6.3 Angular Momentum and Angular Impulse

1. AThey are equal, since the masses and angular speeds are equal.
   A student who thinks rotational inertia depends only on mass picks this. Rotational inertia also depends on how the mass is distributed: the hoop's mass is farther from the axis, so at the same ω the hoop has more angular momentum.
2. BThe hoop's is greater, since its mass is all at the rim. Correct
   L = Iω and the angular speeds are equal, so the comparison is decided by rotational inertia. All of the hoop's mass is a distance R from the axis, while much of the disk's mass is closer to it, so the hoop has the greater I and the greater angular momentum.
3. CThe disk's is greater, since the disk is solid all the way through.
   A student who thinks a solid object is harder to turn than a hollow one of the same mass picks this. The masses are equal; what matters is where the mass is. The disk's mass is on average closer to the axis, so its rotational inertia, and its angular momentum at the same ω, is smaller.
4. DBoth are zero, since neither object's center is moving.
   A student who thinks an object spinning in place has no angular momentum picks this. Each object's LINEAR momentum is zero because its center of mass is at rest, but each has angular momentum L = Iω about its axis because it is spinning.

CED 6.3.A.1 · Read this in Fix

1. A3.6 kg·m²/s Correct
   L = rmv sin θ = (3.0 m)(0.50 kg)(4.0 m/s)(sin 37°) = 6.0 × 0.60 = 3.6 kg·m²/s. The angle between the line from P and the velocity is 37°, so only r sin 37° = 1.8 m, the perpendicular distance from P to the puck's path, counts.
2. B6.0 kg·m²/s
   A student who uses L = rmv with the full distance, (3.0)(0.50)(4.0), picks this. The velocity is not perpendicular to the line from P, so the factor sin 37° = 0.60 is needed.
3. C4.8 kg·m²/s
   A student who uses cos 37° = 0.80 instead of sin 37° picks this. θ is the angle between the line from P and the velocity, and the formula uses its sine: if the puck moved straight away from P, its angular momentum about P would be zero.
4. D2.0 kg·m²/s
   A student who gives the puck's linear momentum, mv = (0.50)(4.0), picks this. That is in kg·m/s and does not depend on P; angular momentum about P also needs the distance factor r sin θ.

Working L = rmv sin θ = 3.0 m × 0.50 kg × 4.0 m/s × sin 37° = 6.0 × 0.60 = 3.6 kg·m²/s. Errors: no sin θ → 6.0; cos 37° → 4.8; mv → 2.0.

CED 6.3.A.2 · Read this in Fix

1. A1.2 N·m·s Correct
   The torque about the hinges is τ = rF sin θ = (0.60 m)(20 N)(sin 30°) = 6.0 N·m. The angular impulse is τΔt = (6.0 N·m)(0.20 s) = 1.2 N·m·s.
2. B2.4 N·m·s
   A student who takes the torque as rF = (0.60)(20) = 12 N·m, ignoring the 30° angle, picks this. Only the component of the force perpendicular to the door, F sin 30° = 10 N, produces torque about the hinges.
3. C6.0 N·m·s
   A student who finds the torque correctly, 6.0 N·m, but gives it as the angular impulse picks this. Angular impulse is the torque multiplied by the time it acts: 6.0 N·m × 0.20 s.
4. D4.0 N·m·s
   A student who multiplies the force by the time, (20 N)(0.20 s), picks this. That is the linear impulse of the force, in N·s; angular impulse uses the torque, which depends on the distance from the hinges and the angle.

Working τ = rF sin θ = 0.60 m × 20 N × sin 30° = 6.0 N·m. Angular impulse = τΔt = 6.0 N·m × 0.20 s = 1.2 N·m·s. Errors: no sin θ → 12 × 0.20 = 2.4; τ only → 6.0; FΔt → 4.0.

CED 6.3.B.1 · Read this in Fix

1. A−6.0 N·m·s Correct
   Angular impulse has the same direction as the torque that delivers it. The brake's torque is clockwise, which is negative here, so the brake's angular impulse is (−3.0 N·m)(2.0 s) = −6.0 N·m·s, even though the wheel turns counterclockwise and speeds up.
2. B+6.0 N·m·s
   A student who gives the angular impulse the direction in which the wheel turns picks this. The wheel turns counterclockwise, but the brake's torque is clockwise, and the angular impulse has the sense of the torque.
3. C−3.0 N·m·s
   A student who takes the torque itself as the angular impulse picks this. The torque acts for 2.0 s, so the angular impulse is τΔt = (−3.0 N·m)(2.0 s).
4. D+4.0 N·m·s
   A student who uses the net torque on the wheel, (5.0 N·m − 3.0 N·m)(2.0 s) = +4.0 N·m·s, picks this. That is the net angular impulse from the motor and the brake together. The brake's angular impulse comes from the brake's torque alone: (−3.0 N·m)(2.0 s) = −6.0 N·m·s.

Working Angular impulse from the brake = τ_brake Δt with τ_brake = −3.0 N·m (clockwise, counterclockwise positive): (−3.0 N·m)(2.0 s) = −6.0 N·m·s. Errors: sense of rotation → +6.0; τ only → −3.0; net torque (5.0 − 3.0)(2.0) → +4.0.

CED 6.3.B.2 · Read this in Fix

1. A48 N·m·s
   A student who multiplies the maximum torque by the whole time, (40 N·m)(1.2 s), picks this. The torque is 40 N·m only at t = 0.8 s; the area of the triangle is half of that rectangle.
2. B40 N·m·s
   A student who reads the highest value on the graph as the angular impulse picks this. That value is the maximum torque, 40 N·m, at one instant; the angular impulse is the area under the graph.
3. C50 N·m·s
   A student who finds the slope of the rising part, (40 N·m)/(0.8 s), picks this. The slope of a torque–time graph is the rate at which the torque changes, not the angular impulse, which is the area.
4. D24 N·m·s Correct
   The angular impulse is the area under the torque–time graph. The graph is a triangle of base 1.2 s and height 40 N·m: (1/2)(1.2 s)(40 N·m) = 24 N·m·s.

Working Area of triangle = (1/2) × 1.2 s × 40 N·m = 24 N·m·s. Errors: peak × time → 40 × 1.2 = 48; peak value → 40; slope of rising part → 40/0.8 = 50.

CED 6.3.B.3 · Read this in Fix

1. A+2Iω₀
   A student who subtracts the magnitudes, 3Iω₀ − Iω₀, ignoring that the rotation reverses, picks this. Once the sense of rotation changes, signs must be kept: the angular momentum goes from +Iω₀ through zero to −3Iω₀.
2. B−4Iω₀ Correct
   ΔL = L − L₀. With counterclockwise positive, L₀ = +Iω₀ and L = −3Iω₀, so ΔL = −3Iω₀ − (+Iω₀) = −4Iω₀. The rotation reverses, so the change is larger than either magnitude. Comparing magnitudes, as in 3Iω₀ − Iω₀, is valid only while the sense of rotation is unchanged.
3. C−3Iω₀
   A student who gives the final angular momentum as the change picks this. The turntable started with +Iω₀, so the change is L − L₀ = −3Iω₀ − Iω₀.
4. D+4Iω₀
   A student who subtracts in the order initial minus final, Iω₀ − (−3Iω₀), picks this. Δ means final minus initial, and the change here is in the clockwise (negative) sense.

Working L₀ = +Iω₀; L = −3Iω₀. ΔL = L − L₀ = −3Iω₀ − Iω₀ = −4Iω₀. Errors: magnitudes 3 − 1 = +2; final only −3; L₀ − L = +4.

CED 6.3.C.1 · Read this in Fix

1. AIts angular momentum is the same at the end as at the start. Correct
   The change in angular momentum equals the net angular impulse. A net angular impulse of zero means ΔL = 0, so the wheel ends the interval with the same angular momentum it began with, whatever torques acted along the way.
2. BIts angular momentum at the end of the interval is zero.
   A student who takes the angular impulse to be the final angular momentum picks this. Zero net angular impulse means zero CHANGE: a wheel that was spinning is still spinning at the end with the same angular momentum.
3. CNo torque was exerted on the wheel at any time in the interval.
   A student who reads 'zero net' as 'nothing acted' picks this. Torques in opposite senses can act, together or one after another, and their angular impulses can add to zero; only the total is zero.
4. DIts angular speed fell, as no net angular impulse kept it going.
   A student who thinks a spinning object needs a net torque to keep turning picks this. With zero net angular impulse the angular momentum does not change, so a rigid wheel keeps the same angular speed.

CED 6.3.C.2.i · Read this in Fix

1. A7.5 N·m
   A student who reads the height of the graph at t = 5.0 s picks this. The height is the angular momentum, 7.5 kg·m²/s; the net torque is the slope, which is zero on the horizontal section.
2. B2.5 N·m
   A student who thinks the torque that spun the wheel up must continue in order to keep it spinning picks this. The slope of the first section, (7.5 kg·m²/s)/(3.0 s), was the net torque from 0 to 3 s; after 3 s the angular momentum is constant, so the net torque is zero.
3. C0.0 N·m Correct
   The net torque equals the slope of the angular momentum–time graph. From t = 3 s to t = 6 s the graph is horizontal, so its slope, and the net torque at t = 5.0 s, is zero. The wheel keeps a constant angular momentum of 7.5 kg·m²/s.
4. D1.5 N·m
   A student who divides the angular momentum at t = 5.0 s by the time, (7.5)/(5.0), picks this. That is the slope of a line from the origin to the point, an average since t = 0; the net torque at t = 5.0 s is the slope of the graph there, which is zero.

Working Net torque = slope of L–t. For 3 s < t < 6 s, L is constant at 7.5 kg·m²/s, so the slope at t = 5.0 s is 0. Errors: height → 7.5; slope of the 0–3 s section 7.5/3.0 = 2.5; L/t = 7.5/5.0 = 1.5.

CED 6.3.C.3 · Read this in Fix

1. A+4.5 kg·m²/s
   A student who gives the angular impulse, the area under the graph, as the final angular momentum picks this. The area is the CHANGE in angular momentum; the wheel already had +2.0 kg·m²/s at t = 0.
2. B+6.5 kg·m²/s Correct
   The angular impulse is the signed area under the net torque–time graph: (+3.0 N·m)(2.0 s) + (−1.5 N·m)(1.0 s) = +6.0 − 1.5 = +4.5 N·m·s. The final angular momentum is +2.0 + 4.5 = +6.5 kg·m²/s.
3. C+9.5 kg·m²/s
   A student who adds the area below the axis as positive, 2.0 + 6.0 + 1.5, picks this. From 2.0 s to 3.0 s the torque is clockwise (negative), so that area reduces the angular momentum by 1.5 kg·m²/s.
4. D−1.5 kg·m²/s
   A student who reads the value of the graph at t = 3.0 s picks this. That value is the net torque at that instant, −1.5 N·m. The angular momentum is found from the initial value plus the area under the graph.

Working Area = (+3.0)(2.0) + (−1.5)(1.0) = +4.5 N·m·s = ΔL. L(3.0 s) = +2.0 + 4.5 = +6.5 kg·m²/s. Errors: area only → +4.5; all areas positive → 2.0 + 6.0 + 1.5 = +9.5; height at 3.0 s → −1.5.

CED 6.3.C.4 · Read this in Fix

1. AL = 2md²ω
   A student who uses the whole rod length d as the distance of each sphere from the axis gets I = 2md² and picks this. The axis is at the center, so each sphere is only d/2 from it, which makes each contribution four times smaller.
2. BL = mdω
   A student who adds the linear momenta of the spheres, 2 × m(ωd/2), picks this. That is a sum of mv values (and, since the spheres move in opposite directions, not even the system's linear momentum); angular momentum needs the distance from the axis as well: L = Iω.
3. CL = md²ω/2 Correct
   Each sphere is d/2 from the axis, so I = 2m(d/2)² = md²/2 and L = Iω = md²ω/2. (Check: each sphere has L = rmv sin 90° = (d/2)m(ωd/2) = md²ω/4, and two of them give md²ω/2.)
4. DL = md²ω²/4
   A student who uses (1/2)Iω² = (1/2)(md²/2)ω² picks this. That is the system's rotational kinetic energy, in joules, not its angular momentum; angular momentum is Iω, with ω to the first power.

Working I = Σ m r² = m(d/2)² + m(d/2)² = md²/2. L = Iω = md²ω/2. Errors: r = d gives 2md²ω; Σmv = 2m(ωd/2) = mdω; (1/2)Iω² = md²ω²/4.

CED 6.3.A.1 · Read this in Fix

1. AIt has the same value about A as it has about B.
   A student who thinks angular momentum belongs to the puck alone, like its mass, picks this. Angular momentum is always about a chosen point: here r and θ differ for A and B, and so does L.
2. BIt is zero about A and about B, as its path is straight.
   A student who thinks only rotating or circling objects have angular momentum picks this. The puck's straight-line motion gives it zero angular momentum only about points on its line, such as A; about B it is mvd.
3. CIt is larger about A, as A is farther from the puck.
   A student who uses L = rmv with the full distance, ignoring the angle, picks this. A is farther away, but the puck moves straight toward A (θ = 180°), so sin θ = 0 and its angular momentum about A is zero.
4. DIt is zero about A and equal to mvd about B. Correct
   About A, the line from A to the puck lies along the velocity, so θ = 180° and L = rmv sin θ = 0. About B, r sin θ is the perpendicular distance d from B to the path, so L = mvd. The value depends on the chosen point.

CED 6.3.A.2.i · Read this in Fix

1. AIt has angular momentum about P only when it is closest to P.
   A student who looks only at the angle, and sees that sin θ is largest at closest approach, picks this. As the puck moves on, r grows while sin θ shrinks, and r sin θ stays equal to d: the angular momentum about P is the same, mvd, at every instant, not just one.
2. BIts velocity is not along the line from P, so rmv sin θ is not zero. Correct
   L = rmv sin θ depends on the distance, mass, speed and the angle between the line from P and the velocity. Because P is off the path, that angle is never 0 or 180°, so the puck has angular momentum mvd about P at every instant, where d is P's perpendicular distance from the path.
3. CIts angular momentum about P grows as its distance from P grows.
   A student who uses L = rmv with the full distance picks this. The growing distance is offset by a shrinking sin θ; for straight-line motion at constant velocity the angular momentum about P does not change.
4. DIts angular momentum about P equals its linear momentum, mv.
   A student who treats angular momentum as linear momentum picks this. The puck does have linear momentum, but angular momentum about P also depends on how far the path is from P: L = mvd, in kg·m²/s.

CED 6.3.A.2.ii · Read this in Fix

1. A6
   A student who treats angular momentum like kinetic energy, with the speed squared, gets (1/2)(2²)(3) = 6 and picks this. Angular momentum is proportional to the speed, not to its square.
2. B9
   A student who carries over the r² of I = mr² gets (1/2)(2)(3²) = 9 and picks this. For an object moving past a point, L = mvd contains the perpendicular distance only once.
3. C1
   A student who compares only the linear momenta, (1/2)(2) = 1, picks this. Linear momentum does not depend on P; the angular momentum about P also grows in proportion to the distance of the path from P.
4. D3 Correct
   For each puck L = rmv sin θ = mv × (perpendicular distance). L₂/L₁ = (1/2)(2)(3) = 3.

Working L = mvd for straight-line motion (r sin θ = d). L₂/L₁ = (m/2)(2v)(3d)/(mvd) = 3. Errors: v² → (1/2)(4)(3) = 6; d² → (1/2)(2)(9) = 9; mv only → 1.

CED 6.3.A.2.ii · Read this in Fix

1. A2.0
   A student who notices that the torque doubles, but ignores the halved time, picks this. Angular impulse is torque × time, and the second push lasts half as long.
2. B0.5
   A student who treats torque as the force, so that the angular impulse is FΔt, sees the force unchanged and the time halved and picks this. Pushing twice as far from the hinges doubles the torque.
3. C1.0 Correct
   Angular impulse = τΔt = rFΔt for a perpendicular push. Second push: (2r)F(Δt/2) = rFΔt, the same as the first. Doubling the torque exactly offsets halving the time.
4. D4.0
   A student who divides the torque by the time, (2)/(1/2) = 4, picks this. Angular impulse is τ MULTIPLIED by Δt; τ/Δt has no physical meaning here, and ΔL/Δt is the torque.

Working J₁ = rFΔt. J₂ = (2r)F(Δt/2) = rFΔt. Ratio = 2 × 0.5 = 1.0. Errors: torque only → 2.0; FΔt → 0.5; τ/Δt → 2/0.5 = 4.0.

CED 6.3.B.1 · Read this in Fix

1. AThe angular impulse delivered to B is double that to A.
   A student who treats B's torque as if it were 6.0 N·m for the whole 2.0 s gets 12 N·m·s and picks this. B reaches 6.0 N·m only at t = 1.0 s; the area of its triangle is 6.0 N·m·s, the same as A's rectangle.
2. BThe angular impulse delivered to B is three times A's.
   A student who compares only the largest torques, 6.0 N·m and 2.0 N·m, picks this. Angular impulse depends on how long each torque acts too: A's smaller torque acts for longer, and the areas come out equal.
3. CThe angular impulses delivered to A and B are equal. Correct
   Angular impulse is the area under each graph. A: rectangle, (2.0 N·m)(3.0 s) = 6.0 N·m·s. B: triangle, (1/2)(2.0 s)(6.0 N·m) = 6.0 N·m·s. The areas, and so the angular impulses, are equal.
4. DNo angular impulse is delivered to A by its torque.
   A student who takes the angular impulse from the slope of the graph sees A's flat line and picks this. A flat line means a constant torque, not zero angular impulse; the area under A's line is 6.0 N·m·s.

Working A: 2.0 N·m × 3.0 s = 6.0 N·m·s. B: (1/2) × 2.0 s × 6.0 N·m = 6.0 N·m·s. Equal. Errors: B at peak for 2.0 s → 12 (twice); peak torques 6.0/2.0 → three times; slope of A = 0 → zero.

CED 6.3.B.3 · Read this in Fix

1. A0.75
   A student who compares rotational kinetic energies, (3I)(ω/2)²/(Iω²) = 0.75, picks this. A torque acting for a time delivers angular impulse, which changes angular momentum Iω, not kinetic energy; the ratio of angular momenta decides the times.
2. B0.50
   A student who compares only the angular speeds, (ω/2)/ω, picks this. The stopping time depends on the angular momentum to be removed, Iω, and Y's three times larger rotational inertia outweighs its halved speed.
3. C1.50 Correct
   Each brake delivers an angular impulse equal to the wheel's change in angular momentum: τΔt = Iω − 0. With equal torques, the stopping time is proportional to the initial angular momentum. L_Y/L_X = (3I)(ω/2)/(Iω) = 1.50.
4. D0.67
   A student who writes the angular impulse as τ/Δt, so that Δt = τ/ΔL, gets the ratio the wrong way up, L_X/L_Y, and picks this. Angular impulse is τΔt, so for equal torques a larger ΔL needs a LONGER time.

Working τΔt = ΔL, so Δt = Iω_initial/τ. t_Y/t_X = (3I × ω/2)/(I × ω) = 1.5 = 1.50. Errors: K ratio 3 × 0.25 = 0.75; ω ratio 0.50; inverted 1/1.5 = 0.67.

CED 6.3.C.2.i · Read this in Fix

1. AL_B is half of L_A.
   A student who treats angular momentum as angular speed picks this. B does gain half of A's angular speed, but its rotational inertia is twice A's, so its angular momentum Iω is the same.
2. BL_B is two times L_A.
   A student who reasons 'more rotational inertia, more angular momentum' picks this. The change in angular momentum is set by the angular impulse, τΔt, which does not depend on the wheel's rotational inertia.
3. CL_B is √2 times L_A.
   A student who assumes the equal torques give the wheels equal kinetic energies gets ω_B = ω_A/√2 and L_B = √2 L_A, and picks this. Equal torques acting for equal TIMES give equal angular impulses, and so equal angular momenta; the kinetic energies are not equal (B's is half of A's).
4. DL_B is the same as L_A. Correct
   Each wheel's change in angular momentum equals the angular impulse delivered, τΔt, which is the same for both. Both started at rest, so L_B = L_A. Wheel B, with twice the rotational inertia, ends with half of A's angular speed.

Working ΔL = τΔt is equal for A and B; both start at rest, so L_B = L_A. (ω_B = ω_A/2; K_B = K_A/2.) Errors: L ↔ ω → half; 'more I, more L' → twice; equal K → (1/2)(2I)ω_B² = (1/2)Iω_A² gives ω_B = ω_A/√2 and L_B = √2 L_A.

CED 6.3.C.2.i · Read this in Fix

1. Aτ/(Iω₀)
   A student who writes the angular impulse as τ/Δt, so that ΔL = τ/Δt and Δt = τ/ΔL, picks this. The change in angular momentum is τ MULTIPLIED by Δt; a larger change needs a longer time, not a shorter one.
2. BIω₀/τ Correct
   For constant rotational inertia, τ_net = ΔL/Δt = IΔω/Δt. Here Δω = 2ω₀ − ω₀ = ω₀, so Δt = IΔω/τ = Iω₀/τ.
3. C3Iω₀²/(2τ)
   A student who sets τΔt equal to the change in rotational kinetic energy, (1/2)I(4ω₀² − ω₀²) = (3/2)Iω₀², picks this. A torque acting for a time changes angular momentum, Iω; kinetic energy is changed by work, τΔθ. The units also fail: 3Iω₀²/(2τ) is in radians, not seconds.
4. D2Iω₀/τ
   A student who takes the angular impulse to be the final angular momentum, I(2ω₀), and leaves out the initial angular momentum picks this. The angular impulse equals the CHANGE in angular momentum, I(2ω₀) − Iω₀ = Iω₀.

Working τ_net = ΔL/Δt = I(2ω₀ − ω₀)/Δt, so Δt = Iω₀/τ. Errors: final angular momentum taken as the change, τΔt = I(2ω₀) → 2Iω₀/τ; ΔL = τ/Δt → Δt = τ/(Iω₀); τΔt = ΔK = (1/2)I(4 − 1)ω₀² → 3Iω₀²/(2τ).

CED 6.3.C.2.ii · Read this in Fix

1. AIt is not zero, since a net torque is needed to keep the wheel spinning.
   A student who thinks steady rotation needs a net torque picks this. A net torque changes angular momentum; a wheel with zero net torque keeps spinning at a constant rate. Real wheels slow down only because friction exerts a net torque.
2. BIt is not zero, since the axle and Earth exert forces on the wheel.
   A student who thinks every force on a rotating object exerts a torque picks this. The axle's force and the gravitational force both act at the center, on the axis, so they have zero lever arm and exert no torque about it.
3. CIt is zero, since a wheel spinning in place has no angular momentum.
   A student who thinks an object spinning in place has no angular momentum picks this. The wheel has angular momentum Iω about its axle; the net torque is zero because that angular momentum is constant, not because it is zero.
4. DIt is zero, since the wheel's angular momentum is not changing. Correct
   τ_net = ΔL/Δt. Constant angular velocity with constant rotational inertia means constant angular momentum, so the net torque is zero. Any torque from a motor or from friction must be balanced by others.

CED 6.3.C.2.ii · Read this in Fix

1. AIncorrect: the graph's slope at t = 2 s is −3 N·m, as at all other times shown. Correct
   The net torque is the slope of the L–t graph. The graph is a straight line from +6 kg·m²/s at t = 0 to −6 kg·m²/s at t = 4 s, so its slope is (−12 kg·m²/s)/(4 s) = −3 N·m everywhere, including at t = 2 s, where L happens to be zero.
2. BCorrect: the angular momentum is zero at t = 2 s, so the net torque is zero.
   A student who thinks a zero angular momentum means a zero net torque picks this. At t = 2 s the wheel is momentarily not turning, but its angular momentum is still changing at the same rate, −3 N·m, as at every other instant.
3. CIncorrect: the net torque turns from counterclockwise to clockwise at t = 2 s.
   A student who thinks the torque acts in the direction the wheel turns picks this. The wheel reverses at t = 2 s, but the torque does not: the slope is −3 N·m throughout, so the net torque is clockwise before and after t = 2 s.
4. DIncorrect: the area under the graph from t = 0 to t = 2 s is not zero.
   A student who takes the net torque from the area under an angular momentum–time graph picks this. Net torque is the SLOPE of the L–t graph; the area under it has no meaning as a torque. The claim is wrong, but because the slope at t = 2 s is −3 N·m.

Working Slope = ΔL/Δt = (−6 − 6) kg·m²/s ÷ (4 − 0) s = −3 N·m, constant, so τ_net(2 s) = −3 N·m ≠ 0.

CED 6.3.C.3 · Read this in Fix