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Unit 5 / Torque and Rotational Dynamics

Unit 5: worked solutions

Compare the model and reasoning at each step. A matching final number is useful; a reason for every step is what lets you solve the next problem.

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Question 1 · AP topic 5.1

1. Rotational Kinematics · 4

  1. Use ω_f = ω_i + αt and set ω_f = 0.

    Why this step: The question supplies constant angular acceleration until the stop.

  2. 0 = 12 − 3.0t gives t = 4.0 s.

    Why this step: The negative acceleration reduces the initial positive angular velocity.

  3. Δθ = ½(12 + 0)(4.0) = +24 rad.

    Why this step: The positive triangular area under ω versus t is angular displacement before the stop.

Use ω_f = ω_i + αt and set ω_f = 0. 0 = 12 − 3.0t gives t = 4.0 s. Δθ = ½(12 + 0)(4.0) = +24 rad.

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Question 2 · AP topic 5.2

2. Connecting Linear and Rotational Motion · 4

  1. Use the no-slip string relation s = rΔθ.

    Why this step: String travel matches axle surface arc only when contact does not slip.

  2. Δθ = 0.60/0.040 = 15 rad.

    Why this step: The relevant radius is the axle where the string contacts the system.

  3. Use a_t = rα to obtain |α| = 0.20/0.040 = 5.0 rad/s².

    Why this step: Tangential acceleration at that radius connects the string and rotational motions.

Use the no-slip string relation s = rΔθ. Δθ = 0.60/0.040 = 15 rad. Use a_t = rα to obtain |α| = 0.20/0.040 = 5.0 rad/s².

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Question 3 · AP topic 5.3

3. Torque · 4

  1. Choose counterclockwise as positive and keep both force positions in the diagram.

    Why this step: Equal force directions can cause opposite turning senses on opposite sides.

  2. The left contribution is +6.0(0.30) = +1.8 N·m.

    Why this step: A downward force left of the pivot tends counterclockwise.

  3. The right contribution is −4.0(0.20) = −0.80 N·m.

    Why this step: A downward force right of the pivot tends clockwise.

  4. Add to get τ_net = +1.0 N·m, tending counterclockwise.

    Why this step: Net torque is the signed sum about the stated pivot.

Choose counterclockwise as positive and keep both force positions in the diagram. The left contribution is +6.0(0.30) = +1.8 N·m. The right contribution is −4.0(0.20) = −0.80 N·m. Add to get τ_net = +1.0 N·m, tending counterclockwise.

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Question 4 · AP topic 5.4

4. Rotational Inertia · 4

  1. At the center each mass has r = 0.40 m.

    Why this step: The center axis bisects the massless 0.80 m rod.

  2. I_cm = 2(0.50)(0.40²) = 0.160 kg·m².

    Why this step: Add the two point-object contributions about the same axis.

  3. Use M = 1.0 kg and d = 0.40 m: I_end = I_cm + Md² = 0.320 kg·m².

    Why this step: The end axis is parallel to the center-of-mass axis.

  4. Check directly: 0.50(0²) + 0.50(0.80²) = 0.320 kg·m².

    Why this step: Agreement between the theorem and the point sum verifies the axis distances.

At the center each mass has r = 0.40 m. I_cm = 2(0.50)(0.40²) = 0.160 kg·m². Use M = 1.0 kg and d = 0.40 m: I_end = I_cm + Md² = 0.320 kg·m². Check directly: 0.50(0²) + 0.50(0.80²) = 0.320 kg·m².

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Question 5 · AP topic 5.5

5. Rotational Equilibrium and Newton’s First Law in Rotational Form · 4

  1. Locate the board’s center 1.0 m from the left, or 0.20 m right of the pivot.

    Why this step: Uniform-board weight acts at the center of mass, which is not at the pivot here.

  2. Balance torques: 100(0.80) = 60(0.20) + W(1.20).

    Why this step: The left load turns counterclockwise; board weight and the right load turn clockwise.

  3. Solve W = (80 − 12)/1.20 = 56.7 N.

    Why this step: The unknown must supply the remaining clockwise torque.

  4. Force balance gives F_pivot = 100 + 60 + 56.7 = 216.7 N upward.

    Why this step: Static equilibrium also needs zero net vertical force.

Locate the board’s center 1.0 m from the left, or 0.20 m right of the pivot. Balance torques: 100(0.80) = 60(0.20) + W(1.20). Solve W = (80 − 12)/1.20 = 56.7 N. Force balance gives F_pivot = 100 + 60 + 56.7 = 216.7 N upward.

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Question 6 · AP topic 5.6

6. Newton’s Second Law in Rotational Form · 4

  1. Use α = τ_net/I with each fixed configuration.

    Why this step: Both trials have the same torque but different inertias.

  2. Initially α = 0.40/0.050 = 8.0 rad/s², so Δω = 8.0(2.0) = 16 rad/s.

    Why this step: Constant torque and inertia give constant angular acceleration.

  3. Outward masses give α = 0.40/0.080 = 5.0 rad/s² and Δω = 10 rad/s.

    Why this step: Greater inertia reduces the response without changing the net torque.

Use α = τ_net/I with each fixed configuration. Initially α = 0.40/0.050 = 8.0 rad/s², so Δω = 8.0(2.0) = 16 rad/s. Outward masses give α = 0.40/0.080 = 5.0 rad/s² and Δω = 10 rad/s.

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Question 7 · AP topic 5.1

7. Unwrap the filmed turn

  1. Unwrap the positions as 350°, 390°, 450°, and 530°. Relative to the first frame, Δθ values are 0°, 40°, 100°, and 180°, or 0, 2π/9, 5π/9, and π rad.

    Why this step: The stated turning sense and less-than-one-turn interval resolve the wrap at 360°. Subtracting the first position sets a convenient angular origin.

  2. The average angular velocities over the three one-second intervals are 2π/9, π/3, and 4π/9 rad/s. Plot them at the interval midpoints 0.5, 1.5, and 2.5 s; their increase is π/9 rad/s each second.

    Why this step: For constant angular acceleration, midpoint interval averages follow a straight ω–t trend. These data are consistent with α = π/9 rad/s², about 0.349 rad/s².

  3. Call constant α a reasonable model over the measured interval, rather than a proven instantaneous behavior. Obtain more frames or repeat the run to test it.

    Why this step: Four position samples can hide changing acceleration between frames, and angle-reading uncertainty can affect the inferred slopes.

Unwrapped relative angles: 0, 2π/9, 5π/9, π rad. The data fit α = π/9 rad/s², subject to limited sampling and angle uncertainty.

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Question 8 · AP topic 5.2

8. Check two measurements against rigidity

  1. Infer angular speeds from v_t/r: the inner mark gives 4.0 rad/s and the outer gives 3.0 rad/s.

    Why this step: Points fixed to a rigid disk must share angular velocity even though their tangential speeds differ.

  2. The allowed speed intervals give angular-speed intervals 3.7–4.3 rad/s and 2.9–3.1 rad/s. They do not overlap.

    Why this step: The stated speed errors are too small to reconcile the reported measurements with a single angular speed.

  3. For example, recheck that both arc travels were measured over the same timestamps and that the camera views along the axis; time several complete turns independently.

    Why this step: Mismatched time intervals or perspective distortion can produce a systematic speed error. An independent turn-period measurement checks ω directly before discarding the rigid-disk model.

The inferred intervals do not overlap, so the measurements are inconsistent with one rigid-disk angular speed. Check synchronization and camera geometry.

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Question 9 · AP topic 5.3

9. Decide which pull turns harder

  1. Calculate τ_A = 0.40(6.0)sin90° = 2.4 N·m and τ_B = 0.30(12)sin30° = 1.8 N·m.

    Why this step: Torque depends on force and the perpendicular geometry, so force magnitude alone cannot rank the turning effects.

  2. The claim is false: A has the larger torque. To match A, solve 2.4 = r(12)(0.5), giving r = 0.40 m.

    Why this step: Keeping force and angle fixed makes torque proportional to application distance from the pivot.

  3. Measure from the pivot to the application point for r, and verify the angle between that position vector and the force. Alternatively measure the perpendicular lever arm, which must be 0.20 m for the matched B trial.

    Why this step: The lever arm is not generally the pivot-to-point distance; the two geometric methods must describe the same line of action.

A gives 2.4 N·m and B 1.8 N·m. Move B to r = 0.40 m, corresponding to a 0.20 m perpendicular lever arm.

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Question 10 · AP topic 5.4

10. Choose a graph that can test the inertia model

  1. Write I_total = 0.010 + 2(0.20)r² = 0.010 + 0.40r², then α = 0.040/I_total.

    Why this step: The frame and two point masses all resist the same angular acceleration about the stated axis.

  2. Rearrange to 1/α = I_total/0.040 = 0.25 + 10r². Choose 1/α vertically against r² horizontally, with slope 10 s²/m² and intercept 0.25 s², treating radians as dimensionless.

    Why this step: Reciprocating acceleration converts the inverse inertia response into a linear relation. In contrast, α versus r² is curved.

  3. Retain the nonzero intercept. Ignoring the frame predicts a line through zero and overestimates acceleration, especially at small r.

    Why this step: The point masses’ contribution shrinks near the axis, while the frame’s own inertia remains.

Plot 1/α against r²: slope 10 s²/m², intercept 0.25 s². Ignoring the frame removes a real positive intercept and overpredicts α.

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Question 11 · AP topic 5.5

11. Use the off-center balance

  1. At the 0.40 m pivot, the known left load has lever arm 0.30 m, the right load 0.50 m, and the meterstick’s center 0.10 m to the right.

    Why this step: A uniform stick’s weight acts at its 0.50 m center of mass, not automatically at its support.

  2. Balance counterclockwise and clockwise torques: 0.20g(0.30) = 0.10g(0.50) + mg(0.10). Cancel g and solve m = 0.10 kg.

    Why this step: Gravity supplies all three downward forces, so its common factor cancels while the lever arms remain.

  3. With the pivot at 0.50 m, the stick’s own lever arm is zero. The left and right load torques have magnitudes 0.20g(0.40) and 0.10g(0.40); the left wins, so the left end initially moves downward, counterclockwise as viewed with the left end to the left.

    Why this step: Equal new lever arms make the unequal hanging masses decisive, and the stick’s center force acts through the new pivot.

The meterstick mass is 0.10 kg. At the center pivot, the left end initially goes down; the stick’s own weight then exerts zero torque.

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Question 12 · AP topic 5.6

12. Separate the bearing torque from the inertia

  1. Plot applied driving torque vertically against angular acceleration horizontally. The data have slope (0.60 − 0.20)/(5.0 − 1.0) = 0.10 kg·m².

    Why this step: The equation is τ_applied − τ_bearing = Iα, so τ_applied = Iα + τ_bearing. Its slope measures inertia even when the constant resistance is unknown.

  2. Use any point: τ_bearing = 0.20 − 0.10(1.0) = 0.10 N·m.

    Why this step: The vertical intercept is the applied torque needed to balance the bearing torque when α = 0, not an additional inertia.

  3. Subtract 0.10 N·m from each applied value to obtain true net torques 0.10, 0.30, and 0.50 N·m. Their graph is τ_net = 0.10α and passes through the origin.

    Why this step: Simply relabeling applied torque leaves the resistance unaccounted for. An ideal constant-inertia true-net graph has zero intercept.

I = 0.10 kg·m² and opposing bearing torque = 0.10 N·m. Applied-torque intercept is 0.10 N·m; true-net-torque intercept is zero.

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Question 13 · AP topic 5.1

13. Rotational Kinematics · 1

  1. Use the stated counterclockwise-positive convention: negative ω is clockwise.

    Why this step: The sign of angular velocity gives turning sense.

  2. Positive α moves −6 rad/s toward zero, so the wheel is slowing at this instant.

    Why this step: Opposite signs of ω and α reduce angular speed until the wheel reaches rest.

It turns clockwise and slows down.

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Question 14 · AP topic 5.2

14. Connecting Linear and Rotational Motion · 1

  1. Rigid-body points sweep the same angle in the same time.

    Why this step: Rigidity keeps their angular positions locked relative to each other.

  2. They share ω, while v_t = r|ω| and a_r = rω² change with radius.

    Why this step: The outer point has larger linear speed and radial acceleration, not a larger angular velocity.

Angular velocity.

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Question 15 · AP topic 5.3

15. Torque · 1

  1. Extend the force’s line of action toward the pivot.

    Why this step: Lever arm is perpendicular distance from the axis to that line.

  2. The line passes through the pivot, so ℓ = 0 and τ = Fℓ = 0.

    Why this step: A nonzero distance to the application point does not guarantee a turning effect.

Zero.

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Question 16 · AP topic 5.4

16. Rotational Inertia · 1

  1. Write the original contribution as I = mr².

    Why this step: A point-object contribution weights the squared perpendicular distance.

  2. At 2r, I_new = m(2r)² = 4I.

    Why this step: Squaring the radius gives a factor of four while mass is unchanged.

It becomes four times as large.

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Question 17 · AP topic 5.5

17. Rotational Equilibrium and Newton’s First Law in Rotational Form · 1

  1. Rotational equilibrium means Στ = 0.

    Why this step: The torques balance about the stated axis.

  2. With constant I, α = 0 and ω stays constant, including a possible nonzero value.

    Why this step: Equilibrium does not require rest or translational equilibrium.

Its angular velocity is constant.

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Question 18 · AP topic 5.6

18. Newton’s Second Law in Rotational Form · 1

  1. Use α = τ_net/I for the fixed rigid configuration.

    Why this step: Inertia and torque refer to the same rotation axis.

  2. Replacing I with 2I gives α_new = α/2.

    Why this step: At fixed net torque the relationship is inverse.

It is halved.

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Question 19 · AP topic 5.1

19. Rotational Kinematics · 2

  1. Multiply the graph units: (rad/s) × s = rad.

    Why this step: This identifies angular displacement rather than angular acceleration.

  2. Add areas with their signs to obtain Δθ.

    Why this step: Clockwise portions below zero offset counterclockwise portions above zero.

Angular displacement.

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Question 20 · AP topic 5.2

20. Connecting Linear and Rotational Motion · 2

  1. Constant ω means α = 0 and therefore a_t = rα = 0.

    Why this step: The speed is not changing.

  2. Velocity still changes direction, requiring a_r = rω² inward.

    Why this step: Constant speed is compatible with nonzero acceleration on a curved path.

Points away from the axis have inward acceleration but zero tangential acceleration.

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Question 21 · AP topic 5.3

21. Torque · 2

  1. Find the perpendicular component: F_perp = 20sin30° = 10 N.

    Why this step: Only the component perpendicular to the wrench produces torque.

  2. τ = rF_perp = 0.40(10) = 4.0 N·m.

    Why this step: The angle is between the force and the wrench position vector.

4.0 N·m

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Question 22 · AP topic 5.4

22. Rotational Inertia · 2

  1. Identify a center-of-mass axis and a parallel displaced axis.

    Why this step: The parallel-axis theorem requires this specific geometry.

  2. Use I = I_cm + Md² with the total mass and axis separation.

    Why this step: An arbitrary known off-center inertia cannot replace I_cm in the formula.

The two axes are parallel and one passes through the center of mass.

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Question 23 · AP topic 5.5

23. Rotational Equilibrium and Newton’s First Law in Rotational Form · 2

  1. The equal opposite forces cancel in ΣF.

    Why this step: Their vector sum controls the center-of-mass acceleration.

  2. Separated lines of action can give turning effects that add.

    Why this step: Their locations determine torques, so zero net force does not force zero net torque.

Net force can be zero while net torque is nonzero.

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Question 24 · AP topic 5.6

24. Newton’s Second Law in Rotational Form · 2

  1. For the mass, write mg − T = ma with downward positive.

    Why this step: The falling object needs a positive net downward force.

  2. Since acceleration is downward, T is less than mg.

    Why this step: Tension both opposes the fall and drives the pulley’s angular acceleration.

T is smaller than mg.

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Question 25 · AP topic 5.1

25. Rotational Kinematics · 3

  1. One complete turn sweeps 360°, or 2π rad.

    Why this step: Radians measure the angle traveled, not just the final orientation.

  2. Counterclockwise is positive, so the unwrapped Δθ is +2π rad.

    Why this step: Matching initial and final orientation does not erase the complete turn.

+2π rad.

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Question 26 · AP topic 5.2

26. Connecting Linear and Rotational Motion · 3

  1. Use s = rΔθ with angle in radians.

    Why this step: Arc length, not straight displacement, is the length related to the angle.

  2. Δθ = 0.60/0.20 = 3.0 rad.

    Why this step: The length units cancel, leaving the radian ratio.

3.0 rad

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Question 27 · AP topic 5.3

27. Torque · 3

  1. For the same force magnitude, τ = Fℓ.

    Why this step: Use the same reference axis for both comparisons.

  2. A greater perpendicular lever arm gives greater torque.

    Why this step: The center of mass is not automatically the selected torque axis.

The one with the greater perpendicular lever arm.

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Question 28 · AP topic 5.4

28. Rotational Inertia · 3

  1. Each mass contribution grows with squared distance from the axis.

    Why this step: Compare distributions about the same center axis.

  2. A hoop puts its mass at the outer radius, so its inertia exceeds the disk’s.

    Why this step: Some disk mass lies closer to the axis and contributes less per kilogram.

The hoop, because more mass is farther from the axis.

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Question 29 · AP topic 5.5

29. Rotational Equilibrium and Newton’s First Law in Rotational Form · 3

  1. Choose the axis through the support force’s application point.

    Why this step: That makes the perpendicular distance to its line of action zero.

  2. The support contributes no torque about that point but remains in ΣF.

    Why this step: Choosing a convenient axis removes an unknown torque term without removing the physical force.

Its force has zero torque about that point.

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Question 30 · AP topic 5.6

30. Newton’s Second Law in Rotational Form · 3

  1. Write τ_net = Iα in the graph form y = slope × x.

    Why this step: Torque is on the vertical axis and angular acceleration on the horizontal axis.

  2. The slope is I, with units kg·m².

    Why this step: This graph estimates inertia from several torque/acceleration trials.

Rotational inertia.

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Explain and apply: worked solutions

Give credit for equivalent correct reasoning. Use the rubric to find the next step to improve, even when your final number matches.

Match models and the full equation set

A rigid wheel rotates about a fixed central axle. Its initial angle is θ₀ = 0.50 rad, its initial angular velocity is +3.0 rad/s, and its constant angular acceleration is +2.0 rad/s² for 4.0 s. Counterclockwise is positive.

Part a

ω = +11 rad/s; θ = 28.5 rad; Δθ = +28 rad. Squared-speed check: 11² = 3² + 2(2)(28) = 121. All three assume constant α over the stated interval and a single defined axis/sign convention.

  1. Use 3 + 2×4 for ω.
  2. Use 0.50 + 3×4 + ½×2×4² for θ.
  3. Use the difference θ−θ₀, not θ itself, in the third equation.
Scoring criteria
  • Credit ω 11 rad/s and θ 28.5 rad/Δθ 28 rad.
  • Credit an independent squared-speed substitution giving 121 on both sides.
  • Credit constant α and consistent axis/sign domain.

Part b

Partners: v = v₀ + at; x = x₀ + v₀t + ½at²; v² = v₀² + 2a(x−x₀). Pairs are Δx↔Δθ, v↔ω and a↔α. The ω–t line runs from (0,3) to (4,11), slope +2 rad/s² and trapezoid area ½(3+11)×4 = +28 rad.

  1. Replace x, v and a with θ, ω and α without changing equation structure.
  2. Slope is the change in ω divided by time; area accumulates angular displacement.
Scoring criteria
  • Credit all three paired equations and the three variable correspondences.
  • Credit labeled graph endpoints, slope 2 rad/s² and area 28 rad.

Part c

The wheel holds shape but opposite rim points have different, opposite tangential velocity directions; a stationary center point hides that spin and the rim motion. The extended rigid model preserves these details. For Earth’s orbital path, a center-of-mass object can preserve revolution when axial spin is irrelevant; it omits the daily rotation and surface-point motions, so it is inadequate if those are asked about.

  1. Rigidity fixes relative positions, not identical linear velocities of all points.
  2. Select the model for the requested motion: spin requires extension; center-of-mass orbit may not.
Scoring criteria
  • Credit fixed shape and differing opposite-rim motion.
  • Credit specific Earth revolution/spin distinction and both model limits.

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Compare both angular rates on a rigid disk

Two marks are fixed to the same rigid disk at radii 0.10 m and 0.30 m. During the same 2.0 s interval the disk’s angular velocity rises from +1.0 to +5.0 rad/s at constant rate.

Part a

Both marks have ω = +5.0 rad/s and α = +2.0 rad/s². Their tangential acceleration magnitudes are 0.20 and 0.60 m/s².

  1. Rigidity fixes angular separation, so both marks sweep equal angle changes at equal times.
  2. Compute α = (5−1)/2 = 2 rad/s² for both.
  3. Use a_t = rα separately at the two radii.
Scoring criteria
  • Credit the choice stating equal ω and α while a_t scales with radius.

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Construct an extended force diagram

A rigid horizontal 1.0 m bar has a marked reference axis perpendicular to the page through its left end. At this instant force A is 6.0 N downward at x = 0.20 m, force B is 9.0 N upward at x = 0.60 m, and force C is 3.0 N downward at x = 1.0 m. There are no other forces; take counterclockwise positive for the one-plane turning calculation.

Part a

Draw a horizontal 1.0 m bar, axis at x=0, downward A at 0.20 m, upward B at 0.60 m, and downward C at 1.0 m. Use arrow-length ratio A:B:C = 2:3:1; label 6, 9, 3 N and distances. Both diagrams represent applied force magnitudes/directions; the extended diagram additionally preserves where they act relative to the axis. A single dot erases lever-arm geometry.

  1. Choose any consistent scale, for example 1 cm per 3 N, so A/B/C arrows are 2/3/1 cm.
  2. Place each arrow tail at its specified contact point on the bar.
  3. Keep the axis and all distances visible before analyzing turning effects.
Scoring criteria
  • Credit bar outline and correctly marked axis.
  • Credit all three application points and distance labels.
  • Credit correct arrow directions, labels and proportional lengths.
  • Credit comparison to free-body representation and the lost lever-arm information.

Part b

Torques are τ_A = −1.2 N·m, τ_B = +5.4 N·m, τ_C = −3.0 N·m; net is +1.2 N·m. Net vertical force is −6 + 9 − 3 = 0 N. Force sum cancels, but different locations weight the forces differently in torque.

  1. All forces are perpendicular to the horizontal position vectors, so each magnitude is xF.
  2. Assign signs from the turning tendency about the left-end axis.
  3. Add forces without lever-arm factors for the translational sum.
Scoring criteria
  • Credit all signed torque contributions and +1.2 N·m sum.
  • Credit zero net force and explanation based on differing application positions.

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Prove the center-axis minimum

A rigid planar assembly has total mass 3.0 kg and provided inertia I_cm = 0.12 kg·m² about a center-of-mass axis perpendicular to its plane. Every axis in this task is parallel to that axis. Let signed offset d measure axis position relative to the center-of-mass axis.

Part a

Since M = 3.0 kg > 0 and d² ≥ 0, I(d) ≥ I_cm = 0.12 kg·m² for every real d, with equality only at d = 0. The axis direction is fixed; the parallel-axis theorem does not rank arbitrary differently oriented axes.

  1. Express every allowed axis using the same I_cm.
  2. The additive Md² term can never be negative.
  3. Identify zero offset as the only zero added term for positive mass.
Scoring criteria
  • Credit nonnegativity argument covering every parallel axis.
  • Credit equality at d=0 and fixed-axis-direction qualification.

Part b

I(±0.20) = 0.24 kg·m² and I(+0.40) = 0.60 kg·m². The parabola I = 0.12 + 3.0d² has vertex (0 m, 0.12 kg·m²). Equal opposite offsets give the same squared separation and inertia.

  1. Add 3(0.20²) = 0.12 to I_cm for either sign.
  2. For 0.40 m add 3(0.40²) = 0.48.
  3. Plot signed offset horizontally and inertia vertically; the minimum is the vertex.
Scoring criteria
  • Credit all three values with kg·m².
  • Credit symmetric upward parabola with labeled vertex and points.

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Separate translation and rotation

A freely moving bar has mass 2.0 kg and provided I_cm = 0.20 kg·m². Ignore all other forces in this ideal comparison. In arrangement A, two 5.0 N forces both point down, one 0.30 m left of the center and one 0.30 m right. In arrangement B, the left 5.0 N force points down and the right 5.0 N force points up. Upward and counterclockwise are positive.

Part a

The dot diagram has two equal downward arrows, net F = −10 N and a_cm = −5.0 m/s². The extended diagram locates them at ±0.30 m and gives torques +1.5 and −1.5 N·m, total zero; α=0. The bar is in rotational equilibrium and not translational equilibrium. ω stays +1.0 rad/s in this arrangement.

  1. Sum forces to determine center-of-mass acceleration.
  2. Include application positions when summing torques about CM.
  3. Zero torque means zero angular acceleration, not necessarily zero ω.
Scoring criteria
  • Credit correct two diagrams with both positions in the extended view.
  • Credit −10 N/−5.0 m/s² and zero net torque/α.
  • Credit both equilibrium judgments and constant +1.0 rad/s.

Part b

F_net = −5 + 5 = 0, so a_cm = 0 and translational equilibrium holds. The two torque contributions are +1.5 N·m each, net +3.0 N·m; α = 3.0/0.20 = +15 rad/s². Rotational equilibrium fails and ω must change. With that net torque maintained for 0.10 s, Δω = +1.5 rad/s.

  1. The opposite forces cancel as vectors but their separated locations give the same turning tendency.
  2. Use α = τ_net/I_cm with the provided constant inertia.
  3. Use Δω = αΔt only under the stated maintained-torque condition.
Scoring criteria
  • Credit zero force and +3.0 N·m torque from two signed contributions.
  • Credit translational yes/rotational no and changing ω.
  • Credit initial α +15 rad/s² and conditional Δω +1.5 rad/s.

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Couple a falling mass and pulley

A 0.80 kg mass falls while a light inextensible string unwinds without slipping from a fixed central axle pulley. Pulley radius is 0.10 m and provided inertia about that axle is 0.0040 kg·m². Bearing torque is negligible. Use g = 10 m/s²; the mass starts from rest. Downward mass motion and the corresponding pulley turn are positive.

Part a

Mass diagram: mg downward, T upward. Pulley diagram: T tangential at radius R and support force at the central axle (pulley weight also acts through its center if included). Equations: mg−T = ma; TR = Iα; a = Rα. Axle/support and central weight have zero lever arms about the axle. Downward acceleration requires mg−T > 0, so T < mg.

  1. Keep mass and pulley as separate objects for the two balances.
  2. Use the actual rim tension to create torque, not the hanging weight.
  3. Use the string constraint to connect a and α only after writing the independent equations.
Scoring criteria
  • Credit both diagrams and all relevant force application locations.
  • Credit both balances and a=Rα with consistent positive senses.
  • Credit zero axle torque and T<mg reason.

Part b

a = mg/(m+I/R²) = 8.0/(0.80+0.40) = 6.67 m/s². T = I a/R² = 2.67 N; α = 66.7 rad/s². v² = 2a(0.30) = 4.0 m²/s², so v = 2.0 m/s and ω = 20 rad/s. K_mass = 1.6 J, K_pulley = 0.80 J, sum 2.4 J = mgs.

  1. Eliminate α using a/R to obtain T = Ia/R².
  2. Substitute tension into mg−T=ma before computing the rates.
  3. Use constant a for v²=2as and the no-slip speed relation for ω.
  4. Compare the two kinetic-energy terms with the gravitational potential decrease of mass plus Earth.
Scoring criteria
  • Credit a 6.67 m/s², T 2.67 N and α 66.7 rad/s².
  • Credit v 2.0 m/s and ω 20 rad/s.
  • Credit separate energies 1.6 and 0.80 J and their 2.4 J gravitational check.

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