3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A merry-go-round is given a push and then left to turn on its own. A student observes that it slows down steadily and claims that a nonzero net torque is being exerted on it. Which reasoning correctly supports the claim?
Answer and reasoning
AThe push left a torque in its direction of rotation that is still acting, though it is weakening A student who thinks the push leaves a torque behind that gradually dies away picks this. Once the push ends, no torque in the direction of rotation remains; the net torque that is acting is opposite to the rotation.
BIts angular velocity is changing, which happens only when the torques on it are unbalancedCorrect Angular velocity changes when, and only when, the net torque on a rigid system is not zero. The merry-go-round's angular velocity is decreasing, so a net torque, opposite to its rotation, must be acting (here, friction at the axle and air resistance).
CIt is still rotating, and any rotating object must have a net torque exerted on it to keep turning A student who thinks rotation itself requires a net torque picks this. A merry-go-round turning at constant angular velocity would have zero net torque; the evidence for a net torque is that ω is changing.
DFriction acts on it, and any force exerted on a rotating object produces a torque on it A student who thinks every force produces a torque, whatever its lever arm, picks this. A force whose line of action passes through the axis exerts no torque, so the existence of a force alone does not establish a net torque.
A grinding wheel has a rotational inertia of 0.20 kg·m² about its axle. A belt exerts a force of 12 N tangent to the wheel at a radius of 0.25 m, while friction at the axle exerts a torque of 1.0 N·m opposing the wheel's rotation. What is the magnitude of the wheel's angular acceleration?
Answer and reasoning
A10 rad/s²Correct The belt's torque is (0.25 m)(12 N) = 3.0 N·m; friction opposes it, so τnet = 2.0 N·m. α = τnet/I = 2.0/0.20 = 10 rad/s².
B15 rad/s² A student who uses the belt's torque alone gets 3.0/0.20 = 15 rad/s². The angular acceleration depends on the net torque, which includes the opposing frictional torque.
C20 rad/s² A student who adds the frictional torque to the belt's gets 4.0/0.20 = 20 rad/s². Friction opposes the rotation, so its torque is subtracted.
D55 rad/s² A student who uses the 12 N force as if it were a torque, ignoring the 0.25 m lever arm, gets (12 − 1.0)/0.20 = 55. The belt's torque is rF = 3.0 N·m.
A rod of mass M lies at rest on a horizontal frictionless surface. A small thruster fixed to one end of the rod then exerts on it a force of constant magnitude F, always parallel to the surface and perpendicular to the rod. Which claim about the rod's motion while the thruster acts is correct?
Answer and reasoning
AIts center of mass accelerates at less than F/M, as it starts rotating A student who thinks part of the force is used up in rotation picks this. The center-of-mass acceleration is Fnet/M = F/M whatever point the force acts at; the rotation does not reduce it.
BIts center of mass accelerates at F/M, and its rotation speeds upCorrect The center of mass responds to the net external force, wherever it acts: acm = F/M in magnitude. Separately, F has a lever arm of half the rod's length about the center of mass, so the net torque is constant and not zero, and the angular velocity keeps increasing. The two analyses are independent.
CIts center of mass stays at rest, and it starts rotating about its center A student who pictures the rod pivoting about its center, as if on an axle, picks this. No axle holds the rod, so the net force F accelerates its center of mass.
DIts center of mass accelerates at F/M, and it turns at a steady rate A student who thinks a constant torque produces a constant angular velocity picks this. A constant net torque produces a constant angular acceleration, so the rod's rotation keeps speeding up.
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 Changing angular velocity Fix
Changing angular velocity
A rigid system's angular velocity changes (it speeds up, slows down or reverses) only when the net torque exerted on it is not zero.
Students often think A slowing object still has a torque in its direction of rotation, given to it when it was pushed, which gradually dies away. In fact No. While the object slows, its angular velocity is decreasing, so the net torque on it points opposite to its rotation; there is no leftover torque in the direction of rotation.
Students often think A rotating object needs a net torque to have an angular velocity, and the size of the torque sets how fast it turns, so a constant torque produces a steady rotation. In fact No. A net torque changes the angular velocity; it is not needed to maintain it. A constant net torque gives a constant angular acceleration, not a constant angular velocity.
5.6.A.2 Newton's second law in rotational form Fix
Newton's second law in rotational form
αsys = Στ/Isys = τnet/Isys: the angular acceleration of a rigid system about an axis is directly proportional to the net torque about that axis, in the same direction, and inversely proportional to the system's rotational inertia about that axis.
Angular acceleration (α)
The rate of change of angular velocity, α = dω/dt (SI unit: rad/s²). In Newton's second law in rotational form it points in the same rotational sense as the net torque.
Net torque versus applied torque
τnet is the sum of all torques about the axis, including opposing ones such as axle friction; using a single applied torque in α = τ/I gives the wrong α whenever other torques act.
Students often think The turning effect of a force depends only on its size, so any force on an object produces a torque equal to it, wherever it is exerted. In fact No. The torque of a force is rF sin θ: it depends on the lever arm, and a force whose line of action passes through the axis exerts no torque.
Students often think Rotational inertia depends only on mass, so objects of equal mass respond equally to equal torques and doubling the mass doubles the rotational inertia. In fact No. Rotational inertia depends on how the mass is distributed as well as on how much there is: a hoop (MR²) has twice the rotational inertia of a uniform disk (½MR²) of the same mass and radius, and a longer rod of the same mass has a larger rotational inertia.
5.6.A.3 Independent linear and rotational analyses Fix
Independent linear and rotational analyses
For a system with parts that translate and parts that rotate, or for an object that both translates and rotates, apply ΣF = ma to the translating parts or the center of mass and Στ = Iα to the rotating parts separately, then connect them with constraints such as a = rα for a string that does not slip.
Massive pulley
A pulley with rotational inertia I. When it has an angular acceleration, the string tensions on its two sides differ, because their torques must provide the net torque Iα.
Center-of-mass motion of a free rigid body
The acceleration of a rigid body's center of mass is acm = Fnet/M whatever points the forces are exerted at; the forces' torques about the center of mass separately determine its angular acceleration.
Students often think A pulley only changes the direction of a string, so the tension is the same on both sides and the pulley's mass does not affect the motion. In fact No. If the pulley has rotational inertia and an angular acceleration, the two tensions must differ: (T₁ − T₂)R = Iα. Equal tensions hold only for a massless (or non-accelerating, frictionless) pulley.
Students often think The tension in a string holding a hanging block equals the block's weight, even while the block accelerates. In fact No. Tension equals weight only when the block's acceleration is zero. For a block accelerating downward, mg − T = ma, so T < mg.
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 three wheels, each of mass M and radius R, that can turn on frictionless axles through their centers. Wheels 1 and 3 are uniform disks and wheel 2 is a thin hoop. Each is at rest when the force shown starts to be exerted on it, tangent to the circle at the point shown. Which ranking of the wheels' angular accelerations, α₁, α₂ and α₃, is correct?
Answer and reasoning
Aα₃ > α₁ > α₂ A student who thinks torque depends only on the size of the force gives wheel 3, with 2F, twice the torque of wheel 1. Its force acts at R/2, so its torque, 2F × R/2 = FR, equals wheel 1's.
Bα₁ = α₂ = α₃ A student who thinks rotational inertia depends only on mass gives all three wheels the same I; with equal torques of FR, all three α would be equal. The hoop's mass is all at the rim, so its rotational inertia is twice a disk's.
Cα₂ > α₁ = α₃ A student who thinks mass at the rim gives the hoop more leverage, making it easier to spin, picks this. Mass far from the axis increases the rotational inertia, so the hoop has the smallest angular acceleration.
Dα₁ = α₃ > α₂Correct Wheels 1 and 3 have equal torques, FR and 2F × R/2 = FR, and equal rotational inertias, ½MR², so α₁ = α₃ = 2F/(MR). The hoop has the same torque but twice the rotational inertia, MR², so α₂ = F/(MR) is half as large.
A uniform rod pivoted at one end on a frictionless horizontal axle is held horizontal and released from rest. It is replaced by a uniform rod of the same material and thickness but twice as long, which is released in the same way. By what factor does the rod's angular acceleration just after release change?
Answer and reasoning
A×4 A student who attends only to the torque, which grows by a factor of 4, picks this. The rotational inertia grows by a factor of 8, so α decreases.
B×2 A student who thinks rotational inertia depends only on mass takes I to double while the torque quadruples, giving ×2. The doubled length also puts mass twice as far from the pivot, multiplying I by 8 in all.
C×½Correct Doubling the length doubles the mass. The weight's torque, MgL/2, grows by 2 × 2 = 4; the rotational inertia, ML²/3, grows by 2 × 2² = 8. So α = τ/I changes by 4/8 = ½, consistent with α = 3g/(2L).
D×1 A student who reasons that everything falls with the same acceleration picks this. For rotation about a pivot, α = 3g/(2L) depends on the length, so the longer rod's angular acceleration is half as large.
Working Same material and thickness: mass doubles, M → 2M; L → 2L. Torque of the weight about the pivot τ = MgL/2 → ×4. I = ML²/3 → ×2 × 4 = ×8. α = τ/I → ×4/8 = ×½. (α = 3g/(2L) directly.)
The diagram shows a uniform rod of mass M and length L that can turn without friction about a horizontal axle through point P, perpendicular to the page. The rod is held horizontal and released from rest. What is the magnitude of its angular acceleration just after release?
Answer and reasoning
A3.00 g/L A student who uses ML²/12, the rotational inertia about the center, gets (MgL/4)/(ML²/12) = 3.00 g/L. The rod turns about P, so the parallel axis term M(L/4)² must be added.
B0.75 g/L A student who uses ML²/3, the result for a pivot at the end, gets (MgL/4)/(ML²/3) = 0.75 g/L. P is not at the end; about P, I = (7/48)ML².
C3.43 g/L A student who measures the weight's lever arm from the end of the rod, L/2, gets (MgL/2)/((7/48)ML²) ≈ 3.43 g/L. The lever arm is measured from the axle at P: L/2 − L/4 = L/4.
D1.71 g/LCorrect The weight acts at the center, L/4 from P, so τ = MgL/4. About P, I = ML²/12 + M(L/4)² = (7/48)ML². α = τ/I = (12/7)(g/L) ≈ 1.71 g/L.
Working The center of mass is L/2 − L/4 = L/4 from P, so τ = Mg(L/4). IP = ML²/12 + M(L/4)² = ML²/12 + ML²/16 = (7/48)ML². α = τ/IP = (L/4)/(7L²/48) × g = (12/7)(g/L) ≈ 1.71 g/L.
Two blocks hang from a light string that passes over a pulley, a uniform disk that turns on a frictionless axle, as shown in the diagram with the masses labeled. The string does not slip on the pulley. The blocks are released from rest. What is the magnitude of the blocks' acceleration, in terms of g?
Answer and reasoning
A0.33g A student who gives the string the same tension on both sides gets mg = 3ma, a ≈ 0.33g, as for a massless pulley. The pulley needs a net torque to speed up, so T₁ > T₂.
B0.20g A student who uses I = MR² = 2mR² for the pulley, as if its mass were at the rim, gets mg = 5ma, a = 0.20g. A uniform disk has I = ½MR² = mR².
C0.25gCorrect Analyse each part separately: 2mg − T₁ = 2ma for the left block, T₂ − mg = ma for the right block, and (T₁ − T₂)R = (½ × 2m × R²)(a/R) for the pulley, so T₁ − T₂ = ma. Adding the three equations: mg = 4ma, a = 0.25g.
D1.00g A student who sets each tension equal to its block's weight gives the pulley a torque (2mg − mg)R = mgR, so α = mgR/(mR²) = g/R and a = Rα = g. The blocks accelerate, so neither tension equals its block's weight.
Working Pulley: I = ½(2m)R² = mR². Left block: 2mg − T₁ = 2ma. Right block: T₂ − mg = ma. Pulley: (T₁ − T₂)R = Iα = mR²(a/R), so T₁ − T₂ = ma. Adding: 2mg − mg = (2m + m + m)a, so a = g/4 = 0.25g.
A uniform rod of mass 0.50 kg and length 0.60 m lies at rest on a horizontal frictionless surface. A horizontal force of 2.0 N, perpendicular to the rod, is exerted at one end. What is the magnitude of the rod's angular acceleration about its center of mass at that instant?
Answer and reasoning
A40 rad/s²Correct About the center of mass, the force's lever arm is L/2 = 0.30 m, so τ = 0.60 N·m, and Icm = ML²/12 = 0.015 kg·m². α = τ/Icm = 40 rad/s². The center of mass's acceleration, F/M, is found separately and does not enter this calculation.
B10 rad/s² A student who uses ML²/3 for the rod gets 0.60/0.060 = 10 rad/s². The angular acceleration is about the center of mass, so the rotational inertia must be about that axis, ML²/12.
C80 rad/s² A student who measures the lever arm from the far end of the rod, 0.60 m, gets τ = 1.2 N·m and α = 80 rad/s². The lever arm is measured from the axis through the center of mass: 0.30 m.
D13 rad/s² A student who links the motions with a = rα finds acm = F/M = 4.0 m/s² and then α = 4.0/0.30 ≈ 13 rad/s². No constraint ties the free rod's translation to its rotation, so α must come from τ/Icm.
Working Torque about the center of mass: τ = F(L/2) = (2.0 N)(0.30 m) = 0.60 N·m. Icm = ML²/12 = (0.50 kg)(0.60 m)²/12 = 0.015 kg·m². α = τ/Icm = 0.60/0.015 = 40 rad/s². (Separately, acm = F/M = 4.0 m/s².)
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
