3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A rigid wheel turns on a fixed axle. The net torque exerted on the wheel is not zero. Which statement about the wheel must be true?
Answer and reasoning
AIts angular velocity is changing with time.Correct A nonzero net torque means the torques on the wheel are unbalanced, so its angular velocity changes: it may speed up, slow down or reverse. Only the change is certain; the sense of rotation, and whether the wheel speeds up or slows down, depend on how the net torque compares with the rotation.
BIt turns in the same sense as the net torque. A student who thinks a wheel turns in the sense of the net torque picks this. The net torque sets the sense of the CHANGE in angular velocity: a wheel turning counterclockwise with a clockwise net torque on it keeps turning counterclockwise for a while as it slows down.
CIts angular speed is increasing with time. A student who thinks a net torque always speeds a rotation up picks this. If the net torque is opposite to the rotation, as when friction or a brake acts on the wheel, the angular speed decreases.
DIt turns at a steady rate set by the torque. A student who thinks a torque produces a steady rotation picks this. A steady rate of rotation needs a zero net torque; a nonzero net torque makes the angular velocity change.
A solid disk of radius 0.50 m and rotational inertia 0.25 kg·m² turns on a fixed axle. A string wrapped around its rim is pulled with a constant force of 2.0 N, tangent to the rim, in the sense in which the disk is turning. Friction at the axle exerts a constant torque of 0.25 N·m that opposes the rotation. What is the magnitude of the disk's angular acceleration?
Answer and reasoning
A4.0 rad/s² A student who uses the pull's torque as the net torque picks this: 1.0 N·m ÷ 0.25 kg·m² = 4.0 rad/s². Newton's second law in rotational form uses the NET torque, so the frictional torque must be subtracted first.
B5.0 rad/s² A student who adds the two torques, 1.0 N·m + 0.25 N·m, picks this. Friction opposes the rotation, so its torque has the opposite sign: the net torque is 1.0 N·m − 0.25 N·m = 0.75 N·m.
C8.0 rad/s² A student who divides the force by the rotational inertia, 2.0 N ÷ 0.25 kg·m², picks this. Angular acceleration depends on torque, not force: the pull's torque is rF = 1.0 N·m, and the net torque is 0.75 N·m. (F/I does not even have the units of rad/s².)
D3.0 rad/s²Correct The pull exerts a torque rF = (0.50 m)(2.0 N) = 1.0 N·m in the sense of rotation, and friction exerts 0.25 N·m in the opposite sense, so τnet = 0.75 N·m. Then α = τnet/I = 0.75 N·m ÷ 0.25 kg·m² = 3.0 rad/s².
Working Torque of the pull: τ = rF = (0.50 m)(2.0 N) = 1.0 N·m, in the sense of rotation. Frictional torque: 0.25 N·m, opposite. τnet = 1.0 N·m − 0.25 N·m = 0.75 N·m. α = τnet/I = 0.75 N·m ÷ 0.25 kg·m² = 3.0 rad/s².
Two identical uniform pucks rest on a horizontal surface with negligible friction. A string is tied to a peg at the center of puck 1, and a string is wrapped many times around the rim of puck 2. Each string is pulled horizontally with the same constant force F; puck 2's string unwinds as puck 2 spins. How do the accelerations of the pucks' centers of mass compare while the strings are pulled?
Answer and reasoning
APuck 1's is greater, since puck 2 uses some of F to spin. A student who thinks the force is shared between moving and spinning picks this. Newton's second law for the center of mass uses the whole net force, F, for both pucks; the torque that spins puck 2 belongs to a separate analysis and takes nothing away from F.
BPuck 1's equals puck 2's: each net force is F.Correct Translational analysis: the only horizontal force on each puck is F, so each center of mass has acceleration F/m, wherever the string is attached. Rotational analysis, done separately: F exerts a torque about puck 2's center, so puck 2 also has an angular acceleration, while puck 1 does not spin. The spin does not reduce puck 2's center-of-mass acceleration.
CPuck 2's is greater, since its string is pulled farther. A student who reasons that the hand pulling puck 2's string moves farther, so puck 2 must move faster, picks this. The hand does move farther, because the string unwinds as puck 2 spins, but the center-of-mass acceleration depends only on the net force and the mass: F/m for both pucks.
DPuck 2's is zero, since F makes puck 2 spin instead. A student who thinks a force that does not point through the center only spins an object picks this. Every external force contributes fully to the center-of-mass acceleration, wherever it is exerted; the force on puck 2 both accelerates its center of mass at F/m and spins it.
Working Translational analysis for each puck: the only horizontal external force is F, so acm = F/m for puck 1 and for puck 2. Rotational analysis about each center: puck 1's string pulls through its center (zero torque, no spin); puck 2's string pulls at the rim (torque FR, angular acceleration FR/I). The rotational result does not change the translational one, so the center-of-mass accelerations are equal.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.6.A.1 Net torque (τnet, unit: N·m) Fix
Net torque (τnet, unit: N·m)
The sum of the torques exerted on an object or rigid system about one axis, with torques in one sense (clockwise or counterclockwise) positive and the other negative: τnet = Σ τ. Only the net torque, not any single torque, determines whether and how the angular velocity changes.
Change in angular velocity
The angular velocity of an object or rigid system changes whenever the net torque on it is not zero. The change can be speeding up, slowing down or reversing the sense of rotation; with a zero net torque the angular velocity stays constant.
Students often think An object rotates in the sense of the net torque exerted on it, so when the net torque reverses, the rotation reverses at once. In fact No. The net torque sets the sense of the CHANGE in angular velocity, not the sense of rotation. A wheel turning counterclockwise with a clockwise net torque on it keeps turning counterclockwise while it slows down.
Students often think A nonzero net torque always increases an object's angular speed; slowing down is not caused by a net torque. In fact No. A net torque opposite to the rotation reduces the angular speed; only a net torque in the same sense as the rotation increases it. Brakes and friction slow rotations by exerting such torques.
The rate at which the angular velocity of a rigid system changes. For a rigid system it has the same sense as the net torque: if that sense is the same as the rotation, the system speeds up; if it is opposite, the system slows down.
Rotational inertia (I, unit: kg·m²)
A rigid system's resistance to changes in its rotation about an axis. It depends on the mass and on how far the mass is from the axis: for a small object, I = mr², so doubling r multiplies I by 4. The larger I is, the smaller the angular acceleration a given net torque produces.
Newton's second law in rotational form
αsys = Σ τ/Isys = τnet/Isys. The angular acceleration of a rigid system is directly proportional to the net torque exerted on it, in the same sense as that net torque, and inversely proportional to its rotational inertia about the axis.
Graph of α against torque
For a rigid system of fixed rotational inertia, a graph of angular acceleration against net torque is a straight line through the origin with slope 1/I. If a constant frictional torque τf opposes the rotation, a graph against the applied torque has the same slope but crosses the torque axis at τf.
Students often think A force and its torque are interchangeable: equal forces exert equal torques, and a force's size alone determines its turning effect. In fact No. Angular acceleration depends on the net torque, τ = rF⊥, so the same force produces a larger angular acceleration when it is exerted farther from the axis. Equal forces at different distances exert different torques.
Students often think The torque someone applies is the net torque, so frictional torques at an axle can be left out of α = τ/I. In fact Only if no other torque acts. α = τnet/I uses the NET torque, so an opposing frictional torque must be subtracted from the applied torque first.
5.6.A.3 Translational analysis of a rigid system Fix
Translational analysis of a rigid system
Newton's second law applied to the system's center of mass: a⃗cm = F⃗net/msys. It uses every external force at its full size, wherever on the system the force is exerted.
Rotational analysis of a rigid system
Newton's second law in rotational form applied about an axis: α = τnet/I. It uses the torques of the forces, so where each force is exerted matters. It is carried out separately from, and in addition to, the translational analysis.
Linking the two analyses
When the separate analyses must be combined, a physical constraint links them. For a string that unwinds from a pulley of radius R without slipping, the string's acceleration equals the tangential acceleration of the rim: a = Rα (from aT = rα).
Students often think A force exerted off-center is shared between making the object move and making it spin, so the center of mass accelerates less than it would if the force acted through the center. In fact No. The center-of-mass acceleration is Fnet/m, wherever the force is exerted. The spinning is described by a separate rotational analysis; it does not use up part of the force.
Students often think The farther the point where a force is exerted moves, the greater the acceleration of the object's center of mass. In fact Not necessarily. The center-of-mass acceleration depends only on the net force and the mass. A string that unwinds as an object spins moves farther than the object, without changing acm.
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
The diagram shows a disk that can rotate about a fixed axle through its center. Two forces, each of magnitude F, are exerted on the disk in the directions shown: one at the rim, a distance R from the axle, and one a distance R/2 from the axle. At the instant the forces begin to be exerted, the disk is rotating counterclockwise. Which describes the disk's motion just after this instant?
Answer and reasoning
AIt keeps turning counterclockwise at a constant rate. A student who compares the forces instead of their torques picks this. The forces are equal, but the force at the rim has twice the lever arm, so its clockwise torque is twice the counterclockwise torque and the net torque is not zero.
BIt keeps turning counterclockwise and speeds up. A student who thinks a net torque always speeds a rotation up picks this. The net torque is clockwise while the disk turns counterclockwise, so it reduces the angular speed; a net torque speeds the disk up only when it is in the same sense as the rotation.
CIt keeps turning counterclockwise but slows down.Correct The force at the rim exerts a clockwise torque FR about the axle; the force at R/2 exerts a counterclockwise torque F(R/2). The net torque is FR/2 clockwise, opposite to the rotation, so the counterclockwise angular velocity decreases. The angular velocity changes gradually, so the disk is still turning counterclockwise just after the forces begin.
DIt stops at once and begins turning clockwise. A student who thinks an object turns in the sense of the net torque picks this. The clockwise net torque changes the angular velocity gradually: the disk slows down while still turning counterclockwise, and only after it has stopped would it begin to turn clockwise.
A light rod can rotate in a horizontal plane about a vertical axle through its center, with negligible friction. Two small spheres, each of mass m, are fixed to the rod, one on each side of the axle at distance r from it. A constant force F exerted perpendicular to the rod at distance d from the axle gives the rod an angular acceleration α₀. The spheres are then moved to distance 2r from the axle, and the same force F is exerted at distance 2d. The new angular acceleration is how many times α₀?
Answer and reasoning
A2 A student who thinks rotational inertia depends only on mass picks this: the torque doubles and I is taken as unchanged. Moving the same mass farther from the axle increases I, here by a factor of 4, since I = mr².
B½Correct Doubling the lever arm doubles the torque: τ = F(2d) = 2Fd. Doubling the spheres' distance multiplies the rotational inertia by 2² = 4: I = 2m(2r)² = 4 × 2mr². So α = τnet/I changes by a factor of 2/4 = ½.
C1 A student who takes rotational inertia to be proportional to r instead of r² picks this: the torque doubles and I is taken to double, so α seems unchanged. With I = mr², doubling r multiplies I by 4.
D¼ A student who treats the same force as giving the same torque picks this: I is correctly multiplied by 4, but the torque is taken as unchanged. Exerting the same force at twice the distance doubles the torque.
Working Originally τ = Fd and I = 2mr², so α₀ = Fd/(2mr²). Afterwards τ = F(2d) = 2Fd and I = 2m(2r)² = 8mr² = 4 × 2mr². α = 2Fd/(8mr²) = (2/4) × Fd/(2mr²) = ½α₀.
A uniform rod of mass M and length L can rotate about a frictionless horizontal axle through one end. The rod's rotational inertia is (1/12)ML² about its center and (1/3)ML² about the axle. The rod is held horizontal and released from rest. Which expression gives the magnitude of the rod's angular acceleration just after it is released?
Answer and reasoning
A1.5g/LCorrect About the axle, only the gravitational force exerts a torque. It acts at the center of mass, L/2 from the axle, so τnet = MgL/2. With the rotational inertia about the axle, α = (MgL/2)/((1/3)ML²) = 3g/(2L) = 1.5g/L.
B3.0g/L A student who takes the lever arm of the gravitational force to be the rod's whole length gets α = MgL/((1/3)ML²) = 3.0g/L and picks this. The gravitational force on the uniform rod acts at its center of mass, L/2 from the axle.
C6.0g/L A student who uses the rotational inertia about the rod's center, (1/12)ML², gets (MgL/2)/((1/12)ML²) = 6.0g/L and picks this. The rod turns about the axle at its end, so the rotational inertia about that axle, (1/3)ML², must be used.
D1.0g/L A student who assumes the free end falls with acceleration g, as a dropped object would, sets aT = g at distance L and gets α = g/L. The axle also exerts a force on the rod, so the angular acceleration must come from α = τnet/I.
Working About the axle, the axle's force exerts no torque, and the gravitational force Mg acts at the center of mass, a lever arm L/2 from the axle (the rod is horizontal). τnet = Mg(L/2). α = τnet/I = (MgL/2)/((1/3)ML²) = 3g/(2L) = 1.5g/L. Errors: lever arm L → MgL/((1/3)ML²) = 3.0g/L; I about the center → (MgL/2)/((1/12)ML²) = 6.0g/L; free end taken to fall at g → α = g/L = 1.0g/L.
A student exerts different constant torques on a turntable that turns on an axle and measures the turntable's angular acceleration α for each. Friction at the axle exerts a constant torque on the turning turntable that opposes its rotation. The graph shows α as a function of the torque the student applies. What is the rotational inertia of the turntable?
Answer and reasoning
A4.0 kg·m² A student who divides the applied torque at the marked point by α, 4.0 N·m ÷ 1.0 rad/s², picks this. That treats the applied torque as the net torque. The line does not pass through the origin: 1.0 N·m of the applied torque only balances friction, so the net torque at that point is 3.0 N·m.
B5.0 kg·m² A student who adds the frictional torque to the applied torque, (4.0 + 1.0) N·m ÷ 1.0 rad/s², picks this. Friction opposes the rotation, so its torque is subtracted: the net torque at that point is 4.0 N·m − 1.0 N·m = 3.0 N·m.
C1.5 kg·m² A student who finds the area under the line, ½ × 3.0 N·m × 1.0 rad/s², picks this. The rotational inertia comes from the slope of an α–τ graph, which is 1/I, not from the area under it.
D3.0 kg·m²Correct The net torque is the applied torque minus the frictional torque, so α = (τapplied − τf)/I: a straight line of slope 1/I. The line crosses the torque axis at 1.0 N·m, the frictional torque, and reaches 1.0 rad/s² at 4.0 N·m. Its slope is 1.0 rad/s² per 3.0 N·m, so I = 3.0 N·m ÷ 1.0 rad/s² = 3.0 kg·m².
Working The net torque is τapplied − τf, so α = (τapplied − τf)/I: a straight line of slope 1/I that meets the torque axis at τf. From the graph, τf = 1.0 N·m and the line passes through (4.0 N·m, 1.0 rad/s²). Slope = (1.0 − 0) rad/s² ÷ (4.0 − 1.0) N·m, so I = 3.0 N·m ÷ 1.0 rad/s² = 3.0 kg·m². Check: at 4.0 N·m, τnet = 3.0 N·m and α = 3.0 N·m ÷ 3.0 kg·m² = 1.0 rad/s².
A block of mass m hangs from a light string wrapped around a pulley of radius R and rotational inertia I, which turns on a frictionless axle. The block is released from rest, and the string unwinds without slipping. Which expression gives the magnitude of the block's acceleration?
Answer and reasoning
Amg/(m+I/R) A student who writes the pulley equation as T = Iα, using the string's force in place of its torque TR, picks this. The torque about the axle is TR, so TR = Iα. The expression picked is not even dimensionally consistent: I/R has units of kg·m, not kg.
Bmg/(I/R²) A student who takes the tension to equal the block's weight picks this: mgR = Iα and a = Rα give a = mgR²/I. The block accelerates downward, so the net force on it is downward and the tension is less than mg; the block's own equation, mg − T = ma, is needed as well.
Cmg/(m+I/R²)Correct Two analyses, one for each object. Block: mg − T = ma. Pulley: TR = Iα, with α = a/R because the string does not slip, so T = Ia/R². Combining: a = mg/(m + I/R²). As a check, a pulley of negligible rotational inertia gives a = g.
Dmg/(m+I) A student who writes α = aR instead of α = a/R picks this: TR = IaR gives T = Ia and a = mg/(m + I). The string's acceleration equals the rim's tangential acceleration, a = Rα, so α = a/R; the expression picked adds a mass to a rotational inertia, which have different units.
Working Block (downward positive): mg − T = ma. Pulley: τ = TR = Iα. No slipping: a = Rα, so α = a/R and T = Iα/R = Ia/R². Substituting: mg − Ia/R² = ma, so a = mg/(m + I/R²). Checks: I → 0 gives a = g; a very large I gives a → 0.
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
