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 drum of radius R that turns on a fixed axle. Starting from rest, the drum turns with constant angular acceleration α, and the string unwinds without slipping. What length of string unwinds in time t?
Answer and reasoning
Aαt²/2 A student who takes the angle turned as the length of string picks this. (1/2)αt² is an angle in radians; the length unwound is R times that angle.
BRαt²/2Correct The drum turns through Δθ = (1/2)αt² from rest, and the string unwinds the arc length at the rim, Δs = RΔθ = Rαt²/2.
CRαt A student who gives the speed of the rim at time t picks this. Rαt is the rate at which string unwinds, in m/s; the length unwound grows as t², not t.
DRαt² A student who multiplies the final rim speed, Rαt, by the time picks this. The rim speeds up from zero, so its average speed is half the final speed and the length is Rαt²/2.
Working Angle turned from rest: Δθ = (1/2)αt². The string unwinds the arc length at the rim: Δs = RΔθ = (1/2)Rαt² = Rαt²/2. (Radius left out: αt²/2. Rim speed at time t: Rαt. Final rim speed × t: Rαt².)
A merry-go-round speeds up uniformly from 0.50 rad/s to 2.0 rad/s in 5.0 s. What is the tangential acceleration of a point on it 0.60 m from its axis?
Answer and reasoning
A0.30 m/s² A student who takes the angular acceleration as the point's tangential acceleration picks this. α = 0.30 rad/s² must be multiplied by the point's distance from the axis: aT = rα = 0.18 m/s².
B0.24 m/s² A student who divides the final speed by the time picks this: (0.60 m)(2.0 rad/s)/(5.0 s) = 0.24 m/s². The merry-go-round was already turning at 0.50 rad/s, so only the change in speed counts.
C0.90 m/s² A student who gives the change in the point's speed as its acceleration picks this: (0.60 m)(1.5 rad/s) = 0.90 m/s. That change takes 5.0 s, so the acceleration is (0.90 m/s)/(5.0 s) = 0.18 m/s².
A rigid disk speeds up as it turns about a fixed axle. Point A on the disk is near its rim and point B is near the axle. Which quantity is greater for A than for B?
Answer and reasoning
AThe point's angular velocity about the axle A student who thinks points farther from the axle rotate faster picks this. Every point of a rigid disk turns through the same angle in the same time, so A and B have the same angular velocity.
BThe point's angular acceleration about the axle A student who identifies angular acceleration with tangential acceleration picks this. A's tangential acceleration, rα, is greater than B's, but both points share the disk's angular acceleration α.
CThe point's time to make one full turn A student who thinks all points move at the same speed picks this: A's longer path would then take longer. All points complete a turn together, because they share one angular velocity.
DThe point's speed along its circular pathCorrect Both points share the disk's angular velocity, and v = rω, so A, being farther from the axle, moves faster along its path.
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 = rΔθ Fix
Arc length, Δs = rΔθ
The distance traveled along its circular path by a point a distance r from a fixed axis while the system rotates through Δθ. Δθ must be in radians. SI unit: m.
Students often think A point's speed and the distance it has traveled (or ω and θ) are treated as the same quantity, so an expression for one is used for the other or is assumed to grow in the same way. In fact No. Distance builds up over time, s = ∫v dt, and has different units from speed. Their time dependence also differs: from rest with constant α, v grows in proportion to t while s grows as t².
Students often think The distance traveled while the speed changes equals the final speed multiplied by the elapsed time. In fact No. That is true only if the point moves at its final speed the whole time. From rest with constant α, the average speed is half the final speed, so s = (1/2)vt.
5.2.A.2 Linear speed of a point, v = rω Fix
Linear speed of a point, v = rω
The speed of a point a distance r from the axis; its velocity is tangent to the point's circle. For a given ω, points farther from the axis move faster. SI unit: m/s.
Tangential acceleration, aT = rα
The component of a point's acceleration along its circular path, equal to the rate at which the point's speed changes. SI unit: m/s².
Acceleration of a point on a rotating rigid system
The vector sum of the tangential component aT = rα and the centripetal component ac = v²/r = rω². The two components are perpendicular, so the magnitude is √(aT² + ac²).
Students often think Linear and angular quantities are the same thing, so a point's arc length, speed or tangential acceleration equals the system's θ, ω or α without a factor of r. In fact No. They are linked through the point's distance from the axis: s = rθ, v = rω and aT = rα. The numbers agree only when r is 1 m, and even then the units differ (m/s and rad/s, for example).
Students often think The acceleration of a point on a rotating system is just aT = rα; the point has no other acceleration. In fact No. A point moving on a circle with ω ≠ 0 also has a centripetal component, ac = rω², directed toward the axis. aT = rα is only the component along the path.
5.2.A.3 Shared angular velocity and angular acceleration Fix
Shared angular velocity and angular acceleration
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; their linear speeds and tangential accelerations differ because their distances from the axis differ.
Students often think All points of a rotating rigid system move with the same linear speed, so points nearer the axis must have a greater angular velocity and points farther out take longer to go around. In fact No. They share one angular velocity, so each point's speed, v = rω, is proportional to its distance from the axis: a point twice as far out moves twice as fast.
Students often think Points farther from the axis rotate faster: a point's angular velocity increases in proportion to its distance from the axis. In fact No. Every point of a rigid system has the same ω. A point farther out moves faster because v = rω, not because it rotates faster.
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 figure shows a rigid disk rotating about a fixed axle at O, and two points on it, P and Q, at the distances from O shown. Which gives the ratios of the angular velocities and of the linear speeds of Q and P?
Answer and reasoning
AωQ/ωP = ½ and vQ/vP = 1 A student who thinks all points of the disk move at the same speed picks this; ω = v/r then makes Q's angular velocity half of P's. The points share one angular velocity, and it is their speeds that differ.
BωQ/ωP = 2 and vQ/vP = 4 A student who thinks a point farther out rotates faster, in proportion to r, picks this; v = rω then gives vQ = (2r)(2ωP) = 4vP. Every point of a rigid disk has the same ω, so vQ/vP = 2.
CωQ/ωP = 1 and vQ/vP = 2Correct All points of a rigid disk turn through the same angle in the same time, so ωQ = ωP. Speed is v = rω, and Q is twice as far from O as P, so Q moves twice as fast.
DωQ/ωP = 1 and vQ/vP = 1 A student who thinks every point of a rigid disk moves in exactly the same way picks this. The points share ω, but Q, twice as far from O, travels twice as far in each turn, so its speed is twice P's.
Working All points of a rigid disk share ω, so ωQ/ωP = 1. v = rω, so vQ/vP = (2r)/r = 2.
A wheel of radius R starts from rest and turns about a fixed axle with constant angular acceleration α. The magnitude of the acceleration of a point on the rim at time t is Rα multiplied by which factor?
Answer and reasoning
A√(1+α²t⁴)Correct The rim point has a tangential component aT = Rα and, since ω = αt, a centripetal component ac = Rω² = Rα²t². These are perpendicular, so a = √(aT² + ac²) = Rα√(1 + α²t⁴).
B1 A student who takes aT = Rα as the whole acceleration picks this. The point also moves on a circle with ω = αt, so it has a centripetal component, Rα²t², that grows with time.
C√(1+α²t⁴/16) A student who takes ω at time t to be θ/t = αt/2 gets ac = R(αt/2)² = Rα²t²/4 and a = √((Rα)² + (Rα²t²/4)²) = Rα√(1+α²t⁴/16). The angular velocity at time t is αt, so ac = Rα²t².
D1+αt² A student who adds the magnitudes of the two components picks this: Rα + Rα²t². The components are perpendicular, so they combine as √(aT² + ac²).
Working Tangential component: aT = Rα. ω = αt, so the centripetal component is ac = v²/R = Rω² = Rα²t². They are perpendicular: a = √((Rα)² + (Rα²t²)²) = Rα√(1 + α²t⁴). (Tangential only: Rα. ω taken as θ/t = αt/2: ac = Rα²t²/4, a = Rα√(1 + α²t⁴/16). Magnitudes added: Rα(1 + αt²).) The factor multiplying Rα is √(1 + α²t⁴).
The figure shows two pulleys on fixed axles connected by a belt that does not slip. The small pulley turns at 20.0 rad/s. What is the angular velocity of the large pulley?
Answer and reasoning
A20.0 rad/s A student who treats the two pulleys as one rigid system with a single angular velocity picks this. They are separate objects linked by the belt; what they share is the rim speed, so the larger pulley turns more slowly.
B8.00 rad/sCorrect The belt does not slip, so both rims move at the belt's speed, v = rsωs = (0.0600 m)(20.0 rad/s) = 1.20 m/s. For the large pulley, ω = v/r = (1.20 m/s)/(0.150 m) = 8.00 rad/s.
C50.0 rad/s A student who thinks the pulley with the larger radius turns faster, in proportion to its radius, picks this: 20.0 rad/s × (0.150 m/0.0600 m). With a shared rim speed, ω = v/r is smaller for the larger radius.
D1.20 rad/s A student who takes the belt's speed, 1.20 m/s, as the large pulley's angular velocity picks this. The speed must be divided by the large pulley's radius: (1.20 m/s)/(0.150 m) = 8.00 rad/s.
Working The belt does not slip, so both rims move at the belt's speed: v = rsωs = (0.0600 m)(20.0 rad/s) = 1.20 m/s. Large pulley: ωL = v/rL = (1.20 m/s)/(0.150 m) = 8.00 rad/s. (Same ω: 20.0 rad/s. Ratio inverted: 20.0 × 0.150/0.0600 = 50.0 rad/s. Belt speed taken as ω: 1.20.)
A turntable starts from rest and turns with constant angular acceleration. At time t, point P, 0.10 m from the axis, has a speed of 0.60 m/s. What is the speed of point Q, 0.30 m from the axis, at time 2t?
Answer and reasoning
A1.2 m/s A student who thinks every point of the turntable moves at the same speed picks this, doubling P's speed for the doubled time. Q is three times as far from the axis, so at any instant it moves three times as fast as P.
B1.8 m/s A student who thinks constant angular acceleration means constant angular velocity picks this, scaling only for Q's larger radius. ω = αt doubles between t and 2t, so the speed doubles too.
C3.6 m/sCorrect v = rω and, from rest with constant α, ω = αt, so v is proportional to rt. Q is three times as far from the axis and the time is twice as long: (0.60 m/s)(3)(2) = 3.6 m/s.
D7.2 m/s A student who lets ω grow as t², the way the angle turned grows, picks this: (0.60 m/s)(3)(4). The angle grows as t², but ω = αt grows only in proportion to t.
Working v = rω and, from rest with constant α, ω = αt, so v ∝ rt. Q is 3 times as far out and the time is doubled: vQ = (0.60 m/s)(3)(2) = 3.6 m/s. (Same speed for all points, time doubled: 1.2 m/s. ω constant: 1.8 m/s. ω taken to grow as t²: 7.2 m/s.)
A student watching a spinning fan claims: 'A point at the tip of a blade has a greater angular velocity than a point near the hub, because the tip moves faster.' Which statement correctly evaluates the claim?
Answer and reasoning
AIncorrect: both points turn through the same angle in any time, so they share one ω.Correct The blade is rigid: a straight line along it stays straight as it turns, so the tip and the point near the hub sweep through the same angle in any time and share one ω. The tip moves faster because v = rω and its r is larger.
BCorrect: the tip moves faster, and a point that moves faster turns at a greater ω. A student who thinks a point that moves faster must rotate faster picks this. Speed depends on the distance from the axis as well as on ω: the tip moves faster only because it is farther out.
CIncorrect: the point near the hub has the greater ω, since its r in ω = v/r is smaller. A student who thinks every point of the blade moves at the same speed picks this; ω = v/r is then larger near the hub. The points share one ω, and the tip's speed is greater because v = rω.
DCorrect: ω and speed are the same quantity, so the faster tip has the greater ω. A student who thinks angular velocity and linear speed are the same quantity picks this. They are related by v = rω; the faster tip has a larger r, not a larger ω.
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
