3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A light string is wrapped around a pulley of radius 0.0600 m that turns on a fixed axle. The string is pulled so that it unwinds without slipping while the pulley turns through 20.0 rad. What length of string unwinds from the pulley?
Answer and reasoning
A2.40 m A student who uses the diameter, 0.120 m, in place of the radius picks this. The string leaves the rim, one radius (0.0600 m) from the axle, so Δs = rΔθ with r = 0.0600 m.
B1.20 mCorrect Each point on the rim travels an arc length Δs = rΔθ, and the string unwinds by that length: (0.0600 m)(20.0 rad) = 1.20 m. The equation applies directly because the angle is in radians.
C7.54 m A student who treats the 20.0 rad as 20.0 revolutions and multiplies by 2π picks this. The angle is already in radians, so Δs = rΔθ needs no conversion.
D68.8 m A student who converts 20.0 rad to degrees (about 1146°) before using Δs = rΔθ picks this. The equation holds only with the angle in radians; with degrees the result is 57.3 times too large.
Working Δs = rΔθ = (0.0600 m)(20.0 rad) = 1.20 m. Distractors: diameter 0.120 m → 2.40 m; ×2π → 7.54 m; degrees 1145.9° × 0.0600 m → 68.8 m.
A wheel rotates counterclockwise about a fixed axle with constant angular velocity ω. Point P is on the rim, a distance r from the axle. Which statement about P's velocity at any instant is correct?
Answer and reasoning
AIt points toward the axle, the direction in which P accelerates. A student who thinks velocity points the same way as acceleration picks this. P's acceleration does point toward the axle, but its velocity is along its path, tangent to the rim; the two are at right angles here.
BIt points away from the axle, since P tends to fly outward. A student who thinks points on a rotating object are thrown outward picks this. P stays on its circle; at each instant it moves along the tangent, and the rest of the wheel exerts forces on it that keep it on the circle.
CIt stays the same at every instant, since P's speed rω is constant. A student who treats speed and velocity as the same quantity picks this. P's speed, rω, is constant, but velocity includes direction, and the direction of P's motion turns continuously as P goes around the axle, so the velocity changes.
DIt is tangent to the rim, so its direction keeps changing.Correct P moves along a circle about the axle, so at any instant its velocity is along the tangent to that circle (in the counterclockwise sense of rotation), with magnitude v = rω. As P goes around, the tangent turns, so the velocity's direction changes continuously even though its magnitude stays rω.
The diagram shows a disk that rotates counterclockwise about a fixed axle through its center O and is speeding up. Points P and Q lie on the same radius, at the distances from O shown. At the instant shown, P's tangential acceleration is 1.2 m/s². What is Q's angular acceleration at that instant?
Answer and reasoning
A8.0 rad/s² A student who finds P's angular acceleration, 4.0 rad/s², and then doubles it because Q is twice as far from O picks this. It is Q's TANGENTIAL acceleration that is twice P's; the angular acceleration is the same at every point of the disk.
B4.0 rad/s²Correct P's angular acceleration is α = aT/r = (1.2 m/s²)/(0.30 m) = 4.0 rad/s². The disk is a rigid system, so every point, Q included, has the same angular acceleration: 4.0 rad/s². Q's tangential acceleration, (0.60 m)(4.0 rad/s²) = 2.4 m/s², is twice P's.
C2.0 rad/s² A student who assumes Q has the same tangential acceleration as P, 1.2 m/s², and divides by Q's distance, 0.60 m, picks this. Points of a rigid system share angular quantities, not linear ones: Q's tangential acceleration is twice P's.
D1.2 rad/s² A student who gives P's tangential acceleration, 1.2 m/s², as the disk's angular acceleration picks this. Tangential acceleration (m/s²) and angular acceleration (rad/s²) are different quantities, related by aT = rα, so α = aT/r.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.2.A.1 Arc length (Δs) Fix
Arc length (Δs)
The distance a point travels along its circular path about a fixed axis. For a point a distance r from the axis, when the system rotates through Δθ, Δs = rΔθ. SI unit: m.
Radian measure in Δs = rΔθ
Δs = rΔθ is the definition of the radian (an arc equal to the radius subtends 1 rad) rearranged, so the angle must be in radians. An angle in revolutions is first multiplied by 2π; an angle in degrees by π/180.
Students often think The size of a wheel or pulley used in Δs = rΔθ, v = rω or aT = rα can be its diameter, since both describe how big it is. In fact r is the distance from the axis of rotation to the point, the radius of the circle the point moves on. Using the diameter doubles the result.
Students often think Radians and revolutions can be swapped: an angle given in radians can be treated as a number of revolutions and multiplied by 2π, or a revolution can be taken as 1 rad. In fact No. One revolution is 2π rad. An angle already in radians goes straight into Δs = rΔθ; only an angle in revolutions must first be multiplied by 2π.
5.2.A.2 Linear (tangential) velocity of a point Fix
Linear (tangential) velocity of a point
A point a distance r from the axis of a rotating rigid system moves along the tangent to its circular path with speed v = rω. Its speed is proportional to its distance from the axis. SI unit: m/s.
Tangential acceleration (aT)
The component of a point's acceleration along its path, aT = rα. It is the rate at which the point's speed changes, so it is zero when the angular velocity is constant. SI unit: m/s².
Tangential and centripetal acceleration of a point
A point on a rotating system can have two perpendicular components of acceleration: tangential, aT = rα, which changes its speed, and centripetal, ac = v²/r, directed toward the axis, which changes the direction of its velocity. A point on a wheel turning at constant ω has aT = 0 but a nonzero ac.
Students often think Angular and linear quantities are interchangeable: a system's angular velocity or angular acceleration can be given as a point's speed or tangential acceleration, or the reverse, without the factor r. In fact No. Angular velocity (rad/s) and speed (m/s), or angular acceleration (rad/s²) and tangential acceleration (m/s²), are different quantities linked by the distance from the axis: v = rω and aT = rα.
Students often think The tangential acceleration of a point on a rotating object is its centripetal acceleration, ω²r (or v²/r). In fact No. They are perpendicular components. The tangential acceleration, aT = rα, is along the path and changes the point's speed. The centripetal acceleration, ac = v²/r = ω²r, points toward the axis and changes the direction of the velocity.
5.2.A.3 Shared angular quantities of a rigid system Fix
Shared angular quantities of a rigid system
Every point of a rigid system turns through the same angle in the same time, so all points have the same angular velocity and the same angular acceleration. Linear quantities (arc length, speed, tangential acceleration) differ from point to point in proportion to r.
Students often think All points of a rotating rigid system move with the same speed and have the same tangential acceleration, because they go around together. In fact No. All points share the same angular velocity and angular acceleration, but their speeds, v = rω, and tangential accelerations, aT = rα, are proportional to their distances from the axis.
Students often think Points farther from the axis of a rotating rigid system have a greater angular velocity and a greater angular acceleration, because they move faster and gain speed faster. In fact No. Every point of a rigid system turns through the same angle in the same time, so all points have the same ω and the same α. Only linear quantities (speed, tangential acceleration) grow with distance from the axis.
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 graph shows the angular velocity ω of a wheel rotating about a fixed axle as a function of time t. Point P on the wheel is 0.50 m from the axle. What is the magnitude of P's tangential acceleration at t = 1.0 s?
Answer and reasoning
A1.0 m/s²Correct The angular acceleration is the slope of the ω–t graph, (9 − 1) rad/s ÷ 4 s = 2.0 rad/s², the same at every instant. P's tangential acceleration is aT = rα = (0.50 m)(2.0 rad/s²) = 1.0 m/s².
B2.0 m/s² A student who gives the angular acceleration, 2.0 rad/s², as the tangential acceleration picks this. They are different quantities: a point's tangential acceleration is its distance from the axle times α, aT = rα.
C4.5 m/s² A student who calculates ω²r = (3 rad/s)²(0.50 m) at t = 1.0 s picks this. That is P's centripetal acceleration, directed toward the axle; the tangential acceleration, along P's path, depends on α, not on ω.
D1.5 m/s² A student who multiplies the angular velocity at t = 1.0 s, 3 rad/s, by 0.50 m picks this. rω = 1.5 m/s is P's SPEED at that instant; the tangential acceleration is the rate at which that speed changes, rα.
Working α = slope = (9 − 1) rad/s ÷ (4 − 0) s = 2.0 rad/s². aT = rα = (0.50 m)(2.0 rad/s²) = 1.0 m/s². At t = 1.0 s, ω = 3 rad/s: rω = 1.5 m/s (speed); ω²r = 4.5 m/s² (centripetal).
The diagram shows a top view of a merry-go-round rotating counterclockwise at a constant rate about its center O. Children A and B sit at the distances from O shown. Which statement correctly compares the children's motions?
Answer and reasoning
AB's speed is equal to A's, since the children go around together. A student who thinks all points of a rotating system move at the same speed picks this. The children go around in the same time, but B's circle is twice as long, so B moves twice as fast.
BB's angular velocity is twice A's, since B is twice as far from O. A student who thinks angular velocity grows with distance from the axis picks this. Both children turn through 2π rad in the same time, so their angular velocities are equal; it is their speeds that differ.
CB's speed is twice A's, since B travels twice as far in each revolution.Correct Both children turn through the same angle in the same time, so they have the same angular velocity. Their speeds are v = rω, so B, twice as far from O, moves twice as fast: in each revolution B travels 2π(2.0 m), twice A's 2π(1.0 m).
DB's speed is four times A's, since B's circle has four times as much area. A student who scales the speed by the area of each child's circle (πr²) picks this. The distance traveled in one revolution is the circumference, 2πr, which is proportional to r, so B's speed is twice A's.
Working Same ω for both. v = rω: vB/vA = 2.0 m/1.0 m = 2. Path per revolution: 2π(1.0) = 6.3 m for A, 2π(2.0) = 12.6 m for B.
A wheel rotates counterclockwise about a fixed axle with constant angular velocity. Which statement about a point on the wheel's rim is correct?
Answer and reasoning
AIts tangential acceleration is zero, because α is zero.Correct With constant angular velocity, α = 0, so the tangential acceleration aT = rα is zero and the point's speed does not change. The point still has a centripetal acceleration, v²/r, toward the axle, because the direction of its velocity changes.
BIts acceleration is zero, as its speed is constant. A student who thinks an object moving at constant speed cannot be accelerating picks this. The point's speed is constant, but the direction of its velocity changes continuously, so it has a centripetal acceleration toward the axle.
CIts speed changes, since its direction of motion keeps changing. A student who treats speed and velocity as the same quantity picks this. The DIRECTION of the velocity changes, but the speed, rω, stays constant because r and ω are constant.
DIts acceleration points along its path, the way it moves. A student who thinks acceleration points the same way as velocity picks this. At constant angular velocity the point's speed does not change, so it has no acceleration along its path; its only acceleration is centripetal, toward the axle, at right angles to its velocity.
A turntable speeds up with constant angular acceleration. A student claims that a dot near the edge of the turntable has a greater angular acceleration than a dot near the center, because the edge dot's speed increases by more each second. Which statement correctly evaluates the claim?
Answer and reasoning
ACorrect: since aT = rα, the dot farther from the axis has the greater angular acceleration. A student who thinks angular quantities grow with distance from the axis reads aT = rα this way and picks this. The equation shows the opposite: α is shared, so aT grows with r.
BIncorrect: both dots gain speed at the same rate, so both have the same angular acceleration. A student who thinks all points of a rotating system have the same linear motion picks this. The conclusion about α is right but the reason is false: the edge dot, farther from the axis, gains speed faster, since aT = rα.
CCorrect: the edge dot moves faster, and a faster dot must have the greater angular acceleration. A student who links a greater speed to a greater acceleration picks this. How fast a dot moves says nothing about how fast its motion changes; both dots have the same α, whatever their speeds.
DIncorrect: each dot gains speed in proportion to its distance from the axis, so both have the same α.Correct Both dots are on one rigid system, so they share the same angular acceleration α. Each dot's tangential acceleration is aT = rα, so the edge dot's speed does increase by more each second, in exact proportion to its greater r; dividing each gain by r gives the same α.
Working Rigid system: α the same for both dots. aT = rα, so a dot at 3r gains speed 3 times as fast as a dot at r, while Δv/(rΔt) = α is the same for both.
A belt runs without slipping around two wheels that turn on fixed axles: a small wheel of radius r and a large wheel of radius R. The small wheel turns with constant angular speed ω. Which expression gives the angular speed of the large wheel?
Answer and reasoning
Aω A student who thinks linked wheels turn together at the same angular speed picks this. The belt fixes the speed of the rims, not the angular speed: the large wheel's rim must cover the same length of belt per second with a larger radius, so it turns more slowly.
Brω A student who finds the belt's speed, rω, and gives it as the large wheel's angular speed picks this. rω is a linear speed in m/s; the large wheel's angular speed is that speed divided by its own radius, rω/R.
Crω/RCorrect The belt does not slip, so both rims move with the belt's speed. For the small wheel that speed is v = rω; for the large wheel v = RωL. Setting RωL = rω gives ωL = rω/R, smaller than ω because R is greater than r.
Dr²ω/R² A student who thinks a rim point's speed scales with the area of its circle, r², sets r²ω = R²ωL and picks this. The distance a rim point travels in one revolution is 2πr, proportional to r, so the speeds are rω and RωL.
Working The belt does not slip, so each point of the belt in contact with a rim moves with that rim point's speed, and the belt has one speed v everywhere. Small wheel: v = rω. Large wheel: v = RωL. So RωL = rω and ωL = rω/R. Errors: same angular speed for linked wheels → ω; belt speed rω given as the angular speed (no division by R) → rω; rim speed taken to scale with r² → r²ω = R²ωL → r²ω/R².
A turntable rotates about a fixed axis through its center. Point P is a distance r from the axis, and point Q is a distance 3r from the axis. At time t1, the turntable's angular velocity has magnitude ω and its angular acceleration has magnitude α. At a later time t2, its angular velocity has magnitude 2ω and its angular acceleration has magnitude α/2. The magnitude of Q's tangential acceleration at t2 is how many times the magnitude of P's tangential acceleration at t1?
Answer and reasoning
A0.5 A student who thinks every point of a rotating rigid object has the same tangential acceleration picks this, keeping only the halving of α. By aT = rα, a point 3 times as far from the axis has 3 times the tangential acceleration, so the factor is 3 × 1/2 = 1.5.
B4.5 A student who thinks a point farther from the axis also has a greater angular acceleration picks this, giving Q 3 times the α as well as 3 times the radius: 3 × 3 × 1/2 = 4.5. All points of a rigid object share one α, so the radius enters only once: 3 × 1/2 = 1.5.
C1.5Correct Tangential acceleration is aT = rα, with α the same for every point of the turntable at a given instant. Q is 3 times as far from the axis, and at t2 the angular acceleration is half as large: 3 × 1/2 = 1.5. The angular velocity does not appear in aT.
D6.0 A student who finds tangential acceleration from angular velocity, as aT = rω, picks this: 3 × 2 = 6.0. Tangential acceleration depends on how fast the angular velocity changes, aT = rα, and α was halved: 3 × 1/2 = 1.5.
Working aT = rα, and every point of the rigid turntable has the same α at a given instant. P at t1: aT = rα. Q at t2: aT = (3r)(α/2) = 1.5rα. Ratio = 3 × 1/2 = 1.5. The angular velocity does not enter aT.
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
