6.3 Angular Momentum and Angular Impulse

1. A3.0 kg·m²/s
   A student who treats all the disk's mass as if it were at the rim picks this: I = MR² = 0.50 kg·m², so Iω = 3.0 kg·m²/s. Much of a solid disk's mass is closer to the axle, so I = (1/2)MR².
2. B1.5 kg·m²/s Correct
   For a uniform solid disk about its central axis, I = (1/2)MR² = (1/2)(2.0 kg)(0.50 m)² = 0.25 kg·m². Then L = Iω = (0.25 kg·m²)(6.0 rad/s) = 1.5 kg·m²/s.
3. C4.5 kg·m²/s
   A student who calculates the rotational kinetic energy, (1/2)Iω² = (1/2)(0.25)(6.0)² = 4.5, in place of the angular momentum picks this. Angular momentum is Iω, proportional to ω, not to ω².
4. D6.0 kg·m²/s
   A student who takes angular momentum to be the linear momentum of the rim picks this: Mv = (2.0 kg)(0.50 m × 6.0 rad/s) = 6.0. Angular momentum about the axle is Iω, which accounts for how each part's mass is distributed about the axle.

Working I = (1/2)MR² = (1/2)(2.0)(0.50)² = 0.25 kg·m². L = Iω = (0.25)(6.0) = 1.5 kg·m²/s.

CED 6.3.A.1 · Read this in Fix

1. A0.60 kg·m²/s Correct
   L = rmv sin θ, where r sin θ is the perpendicular distance from O to the puck's line of motion. The puck moves horizontally, so that distance is the vertical leg, 0.30 m: L = (0.30 m)(0.50 kg)(4.0 m/s) = 0.60 kg·m²/s.
2. B1.00 kg·m²/s
   A student who multiplies mv by the full distance from O, 0.50 m, picks this: (0.50 m)(0.50 kg)(4.0 m/s) = 1.00. The velocity is not perpendicular to the line from O, so only r sin θ = 0.30 m counts.
3. C0.80 kg·m²/s
   A student who uses the horizontal leg, 0.40 m, the distance measured along the direction of motion, picks this: (0.40)(0.50)(4.0) = 0.80. That is r cos θ; the perpendicular distance from O to the line of motion is the vertical leg, 0.30 m.
4. D0.00 kg·m²/s
   A student who thinks an object moving in a straight line has no angular momentum picks this. The puck's line of motion passes 0.30 m from O, so it has angular momentum about O.

Working Perpendicular distance from O to the line of motion = 0.30 m (sin θ = 0.30/0.50 = 0.60). L = rmv sin θ = (0.50)(0.50)(4.0)(0.60) = 0.60 kg·m²/s.

CED 6.3.A.2 · Read this in Fix

1. A8.1 N·m·s
   A student who treats the torque as constant at its greatest value, (2.4)(1.5)² = 5.4 N·m at t = 1.5 s, picks this: (5.4 N·m)(1.5 s) = 8.1 N·m·s. The torque is less than that throughout the interval, so the area under the τ–t graph is smaller.
2. B7.2 N·m·s
   A student who uses the slope of the torque–time graph at t = 1.5 s, dτ/dt = 2(2.4)(1.5) = 7.2, picks this. Angular impulse is the area under the τ–t graph, ∫τ dt, not its slope.
3. C2.7 N·m·s Correct
   Angular impulse = ∫τ dt = ∫₀^1.5 (2.4)t² dt = (2.4)(1.5)³/3 = 2.7 N·m·s. The torque is small for most of the interval and reaches its greatest value, 5.4 N·m, only at the end.
4. D5.4 N·m·s
   A student who reads the torque at the end of the interval, 5.4 N·m, as the angular impulse picks this. The height of a τ–t graph is the torque; the angular impulse is the area under it.

Working ∫₀^1.5 2.4t² dt = [0.80t³]₀^1.5 = (0.80)(3.375) = 2.7 N·m·s.

CED 6.3.B.1 · Read this in Fix

1. ACounterclockwise, the same sense as the wheel's rotation during the interval
   A student who thinks an impulse acts in the direction of the motion picks this. The angular impulse has the sense of the torque, which is clockwise here; it opposes the rotation and slows the wheel.
2. BIt has no sense, as an angular impulse has a size but no direction
   A student who treats angular impulse as a plain amount picks this. Angular impulse has the direction of the torque that delivers it; here it is clockwise.
3. CTo the right, along the force, like the linear impulse of the force
   A student who carries over the direction rule for linear impulse picks this. The force's linear impulse is to the right, but its angular impulse about the axle has the sense of its torque: clockwise.
4. DClockwise, the same sense as the torque the force exerts about the axle Correct
   At the top of the rim, a force to the right turns the wheel clockwise about the axle, so its torque is clockwise. Angular impulse has the direction of the torque that delivers it, so it is clockwise, opposite to the wheel's rotation, and it reduces the wheel's angular momentum.

CED 6.3.B.2 · Read this in Fix

1. A1.60 N·m·s
   A student who adds the areas above and below the time axis as positive picks this: 1.2 + 0.40 = 1.6 N·m·s. From 0.40 s to 0.60 s the torque is negative, so its angular impulse is negative.
2. B3.60 N·m·s
   A student who treats the torque as constant at its greatest value picks this: (6.0 N·m)(0.60 s) = 3.6 N·m·s. The torque varies and is even negative for part of the interval, so the area is much smaller.
3. C0.80 N·m·s Correct
   The angular impulse is the area under the τ–t graph, counted negative below the axis. The triangle from 0 to 0.40 s has area (1/2)(0.40 s)(6.0 N·m) = 1.2 N·m·s; the rectangle from 0.40 s to 0.60 s has area (0.20 s)(−2.0 N·m) = −0.40 N·m·s. The net angular impulse is 0.80 N·m·s.
4. D6.00 N·m·s
   A student who reads the greatest height of the graph, 6.0 N·m, as the angular impulse picks this. The height of a τ–t graph is the torque; the angular impulse is the area under it.

Working Area 0–0.40 s: (1/2)(0.40)(6.0) = 1.2 N·m·s. Area 0.40–0.60 s: (0.20)(−2.0) = −0.40 N·m·s. Net: 0.80 N·m·s.

CED 6.3.B.3 · Read this in Fix

1. A1.0 kg·m²/s
   A student who uses the final angular momentum in place of the change picks this: (0.50 kg·m²)(2.0 rad/s) = 1.0 kg·m²/s. The flywheel did not start from rest, so the change is the final value minus the initial value, 1.0 − 3.0 = −2.0 kg·m²/s.
2. B2.0 kg·m²/s Correct
   The initial and final angular momenta, (0.50 kg·m²)(6.0 rad/s) = 3.0 kg·m²/s and (0.50 kg·m²)(2.0 rad/s) = 1.0 kg·m²/s, are in the same sense, so the magnitude of the change is the difference of their magnitudes: |ΔL| = 3.0 − 1.0 = 2.0 kg·m²/s.
3. C8.0 kg·m²/s
   A student who finds the change in rotational kinetic energy in place of the change in angular momentum picks this: (1/2)(0.50)(6.0² − 2.0²) = 8.0. That is an energy in joules; angular momentum is Iω, so its change involves 6.0 − 2.0 rad/s, not 6.0² − 2.0².
4. D4.0 kg·m²/s
   A student who thinks angular momentum depends only on angular speed picks this: 6.0 − 2.0 = 4.0. L = Iω, so the change in angular momentum is the rotational inertia, 0.50 kg·m², times the change in angular speed.

Working L₀ = (0.50)(6.0) = 3.0 kg·m²/s; L = (0.50)(2.0) = 1.0 kg·m²/s, same sense; |ΔL| = 3.0 − 1.0 = 2.0 kg·m²/s.

CED 6.3.C.1 · Read this in Fix

1. AWheel A's is greater, as the torque on wheel A is greater
   A student who judges the effect by the size of the torque alone picks this. The torque on B is smaller but acts four times as long, so the angular impulses, and the changes in angular momentum, are equal.
2. BWheel B's is greater, as the torque on wheel B acts for longer
   A student who judges the effect by the duration alone picks this. The torque on A acts for a quarter of the time but is four times as large, so the angular impulses are equal.
3. CEqual, as the two wheels receive equal angular impulses Correct
   Each wheel receives an angular impulse τΔt = 4.0 N·m·s (8.0 × 0.50 and 2.0 × 2.0). The angular impulse equals the change in angular momentum, and both wheels start from rest, so they end with equal angular momenta.
4. DEqual only if the two wheels turn through equal angles
   A student who links the change in angular momentum to the angle turned, as for work, picks this. The angular impulse is ∫τ dt over time; it does not depend on the angle through which the wheel turns.

Working A: (8.0 N·m)(0.50 s) = 4.0 N·m·s. B: (2.0 N·m)(2.0 s) = 4.0 N·m·s. ΔL equal; both start from rest, so L_A = L_B.

CED 6.3.C.2 · Read this in Fix

1. AIt is zero, as the angular momentum does not change in that interval. Correct
   The net torque equals the slope of the L–t graph. From 1.0 s to 3.0 s the graph is horizontal, so the slope, and the net torque, is zero, even though the wheel is rotating with L = 4.0 kg·m²/s.
2. BIt is constant and nonzero, as the wheel continues to rotate at that time.
   A student who thinks a wheel needs a continuing torque to keep rotating picks this. With zero net torque, the wheel's angular momentum stays the same and it keeps turning.
3. CIt is at its greatest, as the angular momentum is at its greatest.
   A student who reads the height of the graph as the torque picks this. The net torque is the slope of the L–t graph, which is zero where the graph is flat at its greatest value.
4. DIt is equal to the area under the graph between t = 1.0 s and t = 3.0 s.
   A student who takes the net torque from the area under the L–t graph picks this. The net torque is the slope of the L–t graph; area under a τ–t graph, not an L–t graph, gives an angular impulse.

CED 6.3.C.3 · Read this in Fix

1. AWheel A's is greater, as the torque on it reaches a larger value
   A student who judges by the greatest torque picks this. Wheel A's torque peaks at 8.0 N·m but acts for a shorter time; the areas, and so the changes in angular momentum, are equal.
2. BWheel B's is greater, as its torque acts for a longer time
   A student who judges by the duration alone picks this. Wheel B's torque lasts longer but is smaller; both areas are 2.4 N·m·s.
3. CBoth are zero, as neither torque is still being exerted
   A student who thinks a wheel needs a continuing torque to keep turning picks this. Once the torques end, each wheel keeps the angular momentum it has gained and continues to rotate.
4. DEqual, as the areas under the two torque–time graphs are equal Correct
   The change in angular momentum equals the area under the net external torque–time graph. A: (1/2)(0.60 s)(8.0 N·m) = 2.4 N·m·s. B: (0.80 s)(3.0 N·m) = 2.4 N·m·s. Identical wheels starting from rest with equal changes in angular momentum end with equal angular speeds.

Working Area A = (1/2)(0.60)(8.0) = 2.4 N·m·s; area B = (0.80)(3.0) = 2.4 N·m·s. ΔL_A = ΔL_B, same I, both from rest → ω_A = ω_B.

CED 6.3.C.4 · Read this in Fix

1. A0.50
   A student who takes rotational inertia to depend only on mass picks this: with I unchanged, only the halving of ω counts. The new disk's mass is farther from the axis, and I ∝ R² for a given mass and shape.
2. B1.50
   A student who takes rotational inertia to be proportional to the radius picks this: 3 × 1/2 = 1.50. Rotational inertia depends on the square of the distance, so tripling R multiplies I by 9.
3. C4.50 Correct
   For a uniform disk, I = (1/2)MR², so tripling R with M unchanged multiplies I by 9. Halving ω multiplies Iω by 1/2. The angular momentum changes by 9 × 1/2 = 4.50.
4. D2.25
   A student who treats angular momentum like kinetic energy, proportional to Iω², picks this: 9 × (1/2)² = 2.25. Angular momentum is Iω, proportional to the first power of ω.

Working L ∝ MR²ω. L_new/L_old = 3² × (1/2) = 4.5.

CED 6.3.A.1 · Read this in Fix

1. AZero at all three, as the puck moves in a straight line and not around O
   A student who thinks only objects moving around a point have angular momentum about it picks this. The puck's path does not pass through O, so it has angular momentum rmv sin θ about O.
2. BGreatest at 1 and least at 2, as the puck is farthest from O at position 1
   A student who multiplies mv by the full distance from O picks this. As the puck moves farther away, the angle between r⃗ and v⃗ shrinks, and r sin θ stays the same.
3. CGreatest at 2 and least at 1, as v is perpendicular to the line from O at 2
   A student who looks only at sin θ, which is largest (equal to 1) at position 2, picks this. Where sin θ is smaller, r is larger, and the product r sin θ, the perpendicular distance to the path, is the same at all three positions.
4. DEqual at all three, as the perpendicular distance from O to the path stays the same Correct
   L = rmv sin θ, and r sin θ is the perpendicular distance from O to the line of motion, which is the same at every position on a straight path. With m and v constant, L is the same, and not zero, at all three positions.

Working L = m v (r sin θ) = m v d, where d is the perpendicular distance from O to the path, the same at 1, 2 and 3; so L₁ = L₂ = L₃ ≠ 0.

CED 6.3.A.2.ii · Read this in Fix

1. ANonzero and the same about X, Y and Z, as angular momentum belongs to the puck itself.
   A student who thinks angular momentum belongs to the object alone, like its mass, picks this. Angular momentum is always about a chosen point; here it is zero about Y and has opposite senses about X and Z.
2. BZero about Y, and opposite in sense about X and Z, as the puck passes them on opposite sides. Correct
   About Y, which is on the puck's line of motion, r⃗ and v⃗ are parallel, so L = rmv sin θ = 0. The puck passes above X moving right, which is clockwise around X, and below Z moving right, which is counterclockwise around Z. The angular momentum depends on the point chosen.
3. CZero about Y, and the same sense about X and Z, as the puck moves in one direction.
   A student who judges the sense of angular momentum by the direction of the velocity picks this. The puck goes clockwise around X, below its path, and counterclockwise around Z, above its path. Its angular momentum about Y, on its line of motion, is zero, as this option says, but the senses about X and Z are opposite.
4. DZero about X, Y and Z, as the puck moves in a straight line, not around a point.
   A student who thinks a straight-moving object has no angular momentum picks this. About X and Z, which are off the path, the puck has angular momentum rmv sin θ ≠ 0; only about Y, on the path, is it zero.

CED 6.3.A.2.i · Read this in Fix

1. AMℓ²ω/12
   A student who uses the rotational inertia about the center for rotation about the end picks this. The rotational inertia, and so the angular momentum, depends on the axis: about the end, I = (1/3)Mℓ².
2. BMℓ²ω
   A student who puts all of the rod's mass at its far end picks this: I = Mℓ². The mass is spread along the rod, so I about the end is (1/3)Mℓ².
3. CMℓω/2
   A student who takes angular momentum to be the linear momentum of the rod's center of mass picks this: Mv_cm = M(ωℓ/2). Angular momentum about the axis is Iω, with units kg·m²/s.
4. DMℓ²ω/3 Correct
   About the end, the parallel axis theorem with d = ℓ/2 gives I = (1/12)Mℓ² + M(ℓ/2)² = (1/3)Mℓ², so L = Iω = Mℓ²ω/3. The angular momentum depends on the axis about which it is found.

Working I_end = (1/12)Mℓ² + M(ℓ/2)² = (1/3)Mℓ². L = I_end ω = Mℓ²ω/3.

CED 6.3.A.2.i · Read this in Fix

1. A3.0 Correct
   For a straight path, L = mv(r sin θ) = mvd, with d the perpendicular distance from O to the path. The ratio is 2 × 3 × 1/2 = 3.0.
2. B6.0
   A student who takes angular momentum to be the linear momentum, mv, picks this: 2 × 3 = 6.0. The angular momentum about O also depends on the perpendicular distance from O to the path, which is halved.
3. C1.5
   A student who squares the distance, as in I = mr², picks this: 2 × 3 × (1/2)² = 1.5. A particle's angular momentum, rmv sin θ, contains the distance to the first power.
4. D9.0
   A student who treats angular momentum like kinetic energy, proportional to v², picks this: 2 × 3² × 1/2 = 9.0. Angular momentum is proportional to the speed, not its square.

Working L = m v d. L₂/L₁ = (2)(3)(1/2) = 3.0.

CED 6.3.A.2.ii · Read this in Fix

1. A1.5 Correct
   Angular impulse = τΔt, with τ = rF sin θ. Trial 1: dFΔt. Trial 2: (d/2)(2F)(sin 30°)(3Δt) = 1.5 dFΔt. The ratio is 1.5.
2. B3.0
   A student who leaves out sin θ and uses the full distance d/2 as the lever arm picks this: (1/2)(2)(3) = 3.0. Only the force component perpendicular to the door, 2F sin 30° = F, produces torque.
3. C2.6
   A student who uses cos 30° in place of sin 30° picks this: (1/2)(2)(0.87)(3) ≈ 2.6. The angle given is between the force and the door, which lies along r⃗, so the torque is rF sin θ.
4. D0.5
   A student who takes the angular impulse to be the torque itself, ignoring the time interval, picks this: (1/2)(2)(sin 30°) = 0.5. Angular impulse is the torque multiplied by the time for which it acts, three times as long in trial 2.

Working Ratio = (1/2)(2)(sin 30°)(3) = 1.5.

CED 6.3.B.1 · Read this in Fix

1. Abt₁²/I
   A student who treats the torque as constant at its final, greatest value, bt₁, picks this: angular impulse (bt₁)(t₁). The torque grows from zero, so the angular impulse is half that, bt₁²/2.
2. Bt₁√(b/I)
   A student who sets the angular impulse equal to the rotational kinetic energy, (1/2)Iω² = bt₁²/2, picks this. Angular impulse equals the change in angular momentum, Iω, not a change in kinetic energy.
3. Cb/I
   A student who uses the slope of the torque–time graph, b, as the change in angular momentum picks this. The change in angular momentum is the area under the τ–t graph, ∫τ dt.
4. Dbt₁²/(2I) Correct
   The angular impulse is ∫₀^t₁ bt dt = bt₁²/2, the area of the triangle under the τ–t graph. It equals the change in angular momentum, Iω − 0, so ω = bt₁²/(2I).

Working ΔL = ∫₀^t₁ bt dt = bt₁²/2 = Iω → ω = bt₁²/(2I).

CED 6.3.C.2.i · Read this in Fix

1. A4.0 s Correct
   The kinetic friction force on the rim is μ_kFN = (0.30)(20 N) = 6.0 N, tangent to the rim, so its torque about the axle is (0.40 m)(6.0 N) = 2.4 N·m. The angular impulse must remove L = Iω = (2.4)(4.0) = 9.6 kg·m²/s, so Δt = 9.6/2.4 = 4.0 s.
2. B8.0 s
   A student who sets the angular impulse equal to the wheel's kinetic energy picks this: (1/2)(2.4)(4.0)² = 19.2 J, and 19.2/2.4 = 8.0 s. Angular impulse equals the change in angular momentum, Iω = 9.6 kg·m²/s.
3. C1.2 s
   A student who takes the 20 N normal force as the force producing the torque picks this: τ = (0.40)(20) = 8.0 N·m and 9.6/8.0 = 1.2 s. The normal force points toward the axle, so its line of action passes through the axle and its torque is zero; only friction, tangent to the rim, has a torque.
4. D1.6 s
   A student who leaves the lever arm out and sets the friction force times time equal to the angular momentum picks this: 9.6/6.0 = 1.6 s. The torque of the friction force is the force times its lever arm, (0.40 m)(6.0 N) = 2.4 N·m.

Working f = μ_kFN = (0.30)(20 N) = 6.0 N. τ = rf = (0.40 m)(6.0 N) = 2.4 N·m. L₀ = Iω = (2.4)(4.0) = 9.6 kg·m²/s. τΔt = ΔL → Δt = 9.6/2.4 = 4.0 s.

CED 6.3.C.2.i · Read this in Fix

1. A75 N·m
   A student who takes the net torque to be proportional to the angular speed picks this: Iω = (1.5)(50) = 75. The net torque equals the rate of change of the angular momentum, I dω/dt, not Iω.
2. B30 N·m Correct
   With constant rotational inertia, τ_net = dL/dt = I dω/dt. Here dω/dt = (4.0 rad/s³)t = 20 rad/s² at t = 5.0 s, so τ_net = (1.5 kg·m²)(20 rad/s²) = 30 N·m.
3. C15 N·m
   A student who divides the angular momentum by the time picks this: (1.5)(50)/5.0 = 15. That is the average rate of change from t = 0; at t = 5.0 s the instantaneous rate, I dω/dt, is twice as large.
4. D20 N·m
   A student who leaves out the rotational inertia picks this: dω/dt = 20 rad/s². The net torque is I dω/dt; the rotational inertia multiplies the angular acceleration.

Working ω = 2.0t²; dω/dt = 4.0t = 20 rad/s² at 5.0 s. τ_net = I dω/dt = (1.5)(20) = 30 N·m.

CED 6.3.C.2.ii · Read this in Fix

1. AThe net torque is proportional to the angular momentum that the wheel has.
   A student who links torque to the rotation itself rather than to its change picks this. A wheel turning steadily on a frictionless axle has angular momentum but zero net torque.
2. BThe net torque is zero at any instant when the wheel's angular momentum is zero.
   A student who thinks zero angular momentum means zero torque picks this. A wheel reversing its rotation has L = 0 for an instant while a net torque is still changing its angular momentum.
3. CA net torque is needed to keep any nonzero angular momentum going.
   A student who thinks rotation needs a continuing torque picks this. With zero net torque, the angular momentum does not change; a net torque is needed only to change it.
4. DThe net torque sets how fast the angular momentum changes, not its value. Correct
   Newton's second law in rotational form is τ_net = dL/dt: the net torque is the rate of change of the angular momentum, which for constant rotational inertia is Iα. The angular momentum itself can have any value when the net torque is zero.

CED 6.3.C.2.ii · Read this in Fix

1. A4.5 N·m
   A student who reads the height of the graph at t = 4.0 s, L = 4.5 kg·m²/s, as the torque picks this. The height of an L–t graph is the angular momentum; the net torque is its slope.
2. B1.1 N·m
   A student who divides the angular momentum at t = 4.0 s by the time, 4.5/4.0 ≈ 1.1, picks this. That is the slope of a line from the origin, an average from t = 0; the net torque at 4.0 s is the slope of the graph at that instant.
3. C1.5 N·m Correct
   The net torque is the slope of the L–t graph. From 3.0 s to 5.0 s, L falls from 6.0 to 3.0 kg·m²/s along a straight line, so at t = 4.0 s the slope is (3.0 − 6.0)/(2.0 s) = −1.5 N·m, a magnitude of 1.5 N·m.
4. D9.0 N·m
   A student who takes the area under the L–t graph from 3.0 s to 5.0 s, (1/2)(6.0 + 3.0)(2.0) = 9.0, as the torque picks this. The net torque is the slope of the L–t graph, not the area under it.

Working Slope from 3.0 s to 5.0 s: (3.0 − 6.0)/(5.0 − 3.0) = −1.5 N·m; |τ_net| = 1.5 N·m at t = 4.0 s.

CED 6.3.C.3 · Read this in Fix

1. AL₀/e
   A student who takes the value of the angular momentum at t = T, L₀e⁻¹, as the torque picks this. The net torque is the rate of change of L, which has units of N·m, not kg·m²/s.
2. BL₀/(eT) Correct
   The net torque is the slope of the L–t graph: dL/dt = −(L₀/T)e^−t/T. At t = T, its magnitude is (L₀/T)e⁻¹ = L₀/(eT).
3. C(1−1/e)L₀/T
   A student who uses the average rate of change from t = 0 to t = T, (L₀ − L₀/e)/T, picks this. The net torque at t = T is the instantaneous rate, the derivative of L at that instant.
4. DL₀T(1−1/e)
   A student who integrates L over time from 0 to T, the area under the L–t graph, picks this. The net torque is the slope of the L–t graph, found by differentiating, not by integrating.

Working τ_net = dL/dt = −(L₀/T)e^−t/T; at t = T, |τ_net| = L₀/(eT). Distractors: L(T) = L₀/e; [L(0) − L(T)]/T = (1 − 1/e)L₀/T; ∫₀^T L dt = L₀T(1 − 1/e).

CED 6.3.C.3 · Read this in Fix

1. Aω₀−τ₀T/(2I) Correct
   Taking the original sense of rotation as positive, the torque is −τ₀(1 − t/T). The angular impulse is the area under the τ–t graph, a triangle below the axis: ∫₀^T −τ₀(1 − t/T) dt = −τ₀T/2. It equals the change in angular momentum, I(ω − ω₀), so ω = ω₀ − τ₀T/(2I).
2. Bω₀−τ₀T/I
   A student who treats the torque as constant at its greatest magnitude, τ₀, for the whole interval picks this: angular impulse −τ₀T. The torque falls linearly to zero, so the angular impulse is half of that, −τ₀T/2.
3. Cω₀+τ₀/(IT)
   A student who takes the slope of the τ–t graph as the change in angular momentum picks this: the torque is negative and rises linearly to zero, so the slope is +τ₀/T, giving I(ω − ω₀) = τ₀/T. The change in angular momentum is the area under the graph, −τ₀T/2, not its slope.
4. Dω₀−τ₀/I
   A student who takes the angular impulse to be the initial torque itself, −τ₀, picks this. Angular impulse is torque integrated over time, ∫τ dt, with units N·m·s.

Working Take the original sense of rotation as positive: τ = −τ₀(1 − t/T). ΔL = ∫₀^T −τ₀(1 − t/T) dt = −(τ₀T − τ₀T/2) = −τ₀T/2 = I(ω − ω₀) → ω = ω₀ − τ₀T/(2I), positive because the wheel still turns in its original sense.

CED 6.3.C.4 · Read this in Fix

1. AThe magnitude of r⃗ × p⃗ is rp, whatever the angle, and Iω reduces to the same product.
   A student who multiplies r by p whatever the angle between them picks this. |r⃗ × p⃗| = rp sin θ; it equals rp here only because the velocity is tangent, perpendicular to r⃗.
2. BFor a point mass, I is mr² and ω is v/r, and v⃗ is tangent, perpendicular to r⃗. Correct
   For a small ball, I = mr² about the center and ω = v/r, so Iω = mvr. The velocity in circular motion is tangent to the circle, perpendicular to the radius, so sin θ = 1 and the cross product has magnitude rmv = mvr. The two methods agree.
3. CAngular momentum is the linear momentum mv, changed into rotational units by multiplying it by r.
   A student who takes angular momentum to be linear momentum in other units picks this. The factor is r sin θ, the perpendicular distance from the center to the line of motion; it equals r here only because v⃗ is perpendicular to r⃗.
4. DThe tension along the string exerts a torque about the center that keeps L equal to mvr.
   A student who thinks a force toward the center exerts a torque about it picks this. The tension's line of action passes through the center, so its torque about the center is zero; that is why L stays constant, but it does not show that L = mvr.

Working I = mr², ω = v/r → Iω = mvr. |r⃗ × p⃗| = rmv sin 90° = mvr.

CED 6.3.A.2 · Read this in Fix

1. AThe final angular momentum is set by the torque and time alone, so the change in it is too.
   A student who confuses the final angular momentum with the change picks this. The final angular momentum is L₀ + τΔt, which does depend on the initial value; only the change, τΔt, does not.
2. BThe torque does the same work on the wheel whatever angular speed the wheel starts with.
   A student who links the change in angular momentum to the work done picks this. Work is τΔθ, and a faster wheel turns through a larger angle in 3.0 s, so the work does depend on the initial angular speed; it is the angular impulse, τΔt, that does not.
3. CThe angular impulse τΔt equals ΔL, and τΔt does not depend on how fast the wheel is turning. Correct
   By the angular impulse–angular momentum theorem, ΔL = ∫τ dt = τΔt = (2.0 N·m)(3.0 s) = 6.0 kg·m²/s. Neither the torque nor the time interval depends on the initial angular speed, so neither does ΔL.
4. DThe wheel's kinetic energy changes by the same amount whatever its initial angular speed.
   A student who treats kinetic energy and angular momentum as the same kind of quantity picks this. The change in kinetic energy, (1/2)I(ω² − ω₀²), does depend on ω₀; the change in angular momentum, IΔω = τΔt, does not.

Working ΔL = τΔt = (2.0 N·m)(3.0 s) = 6.0 kg·m²/s, independent of ω₀.

CED 6.3.C.2.i · Read this in Fix

1. Amgv₀t²/2 Correct
   Gravity's torque about P is mg times the horizontal distance v₀t, so it grows linearly from zero. The angular impulse is ∫₀ᵗ mgv₀t′ dt′ = mgv₀t²/2, and since the angular momentum about P is zero at the throw, this is L. The same result comes from |r⃗ × p⃗| = |xp_y − yp_x|.
2. Bmgv₀t²
   A student who multiplies gravity's torque about P at time t, mgv₀t, by the whole time t picks this. That torque grows from zero at the throw, so the angular impulse is the area under a rising line, half of mgv₀t².
3. C0
   A student who thinks only spinning objects, or objects moving in a circle, have angular momentum picks this, since the ball neither spins nor circles P. Angular momentum about a point is r⃗ × p⃗, which is nonzero here because the ball's momentum does not point along the line from P.
4. Dm√(v₀²+g²t²)
   A student who takes the ball's angular momentum to be its linear momentum, m times its speed √(v₀²+g²t²), picks this. Angular momentum about P is r⃗ × p⃗ and depends on where the ball is relative to P; this expression does not even have the units of angular momentum.

Working Take P as the origin, x horizontal in the direction of the throw and y upward. At time t: r⃗ = (v₀t, −gt²/2) and p⃗ = (mv₀, −mgt). L_z = xp_y − yp_x = (v₀t)(−mgt) − (−gt²/2)(mv₀) = −mgv₀t² + mgv₀t²/2 = −mgv₀t²/2, so |L⃗| = mgv₀t²/2. Check by angular impulse: the ball starts at P, so L₀ = 0; gravity's torque about P has magnitude mg(v₀t), its lever arm being the horizontal distance v₀t; L = ∫₀ᵗ mgv₀t′ dt′ = mgv₀t²/2. Distractors: (mgv₀t)·t = mgv₀t² (torque at time t times the whole time); 0 (no angular momentum without spin or circular motion); |p⃗| = m√(v₀²+g²t²) (linear momentum taken as angular momentum).

CED 6.3.A.2 · Read this in Fix

1. A0.43Mℓ√(gℓ)
   A student who treats the rod as a particle of mass M at its center of mass picks this: with ω = √(3g/ℓ), L = M(ℓ/2)(ℓ/2)ω = (√3/4)Mℓ√(gℓ) ≈ 0.43Mℓ√(gℓ). The rod's rotational inertia about the axle is (1/3)Mℓ², not M(ℓ/2)².
2. B0.67Mℓ√(gℓ)
   A student who writes the rod's kinetic energy as (1/2)Mv_cm² picks this: v_cm = √(gℓ), so ω = v_cm/(ℓ/2) = 2√(g/ℓ) and L = (2/3)Mℓ√(gℓ). The rod also turns about its center of mass; its kinetic energy is (1/2)Iω² with I about the axle.
3. C0.58Mℓ√(gℓ) Correct
   The rod's rotational inertia about its end is (1/3)Mℓ², and its center of mass falls ℓ/2, so Mgℓ/2 = (1/2)(1/3)Mℓ²ω² and ω = √(3g/ℓ). The angular momentum about the axle is L = Iω = (1/3)Mℓ²√(3g/ℓ) = Mℓ√(gℓ/3) ≈ 0.58Mℓ√(gℓ).
4. D1.00Mℓ√(gℓ)
   A student who takes the rod's rotational inertia about the axle to be Mℓ², as if all its mass were at the far end, picks this: Mgℓ/2 = (1/2)Mℓ²ω², ω = √(g/ℓ) and L = Mℓ√(gℓ). For a thin uniform rod about one end, I = ∫x² dm = (1/3)Mℓ².

Working Rotational inertia about the end: I = ∫₀^ℓ x²(M/ℓ) dx = (1/3)Mℓ². Energy (rod–Earth system, no friction): the center of mass falls ℓ/2, so Mg(ℓ/2) = (1/2)(1/3)Mℓ²ω², giving ω² = 3g/ℓ and ω = √(3g/ℓ). Then L = Iω = (1/3)Mℓ²√(3g/ℓ) = Mℓ√(gℓ/3) ≈ 0.58Mℓ√(gℓ). Distractors: rod treated as a particle of mass M at its center of mass, L = M(ℓ/2)(ℓ/2)ω with ω = √(3g/ℓ): L = (√3/4)Mℓ√(gℓ) ≈ 0.43Mℓ√(gℓ); K = (1/2)Mv_cm²: v_cm = √(gℓ), ω = v_cm/(ℓ/2) = 2√(g/ℓ), L = (2/3)Mℓ√(gℓ) ≈ 0.67Mℓ√(gℓ); I = Mℓ² (all mass at the far end): Mgℓ/2 = (1/2)Mℓ²ω², ω = √(g/ℓ), L = Mℓ√(gℓ) = 1.00Mℓ√(gℓ).

CED 6.3.A.1 · Read this in Fix

1. A2.0ω₁R/g
   A student who takes the string's tension to be the block's weight, mg, picks this: mgR t = 2mR²ω₁. While the block accelerates downward the tension is less than mg, so the torque on the pulley is smaller and the time is longer.
2. B3.0ω₁R/g Correct
   The disk's rotational inertia is (1/2)(4m)R² = 2mR². Newton's second law for the block, mg − F_T = ma, and for the pulley, F_T R = Iα with a = αR, give a = g/3 and F_T = (2/3)mg. The angular impulse of the string's torque equals the pulley's change in angular momentum: F_T R t = 2mR²ω₁, so t = 3.0ω₁R/g.
3. C1.0ω₁R/g
   A student who treats the block as falling freely, with acceleration g, picks this: the string's speed ω₁R is reached when gt = ω₁R. The string must exert a torque to turn the pulley, so it pulls up on the block and the block's acceleration is less than g.
4. D5.0ω₁R/g
   A student who takes the disk's rotational inertia to be (4m)R², as if all its mass were at the rim, picks this: a = g/5 and F_T = (4/5)mg. A uniform solid disk has I = (1/2)MR² about its central axis, so I = 2mR² here.

Working Pulley: I = (1/2)(4m)R² = 2mR². Block: mg − F_T = ma. Pulley: F_T R = Iα, with a = αR. So F_T = Ia/R² = 2ma, mg = 3ma, a = g/3 and F_T = (2/3)mg. Rotational impulse–momentum theorem for the pulley (the axle force and the pulley's weight act through the axle): F_T R t = Iω₁ − 0, so t = 2mR²ω₁/((2/3)mgR) = 3.0ω₁R/g. Check with the pulley–block system about the axle: the only external torque is the block's weight, mgR; the system's angular momentum is Iω₁ + m(ω₁R)R = 3mR²ω₁, so mgR t = 3mR²ω₁, t = 3ω₁R/g. Distractors: F_T = mg: mgR t = 2mR²ω₁, t = 2.0ω₁R/g; block falls with acceleration g: gt = ω₁R, t = 1.0ω₁R/g; I = (4m)R²: mg = 5ma, F_T = 4mg/5, t = 4mR²ω₁/((4/5)mgR) = 5.0ω₁R/g.

CED 6.3.C.2 · Read this in Fix