5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Wheel A, with rotational inertia I, spins at angular speed 2ω on a fixed axle with negligible friction. Wheel B, with rotational inertia 2I, spins at angular speed ω in the opposite sense on the same axle, without touching wheel A. Which statement about the total angular momentum of the two-wheel system about the axle is correct?
Answer and reasoning
AIt is 4Iω, as each wheel contributes 2Iω and the two contributions add A student who adds magnitudes regardless of sense picks this. Each wheel does have 2Iω, but they turn in opposite senses, so one is positive and the other negative and they cancel.
BIt is in wheel A's sense, as wheel A has the greater angular speed A student who judges angular momentum by angular speed alone picks this. Wheel A spins twice as fast, but wheel B has twice the rotational inertia, so their values of Iω are equal and the total is zero.
CIt is zero, as the wheels' angular momenta are equal and oppositeCorrect The total is the sum of the parts' angular momenta about the axle, with signs. Wheel A has I(2ω) = 2Iω in one sense and wheel B has (2I)ω = 2Iω in the other, so they cancel and the total is zero (although each wheel has kinetic energy).
DIt is in wheel B's sense, as wheel B has the greater rotational inertia A student who judges angular momentum by rotational inertia alone picks this. Wheel B has twice the rotational inertia, but it spins at half the angular speed, so its Iω equals wheel A's and the total is zero.
Working Take wheel A's sense as positive: LA = I(2ω) = +2Iω and LB = (2I)(−ω) = −2Iω. Total = 0.
A space station floats far from other objects. A motor mounted on the station spins up a flywheel whose rotational inertia is much smaller than the station's, about an axis shared with the station. During the spin-up, how does the angular impulse exerted on the flywheel by the station compare with the angular impulse exerted on the station by the flywheel?
Answer and reasoning
ALarger on the flywheel, as the station is far more massive than the flywheel A student who thinks the larger body exerts the larger torque picks this. The torques the two exert on each other form a third-law pair, equal and opposite whatever their sizes, so the angular impulses are equal in magnitude.
BLarger on the station, as the flywheel ends up spinning faster than the station A student who thinks the faster-spinning object exerts the larger torque picks this. The flywheel's larger angular speed comes from its smaller rotational inertia (Δω = ΔL/I), not from a larger angular impulse; the two angular impulses are equal and opposite.
CZero on the station, as the motor turns only the flywheel, not the station A student who thinks only the active agent exerts a torque picks this. The flywheel exerts an equal and opposite torque back on the motor and so on the station, which therefore starts to turn the other way.
DEqual in size and opposite in sense, as the two torques form a third-law pairCorrect The station (through the motor) and the flywheel exert torques on each other about the shared axis. By Newton's third law these torques are equal in magnitude and opposite in sense at every instant, so the angular impulses are too. The flywheel ends up spinning faster only because its rotational inertia is smaller.
A flywheel spins on an axle mounted in a machine frame that is bolted to the floor. A brake pad fixed to the frame presses on the flywheel and brings it to rest. Which statement about the angular momentum that the flywheel loses is correct?
Answer and reasoning
AIt turns into thermal energy in the pad and wheel A student who conflates angular momentum with energy picks this. The flywheel's kinetic energy becomes thermal energy, but angular momentum is a different quantity and cannot become energy; it is transferred to Earth.
BIt runs down, as the spin of every object does in time A student who thinks spin dies away by itself picks this. The flywheel slows because the brake exerts a torque on it; without that torque it would keep spinning, and the angular momentum it loses goes to Earth.
CIt is passed through the frame and the floor to EarthCorrect Every interaction conserves angular momentum. The pad exerts a torque on the flywheel, and the flywheel exerts an equal and opposite torque on the pad, frame and floor, so Earth gains the angular momentum the flywheel loses. Earth's enormous rotational inertia means its rotation changes immeasurably.
DIt is destroyed, as friction is a nonconservative force A student who reads 'nonconservative' as 'does not conserve' picks this. That word describes the path dependence of a force's work; angular momentum is conserved in every interaction, including frictional ones, and passes to Earth.
Disk A, a uniform disk of mass M and radius R, rotates at angular speed ω₀ on a vertical axle with negligible friction. Disk B, a uniform disk of mass M and radius 2R that is not rotating, is dropped onto disk A, centered on the axle, and the two come to a common angular speed. What is that angular speed?
Answer and reasoning
A0.50ω₀ A student who thinks rotational inertia depends only on mass gives the two equal-mass disks the same I and gets ω₀/2. Disk B's mass is spread out to twice the radius, so its rotational inertia is four times disk A's.
B0.33ω₀ A student who thinks rotational inertia grows in proportion to radius takes IB = 2IA and gets ω₀/3. For the same mass, I ∝ R², so doubling the radius makes IB = 4IA.
C0.20ω₀Correct Rotational inertia grows as the square of the radius for disks of the same mass: IB = (1/2)M(2R)² = 4IA. Angular momentum is conserved, so IAω₀ = (IA + 4IA)ω and ω = ω₀/5 = 0.20ω₀.
D0.45ω₀ A student who conserves kinetic energy sets (1/2)IAω₀² = (1/2)(5IA)ω² and gets ω₀/√5 = 0.45ω₀. Friction between the disks dissipates energy; angular momentum is what is conserved.
Working IA = (1/2)MR²; IB = (1/2)M(2R)² = 4IA. L conserved: IAω₀ = 5IAω ⇒ ω = ω₀/5 = 0.20ω₀.
A uniform solid disk of mass M and radius R spins at angular speed ω₀ about a fixed axle through its center, with negligible friction at the axle. A brake pad presses on the flat face of the disk at distance r from the axle (r < R) with normal force FN, and the coefficient of kinetic friction between pad and disk is μ. How long does the disk take to stop?
Answer and reasoning
AMR²ω₀/(2μFN) A student who treats the angular impulse as force × time, with no lever arm, sets μFNΔt = (1/2)MR²ω₀. The friction force acts at distance r from the axle, so the torque is μFN r; this expression does not even have units of time.
BMR²ω₀/(2μFN r)Correct The friction force μFN acts at distance r, so the braking torque is μFN r. It must remove all of the disk's angular momentum, (1/2)MR²ω₀, which it transfers to the brake: (μFN r)Δt = (1/2)MR²ω₀, giving Δt = MR²ω₀/(2μFN r).
CMR²ω₀/(μFN r) A student who uses I = MR² for the disk picks this. A uniform solid disk has I = (1/2)MR², so its angular momentum, and the stopping time, are half as large.
DMRω₀/(2μFN) A student who takes the lever arm of every force on a wheel to be its radius uses μFN R as the torque and gets (1/2)MR²ω₀/(μFN R). The pad presses at distance r from the axle, so the lever arm is r.
Working Friction force μFN acts tangentially at radius r: τ = μFN r. All of the disk's angular momentum, L = Iω₀ = (1/2)MR²ω₀, is transferred to the brake: τΔt = L ⇒ Δt = MR²ω₀/(2μFN r).
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.4.A.1 Total angular momentum of a system Fix
Total angular momentum of a system
The sum of the angular momenta of all the parts of a system about one chosen rotational axis, each counted with a sign for its sense of rotation about that axis (for example, counterclockwise positive). Unit: kg·m²/s.
Angular momentum of a small object about an axis
For an object of mass m moving with speed v, L = r⊥mv, where r⊥ is the perpendicular distance from the axis to the object's line of motion (equivalently L = rmv sin θ, from L⃗ = r⃗ × p⃗). An object moving in a straight line that does not pass through the axis has nonzero angular momentum about it. Unit: kg·m²/s.
Students often think The total angular momentum of a system is the sum of the magnitudes of its parts' angular momenta, whatever their senses of rotation. In fact No. Angular momenta about one axis are added with signs: one sense of rotation is positive and the other negative, so parts with opposite senses partly or wholly cancel.
Students often think Only objects that spin or move in a circle have angular momentum; an object moving in a straight line has none about any axis. In fact Yes. Its angular momentum about the axis has magnitude r⊥mv, where r⊥ is the perpendicular distance from the axis to its line of motion. It is zero only if the line of motion passes through the axis.
6.4.A.2 System and surroundings Fix
System and surroundings
The system is the set of objects chosen for analysis; everything else is its surroundings. Torques that parts of the system exert on each other are internal and cannot change the system's total angular momentum; only torques exerted by the surroundings (external torques) can.
Third-law pair of angular impulses
When two objects interact, the torques they exert on each other about a common axis are equal in magnitude and opposite in sense at every instant, so the angular impulses they deliver to each other over the interaction are equal in magnitude and opposite in sense.
Choice of system for constant angular momentum
Angular momentum is constant for a system chosen so that every object exerting a torque on its parts is inside it (or so that the external torques about the axis cancel). For a person throwing a ball from a turntable, the person, the turntable and the ball together form such a system.
Nonrigid system
A system whose shape can change, such as a diver tucking or beads sliding along a spinning rod. Moving mass toward the axis decreases its rotational inertia I; with no external torque, L = Iω stays constant, so its angular speed ω increases (and the reverse when mass moves outward).
Angular impulse
The integral of torque over time, ∫τ dt, equal to the area under a graph of torque against time. The net external angular impulse on a system equals the change in its angular momentum, ΔL. Unit: N·m·s (the same as kg·m²/s).
Students often think When nothing external acts on a spinning system, its angular speed is what stays constant, even if its shape changes. In fact No. Its angular momentum L = Iω stays the same. If mass moves toward the axis, I decreases and ω increases; if mass moves outward, I increases and ω decreases.
Students often think Rotational inertia depends only on mass, so changing shape or size without changing mass leaves it unchanged. In fact No. Rotational inertia depends on how the mass is distributed relative to the axis (I = Σmr² or ∫r² dm). The same mass farther from the axis gives a larger rotational inertia.
6.4.B.1 Conservation of angular momentum in all interactions Fix
Conservation of angular momentum in all interactions
In every interaction the angular momentum one object or system gains is lost by another, so the total for all the interacting objects together is unchanged. A flywheel stopped by a brake fixed to Earth passes its angular momentum to Earth.
Students often think When friction stops a spinning object, its angular momentum is converted into thermal energy, as its kinetic energy is. In fact No. Angular momentum and energy are different quantities and cannot be converted into each other. The object's kinetic energy becomes thermal energy, while its angular momentum is transferred to whatever exerts the friction torque (ultimately Earth).
Students often think Friction is a nonconservative force, so it destroys angular momentum rather than transferring it. In fact No. Every interaction conserves angular momentum. The terms 'conservative' and 'nonconservative' describe whether a force's work depends on the path; they say nothing about whether angular momentum is conserved.
6.4.B.2 Conservation of angular momentum for a selected system Fix
Conservation of angular momentum for a selected system
If the net external torque on a chosen system about an axis is zero, the system's total angular momentum about that axis is constant, even when external forces act (for example, a frictionless axle's force, whose line of action passes through the axis). Kinetic energy need not be constant.
Students often think A spinning object's rotation naturally dies away over time, even when nothing acts on it. In fact No. With zero net external torque, its angular momentum stays constant indefinitely. Objects on Earth slow down because friction or air resistance exerts a torque on them.
Students often think A rod's rotational inertia is (1/12)Mℓ² whatever axis it turns about. In fact No. (1/12)Mℓ² is the rod's rotational inertia about an axis through its center. About an end it is (1/3)Mℓ², which follows from ∫r² dm or from the parallel-axis theorem: (1/12)Mℓ² + M(ℓ/2)².
6.4.B.3 Transfer of angular momentum to the environment Fix
Transfer of angular momentum to the environment
When the net external torque on a system is nonzero, angular momentum passes between the system and its surroundings at a rate equal to that torque, τnet = dL/dt; the system's loss is the environment's gain.
Students often think If a force does no work on an object, it cannot change the object's angular momentum. In fact Yes. A force can do no work (because it is perpendicular to the velocity) and still exert a torque about an axis. If no other force does work on the object, its kinetic energy is constant while its angular momentum changes.
Students often think A force that pulls an object toward a body exerts no torque about that body's center, wherever its line of action lies. In fact Only if its line of action passes through the center. A string pulling toward the edge of a post (tangent to it) has a lever arm equal to the post's radius about the post's center and exerts a torque about it.
13 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 13
The diagram is a top view of a uniform rod that rotates on a horizontal frictionless table about a fixed vertical axle through its end O, and of a small puck sliding in a straight line on the same table. The rod and the puck do not touch. At the instant shown, what is the magnitude of the total angular momentum of the rod–puck system about the axle?
Answer and reasoning
A0.50 kg·m²/sCorrect The total is the signed sum of the parts' angular momenta about the axle. The rod has (1/3)Mℓ²ω = (0.20 kg·m²)(4.0 rad/s) = 0.80 kg·m²/s counterclockwise. The puck passes above O moving right, which is clockwise about O, with r⊥mv = (0.50)(0.20)(3.0) = 0.30 kg·m²/s. The total is 0.80 − 0.30 = 0.50 kg·m²/s.
B1.10 kg·m²/s A student who adds the magnitudes without regard to sense picks 0.80 + 0.30 = 1.10 kg·m²/s. The puck goes around O clockwise while the rod turns counterclockwise, so the puck's angular momentum is negative and reduces the total.
C0.80 kg·m²/s A student who thinks an object moving in a straight line has no angular momentum counts only the rod. The puck's line of motion passes 0.50 m from O, so it has angular momentum r⊥mv = 0.30 kg·m²/s about the axle, which must be included.
D0.32 kg·m²/s A student who uses the full distance from O to the puck finds rmv = (0.80)(0.20)(3.0) = 0.48 kg·m²/s and gets 0.80 − 0.48 = 0.32 kg·m²/s. Only the perpendicular distance from O to the puck's line of motion, 0.50 m, counts.
Working Take counterclockwise as positive. Rod about its end: I = (1/3)Mℓ² = (1/3)(0.60 kg)(1.0 m)² = 0.20 kg·m², so Lrod = Iω = (0.20)(4.0) = +0.80 kg·m²/s. Puck: it moves to the right along a line 0.50 m above O, which is clockwise about O, so Lpuck = −r⊥mv = −(0.50 m)(0.20 kg)(3.0 m/s) = −0.30 kg·m²/s. Total: 0.80 − 0.30 = +0.50 kg·m²/s (counterclockwise), magnitude 0.50 kg·m²/s.
A diver leaves a springboard rotating slowly about an axis through her center of mass, then pulls into a tight tuck. Air resistance is negligible. A student claims that her angular momentum about her center of mass is the same in the tuck as just after she leaves the board. Which reasoning correctly supports the claim?
Answer and reasoning
AHer angular speed is unchanged by the tuck, and her angular momentum depends on that speed A student who thinks angular speed stays constant without an external torque picks this. Tucking decreases her rotational inertia, so her angular speed increases while Iω stays constant.
BGravity exerts no torque about her center of mass, and her muscles' torques are internalCorrect A change in her angular momentum requires an external torque. The only external force, gravity, acts at her center of mass, so it has no lever arm about it; her muscles exert equal and opposite torques on parts of her own body, which cannot change her total. So L = Iω stays the same while I decreases and ω increases.
CHer mass is the same in the tuck, so her rotational inertia and angular momentum are also A student who thinks rotational inertia depends only on mass picks this. Tucking moves mass toward the axis and decreases her rotational inertia; her angular momentum is constant because no external torque acts, not because I is unchanged.
DHer rotational kinetic energy is unchanged by the tuck, so her angular momentum is unchanged A student who treats kinetic energy and angular momentum as conserved together picks this. Her muscles do work as she pulls in, so her rotational kinetic energy L²/(2I) increases as I decreases; only her angular momentum is constant.
Working Only gravity acts on her from outside, and it acts at her center of mass, so its torque about that point is zero. Her muscles' forces are internal, exerting equal and opposite torques on her body parts. Zero net external torque ⇒ L constant (I decreases, ω increases).
A turntable with rotational inertia I about its vertical axle, which has negligible friction, is at rest. A toy car of mass m sits at rest on the turntable at distance R from the axle. The car is switched on and drives around a circle of radius R centered on the axle, reaching speed v relative to the ground. Treat the car as a point object. What is then the angular speed of the turntable?
Answer and reasoning
Av/R A student who thinks equal and opposite interactions give equal and opposite motions picks this: the car's angular speed about the axle, v/R. The angular momenta are equal and opposite, not the angular speeds; the turntable's ω is mvR divided by its own rotational inertia.
BmvR/(I+mR²) A student who adds the car's mR² to the turntable's rotational inertia, as if the car rode along with it, writes mvR = (I + mR²)ω and picks this. The car moves on its own at speed v; its angular momentum, mvR, is counted once, so the turntable alone must have Iω = mvR.
CmvR/ICorrect Choose the turntable and car as the system: the forces they exert on each other are internal and the axle exerts no torque about the axle, so the total angular momentum stays zero. The car gains mvR in one sense and the turntable must gain Iω = mvR in the other, so ω = mvR/I.
Dmv/I A student who sets the car's linear momentum equal to the turntable's angular momentum picks this. The car's angular momentum about the axle is r⊥mv = mvR; mv/I does not even have units of angular speed.
Working System: turntable + car. The car and turntable exert torques on each other (internal); the axle exerts no torque about the axle. Ltotal = 0 before and after: mvR − Iω = 0 ⇒ ω = mvR/I (the turntable turns opposite to the car).
A student stands on a platform that rotates with negligible friction about a vertical axle. The rotational inertia of the student and platform together about the axle is 2.0 kg·m²; this does not include a 0.50 kg ball that the student holds 0.80 m from the axle. Student, platform and ball rotate together counterclockwise, seen from above, at 2.0 rad/s. The student throws the ball horizontally, tangent to its circle, so that just after release it moves clockwise about the axle at 4.0 m/s relative to the ground. What is the angular speed of the student and platform just after the throw?
Answer and reasoning
A3.1 rad/sCorrect For the student, platform and ball together the throw is internal, so the total angular momentum is constant. Before: (2.0 + 0.50 × 0.80²)(2.0) = 4.64 kg·m²/s counterclockwise. After, the ball has 0.80 × 0.50 × 4.0 = 1.60 kg·m²/s clockwise, so the student and platform have 4.64 + 1.60 = 6.24 kg·m²/s and ω = 6.24/2.0 = 3.1 rad/s.
B2.0 rad/s A student who thinks angular speed is what stays constant picks this. Throwing the ball clockwise gives it clockwise angular momentum, so the student and platform must gain counterclockwise angular momentum and speed up.
C2.8 rad/s A student who leaves the held ball out of the initial rotational inertia starts from 2.0 × 2.0 = 4.0 kg·m²/s and gets (4.0 + 1.60)/2.0 = 2.8 rad/s. Before the throw the ball moves with the platform and contributes 0.50 × 0.80² × 2.0 = 0.64 kg·m²/s.
D1.5 rad/s A student who ignores the sense of the ball's motion treats its 1.60 kg·m²/s as counterclockwise and gets (4.64 − 1.60)/2.0 = 1.5 rad/s. The ball moves clockwise, so its angular momentum is negative and the platform's must increase.
Working System: student + platform + ball (the throw is internal; the axle exerts no torque about the axle). Counterclockwise positive. Before: L = (2.0 + 0.50 × 0.80²)(2.0) = (2.32)(2.0) = 4.64 kg·m²/s. After: ball L = −(0.80)(0.50)(4.0) = −1.60 kg·m²/s. So 2.0ω = 4.64 + 1.60 = 6.24 ⇒ ω = 3.12 ≈ 3.1 rad/s counterclockwise.
The diagram is a top view of a rod that rotates in a horizontal plane about a fixed vertical axle through its center, with negligible friction. The rod's rotational inertia about the axle is 0.024 kg·m². Two small beads, each of mass 0.10 kg, are held by catches at the positions shown by solid circles while the system rotates at 5.0 rad/s. The catches are released, and the beads slide out to the stops at the rod's ends, shown by dashed circles. What is the angular speed of the system once the beads rest against the stops?
Answer and reasoning
A5.0 rad/s A student who thinks the angular speed stays constant without an external torque picks this. The beads move outward, so the rotational inertia increases from 0.026 to 0.042 kg·m², and with Iω constant the angular speed must decrease.
B3.1 rad/sCorrect The forces between beads, catches and stops are internal, and the axle exerts no torque, so Iω is constant. Ii = 0.024 + 2(0.10)(0.10)² = 0.026 kg·m² and If = 0.024 + 2(0.10)(0.30)² = 0.042 kg·m², so ωf = (0.026)(5.0)/0.042 = 3.1 rad/s.
C3.9 rad/s A student who keeps the kinetic energy constant sets (1/2)Iiωi² = (1/2)Ifωf² and gets 5.0 × √(0.026/0.042) = 3.9 rad/s. Kinetic energy is not conserved here (the beads strike the stops); angular momentum is, giving 3.1 rad/s.
D2.6 rad/s A student who takes each bead's rotational inertia as proportional to its distance, mr instead of mr², gets Ii = 0.044 and If = 0.084 (in those wrong units) and ωf = 5.0 × 0.044/0.084 = 2.6 rad/s. A bead contributes mr², so moving it from 0.10 m to 0.30 m multiplies its contribution by nine.
Working No external torque about the axle (the catches and stops are internal). Ii = 0.024 + 2(0.10)(0.10)² = 0.026 kg·m²; If = 0.024 + 2(0.10)(0.30)² = 0.042 kg·m². ωf = Iiωi/If = (0.026)(5.0)/0.042 = 3.10 ≈ 3.1 rad/s.
A figure skater spins on ice with negligible friction and air resistance. She pulls her arms in, and her rotational inertia about her spin axis decreases to one-third of its initial value. How does her rotational kinetic energy change?
Answer and reasoning
AIt stays the same as its initial value A student who thinks kinetic energy is conserved whenever angular momentum is picks this. Her muscles do positive work pulling her arms inward against their tendency to move in a circle, so her rotational kinetic energy increases.
BIt increases to nine times its initial value A student who notes that ω triples and that K depends on ω² multiplies by nine, keeping I fixed. Her rotational inertia falls to one-third at the same time, so K = (1/2)Iω² changes by (1/3)(3²) = 3.
CIt decreases to one-third of its initial value A student who thinks angular speed stays constant picks this: with ω fixed, K = (1/2)Iω² falls in proportion to I. Her angular momentum, not her angular speed, is constant; ω triples and K triples.
DIt increases to three times its initial valueCorrect With no external torque, L = Iω is constant, so a third of the rotational inertia means three times the angular speed. Krot = (1/2)Iω² = L²/(2I), so with L fixed, Krot is inversely proportional to I and triples. The extra energy comes from the work her muscles do pulling her arms in.
Working L = Iω constant ⇒ ωf = 3ωi. K = (1/2)Iω² = L²/(2I) ⇒ Kf/Ki = Ii/If = 3.
The graph shows the net external torque exerted on a wheel as a function of time. At t = 0, the wheel's angular momentum is 1.0 kg·m²/s, in the same sense as the torque. What is the magnitude of the wheel's angular momentum at t = 5.0 s?
Answer and reasoning
A21 kg·m²/s A student who multiplies the largest torque by the whole time takes the angular impulse as (4.0)(5.0) = 20 N·m·s and gets 21 kg·m²/s. For the first 3.0 s the torque is less than 4.0 N·m, so the area is a triangle plus a rectangle, 14 N·m·s.
B15 kg·m²/sCorrect The change in angular momentum equals the angular impulse, the area under the τ–t graph: (1/2)(3.0)(4.0) + (2.0)(4.0) = 6.0 + 8.0 = 14 N·m·s. Adding the initial 1.0 kg·m²/s gives 15 kg·m²/s.
C14 kg·m²/s A student who takes the angular impulse as the final angular momentum picks this. The angular impulse, 14 N·m·s, is the change in angular momentum; the wheel already had 1.0 kg·m²/s, so its final value is 15 kg·m²/s.
D11 kg·m²/s A student who averages the first and last torques, (0 + 4.0)/2 = 2.0 N·m, and multiplies by 5.0 s gets 10 N·m·s and 11 kg·m²/s. The graph is not one straight line, so the area must be found piece by piece: 14 N·m·s.
Working Angular impulse = area under τ–t: triangle (1/2)(3.0 s)(4.0 N·m) = 6.0 N·m·s plus rectangle (2.0 s)(4.0 N·m) = 8.0 N·m·s, total 14 N·m·s. L = L₀ + ΔL = 1.0 + 14 = 15 kg·m²/s.
The graph shows the angular momentum of wheel X and of wheel Y, each turning on its own fixed axle, over the whole interval shown. What is the ratio of the net angular impulse exerted on X to the net angular impulse exerted on Y over this interval?
Answer and reasoning
A1.00Correct The net angular impulse on each wheel equals its change in angular momentum. X changes from 2.0 to 8.0 kg·m²/s and Y from 4.0 to 10.0 kg·m²/s, so both receive 6.0 N·m·s and the ratio is 1.00. How quickly the change happens does not matter.
B4.00 A student who reads angular impulse from the steepness of the L–t graph compares slopes: X rises 6.0 in 1.0 s and Y rises 6.0 in 4.0 s, giving 4.00. The slope is the net torque; the angular impulse is the change in L, the same 6.0 kg·m²/s for each.
C0.80 A student who takes the angular impulse as the final angular momentum compares 8.0 with 10.0 and gets 0.80. The angular impulse is the change, 8.0 − 2.0 for X and 10.0 − 4.0 for Y, which are equal.
D0.39 A student who takes the angular impulse as the area under the L–t graph finds 11 for X and 28 for Y and gets 0.39. The area under an L–t graph is not an angular impulse; angular impulse is the area under a torque–time graph, and it equals the change in L read from this graph.
Working Net angular impulse = ΔL. X: 8.0 − 2.0 = 6.0 kg·m²/s. Y: 10.0 − 4.0 = 6.0 kg·m²/s. Ratio = 1.00.
The diagram shows a uniform rod hanging at rest from a frictionless pivot at its top end and a small lump of clay moving horizontally toward the rod's lower end; the masses, the rod's length and the clay's speed are labeled. The clay strikes the lower end of the rod and sticks to it. What is the angular speed of the rod–clay system just after the collision?
Answer and reasoning
A0.80v/ℓ A student who uses the rod's rotational inertia about its center, (1/12)(3m)ℓ² = 0.25mℓ², gets I = 1.25mℓ² and ω = 0.80v/ℓ. The rod turns about its end, where I = (1/3)(3m)ℓ² = mℓ².
B1.00v/ℓ A student who leaves the clay out of the final rotational inertia uses I = mℓ² for the rod alone and gets v/ℓ. The clay moves with the rod's end after sticking and adds mℓ², doubling the rotational inertia.
C0.71v/ℓ A student who conserves kinetic energy sets (1/2)mv² = (1/2)(2mℓ²)ω² and gets ω = v/(ℓ√2) = 0.71v/ℓ. The clay sticks, so kinetic energy is dissipated; angular momentum about the pivot is what is conserved.
D0.50v/ℓCorrect Angular momentum about the pivot is conserved, since the pivot's force acts at the axis. Before: mvℓ. After: I = (1/3)(3m)ℓ² + mℓ² = 2mℓ². So ω = mvℓ/(2mℓ²) = 0.50v/ℓ.
Working Angular momentum about the pivot is conserved (the pivot's force acts at the axis; gravity's lever arm is zero while the rod hangs vertically). Before: L = mvℓ. After: I = (1/3)(3m)ℓ² + mℓ² = 2mℓ². ω = mvℓ/(2mℓ²) = v/(2ℓ) = 0.50 v/ℓ.
Disk 1, with rotational inertia I₁, spins at angular speed ω₀ on a vertical axle with negligible friction. Disk 2, with rotational inertia I₂ and not rotating, is dropped onto disk 1, centered on the axle, and friction between the disks brings them to a common angular speed. What is the ratio of the kinetic energy of the two disks at the common angular speed to the initial kinetic energy?
Answer and reasoning
A1 A student who thinks kinetic energy is conserved whenever angular momentum is picks this. The disks slide over each other until they move together, so friction dissipates energy, as in a perfectly inelastic collision.
BI₁²/(I₁+I₂)² A student who finds the common angular speed correctly, ω = I₁ω₀/(I₁ + I₂), but takes the kinetic energy to scale with ω² alone gets Kf/Ki = (ω/ω₀)² = I₁²/(I₁ + I₂)². The rotational inertia also grows, from I₁ to I₁ + I₂, so Kf/Ki = (I₁ + I₂)ω²/(I₁ω₀²) = I₁/(I₁ + I₂).
CI₁/(I₁+I₂)Correct Angular momentum is conserved (internal friction, no torque from the axle): ω = I₁ω₀/(I₁ + I₂). Then Kf/Ki = (I₁ + I₂)ω²/(I₁ω₀²) = I₁/(I₁ + I₂); friction dissipates the remaining fraction, I₂/(I₁ + I₂).
D1/2 A student who remembers that half the kinetic energy is lost when two equal objects stick, so half remains, picks this. The ratio is I₁/(I₁ + I₂), which is one-half only when I₂ = I₁.
Working L conserved: I₁ω₀ = (I₁ + I₂)ω ⇒ ω = I₁ω₀/(I₁ + I₂). Ki = (1/2)I₁ω₀²; Kf = (1/2)(I₁ + I₂)ω² = (1/2)I₁²ω₀²/(I₁ + I₂). Kf/Ki = I₁/(I₁ + I₂); the fraction I₂/(I₁ + I₂) is dissipated by friction. Distractors: K conserved: 1; 'half lost when objects stick': 1/2; K scaled by ω² alone: (ω/ω₀)² = I₁²/(I₁ + I₂)².
A child runs along a line tangent to the rim of a merry-go-round that is at rest and is mounted on a fixed vertical axle with negligible friction. She jumps onto the rim and holds on, and the two rotate together. During the landing, which quantity of the child–merry-go-round system is conserved?
Answer and reasoning
AIts linear momentum, as the landing lasts only a very short time compared with the run A student who applies conservation of linear momentum to every short collision picks this. The axle holds the merry-go-round's center in place and exerts a large horizontal impulse during the landing, so the system's linear momentum changes.
BIts kinetic energy, as there is no net external torque on the system about the axle A student who thinks kinetic energy is conserved whenever angular momentum is picks this. The child and rim come to a common motion, like a perfectly inelastic collision, so kinetic energy is dissipated.
CNo such quantity, as the axle exerts an external force on the system during the landing A student who thinks any external force rules out conservation of angular momentum picks this. The axle's force acts at the axis and has no lever arm about it, so it exerts no torque about the axle; angular momentum about the axle is conserved.
DIts angular momentum about the axle, as the axle's force has no lever arm about itCorrect The only horizontal external force during the landing is the axle's, and it acts at the axis, so its torque about the axle is zero (gravity and the vertical support forces exert no torque about the vertical axle). With zero net external torque about the axle, the system's angular momentum about it is constant.
The diagram is a top view of a puck sliding on a horizontal frictionless table. A string tied to the puck is wrapped around a fixed vertical post centered at O, and as the puck moves the string winds onto the post, so the puck's path spirals inward. The string stays taut. Which statement about the puck's angular momentum about O is correct?
Answer and reasoning
AIt decreases, as the string's force has a lever arm about OCorrect The string pulls along a line tangent to the post, not through O, so its lever arm about O is the post's radius and it exerts a torque opposite to the puck's rotation. Angular momentum about O is transferred to the post: L = mv × (free string length) falls as the string winds on, even though the speed stays constant.
BIt stays constant, as the string's force does no work on the puck A student who thinks a force that does no work cannot change angular momentum picks this. The tension is perpendicular to the velocity, so the puck's kinetic energy is constant, but the tension still has a lever arm about O and exerts a torque.
CIt stays constant, as the string pulls the puck toward the post A student who treats any pull toward the post as a pull toward its center picks this. The string's line of action is tangent to the post and misses O by the post's radius, so it exerts a torque about O.
DIt increases, as its angular speed about O increases A student who judges angular momentum by angular speed picks this. At the stage shown the puck does go around O faster, but it is also getting closer to O; angular momentum is mv times the free string length, which decreases.
Working The tension acts along the string, whose line is tangent to the post, so its lever arm about O is the post's radius a: τ = aT, opposite in sense to the puck's rotation about O. Hence L about O decreases (dL/dt = −aT). The tension is perpendicular to v, so it does no work and the speed is constant; L = mv × (free string length) decreases as the string shortens.
A turntable is mounted on a fixed vertical axle with negligible friction; its rotational inertia about the axle is I. A person of mass m stands at distance R from the axle, and the person and turntable are at rest. The person then walks along a circle of radius R centered on the axle, at speed u relative to the turntable's surface. Treat the person as a point object. What is the angular speed of the turntable relative to the ground while the person walks?
Answer and reasoning
AmuR/I A student who uses u as the person's speed relative to the ground picks this: muR = Iω. The surface under the person moves backward at ωR, so the person's ground speed is only u − ωR.
Bu/(2R) A student who expects the person and turntable to end up with equal and opposite angular velocities about the axle picks this, splitting the relative angular speed u/R equally. Their angular momenta, not their angular velocities, are equal and opposite, and the turntable's rotational inertia I differs from the person's, mR².
CmuR/(I+mR²)Correct The axle exerts no torque about the axis, so the system's angular momentum stays zero. If the turntable turns at ω in the opposite sense, the person's speed relative to the ground is u − ωR, and m(u − ωR)R = Iω gives ω = muR/(I + mR²).
D0 A student who writes the system's angular momentum as (I + mR²)ω, as if person and turntable turned together as one rigid body, gets (I + mR²)ω = 0 and picks this. The person moves relative to the turntable, so each part's angular momentum must be found separately; they cancel with the turntable turning.
Working System: person + turntable. The axle's force passes through the axis and friction is negligible, so the net external torque about the axle is zero and the system's angular momentum stays 0. Let the turntable turn with angular speed ω opposite to the person's walking. The surface under the person moves at ωR opposite to the person, so the person's speed relative to the ground is u − ωR. Then m(u − ωR)R − Iω = 0, so ω = muR/(I + mR²). Distractors: u used as the ground speed: muR = Iω, ω = muR/I; equal and opposite angular velocities: the person's angular speed about the axle relative to the ground equals ω, so the relative angular speed 2ω = u/R, ω = u/(2R); system treated as one rigid body: (I + mR²)ω = 0, ω = 0.
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
