5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Two identical small balls, each of mass m, are fixed to the ends of a light rod. The rod rotates in a horizontal plane about a fixed vertical axle through its center, with negligible friction, and each ball moves with speed v in a circle of radius r. Which statement about the total angular momentum of the rod–balls system about the axle is correct?
Answer and reasoning
AThe balls move in opposite directions, so their angular momenta cancel. A student who treats angular momentum like linear momentum picks this. The balls' linear momenta do cancel, but angular momentum depends on the sense of rotation about the axle, and both balls turn in the same sense, so their angular momenta add.
BThe system's center of mass stays at rest, so there is no angular momentum. A student who judges rotation from the motion of the center of mass picks this. A center of mass at rest means zero total linear momentum only; every part of the system turns in the same sense about the axle, so the angular momentum is 2mvr.
CNo net external torque acts on it, so the system's angular momentum is zero. A student who confuses constant with zero picks this. With no net external torque, the angular momentum stays constant; it keeps whatever value it has, here 2mvr, just as an object with no net force keeps its velocity.
DEach ball has angular momentum mvr in the same sense, so the two add together.Correct The system's total is the sum of its parts' angular momenta about the axle. Both balls go around the axle in the same sense, so each contributes +mvr and the total is 2mvr (the light rod contributes nothing). The opposite directions of their velocities do not matter; their sense of rotation does.
An astronaut floats inside a space station, touching nothing and holding nothing, and rotates slowly about an axis through her center of mass. Air resistance is negligible. Which of the following could change her angular momentum about that axis?
Answer and reasoning
APulling her arms and legs in close to her body A student who equates angular momentum with angular speed picks this. Pulling her limbs in reduces her rotational inertia, so she spins faster, but Iω stays the same: nothing outside her exerts a torque.
BTwisting her torso one way and her legs the other A student who thinks internal torques can change the total picks this. Her muscles exert equal and opposite torques on her torso and her legs, so one part gains exactly the angular momentum the other loses and her total is unchanged.
CPushing sideways on a handrail fixed to the stationCorrect A change in her angular momentum requires an interaction with her surroundings. Pushing sideways on the handrail makes the handrail push back on her; a force whose line of action does not pass through her center of mass exerts a torque about the axis and changes her angular momentum.
DWaiting, since spinning slowly dies away when nothing acts A student who believes a spin runs down by itself picks this. With no external torque her angular momentum stays constant indefinitely; objects on Earth slow down only because friction or air resistance exerts a torque on them.
A lump of clay moving horizontally in a straight line strikes and sticks to the outer edge of a door that is initially at rest. The door can rotate about vertical hinges with negligible friction. The collision decreases the kinetic energy of the clay–door system. Which statement about the angular momentum of the clay–door system about the hinge axis is correct?
Answer and reasoning
AIt decreases, since the system's kinetic energy decreases. A student who thinks angular momentum and kinetic energy rise and fall together picks this. Kinetic energy is transformed into internal energy as the clay deforms, but no external torque acts about the hinge axis, so the angular momentum does not change.
BIt is the same just after the collision as it was just before.Correct Angular momentum is conserved in every interaction. The forces between the clay and the door are internal to the system; the hinges exert forces only at the axis and gravity is parallel to it, so neither exerts a torque about the hinge axis. The loss of kinetic energy does not change this.
CIt increases from zero, as the clay moved in a straight line. A student who thinks only rotating objects have angular momentum picks this. Before the collision the clay already has angular momentum rmv sin θ about the hinge axis, because its line of motion does not pass through the hinges; that is what the door and clay share afterward.
DIt changes, because the hinges exert a force on the door. A student who confuses force with torque picks this. The hinges do exert a force during the collision, but it acts at the hinge axis, so its torque about that axis is zero and the system's angular momentum is unchanged.
Disk 1, with rotational inertia 0.40 kg·m², rotates counterclockwise at 5.0 rad/s on a vertical axle. Disk 2, with rotational inertia 0.20 kg·m², rotates clockwise at 4.0 rad/s on the same axle, just above disk 1. Friction at the axle is negligible. Disk 2 drops onto disk 1, and the two disks soon rotate together. What is the angular speed of the two disks when they rotate together?
Answer and reasoning
A2.0 rad/sCorrect No net external torque acts on the two-disk system, so its angular momentum is constant. With counterclockwise positive, L = (0.40)(5.0) − (0.20)(4.0) = 1.2 kg·m²/s, and together the disks have I = 0.60 kg·m², so ω = 1.2/0.60 = 2.0 rad/s.
B4.7 rad/s A student who adds the angular momenta as magnitudes gets (2.0 + 0.80)/0.60 = 4.7 rad/s. The disks rotate in opposite senses, so disk 2's angular momentum is negative and partly cancels disk 1's.
C0.5 rad/s A student who ignores the rotational inertias and averages the angular velocities gets (5.0 − 4.0)/2 = 0.5 rad/s. It is angular momentum, Iω, that is conserved, so each disk's contribution depends on its rotational inertia as well as on its angular velocity.
D3.0 rad/s A student who divides the total angular momentum by disk 1's rotational inertia alone gets 1.2/0.40 = 3.0 rad/s. After they join, both disks rotate together, so the rotational inertia is 0.40 + 0.20 = 0.60 kg·m².
Working System: both disks. The torques between the disks are internal and the frictionless axle exerts no torque, so the net external torque is zero and L is constant. Take counterclockwise as positive: L = (0.40)(5.0) + (0.20)(−4.0) = 2.0 − 0.80 = 1.2 kg·m²/s. Together: ω = L/(I₁ + I₂) = 1.2/0.60 = 2.0 rad/s, counterclockwise.
A flywheel rotates on an axle in a machine. For part of the time shown, a brake pad fixed to the machine's frame, which is bolted to the floor, presses against the flywheel. The graph shows the angular momentum L of the flywheel about its axle as a function of time t. Taking the flywheel alone as the system, which claim about its angular momentum is correct?
Answer and reasoning
AFrom 6.0 s to 8.0 s, a net torque must still act on the flywheel to keep it turning. A student who thinks a spinning object needs a torque to keep it spinning picks this. From 6.0 s to 8.0 s the graph is horizontal, so the angular momentum is constant and the net torque on the flywheel is zero.
BThe net torque on the flywheel is greatest from 0 to 2.0 s, where L is greatest. A student who reads the torque from the height of the graph picks this. The net torque is the slope of the L–t graph; from 0 to 2.0 s the graph is horizontal, so the net torque is zero there, and it is nonzero only from 2.0 s to 6.0 s.
CFrom 2.0 s to 6.0 s, angular momentum passes to the pad, frame and Earth.Correct From 2.0 s to 6.0 s the flywheel's angular momentum falls steadily from 12 to 4.0 kg·m²/s, so a nonzero net external torque acts on it (the slope, −2.0 N·m). That torque transfers 8.0 kg·m²/s from the flywheel to its environment: the brake pad, the frame and Earth.
DFrom 2.0 s to 6.0 s, friction from the brake pad destroys angular momentum. A student who thinks friction destroys angular momentum picks this. The flywheel does lose 8.0 kg·m²/s, but angular momentum is conserved in all interactions: the pad, the frame and Earth gain exactly what the flywheel loses.
Working The net external torque on the flywheel is the slope of its L–t graph. 0 to 2.0 s and 6.0 s to 8.0 s: horizontal, net torque zero, L constant. 2.0 s to 6.0 s: slope = (4.0 − 12) kg·m²/s ÷ (6.0 − 2.0) s = −2.0 N·m, so a nonzero net external torque (from the brake pad) acts and the flywheel's angular momentum decreases by 8.0 kg·m²/s. By conservation, the surroundings (pad, frame, Earth) gain 8.0 kg·m²/s.
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 taken with its sign for its sense of rotation about that axis (for example, counterclockwise positive). Unit: kg·m²/s.
Angular momentum of an object moving in a straight line
An object of mass m moving with speed v has angular momentum about a point of magnitude L = rmv sin θ, where r is its distance from the point and θ is the angle between the line from the point to the object and its velocity. It is zero only if the object moves along a line through the point. Unit: kg·m²/s.
Rotational sense
Whether a part of a system turns, or moves, counterclockwise or clockwise about the chosen axis. In AP Physics 1 one sense is chosen as positive and angular momenta of the other sense are negative, so parts with opposite senses partly cancel in the total.
Students often think Only objects that are spinning or moving in a circle have angular momentum; an object moving in a straight line has none. In fact Yes. Its angular momentum about the axis is L = rmv sin θ, which is zero only if the object moves along a line through the axis. A puck sliding past a turntable's axle, or a lump of clay flying toward the edge of a door, has angular momentum about that axle or hinge.
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: choose one sense as positive, and give the parts that turn or move in the other sense negative angular momentum. Parts with opposite senses partly cancel in the total.
6.4.A.2 System and surroundings (rotational) Fix
System and surroundings (rotational)
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.
Angular impulse
The product of the torque exerted on an object or system and the time interval during which it is exerted, τΔt; for a torque that varies, the area under the torque–time graph. Unit: N·m·s, which is the same as kg·m²/s.
Equal and opposite angular impulses
When two objects exert torques on each other about the same axis, the angular impulse each exerts on the other has the same magnitude and the opposite sense, as a direct result of Newton's third law; this holds whatever their rotational inertias or angular speeds.
Choosing a system with constant angular momentum
Drawing the system boundary so that every object that exerts a torque during the interaction is inside it. The torques are then internal, the net external torque is zero, and the total angular momentum of the system is constant even though its parts exchange angular momentum.
Nonrigid system
A system whose parts can move relative to each other, so that its shape and its rotational inertia can change, such as a skater moving her arms or beads sliding along a rotating rod. With no net external torque its angular momentum Iω is constant, so its angular speed changes inversely with its rotational inertia.
Rotational kinetic energy in a shape change
K = (1/2)Iω², which can be written K = L²/(2I). When a nonrigid system pulls mass toward the axis with L constant, I decreases and K increases; the extra energy comes from work done by forces inside the system (for example, a person's muscles). Unit: J.
Angular impulse–angular momentum theorem for a system
The change in a system's total angular momentum equals the net angular impulse exerted on it by its surroundings: ΔL = τnet Δt, where only external torques count. Internal torques transfer angular momentum between parts but do not change the total.
Students often think A person or machine can change the total angular momentum of a system by exerting torques inside it, for example by twisting or swinging parts of the body. In fact No. Torques that parts of a system exert on each other are internal: by Newton's third law they come in equal and opposite pairs, so they transfer angular momentum between the parts without changing the total.
Students often think Angular momentum depends only on how fast something rotates, so a change in angular speed means a change in angular momentum, and objects rotating together have equal angular momenta. In fact No. Angular momentum is L = Iω, so it depends on the rotational inertia as well as on the angular velocity. A skater who pulls in her arms spins faster with the same L, and two objects turning at the same ω can have very different L.
6.4.B.1 Conservation of angular momentum Fix
Conservation of angular momentum
Angular momentum is never created or destroyed. In every interaction, including those in which kinetic energy is transformed, any angular momentum one object or system loses is gained by the objects it interacts with, so the total for all interacting objects is unchanged.
Students often think A system's angular momentum can be conserved only if no external force at all is exerted on it; any force from outside, even at the axis, changes it. In fact No. Angular momentum about an axis changes only if there is a net external torque about that axis. A force whose line of action passes through the axis, such as the force from a frictionless hinge or axle, exerts zero torque about it.
6.4.B.2 Net external torque Fix
Net external torque
The sum, with signs for sense, of the torques exerted on a selected object or rigid system by objects outside it. A force whose line of action passes through the axis, such as the force from a frictionless axle or hinge, exerts zero torque about that axis.
Students often think When the net torque on a system is zero, the system has no angular momentum. In fact No. Zero net external torque means the angular momentum is constant, not zero. A wheel spinning on a frictionless axle keeps its (nonzero) angular momentum.
Students often think After objects join and rotate together, results can be found with the rotational inertia of only one of the objects, even when the combined value, or the other object's value, is the one that applies. In fact No. Each quantity needs the rotational inertia that belongs to it: the combined system's final angular speed uses the sum I1 + I2, and one object's own angular momentum afterward uses that object's rotational inertia only.
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 selected object or rigid system is not zero, the system's angular momentum changes and exactly that amount is gained by its environment (for example, by a brake, its frame and Earth). The system's angular momentum is not conserved, but the total including the environment is.
11 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 11
The diagram is a top view of a turntable that rotates counterclockwise about a fixed axle through its center O, and a small puck sliding in a straight line across the level floor beside it. The turntable has rotational inertia 0.20 kg·m² about the axle and angular speed 3.0 rad/s. The puck has mass 0.40 kg and speed 2.0 m/s. At the instant shown, the puck is 0.50 m from O and its velocity makes an angle of 37° with the extension of the line from O to the puck (sin 37° = 0.60). What is the magnitude of the total angular momentum of the turntable–puck system about the axle at this instant?
Answer and reasoning
A0.36 kg·m²/sCorrect The total is the sum of the parts' angular momenta about the axle, with signs. The turntable has Iω = (0.20)(3.0) = 0.60 kg·m²/s counterclockwise. The puck moves clockwise about O with rmv sin θ = (0.50)(0.40)(2.0)(0.60) = 0.24 kg·m²/s. The total is 0.60 − 0.24 = 0.36 kg·m²/s.
B0.84 kg·m²/s A student who adds the magnitudes without regard to sense picks 0.60 + 0.24 = 0.84 kg·m²/s. The puck goes around O clockwise while the turntable turns counterclockwise, so the puck's angular momentum is negative and reduces the total.
C0.60 kg·m²/s A student who thinks an object moving in a straight line has no angular momentum counts only the turntable. The puck's line of motion does not pass through O, so it has angular momentum rmv sin θ = 0.24 kg·m²/s about the axle, which must be included.
D0.20 kg·m²/s A student who drops the sin θ factor finds the puck's angular momentum as rmv = (0.50)(0.40)(2.0) = 0.40 kg·m²/s and gets 0.60 − 0.40 = 0.20 kg·m²/s. Only the velocity component perpendicular to the line from O counts, so the puck's value is 0.40 × 0.60 = 0.24 kg·m²/s.
Working Take counterclockwise as positive. Turntable: L = Iω = (0.20 kg·m²)(3.0 rad/s) = +0.60 kg·m²/s. Puck: its velocity has a component perpendicular to the line from O pointing downward in the diagram, so it moves clockwise about O: L = −rmv sin θ = −(0.50 m)(0.40 kg)(2.0 m/s)(0.60) = −0.24 kg·m²/s. Total: 0.60 − 0.24 = +0.36 kg·m²/s, magnitude 0.36 kg·m²/s (counterclockwise).
A student sits at rest on a stool that can rotate with negligible friction about a vertical axle. She holds a bicycle wheel directly above her head with the wheel's axle vertical and in line with the stool's axle, and with her other hand she sets the wheel spinning. As the wheel starts to spin, she and the stool start to turn slowly in the opposite sense. How does the angular impulse exerted on the wheel by the student compare with the angular impulse exerted on the student by the wheel?
Answer and reasoning
AThe one on the student is larger, as the wheel is spinning much faster. A student who thinks the faster-spinning object exerts the larger torque picks this: the wheel seems to push back on her harder than she pushes on it. The torques the student and the wheel exert on each other are equal in magnitude and opposite in sense however fast either is turning, so the angular impulses are equal in magnitude; the wheel spins faster only because its rotational inertia is smaller.
BThey are equal in magnitude and opposite in their rotational sense.Correct The student's hand and the wheel exert torques on each other about the same axis, so by Newton's third law the angular impulses are equal in magnitude and opposite in sense, for every instant of the interaction. The wheel spins faster than the student only because its rotational inertia is smaller.
CThe one by the student is larger, as her rotational inertia is larger. A student who thinks the object with more rotational inertia exerts the larger torque picks this: she seems to deliver more angular impulse to the wheel than it delivers to her. The torques that the student and the wheel exert on each other are equal and opposite, whatever their rotational inertias, and so are the angular impulses.
DThe one on the student is zero, as she alone exerts a torque here. A student who thinks the object being turned cannot turn anything back picks this. The wheel exerts a torque on her hand while she exerts one on the wheel, which is exactly why she and the stool start to turn the other way.
A student stands at rest at the edge of a turntable that is also at rest and can rotate with negligible friction about a fixed vertical axle. The student starts to walk counterclockwise around the edge, and the turntable starts to rotate clockwise. For which system is the total angular momentum about the axle constant while the student starts to walk?
Answer and reasoning
AThe student and turntable together, as their torques on each other are internalCorrect With the student and the turntable both in the system, the frictional torques they exert on each other are internal and cancel in pairs; the axle and gravity exert no torque about the axle. The total stays zero: the student's counterclockwise angular momentum is matched by the turntable's clockwise angular momentum.
BThe turntable alone, since its axle exerts only negligible friction on it A student who thinks a frictionless axle is enough picks this. The student's feet exert a frictional force on the turntable, which is an external torque on the turntable alone, and it changes the turntable's angular momentum from zero to clockwise.
CThe student alone, since she sets herself moving using her own muscles A student who thinks a person can start moving without being pushed picks this. The student starts moving counterclockwise only because the turntable exerts a frictional force on her feet, so her angular momentum changes from zero.
DNo system, since friction acts between the student's feet and the turntable's surface A student who thinks friction destroys angular momentum picks this. The frictional torques on the student and on the turntable are equal and opposite, so they transfer angular momentum between the two without changing the total of the combined system.
Disk X rotates on a vertical axle with negligible friction. Disk Y is held at rest just above X on the same axle. At t = 0.20 s, Y is released onto X, and friction between the disks brings them to a common angular velocity. The graph shows the angular momentum L of each disk about the axle as a function of time t. Which claim about the system of both disks do the data support?
Answer and reasoning
AThe total decreases while the disks slip, since friction between them removes it. A student who thinks friction destroys angular momentum picks this. The data show that every 1 kg·m²/s X loses, Y gains, so the total stays at 6.0 kg·m²/s even while the disks slip. Friction transfers angular momentum between the disks.
BThe disks end with equal angular momenta, as they end with the same angular velocity. A student who equates angular momentum with angular velocity picks this. The graph shows final values of 4.0 and 2.0 kg·m²/s: with the same ω, the disk with the larger rotational inertia has the larger angular momentum (X has twice Y's rotational inertia).
CThe angular momentum X loses, Y gains, so the system's total is constant.Correct From 0.20 s to 0.60 s, X's angular momentum falls from 6.0 to 4.0 kg·m²/s while Y's rises from 0 to 2.0 kg·m²/s. At every instant the two add to 6.0 kg·m²/s: the frictional torques between the disks are internal to the system and only transfer angular momentum from X to Y.
DThe kinetic energy of the system stays constant, as its angular momentum does. A student who thinks kinetic energy is conserved whenever angular momentum is picks this. K = L²/(2I) for each disk: before, X alone has 36/(2IX) = 18/IX; after, 16/(2IX) + 4/(2 × IX/2) = 12/IX. Friction between the slipping disks transforms a third of the kinetic energy into internal energy.
A figure skater spins on ice with negligible friction and air resistance. Her rotational kinetic energy is 240 J when her rotational inertia about her spin axis is 4.5 kg·m². She then pulls her arms in, reducing her rotational inertia to 3.0 kg·m². What is her rotational kinetic energy after she pulls her arms in?
Answer and reasoning
A240 J A student who thinks kinetic energy is conserved along with angular momentum picks this. Only L is constant here; her muscles do work as she pulls her arms inward, so her rotational kinetic energy increases.
B540 J A student who notices that ω increases by a factor of 1.5 and squares it, as if I had stayed 4.5 kg·m², gets 240 J × 2.25 = 540 J. I decreased by a factor of 1.5 at the same time, so K increases by 2.25/1.5 = 1.5 only.
C160 J A student who thinks her angular speed stays the same, since no external torque acts, uses K ∝ I and gets 240 J × 3.0/4.5 = 160 J. With no external torque it is Iω that stays the same, so ω increases as I decreases.
D360 JCorrect Her angular momentum is constant, and K = L²/(2I), so K is inversely proportional to I. Reducing I by a factor of 1.5 increases K by a factor of 1.5: 240 J × 4.5/3.0 = 360 J. The extra energy comes from the work her muscles do pulling her arms in.
Working No external torque, so L = Iω is constant. K = (1/2)Iω² = L²/(2I), so with L constant K is inversely proportional to I: Kf = K0 × I0/If = 240 J × (4.5/3.0) = 240 J × 1.5 = 360 J. (Equivalently ω increases by a factor of 1.5, ω² by 2.25, and I decreases by a factor of 1.5: 2.25/1.5 = 1.5.)
A student sits on a stool that rotates with negligible friction about a vertical axle, holding a dumbbell in each outstretched hand. While the stool rotates, she pulls the dumbbells in toward her chest, and the system of student, stool and dumbbells rotates faster. Which statement about the rotational kinetic energy of the system is correct?
Answer and reasoning
AIt stays the same, as it is conserved along with angular momentum. A student who thinks kinetic energy is conserved whenever angular momentum is picks this. L = Iω is constant, but I decreases and ω increases in the same ratio, so (1/2)Iω² increases in that ratio.
BIt increases, because her arms do work pulling the dumbbells inward.Correct The system's angular momentum is constant, and with smaller rotational inertia K = L²/(2I) is larger. The energy comes from inside the system: her arms pull each dumbbell inward while it moves inward, so they do positive work on it, using energy stored in her body.
CIt increases, because some angular momentum is converted into kinetic energy. A student who thinks angular momentum can change into energy picks this. The system's angular momentum does not change at all; the kinetic energy comes from the work done by the student's muscles.
DIt increases, because the axle must exert a torque to speed up the stool. A student who thinks every increase in angular speed needs an external torque picks this. That rule is for rigid systems. Here the rotational inertia changes, so ω increases with no external torque; the frictionless axle exerts no torque about its own axis.
Two identical small beads can slide along a light rod that rotates in a horizontal plane about a vertical axle through the rod's center, with negligible friction at the axle. A mechanism in the system moves the beads along the rod. The diagram shows three arrangements of the beads, with each bead's distance from the axle. The system has the same angular momentum in all three arrangements. Which correctly ranks the system's angular speeds ω₁, ω₂ and ω₃ in arrangements 1, 2 and 3?
Answer and reasoning
Aω₂ = ω₃ > ω₁ A student who thinks rotational inertia depends on the average distance of the mass from the axle ranks 2 and 3 equal, since both have an average distance of 0.20 m. Squaring makes the bead at 0.30 m count much more: I₂ = 0.10m is larger than I₃ = 0.08m.
Bω₁ > ω₂ > ω₃ A student who thinks mass farther from the axle makes a system spin faster ranks the arrangements in order of how far out the mass is, which is the order of rotational inertia: 1, 2, 3. The beads farther out do move faster for the same ω, but with the same angular momentum a larger rotational inertia means a smaller angular speed, so the order is reversed.
Cω₁ = ω₂ = ω₃ A student who thinks that with no external torque the angular speed stays the same picks this. It is Iω that stays the same; the arrangements have different rotational inertias, so they have different angular speeds.
Dω₃ > ω₂ > ω₁Correct Each bead contributes mr². The rotational inertias are 0.18m (arrangement 1), 0.10m (arrangement 2) and 0.08m (arrangement 3). With the same angular momentum, ω = L/I, so the smallest rotational inertia gives the largest angular speed: ω₃ > ω₂ > ω₁.
Working With bead mass m: I₁ = m(0.30)² + m(0.30)² = 0.18m; I₂ = m(0.10)² + m(0.30)² = 0.10m; I₃ = m(0.20)² + m(0.20)² = 0.08m (in kg·m² with m in kg). L = Iω is the same, so ω ∝ 1/I: ω₃ > ω₂ > ω₁.
Two children sit on a merry-go-round that can rotate with negligible friction about a fixed vertical axle. Starting at t = 0, a parent standing on the ground pushes tangentially on the rim. The graph shows the torque τ exerted on the merry-go-round by the parent as a function of time t. What is the change in the angular momentum of the system of merry-go-round and children from t = 0 to t = 3.0 s?
Answer and reasoning
A9.0 kg·m²/s A student who treats the region under the sloping line as a rectangle gets (3.0)(1.0) + (3.0)(2.0) = 9.0 kg·m²/s. From 1.0 s to 3.0 s the torque falls steadily to zero, so that region is a triangle with area (1/2)(3.0)(2.0) = 3.0 kg·m²/s.
B3.0 kg·m²/s A student who reads the height of the graph as the change in angular momentum picks the largest torque, 3.0 N·m. The change in angular momentum is the angular impulse, the area under the torque–time graph, which also depends on how long the torque acts.
C6.0 kg·m²/sCorrect The parent is the only object outside the system that exerts a torque about the axle, so the system's change in angular momentum equals the parent's angular impulse, the area under the graph: (3.0)(1.0) + (1/2)(3.0)(2.0) = 6.0 kg·m²/s.
D1.5 kg·m²/s A student who takes the slope instead of the area uses the sloping part: 3.0 N·m ÷ 2.0 s = 1.5. That has units of N·m/s; the change in angular momentum is the area under the graph, in N·m·s = kg·m²/s.
Working The parent is outside the system, so the parent's torque is the net external torque (torques between the children and the merry-go-round are internal). ΔL = angular impulse = area under the τ–t graph = rectangle + triangle = (3.0 N·m)(1.0 s) + (1/2)(3.0 N·m)(2.0 s) = 3.0 + 3.0 = 6.0 kg·m²/s.
A merry-go-round of radius R and rotational inertia I about its axle is at rest and can rotate with negligible friction. A child of mass m runs with speed v along a line tangent to the rim and jumps onto the rim, and the child and merry-go-round then rotate together. Model the child as a point object. What is the magnitude of the angular impulse exerted on the merry-go-round by the child?
Answer and reasoning
AImvR/(I+mR²)Correct For the child–merry-go-round system no external torque acts, so mvR = (I + mR²)ωf and ωf = mvR/(I + mR²). The angular impulse on the merry-go-round equals its own change in angular momentum, Iωf − 0 = ImvR/(I + mR²).
BmvR/(I+mR²) A student who treats angular momentum and angular velocity as the same thing gives the final angular speed ωf. The angular impulse equals the change in angular momentum, Iωf, which includes the merry-go-round's rotational inertia.
Cm²vR³/(I+mR²) A student who multiplies ωf by the child's rotational inertia, mR², finds the angular momentum the child keeps. The angular impulse on the merry-go-round equals the merry-go-round's own change in angular momentum, so its rotational inertia I is the one to use: Iωf.
DIv√(m/(I+mR²)) A student who treats the landing as if kinetic energy were conserved sets (1/2)mv² = (1/2)(I + mR²)ω² and multiplies the resulting ω by I. The child and merry-go-round stick together, so kinetic energy is not conserved; only angular momentum is, which gives ωf = mvR/(I + mR²).
Working System child + merry-go-round: no external torque about the axle, so L is constant. Before: L = mvR (child moving tangent to the rim, sin θ = 1); merry-go-round 0. After: (I + mR²)ωf = mvR, so ωf = mvR/(I + mR²). The angular impulse on the merry-go-round equals its change in angular momentum: Iωf − 0 = ImvR/(I + mR²). The child keeps mR²ωf = m²vR³/(I + mR²); the two shares add to mvR.
The diagram shows a wheel that rotates on a fixed axle through its center O, with negligible friction at the axle. Rope 1 is wrapped around a hub of radius 0.25 m and pulls tangentially with a force of 24 N. Rope 2 is attached to the rim at point P, 0.40 m from O, and pulls at 30° to the line OP, as shown (sin 30° = 0.50, cos 30° = 0.87). The arrows show the directions of the pulls only; they are not drawn to scale. What magnitude of force must rope 2 exert for the wheel's angular momentum to stay constant?
Answer and reasoning
A15 N A student who leaves out sin θ uses the full distance OP as the lever arm: 6.0/0.40 = 15 N. Rope 2 pulls at 30° to OP, so only the component F₂ sin 30° exerts a torque, and F₂ must be twice as large.
B17 N A student who uses cos 30° instead of sin 30° gets 6.0/(0.40 × 0.87) = 17 N. The component of rope 2's pull along OP exerts no torque; the perpendicular component, F₂ sin 30°, does.
C24 N A student who thinks equal forces balance each other's turning effect picks 24 N, the same as rope 1. Torque depends on the lever arm and angle as well as the force, so the two torques are equal only if (0.40)F₂ sin 30° = (0.25)(24).
D30 NCorrect L is constant when the net external torque is zero. Rope 1 exerts (0.25)(24) = 6.0 N·m counterclockwise, so rope 2 must exert 6.0 N·m clockwise: (0.40)F₂ sin 30° = 6.0, which gives F₂ = 6.0/(0.40 × 0.50) = 30 N.
Working The angular momentum stays constant if the net external torque is zero. The axle's force acts at O and exerts no torque. Rope 1: τ₁ = rF = (0.25 m)(24 N) = 6.0 N·m, counterclockwise. Rope 2: τ₂ = rF sin θ = (0.40 m)F₂(0.50), clockwise. τ₂ = τ₁ gives F₂ = 6.0 N·m / (0.40 m × 0.50) = 30 N.
A turntable rotates with angular speed ω₀ about a vertical axle with negligible friction. A thin ring of mass M and radius R that is not rotating is dropped onto the turntable with its center on the axle, and friction soon makes the ring rotate with the turntable. All of the ring's mass is at distance R from the axle, and the turntable's rotational inertia about the axle is 3MR². What is the final angular speed of the turntable and ring?
Answer and reasoning
A0.87ω₀ A student who treats the kinetic energy as conserved sets (1/2)(3MR²)ω₀² = (1/2)(4MR²)ω² and gets ω = √(3/4)ω₀ = 0.87ω₀. The ring slips before it moves with the turntable, so friction transforms some kinetic energy; only the angular momentum is conserved.
B0.75ω₀Correct No net external torque acts on the turntable–ring system, so its angular momentum is constant: (3MR²)ω₀ = (3MR² + MR²)ω. The ring increases the rotational inertia by a third, so ω = (3/4)ω₀ = 0.75ω₀.
C1.00ω₀ A student who thinks the angular speed stays the same when no external torque acts picks this. It is Iω that stays the same; adding the ring increases the rotational inertia, so the angular speed decreases.
D3.00ω₀ A student who divides the angular momentum 3MR²ω₀ by the ring's rotational inertia alone, MR², gets 3ω₀. After the ring and turntable rotate together, the rotational inertia is that of both, 4MR².
Working System: turntable + ring. The frictional torques between them are internal and the axle exerts no torque, so L is constant. Ring: Iring = MR² (all its mass at distance R). Before: L = (3MR²)ω₀. After: (3MR² + MR²)ω = 4MR²ω. So ω = (3/4)ω₀ = 0.75ω₀.
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
