AP Physics C Mechanics Quiz: Angular Momentum And Angular Impulse
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Angular Momentum And Angular ImpulseQuestion 1 of 20
A spacecraft with I=80kg\cdotpm2 must reach ω=0.60rad/s from rest; during a short burn, external torques are negligible except for the thrusters, so J=ΔL=IΔω. Assume the torque direction matches the desired rotation. Considering the described system, calculate the angular impulse applied to the spacecraft during the interval.
AP Physics C Mechanics Quiz: Angular Momentum And Angular Impulse
Practice Angular Momentum And Angular Impulse in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
What this quiz covers
This quiz focuses on Angular Momentum And Angular Impulse, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.
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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
All questions
Question 1
A spacecraft with I=80kg\cdotpm2 must reach ω=0.60rad/s from rest; during a short burn, external torques are negligible except for the thrusters, so J=ΔL=IΔω. Assume the torque direction matches the desired rotation. Considering the described system, calculate the angular impulse applied to the spacecraft during the interval.
48 N·m·s (correct answer)
133 N·m·s
48 N·m
0.48 N·m·s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft needs to change from rest (ωi = 0) to ωf = 0.60 rad/s, requiring a change in angular momentum. Choice A is correct because the angular impulse J = ΔL = IΔω = I(ωf - ωi) = 80 × (0.60 - 0) = 48 N·m·s. Choice B is incorrect because it might result from a calculation error or misunderstanding of the initial conditions, possibly assuming a different initial angular velocity. To help students: Emphasize careful reading of initial conditions (starting from rest means ωi = 0). Practice problems with various initial states to build pattern recognition. Watch for: assumptions about initial conditions and ensure students identify when objects start from rest versus already in motion.
Question 2
An ice skater rotates about a vertical axis with Ii=3.0kg\cdotpm2 and ωi=2.0rad/s. The skater pulls in arms to If=1.2kg\cdotpm2 with negligible external torque, so L=Iω is conserved. Considering the described system, what is the final angular velocity of the system after the skater pulls in arms?
0.80 rad/s
5.0 rad/s (correct answer)
3.2 rad/s
7.2 rad/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is conserved in a closed system when no external torque acts, which is the key principle in this ice skater problem. In this problem, the ice skater changes their moment of inertia by pulling in their arms, with negligible external torque, so angular momentum remains constant. Choice B is correct because it accurately applies conservation of angular momentum: Li = Lf, so Iiωi = Ifωf, giving ωf = Iiωi/If = (3.0 × 2.0)/1.2 = 5.0 rad/s. Choice C is incorrect because it appears to use an incorrect calculation, possibly confusing the ratio of moments of inertia. To help students: Emphasize that angular momentum conservation is analogous to linear momentum conservation but involves I and ω. Encourage students to identify when external torque is negligible (like this skating problem). Watch for: confusion about when angular momentum is conserved versus when angular velocity is constant, and ensure understanding that decreasing I increases ω.
Question 3
A wind turbine rotor with I=2.0×104kg\cdotpm2 spins at ωi=0.80rad/s. The wind drops and net torque becomes τ=−2.0×103N\cdotpm for t=4.0s, using ΔL=τt and L=Iω. Considering the described system, what is the final angular velocity of the system after 4.0 seconds?
0.60 rad/s (correct answer)
1.0 rad/s
0.40 rad/s
8.0×103 kg·m$^2$/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine rotor experiences a negative net torque as the wind drops, which decreases its angular momentum and slows its rotation. Choice A is correct because it accurately calculates the final angular velocity: ΔL = τt = -2.0×10³ × 4.0 = -8.0×10³ kg·m²/s, then ωf = (Li + ΔL)/I = (2.0×10⁴ × 0.80 - 8.0×10³)/(2.0×10⁴) = (1.6×10⁴ - 8.0×10³)/(2.0×10⁴) = 0.60 rad/s. Choice B is incorrect because it appears to use an incorrect calculation, possibly making an error with the scientific notation. To help students: Emphasize careful handling of scientific notation in multi-step problems. Encourage students to verify that negative torque reduces angular velocity. Watch for: calculation errors with large numbers and scientific notation, and ensure understanding that wind turbines slow down when wind decreases.
Question 4
A wind turbine rotor with I=1.0×103kg\cdotpm2 spins at ωi=2.0rad/s; the wind drops and the net torque becomes τ=−200N\cdotpm for t=5.0s, so ΔL=τt and L=Iω. Neglect other torques. Based on the scenario above, what is the final angular velocity of the system after 5.0 seconds?
3.0 rad/s
1.0 rad/s (correct answer)
0.0 rad/s
2.2 rad/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine experiences a negative torque of -200 N·m for 5.0 seconds, changing its angular momentum by ΔL = τt = -200 × 5.0 = -1000 N·m·s. Choice B is correct because the initial angular momentum is Li = Iωi = 1000 × 2.0 = 2000 N·m·s, the final angular momentum is Lf = Li + ΔL = 2000 + (-1000) = 1000 N·m·s, and the final angular velocity is ωf = Lf/I = 1000/1000 = 1.0 rad/s. Choice C is incorrect because it would mean the turbine stops completely, which would require exactly -2000 N·m·s of impulse, not -1000 N·m·s. To help students: Practice problems involving partial reduction of angular velocity versus complete stopping. Emphasize checking if the final result makes physical sense. Watch for: assumptions that negative torque always brings objects to rest, rather than just reducing angular velocity.
Question 5
A solid disk with I=0.80kg\cdotpm2 rotates at ωi=12rad/s. A motor applies constant torque τ=3.2N\cdotpm for t=5.0s, so ΔL=τt. Based on the scenario above, what is the final angular velocity of the system after 5.0 seconds?
8.0 rad/s
32 rad/s
20 rad/s (correct answer)
20 N·m
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system, while angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the solid disk experiences a constant torque which changes its angular momentum, as described in the scenario. Choice C is correct because it accurately calculates the final angular velocity using the angular impulse-momentum theorem: ΔL = τt = 3.2 × 5.0 = 16 kg·m²/s, then ωf = (Li + ΔL)/I = (0.80 × 12 + 16)/0.80 = 20 rad/s. Choice B is incorrect because it appears to double the correct answer, possibly by misapplying the formula. To help students: Emphasize the step-by-step application of ΔL = τt and L = Iω. Encourage drawing before/after diagrams showing initial and final states. Watch for: confusion between torque and angular momentum units, and ensure students understand that angular impulse changes the angular momentum, not the angular velocity directly.
Question 6
A rotating disk has I=0.50kg\cdotpm2 and ωi=16rad/s. An external torque τ=1.5N\cdotpm acts for t=4.0s, so ΔL=τt and L=Iω. Based on the scenario above, determine the change in angular momentum of the disk.
0.75 kg·m$^2$/s
6.0 kg·m$^2$/s (correct answer)
6.0 N·m
32 kg·m$^2$/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the rotating disk experiences an external torque which changes its angular momentum by the amount of the angular impulse applied. Choice B is correct because it accurately calculates the change in angular momentum using the angular impulse formula: ΔL = τt = 1.5 × 4.0 = 6.0 kg·m²/s. Choice C is incorrect because it gives the answer in N·m units (torque units) rather than kg·m²/s (angular momentum units), showing unit confusion. To help students: Emphasize that the change in angular momentum equals the angular impulse, regardless of initial conditions. Encourage dimensional analysis to catch unit errors. Watch for: confusion between torque and angular momentum units, and ensure students understand that ΔL is independent of the initial angular velocity for a given impulse.
Question 7
A spacecraft has rotational inertia I=120kg\cdotpm2 about its yaw axis and initially ωi=0; its thrusters provide a constant torque τ=30N\cdotpm for t=4.0s, with angular impulse J=τt=ΔL. Ignore external torques from space. Based on the scenario above, calculate the angular impulse applied to the spacecraft during the interval.
7.5 N·m·s
120 N·m·s (correct answer)
120 N·m
480 N·m·s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft's thrusters provide a constant torque of 30 N·m for 4.0 seconds, creating an angular impulse. Choice B is correct because the angular impulse J = τt = 30 × 4.0 = 120 N·m·s, which equals the change in angular momentum ΔL. Choice C is incorrect because it gives only the torque value (30 N·m) multiplied by 4, confusing torque with impulse and missing the proper units. To help students: Emphasize that angular impulse has units of N·m·s (or kg·m²/s), not just N·m. Practice problems involving both impulse calculations and the resulting motion changes. Watch for: unit confusion between torque and angular impulse, and ensure understanding that impulse is the time integral of torque.
Question 8
A bicycle wheel with moment of inertia I=0.60kg\cdotpm2 starts from rest; a tangential force F=18N is applied at the rim of radius r=0.35m for t=2.0s, so τ=rF and ΔL=τt. Neglect friction and assume the force stays perpendicular to the radius. Considering the described system, determine the change in angular momentum of the wheel.
12.6 kg·m$^2$/s (correct answer)
6.3 kg·m$^2$/s
25.2 kg·m$^2$/s
12.6 N
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the bicycle wheel experiences a tangential force at the rim, creating a torque τ = rF = 0.35 × 18 = 6.3 N·m for 2.0 seconds. Choice A is correct because the change in angular momentum is ΔL = τt = 6.3 × 2.0 = 12.6 kg·m²/s, using the impulse-momentum theorem for rotation. Choice B is incorrect because it only calculates the torque (6.3 N·m) without multiplying by time, a common error when students forget that impulse involves both force (or torque) and time duration. To help students: Reinforce the parallel between linear impulse (Ft) and angular impulse (τt). Encourage careful unit analysis to distinguish between torque (N·m) and angular momentum (kg·m²/s). Watch for: confusion between instantaneous quantities (torque) and time-integrated quantities (impulse).
Question 9
A bicycle wheel with I=0.40kg\cdotpm2 starts from rest; a constant tangential force F=10N is applied at radius r=0.30m for t=5.0s, so τ=rF and ΔL=τt. Ignore friction and keep the force perpendicular to the radius. Based on the scenario above, determine the change in angular momentum of the wheel.
15 kg·m$^2$/s (correct answer)
6.0 kg·m$^2$/s
3.0 kg·m$^2$/s
15 N·m
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the bicycle wheel experiences a tangential force creating a torque τ = rF = 0.30 × 10 = 3.0 N·m for 5.0 seconds. Choice A is correct because the change in angular momentum is ΔL = τt = 3.0 × 5.0 = 15 kg·m²/s, properly applying the angular impulse-momentum theorem. Choice D is incorrect because it gives the answer in N·m instead of kg·m²/s, confusing the units of torque with those of angular momentum. To help students: Stress the importance of dimensional analysis in physics problems. Create unit conversion charts showing the relationships between rotational quantities. Watch for: unit confusion between torque (N·m) and angular momentum (kg·m²/s or N·m·s).
Question 10
An ice skater spins with Ii=3.0kg\cdotpm2 at ωi=2.5rad/s; they pull in their arms to If=1.2kg\cdotpm2 while external torque is negligible, so L=Iω is conserved. Assume the change happens quickly compared with frictional effects. Considering the described system, what is the final angular velocity after pulling in their arms?
1.0 rad/s
6.25 rad/s (correct answer)
3.0 rad/s
2.0 rad/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the ice skater pulls in their arms with negligible external torque, so angular momentum is conserved: Li = Lf. Choice B is correct because using conservation of angular momentum: Iiωi = Ifωf, so 3.0 × 2.5 = 1.2 × ωf, giving ωf = 7.5/1.2 = 6.25 rad/s. Choice C is incorrect because it might result from incorrectly assuming the angular velocity changes by the same ratio as the moment of inertia, a common misconception about rotational dynamics. To help students: Emphasize that angular momentum (not angular velocity) is conserved when no external torque acts. Use analogies with linear momentum conservation in collisions. Watch for: confusion between conserved quantities and ensure students understand why angular velocity increases when moment of inertia decreases.
Question 11
A wind turbine rotor has I=1.0×104kg\cdotpm2 and spins at ω0=2.0rad/s. The wind suddenly drops so external torque is approximately zero for t=20s, with τ=ΔL/Δt and L=Iω. Considering the described system, what is the effect of removing an external force on the system's rotation?
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system when no external torque acts. In this problem, the wind turbine rotor experiences zero external torque for 20 seconds when the wind drops, creating a situation where angular momentum must be conserved. Choice B is correct because with zero external torque, angular momentum L = Iω remains constant, and since the moment of inertia I doesn't change, the angular velocity ω must also remain constant at 2.0 rad/s. Choice A is incorrect because it assumes the rotor will slow down without any opposing torque - this violates conservation of angular momentum and reflects a common misconception that rotating objects naturally come to rest. To help students: Emphasize that objects in rotation continue rotating unless acted upon by an external torque (rotational analog of Newton's first law). Use space-based examples where friction is negligible to reinforce this concept.
Question 12
A wind turbine rotor has I=5.0×103kg\cdotpm2 and spins at ωi=1.5rad/s. A gust produces constant net torque τ=1.0×103N\cdotpm for t=6.0s, with ΔL=τt. Considering the described system, determine the change in angular momentum of the turbine.
3.0×103 kg·m$^2$/s
6.0×103 kg·m$^2$/s (correct answer)
6.0×103 N·m
1.2 rad/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system, while angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine rotor experiences a constant net torque from a gust which changes its angular momentum, as described. Choice B is correct because it accurately calculates the change in angular momentum: ΔL = τt = 1.0×10³ × 6.0 = 6.0×10³ kg·m²/s. Choice C is incorrect because it gives the answer in N·m units (torque units) rather than kg·m²/s (angular momentum units), showing confusion between torque and angular momentum. To help students: Reinforce the relationship between angular impulse and change in angular momentum through practice with various scenarios. Encourage careful attention to scientific notation in calculations. Watch for: unit confusion and ensure students understand that ΔL represents change in angular momentum, not torque or angular velocity.
Question 13
A spacecraft has moment of inertia I=1200kg\cdotpm2 about its yaw axis and initially ωi=0. Its attitude-control thrusters provide a constant torque τ=240N\cdotpm for t=3.0s, so angular impulse J=τt=ΔL. Based on the scenario above, calculate the angular impulse applied to the spacecraft during the interval.
720 N·m·s (correct answer)
240 N·m
0.60 rad/s
360 N·m·s
Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system, while angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft's attitude-control thrusters provide a constant torque which creates an angular impulse, as described in the scenario. Choice A is correct because it accurately calculates the angular impulse using J = τt = 240 × 3.0 = 720 N·m·s (which equals kg·m²/s). Choice B is incorrect because it only gives the torque value without multiplying by time, missing the concept of impulse entirely. To help students: Emphasize that angular impulse is torque multiplied by time, analogous to linear impulse being force times time. Encourage students to write out units to verify their calculations. Watch for: confusion between instantaneous torque and impulse over time, and ensure understanding that N·m·s = kg·m²/s through unit analysis.
Question 14
A uniform disk of radius R and mass M rotates about its center with angular velocity ω0. A second identical disk, initially not rotating, is dropped onto the first disk so that their axes coincide. Due to friction between the surfaces, they eventually rotate together. During this process, the angular impulse experienced by the initially stationary disk is:
41MR2ω0 (correct answer)
21MR2ω0
43MR2ω0
MR2ω0
Explanation: Initially, only the first disk has angular momentum Li=21MR2ω0. After coupling, both disks rotate with the same angular velocity. By conservation of angular momentum: 21MR2ω0=2⋅21MR2ωf, so ωf=2ω0. The initially stationary disk gains angular momentum ΔL=21MR2⋅2ω0=41MR2ω0. This equals the angular impulse it received. Choice B is the initial angular momentum of one disk. Choice C would be the angular impulse on the initially moving disk. Choice D is twice the initial angular momentum of one disk.
Question 15
A uniform sphere of radius R and mass M rolls without slipping down an inclined plane. When the sphere has descended a vertical distance h, its angular momentum about its center of mass is Lcm. What is the angular momentum of the sphere about a point on the inclined surface directly below the sphere's center?
Lcm
Lcm+MvR
57Lcm (correct answer)
512Lcm
Explanation: For a rolling sphere, Lcm=Iω=52MR2ω. The angular momentum about a point on the surface uses the parallel axis theorem: Lsurface=Lcm+MVR⊥, where R⊥=R is the perpendicular distance from the center to the contact point. Since v=Rω for rolling without slipping: Lsurface=52MR2ω+MR(Rω)=52MR2ω+MR2ω=57⋅52MR2ω=57Lcm. Choice A ignores the translation contribution. Choice B has the right idea but wrong coefficient. Choice D uses an incorrect factor.
Question 16
A uniform thin hoop of mass M and radius R rotates freely about a vertical axis through its center with angular velocity ω. A small mass m (where m≪M) slides radially outward along a spoke from the center to the rim. During this process, which statement about the angular impulse on the small mass is correct?
The angular impulse is zero because no external torques act on the system
The angular impulse on the mass equals mR2ω⋅M+mm
The angular impulse on the mass equals mR2ω(1−M+mm)
The angular impulse on the mass equals M+mmMR2ω (correct answer)
Explanation: Initially, only the hoop rotates with angular momentum Li=MR2ω. The small mass has zero angular momentum. By conservation of angular momentum: MR2ω=(MR2+mR2)ωf, so ωf=M+mMω. The final angular momentum of the small mass is Lm,f=mR2ωf=M+mmMR2ω. Since the mass started with zero angular momentum, this equals the angular impulse it received. Choice A incorrectly assumes no internal forces can change individual angular momenta. Choice B has an incorrect factor. Choice C uses the wrong expression for conservation.
Question 17
Two identical uniform rods, each of mass M and length L, are initially at rest. Rod A lies along the x-axis with its center at the origin. Rod B lies along the y-axis with its center also at the origin, forming a cross. A brief torque is applied about the z-axis (perpendicular to both rods) to the entire system, giving it an angular impulse J. What is the final angular momentum of rod A about the origin?
2J (correct answer)
4J
12ItotalML2J where Itotal is the total moment of inertia
4J2
Explanation: Both rods have the same moment of inertia about the z-axis: I=121ML2. The total moment of inertia is Itotal=2⋅121ML2=61ML2. The angular impulse gives the system angular velocity ω=ItotalJ=ML26J. Each rod rotates with this same angular velocity, so rod A's angular momentum is LA=IAω=121ML2⋅ML26J=2J. Choice B would be correct if there were 4 identical rods. Choice C is unnecessarily complicated and doesn't simplify correctly. Choice D introduces an incorrect 2 factor.
Question 18
A rigid body rotates about a fixed axis with moment of inertia I. A constant torque τ acts on the body for time t, changing its angular velocity from ω1 to ω2. If the same angular impulse were applied over a time interval 2t (with appropriately reduced torque), and the body started from rest, what would be its final angular velocity?
2ω2−ω1
ω2−ω1 (correct answer)
2ω2+ω1
2(ω2−ω1)
Explanation: The angular impulse in the first scenario is J=τt=I(ω2−ω1). Angular impulse equals the change in angular momentum regardless of how it's applied. In the second scenario, starting from rest with the same total angular impulse: J=Iωfinal−I(0)=Iωfinal. Therefore: Iωfinal=I(ω2−ω1), giving ωfinal=ω2−ω1. Choice A incorrectly assumes the result should be halved due to the longer time. Choice C incorrectly averages the initial and final angular velocities. Choice D incorrectly doubles the result.
Question 19
A particle of mass m is attached to a string of length ℓ and swings as a pendulum. At the bottom of its swing, the particle has speed v. At this instant, a horizontal impulse J is applied to the particle in the direction perpendicular to its velocity. What is the change in angular momentum about the pivot point?
Jℓ (correct answer)
Jℓsinθ where θ is the angle the impulse makes with the radial direction
Jℓ−mvΔt where Δt is the duration of the impulse
(mvℓ)2+(Jℓ)2−mvℓ
Explanation: The change in angular momentum is ΔL=r×Δp. At the bottom of the swing, the position vector from the pivot has magnitude ℓ and points vertically downward. The impulse J is horizontal and perpendicular to the initial velocity (which is also horizontal). Since the impulse is perpendicular to both the position vector and the initial momentum direction, ∣ΔL∣=ℓ∣J∣=Jℓ. Choice B would apply if the impulse weren't perpendicular to the radius, but the problem states it's horizontal at the bottom of the swing. Choice C incorrectly tries to subtract the original angular momentum component. Choice D attempts vector addition but misapplies it.
Question 20
A figure skater performs a spin with arms extended, having moment of inertia I1 and angular velocity ω1. She then pulls her arms inward, reducing her moment of inertia to I2=3I1. If this process takes time Δt, what is the average torque that her muscles exerted during the arm-pulling motion?
3ΔtI1ω1(3−1)
3Δt2I1ω1
Zero, because angular momentum is conserved (correct answer)
ΔtI1ω1
Explanation: During the arm-pulling process, no external torques act on the skater about the vertical axis. The internal forces (muscle forces) that pull the arms inward do not create a net torque about the rotation axis because they act radially. Angular momentum is conserved: L=I1ω1=I2ω2=3I1ω2, so ω2=3ω1. Since L is constant, dtdL=τexternal=0. The average external torque is zero. Choices A, B, and D incorrectly assume that the change in angular velocity requires an external torque, confusing internal forces with external torques.