of the end values is not its average; the area is 26 J.
D26 JCorrect The work is the area under the τ–θ graph. From 0 to 2.0 rad it is a triangle: (1/2)(2.0 rad)(6.0 N·m) = 6.0 J. From 2.0 to 6.0 rad it is a trapezoid: (1/2)(6.0 + 4.0 N·m)(4.0 rad) = 20 J. The total is 26 J.
Working Area 0–2.0 rad: (1/2)(2.0)(6.0) = 6.0 J. Area 2.0–6.0 rad: (1/2)(6.0 + 4.0)(4.0) = 20 J. W = 26 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 Work done by a torque Fix
Work done by a torque
The energy transferred into or out of a rigid system by a torque exerted while the system rotates through an angular displacement. A torque exerted on a system that does not rotate does no work on it. Unit: joule (J).
Angular displacement, Δθ
The angle, measured in radians, through which a rigid system rotates about an axis: Δθ = θ − θ₀. One revolution is 2π rad.
Students often think The energy a torque (or force) transfers depends on how long it is exerted, so a torque exerted for a longer time does more work. In fact Not in itself. The work done by a torque depends on the angle through which the system turns while the torque is exerted, W = ∫τ dθ, not on the time. A torque exerted for any length of time on an object that does not turn does no work on it.
Students often think A torque that is balanced by an equal and opposite torque does no work. In fact No. Each torque does work τΔθ on its own if the object turns through an angle. Balanced torques make the net work zero, but each individual torque can still transfer energy, as a motor's torque does on a wheel turning steadily against friction.
6.2.A.2 Torque, τ Fix
Torque, τ
The turning effect of a force about an axis, τ⃗ = r⃗ × F⃗, of magnitude rF sin θ, where r is the distance from the axis to the point where the force is exerted and θ the angle between r⃗ and F⃗. Unit: N·m.
Work done by a torque (equation)
W = ∫θ1θ2 τ dθ. For a constant torque this becomes W = τΔθ, with Δθ in radians.
Sign of the work done by a torque
The work done by a torque is positive when the torque is in the same sense as the rotation, transferring energy into the system, and negative when it is in the opposite sense, transferring energy out (as a friction torque at an axle does).
Work–energy theorem for a rotating rigid system
For a rigid system rotating about a fixed axis, the net work done on it by the torques exerted on it equals the change in its rotational kinetic energy: Wnet = ΔKrot = (1/2)Iω² − (1/2)Iω₀².
Students often think The unit of an angle does not matter in W = τΔθ, so a number of revolutions or degrees can be used as Δθ. In fact No. The equation W = τΔθ, like s = rθ, holds with Δθ in radians. One revolution is 2π rad.
Students often think The torque of a force does not depend on where the force is exerted, so the work done by a torque can be found as the force times the angular displacement. In fact No. The torque of a force is rF sin θ, so it depends on the distance from the axis to the point where the force is exerted. The work done by the torque, τΔθ, depends on that distance too.
6.2.A.3 Work from a torque–angular position graph Fix
Work from a torque–angular position graph
The work done by a torque between two angular positions equals the area between the graph of τ against θ and the θ-axis over that interval, counted positive above the axis and negative below it.
Students often think The average of any changing quantity is the mean of its initial and final values, whatever happens in between. In fact It is if the torque changes linearly over the interval, but not in general: the average torque is (1/Δθ)∫τ dθ, which depends on the shape of the graph between the end points.
Students often think The work done by a varying torque is the torque at the final position multiplied by the whole angular displacement. In fact No. The work done is the area under the τ–θ graph over the whole interval. The torque at one position says nothing about its values elsewhere.
5 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 5
A wheel of rotational inertia I starts from rest. The net torque exerted on it depends on its angular position θ as τ = τ₀(1 − θ²/θ₁²), where τ₀ and θ₁ are positive constants, so the torque falls to zero at θ = θ₁. What is the wheel's angular speed at θ = θ₁?
Answer and reasoning
A1.41√(τ₀θ₁/I) A student who treats the torque as constant at its greatest value, τ₀, picks this: W = τ₀θ₁ and (1/2)Iω² = τ₀θ₁, so ω = √(2τ₀θ₁/I) ≈ 1.41√(τ₀θ₁/I). The torque falls to zero, so the work is ∫τ dθ = 2τ₀θ₁/3, less than τ₀θ₁.
B1.15√(τ₀θ₁/I)Correct The work done is W = ∫₀θ₁ τ₀(1 − θ²/θ₁²) dθ = τ₀θ₁ − τ₀θ₁/3 = 2τ₀θ₁/3, the area under the τ–θ curve. Setting (1/2)Iω² = 2τ₀θ₁/3 gives ω = √(4τ₀θ₁/(3I)) = √(4/3)·√(τ₀θ₁/I) ≈ 1.15√(τ₀θ₁/I).
C0.82√(τ₀θ₁/I) A student who finds the work correctly as 2τ₀θ₁/3 but writes the kinetic energy as Iω² picks this: ω² = 2τ₀θ₁/(3I), so ω ≈ 0.82√(τ₀θ₁/I). With Krot = (1/2)Iω², ω² = 4τ₀θ₁/(3I).
D1.00√(τ₀θ₁/I) A student who takes the average torque to be the mean of its values at the two ends, (τ₀ + 0)/2 = τ₀/2, picks this: W = τ₀θ₁/2 and (1/2)Iω² = τ₀θ₁/2, so ω = √(τ₀θ₁/I). The torque does not fall linearly with θ, so its average is (1/θ₁)∫τ dθ = 2τ₀/3, not τ₀/2.
Working W = ∫₀θ₁ τ₀(1 − θ²/θ₁²) dθ = τ₀θ₁ − τ₀θ₁/3 = 2τ₀θ₁/3. Work–energy: (1/2)Iω² − 0 = 2τ₀θ₁/3 → ω = √(4τ₀θ₁/(3I)) = √(4/3)·√(τ₀θ₁/I) ≈ 1.15√(τ₀θ₁/I). Distractors: W = τ₀θ₁ (torque taken as τ₀ throughout) → √(2τ₀θ₁/I) ≈ 1.41√(τ₀θ₁/I); Iω² = 2τ₀θ₁/3 → √(2τ₀θ₁/(3I)) ≈ 0.82√(τ₀θ₁/I); average torque taken as (τ₀ + 0)/2, W = τ₀θ₁/2 → 1.00√(τ₀θ₁/I).
A wheel is rotating in the positive θ direction as it passes θ = 0. The graph shows the net torque τ exerted on the wheel as a function of its angular position θ. At which position is the wheel's rotational kinetic energy greatest?
Answer and reasoning
AAt θ₁, as beyond θ₁ the torque decreases and so the wheel begins to slow down. A student who thinks a decreasing torque slows the wheel picks this. Between θ₁ and θ₂ the net torque is smaller but still in the sense of rotation, so it still does positive work and the wheel keeps speeding up.
BAt θ₃, as by then the wheel has turned through the greatest angle under the torque. A student who adds all the area under the graph as positive picks this. Beyond θ₂ the torque opposes the rotation, so the area there is negative work and the kinetic energy decreases.
CEverywhere from 0 to θ₁, as the torque has its greatest value throughout that range. A student who reads the kinetic energy from the height of the graph picks this. The torque gives the rate of energy transfer per radian; the kinetic energy follows the accumulated area, which is still growing after θ₁.
DAt θ₂, as the graph's area is positive up to θ₂ and negative beyond it.Correct The kinetic energy at any position equals the kinetic energy at θ = 0 plus the net work done since, the area under the τ–θ graph from 0. That area grows while τ is positive, up to θ₂, and shrinks once τ is negative, so the kinetic energy is greatest at θ₂.
A constant net torque turns a wheel, initially at rest, through an angle θ₁; the wheel's angular speed is then ω₁. Starting again from rest, the same torque turns the same wheel through an angle 2θ₁. What is the wheel's new final angular speed divided by ω₁?
Answer and reasoning
A1.41Correct The work done by the constant torque is τΔθ, so doubling the angle doubles the work and the final kinetic energy, (1/2)Iω². Since ω ∝ √K, the angular speed increases by a factor of √2 ≈ 1.41.
B2.00 A student who takes kinetic energy to be proportional to angular speed picks this: twice the work gives twice the angular speed. Krot depends on ω², so twice the energy gives √2 times the angular speed.
C4.00 A student who applies the dependence the wrong way round, squaring the factor instead of taking its square root, picks this: 2² = 4. Since ω² ∝ W, the factor for ω is √2.
D1.00 A student who thinks a given torque produces a given angular speed picks this. A net torque produces angular acceleration, so the longer it acts (here, through twice the angle) the faster the wheel turns.
Working W = τΔθ; (1/2)Iω² = τΔθ → ω ∝ √Δθ. Doubling Δθ: ω/ω₁ = √2 = 1.41.
Motor A exerts a torque that holds a wheel at rest against a spring. Motor B exerts a torque of the same magnitude that keeps an identical wheel turning at constant angular speed against friction at its axle. Each motor exerts its torque for the same time interval. Which motor transfers more energy to its wheel during the interval?
Answer and reasoning
ANeither, as neither wheel's angular speed changes during the time interval. A student who thinks energy is transferred only when speed changes picks this. Motor B does positive work on wheel B and friction does equal negative work, so the net work is zero, but motor B still transfers energy to the wheel.
BThey transfer equal amounts, as equal torques act for equal intervals of time. A student who links the energy transferred by a torque to the time it acts picks this. Work depends on angular displacement: wheel A turns through no angle, so motor A does no work on it.
CMotor B, as only wheel B turns through an angle while the torque is exerted.Correct Work done by a torque is ∫τ dθ. Wheel A does not rotate, so motor A does no work on it. Wheel B turns through an angle, so motor B does positive work on it; friction at the axle does an equal amount of negative work, which is why wheel B's kinetic energy stays the same.
DMotor A, as holding the wheel still against the spring takes more effort. A student who equates effort with energy transferred picks this. A torque that holds an object at rest does no work on it, because the object turns through no angle.
A wheel of rotational inertia I, initially at rest, is turned by a constant net torque of magnitude τ through an angle θ₁. What is the wheel's angular speed at the end of this rotation?
Answer and reasoning
A√(τθ₁/I) A student who writes the kinetic energy as Iω², without the ½, picks this. With (1/2)Iω² = τθ₁, ω² = 2τθ₁/I.
B√(2τθ₁/I)Correct The constant net torque does work W = τθ₁ on the wheel, which by the work–energy theorem equals its gain in kinetic energy: (1/2)Iω² = τθ₁, so ω = √(2τθ₁/I).
C2τθ₁/I A student who takes kinetic energy to be proportional to angular speed, (1/2)Iω = τθ₁, picks this. Krot = (1/2)Iω², so ω is the square root of 2τθ₁/I.
Dτ/I A student who confuses angular acceleration with angular speed picks this. τ/I is the angular acceleration α, in rad/s²; the angular speed reached also depends on the angle through which the torque acts.
Working W = τθ₁ = ΔK = (1/2)Iω² − 0 → ω = √(2τθ₁/I). Check with kinematics: α = τ/I, ω² = 2αθ₁ = 2τθ₁/I.
Compiled from the AP Physics C: Mechanics 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
