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Unit 6 / Energy and Momentum of Rotating Systems

Unit 6: 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 6.1

1. Rotational Kinetic Energy · 4

  1. Rearrange K = ½Iω² to |ω| = √(2K/I).

    Why this step: Both wheels have supplied fixed-axis energy and inertia.

  2. Wheel A: √[2(4.0)/0.020] = 20 rad/s.

    Why this step: Its smaller inertia permits greater angular speed at the same energy.

  3. Wheel B: √[2(4.0)/0.080] = 10 rad/s.

    Why this step: Four times the inertia halves angular speed because energy depends on ω².

Rearrange K = ½Iω² to |ω| = √(2K/I). Wheel A: √[2(4.0)/0.020] = 20 rad/s. Wheel B: √[2(4.0)/0.080] = 10 rad/s.

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Question 2 · AP topic 6.2

2. Torque and Work · 4

  1. Find K_i = ½(0.050)(12²) = 3.6 J and K_f = 0.

    Why this step: The wheel’s stated final condition is rest.

  2. Use W_net = ΔK = −3.6 J.

    Why this step: The opposing torque removes the initial rotational kinetic energy.

  3. Write −0.30Δθ = −3.6 to get Δθ = 12 rad.

    Why this step: Work uses angular displacement; this calculation does not supply stopping time.

Find K_i = ½(0.050)(12²) = 3.6 J and K_f = 0. Use W_net = ΔK = −3.6 J. Write −0.30Δθ = −3.6 to get Δθ = 12 rad.

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Question 3 · AP topic 6.3

3. Angular Momentum and Angular Impulse · 4

  1. Compute L_i = Iω_i = 0.20(5.0) = +1.0 kg·m²/s.

    Why this step: Use the specified axis and positive initial turning sense.

  2. J_angular = (−0.40)(1.5) = −0.60 kg·m²/s.

    Why this step: The braking torque transfers angular momentum out over the stated time.

  3. L_f = 1.0 − 0.60 = +0.40 kg·m²/s; ω_f = L_f/I = +2.0 rad/s.

    Why this step: Subtract the impulse from initial L, and check that the remaining positive value has not reversed the spin.

Compute L_i = Iω_i = 0.20(5.0) = +1.0 kg·m²/s. J_angular = (−0.40)(1.5) = −0.60 kg·m²/s. L_f = 1.0 − 0.60 = +0.40 kg·m²/s; ω_f = L_f/I = +2.0 rad/s.

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Question 4 · AP topic 6.4

4. Conservation of Angular Momentum · 4

  1. L_i = I_iω_i = 2.0(3.0) = +6.0 kg·m²/s.

    Why this step: The initial configuration supplies the initial angular momentum.

  2. Add external angular impulse: L_f = 6.0 − 1.0 = +5.0 kg·m²/s.

    Why this step: The external brake means total L of the selected system is not constant.

  3. ω_f = L_f/I_f = 5.0/1.0 = +5.0 rad/s.

    Why this step: The reduced inertia increases speed even though the system loses angular momentum.

L_i = I_iω_i = 2.0(3.0) = +6.0 kg·m²/s. Add external angular impulse: L_f = 6.0 − 1.0 = +5.0 kg·m²/s. ω_f = L_f/I_f = 5.0/1.0 = +5.0 rad/s.

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Question 5 · AP topic 6.5

5. Rolling · 4

  1. Compute K_trans = ½(1.5)(2.0²) = 3.0 J.

    Why this step: The center-of-mass motion contributes the translational term.

  2. Insert I_cm = MR² and ω = v/R into K_rot.

    Why this step: The supplied hoop inertia and no-slip constraint apply together.

  3. K_rot = ½MR²(v/R)² = ½Mv² = 3.0 J; K_total = 6.0 J.

    Why this step: Radius cancels in this particular model, and the two distinct contributions add.

Compute K_trans = ½(1.5)(2.0²) = 3.0 J. Insert I_cm = MR² and ω = v/R into K_rot. K_rot = ½MR²(v/R)² = ½Mv² = 3.0 J; K_total = 6.0 J.

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Question 6 · AP topic 6.6

6. Motion of Orbiting Satellites · 4

  1. Use center distance r = 8.0 × 10⁶ m and the supplied GM.

    Why this step: Orbital expressions use center-to-center distance, not surface height.

  2. v_c = √(GM/r) = √(4.0 × 10⁷) ≈ 6.32 × 10³ m/s.

    Why this step: Circular motion requires gravity to supply the centripetal force.

  3. v_e = √(2GM/r) ≈ 8.94 × 10³ m/s.

    Why this step: Threshold escape sets total mechanical energy to zero, with zero speed at infinity in the ideal model.

  4. The ratio is √2 and satellite mass cancels from both.

    Why this step: The same gravitational and inertial mass appears in the force or energy balance.

Use center distance r = 8.0 × 10⁶ m and the supplied GM. v_c = √(GM/r) = √(4.0 × 10⁷) ≈ 6.32 × 10³ m/s. v_e = √(2GM/r) ≈ 8.94 × 10³ m/s. The ratio is √2 and satellite mass cancels from both.

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Question 7 · AP topic 6.1

7. Compare a different speed change

  1. Trial A has K_trans = ½(2.0)(3.0²) = 9.0 J and K_rot = ½(0.040)(15²) = 4.5 J, totaling 13.5 J.

    Why this step: Center translation and rotation about the center are distinct contributions, so they add.

  2. Trial B has K_trans = ½(2.0)(6.0²) = 36 J and K_rot = ½(0.040)(7.5²) = 1.125 J, totaling 37.125 J.

    Why this step: Doubling center speed quadruples its energy term, while halving angular speed quarters the other term. These changes do not generally cancel.

  3. Use one bar scale, such as 4.5 J per division: A has bars 2 and 1 divisions; B has bars 8 and 0.25 divisions. Reject the equal-total prediction.

    Why this step: A common scale makes the changed partition visible. The decomposition is valid without a no-slip relationship between the two speeds.

Trial A: 9.0 J translation + 4.5 J rotation = 13.5 J. Trial B: 36 J + 1.125 J = 37.125 J; the totals are unequal.

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Question 8 · AP topic 6.2

8. Choose the work coordinate

  1. Use torque vertically against angle horizontally.

    Why this step: Work is torque–angle area. A torque–time graph would determine angular impulse, and no elapsed-time measurements are supplied here.

  2. Driving work is ½(2.0 + 4.0)(1.0) + 4.0(1.0) = 7.0 J. Bearing work is −0.50(2.0) = −1.0 J, so net work is +6.0 J.

    Why this step: Add the first trapezoid and second rectangle, then include the opposing torque through the same angular displacement.

  3. K_f = K_i + W_net = 1.0 + 6.0 = 7.0 J, assuming the stated torque work accounts for all energy transfers and no other energy storage changes.

    Why this step: The fixed-axis work–energy theorem connects net torque work to rotational kinetic energy only when the ledger includes all relevant work and energy changes.

Use a torque–angle graph. Driving work is 7.0 J, bearing work −1.0 J, and final kinetic energy 7.0 J under the stated complete-work assumption.

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Question 9 · AP topic 6.3

9. Follow a pulse and an opposing torque

  1. The triangular angular impulse is ½(2.0)(2.0) = +2.0 kg·m²/s, so L reaches +2.50 kg·m²/s at 2.0 s.

    Why this step: Signed torque–time area changes L, and the positive triangle adds to the initial value.

  2. During the triangle, the L graph increases with a slope that rises from zero to +2.0 N·m and falls back to zero. L is +1.50 kg·m²/s at 1.0 s. During the last interval it is a straight line of slope −1.0 N·m, ending at L = 2.50 − 1.0(1.5) = +1.00 kg·m²/s.

    Why this step: Torque is the instantaneous slope of L against time. The first two halves have equal triangular areas; the final constant torque gives linear decrease.

  3. The wheel does not reverse: L stays positive, with its smallest value +0.50 initially and +1.00 after braking.

    Why this step: For positive fixed inertia, L and signed angular velocity share a sign. A negative torque can reduce positive L without making it cross zero.

L: +0.50 initially, +1.50 at 1.0 s, +2.50 at 2.0 s, +1.00 at 3.5 s. It never crosses zero, so the wheel does not reverse.

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Question 10 · AP topic 6.4

10. Judge a conservation claim with uncertainty

  1. Initial L has possible values 2.4(1.90) to 2.4(2.10), or 4.56–5.04 kg·m²/s. Final L lies between 0.80(5.40) and 0.80(5.60), or 4.32–4.48 kg·m²/s.

    Why this step: Use the supplied speed intervals rather than assuming the central values are exact; positive constant inertia scales each interval.

  2. The intervals do not overlap. Final L is below initial L even at the closest endpoints, so the data support a negative external angular impulse rather than zero.

    Why this step: The largest possible final value, 4.48, is below the smallest possible initial value, 4.56. Increasing angular speed does not by itself establish constant L.

  3. As a control, record the same stool coasting for the same duration with a fixed mass configuration and similar rotation speeds. Compare the measured L decrease with the shape-change trial.

    Why this step: Keeping configuration fixed separates a background drag trend from the inertia change. Additional uncertainty in real inertia estimates must also be assessed before a real experiment makes the same claim.

The possible L intervals do not overlap; a negative external angular impulse is supported. A fixed-configuration coast-down control can test bearing drag.

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Question 11 · AP topic 6.5

11. Diagnose the contact motion

  1. For the disk, R|ω| = 0.25(8.0) = 2.0 m/s, below v_cm = 3.0 m/s; its bottom slips right at 1.0 m/s. For the hoop, R|ω| = 0.25(10) = 2.5 m/s, matching v_cm.

    Why this step: Translation and clockwise surface motion subtract at the bottom. Matching values support instantaneous no slipping; the disk’s unequal values establish slipping.

  2. Kinetic friction on the disk acts left. With other horizontal forces negligible, it reduces the rightward center speed and produces a clockwise torque that increases clockwise spin; contact sliding dissipates mechanical energy.

    Why this step: Friction opposes the contact point’s motion relative to the floor, and its force and torque have different effects on center translation and rotation.

  3. The data cannot rank total kinetic energies without masses and rotational inertias about the centers.

    Why this step: K_total = ½Mv_cm² + ½I_cmω² depends on those quantities as well as the measured speeds. Radius and shape names alone do not supply all the needed values.

The hoop matches instantaneous no slipping; the disk slips right at contact, so friction acts left, slows translation, increases clockwise spin, and dissipates mechanical energy. Energy ranking needs masses and inertias.

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Question 12 · AP topic 6.6

12. Check an elliptical orbit ledger

  1. At the near and far turning points, velocity is perpendicular to radius. Use r_near v_near = r_far v_far to get v_far = (2.0 × 10⁷)(3.0 × 10³)/(6.0 × 10⁷) = 1.0 × 10³ m/s.

    Why this step: Gravity is radial, so its torque about the central object is zero; the simple mrv form applies at these turning points.

  2. Near the central object, K = ½(100)(3.0 × 10³)² = 4.5 × 10⁸ J and U = −(1.20 × 10¹⁴)(100)/(2.0 × 10⁷) = −6.0 × 10⁸ J.

    Why this step: Kinetic energy is positive, while the infinity-zero gravitational potential energy is negative and lower at smaller radius.

  3. At the far point, K = 5.0 × 10⁷ J and U = −2.0 × 10⁸ J. Both bar charts sum to E = −1.5 × 10⁸ J, so the orbit is bound in the stated gravity-only model.

    Why this step: The decrease in K offsets the increase in U. Negative total energy is below the threshold needed to reach infinity with zero speed.

  4. Draw force toward the central object at both points and show its line of action through that object.

    Why this step: The zero lever arm explains constant angular momentum even while conservative gravity exchanges kinetic and potential energy and changes speed.

Far speed: 1.0 × 10³ m/s. Near energies: K = 4.5 × 10⁸ J, U = −6.0 × 10⁸ J; far: K = 5.0 × 10⁷ J, U = −2.0 × 10⁸ J. E = −1.5 × 10⁸ J, a bound orbit.

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Question 13 · AP topic 6.1

13. Rotational Kinetic Energy · 1

  1. Use K_rot = ½Iω².

    Why this step: Rotational kinetic energy is a scalar with squared angular velocity.

  2. Squaring +10 and −10 gives the same value, so endpoint energies match.

    Why this step: Changing turning sense alone does not change energy at equal angular speed.

They are equal.

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Question 14 · AP topic 6.2

14. Torque and Work · 1

  1. Inspect the horizontal axis: it is angular position.

    Why this step: The choice between work and angular impulse depends on whether the axis is angle or time.

  2. The signed area is τΔθ, which is work in joules.

    Why this step: Torque–time area would instead have units N·m·s.

Work done by the torque.

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Question 15 · AP topic 6.3

15. Angular Momentum and Angular Impulse · 1

  1. Use L = mvb, where b is distance from the reference point to the straight path.

    Why this step: Angular momentum is defined about a chosen point or axis.

  2. When the point is off the path, b can be nonzero and L is nonzero.

    Why this step: The path need not curve, and nonzero angular momentum does not require a current force.

Yes, if the reference point is off the line of motion.

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Question 16 · AP topic 6.4

16. Conservation of Angular Momentum · 1

  1. Negligible external angular impulse gives I_iω_i = I_fω_f.

    Why this step: Internal shape-changing forces do not change the system total L.

  2. Reducing I allows ω to increase while L stays constant.

    Why this step: Internal work can change kinetic energy, so energy need not stay constant.

Angular momentum stays constant while angular speed can increase.

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Question 17 · AP topic 6.5

17. Rolling · 1

  1. For no slipping, v_cm = R|ω|.

    Why this step: The translational and rotational surface motions match at contact.

  2. At the bottom they cancel in the stationary-surface frame.

    Why this step: The center still moves at v_cm and the top adds the two motions, giving 2v_cm.

The instantaneous contact point is at rest in the surface frame.

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Question 18 · AP topic 6.6

18. Motion of Orbiting Satellites · 1

  1. Use U = −GMm/r, with zero at infinity.

    Why this step: The orbital potential energy is for the two-object system and uses center distance.

  2. Smaller positive r makes U more negative.

    Why this step: Inward motion decreases potential energy and can increase kinetic energy at fixed total E.

It becomes more negative.

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Question 19 · AP topic 6.1

19. Rotational Kinetic Energy · 2

  1. Substitute v = rω and I = mr² into ½mv².

    Why this step: Both expressions describe the same point object’s motion about a fixed axis.

  2. The result is ½Iω²; do not add the expressions.

    Why this step: Adding equal descriptions would count the same kinetic energy twice.

They are two descriptions of the same energy.

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Question 20 · AP topic 6.2

20. Torque and Work · 2

  1. The wheel’s angular displacement is zero.

    Why this step: A force or torque can act while its application point has no rotational motion.

  2. W = τΔθ = 0.

    Why this step: Elapsed time alone does not imply work or rotational kinetic-energy transfer.

Zero.

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Question 21 · AP topic 6.3

21. Angular Momentum and Angular Impulse · 2

  1. Angular impulse is signed torque–time area.

    Why this step: The pulse changes angular momentum, not directly an energy value.

  2. J_angular = ½(0.50)(4.0) = +1.0 N·m·s = +1.0 kg·m²/s.

    Why this step: The triangular shape halves the rectangular peak-times-duration value.

+1.0 kg·m²/s

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Question 22 · AP topic 6.4

22. Conservation of Angular Momentum · 2

  1. Select both disks and identify contact friction as internal.

    Why this step: Each frictional angular impulse is accompanied by an equal opposite transfer to the other disk.

  2. With negligible external bearing torque, total L of both disks is constant.

    Why this step: Selecting only one disk leaves the other disk’s friction as an external transfer.

Both disks together.

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Question 23 · AP topic 6.5

23. Rolling · 2

  1. The bottom surface slips left relative to the floor.

    Why this step: Kinetic friction opposes relative motion at contact, not necessarily center motion.

  2. Friction acts right, increasing center speed while reducing clockwise spin.

    Why this step: Its force accelerates the center right and its torque opposes the existing clockwise rotation.

Right.

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Question 24 · AP topic 6.6

24. Motion of Orbiting Satellites · 2

  1. Gravity is conservative and radial about the central object.

    Why this step: A conservative interaction keeps mechanical energy constant, and a radial force has zero torque there.

  2. E and L stay constant while K, U, speed, and radius can change.

    Why this step: An elliptical orbit is not uniform circular motion.

Total mechanical energy and angular momentum.

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Question 25 · AP topic 6.1

25. Rotational Kinetic Energy · 3

  1. Separate motion into center translation and motion relative to the center.

    Why this step: The two contributions must refer to distinct parts of the motion description.

  2. Use I_cm in the rotation term.

    Why this step: An external-axis inertia already includes some center motion and can cause double counting if used with the translation term.

Inertia about the center-of-mass rotation axis.

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Question 26 · AP topic 6.2

26. Torque and Work · 3

  1. Starting from rest gives W_net = ½Iω_f².

    Why this step: There is no initial rotational kinetic energy to subtract.

  2. At fixed I, doubling W multiplies ω_f by √2.

    Why this step: Angular speed depends on the square root of supplied net work.

It becomes √2 times as large.

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Question 27 · AP topic 6.3

27. Angular Momentum and Angular Impulse · 3

  1. Net external torque is the slope of L against t.

    Why this step: Torque measures the rate of change of angular momentum.

  2. A horizontal graph has zero slope, so τ_net = 0.

    Why this step: Its positive height is angular momentum already present, not torque.

Zero.

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Question 28 · AP topic 6.4

28. Conservation of Angular Momentum · 3

  1. At fixed L, replace ω by L/I in K = ½Iω².

    Why this step: This removes angular speed in favor of the conserved quantity.

  2. K = L²/(2I), so halving I doubles K.

    Why this step: The extra energy must come from internal work or another energy source during the shape change.

It doubles.

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Question 29 · AP topic 6.5

29. Rolling · 3

  1. Use Mgh = ½Mv² + ½I(v/R)².

    Why this step: Both start from rest with the same height change and roll without slipping.

  2. The hoop has greater I/(MR²), so less energy is available for translation at a given total.

    Why this step: The disk reaches a larger center speed; this is an energy partition effect.

The disk.

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Question 30 · AP topic 6.6

30. Motion of Orbiting Satellites · 3

  1. Circular motion gives v_c² = GM/r; threshold escape energy gives v_e² = 2GM/r.

    Why this step: The circular condition comes from force, while escape comes from E = 0.

  2. Taking the ratio and positive square root gives v_e/v_c = √2.

    Why this step: The two speeds are compared at the same starting radius and under the same gravity-only model.

It is √2 times as large.

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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.

Explain energy with a stationary center

Two 0.40 kg masses are fixed to opposite ends of a massless rigid arm, each 0.25 m from a fixed axle through their center of mass. The arm rotates at angular speed 8.0 rad/s.

Part a

Each speed is rω = 2.0 m/s, and each K = ½(0.40)(2.0²) = 0.80 J; total 1.6 J. The masses move tangentially in opposite directions while the CM stays fixed. Their scalar kinetic energies add rather than cancel like velocity or momentum components.

  1. Compute speeds relative to the fixed axle.
  2. Apply ordinary particle kinetic energy to each mass.
  3. Opposite directions can cancel momentum but K depends on speed squared.
Scoring criteria
  • Credit both speeds and 0.80+0.80=1.6 J.
  • Credit moving-material explanation despite stationary CM and scalar addition.

Part b

I = 2(0.40)(0.25²) = 0.050 kg·m². K_rot = ½(0.050)(8²) = 1.6 J. The CM translation term is zero. The individual energies and ½Iω² are the same motion sum, so they must not be added again.

  1. Sum mr² about the same axle.
  2. Use common angular speed for the rigid assembly.
  3. The rotational formula compactly replaces the particle-energy sum.
Scoring criteria
  • Credit I 0.050 kg·m² and K_rot 1.6 J.
  • Credit zero CM translation and no duplicate energy addition.

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Use four factors and derive angular impulse

For a straight-moving puck, m = 0.50 kg and v = 4.0 m/s. At one instant relative to a fixed point O, r = 0.80 m and the angle between the radial vector and velocity is 30°. Separately, a wheel with fixed I = 0.30 kg·m² receives constant net torque +0.60 N·m for 0.50 s; counterclockwise is positive.

Part a

Original L = rmv sinθ = 0.80 kg·m²/s. Each listed change doubles it to 1.60 kg·m²/s. On one fixed straight path, r and θ are linked by constant b=r sinθ, so the independent r/θ comparisons refer to changed reference/path arrangements.

  1. Compute 0.80×0.50×4×sin30° = 0.80.
  2. Use linear dependence on r, m and v; sin90°/sin30° = 2.
  3. Distinguish independent factor comparisons from correlated geometric changes along a path.
Scoring criteria
  • Credit original 0.80 and all four new 1.60 kg·m²/s values.
  • Credit fixed-path r sinθ relation and independent-comparison limit.

Part b

b = 0.80sin30° = 0.40 m. At r=0.40 m and θ=90°, L = 0.40×0.50×4 = 0.80 kg·m²/s. A reference point on the path has b=0 and L=0, despite nonzero linear momentum.

  1. Use the perpendicular offset, not r alone, as the straight-path geometry.
  2. The force-free path keeps speed and b constant about fixed O.
Scoring criteria
  • Credit b 0.40 m and unchanged L 0.80 kg·m²/s.
  • Credit zero L for an on-path reference point.

Part c

For this constant torque, α=Δω/Δt. With unchanged I, L=Iω gives ΔL=IΔω. Thus τ_net=IΔω/Δt=ΔL/Δt and multiplying by Δt gives τ_netΔt=ΔL. Angular impulse is +0.30 N·m·s = +0.30 kg·m²/s; Δω = ΔL/I = +1.0 rad/s. For varying torque, use interval averages or signed graph area rather than assuming constant instantaneous torque.

  1. State fixed I before replacing IΔω by ΔL.
  2. Keep the same interval in the acceleration and momentum rates.
  3. Multiply the rate equality by 0.50 s and use the inertia for the spin change.
Scoring criteria
  • Credit starting rotational second law and the full equality chain.
  • Credit constant-I and constant-torque/average scope.
  • Credit +0.30 kg·m²/s and +1.0 rad/s.

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Account for both disks and the room

Two coaxial disks have I_A = 0.30 kg·m², ω_A = +4.0 rad/s and I_B = 0.20 kg·m², ω_B = −1.0 rad/s. They make contact and eventually turn together. Neglect external torque during contact. After contact, a brake connected to the room removes 0.20 kg·m²/s from the pair. All quantities refer to the same fixed axle; counterclockwise is positive.

Part a

Initially L_A = +1.2 and L_B = −0.20 kg·m²/s, total +1.0. The common ω_f = 1.0/(0.30+0.20) = +2.0 rad/s. Final components are +0.60 and +0.40 kg·m²/s, total +1.0.

  1. Multiply each inertia by its signed ω separately.
  2. Conserve the pair sum over contact, then divide by combined inertia.
  3. Check the final sum from the two individual components rather than from combined inertia alone.
Scoring criteria
  • Credit both initial components and +1.0 sum.
  • Credit +2.0 rad/s and both final components/sum.

Part b

ΔL_A = 0.60−1.2 = −0.60 kg·m²/s; ΔL_B = 0.40−(−0.20) = +0.60 kg·m²/s. At each matched contact location the third-law tangential forces are opposite and share the same radius, so their torques about the axle are opposite. Summing over the same contact time gives equal opposite angular impulses. Each disk receives external friction torque from the other when considered alone.

  1. Subtract each component’s initial L from its final L.
  2. Apply τ=rF_t at the same contact radius for each force pair.
  3. Sum paired torque×time contributions; only the enlarged pair internalizes that contact interaction.
Scoring criteria
  • Credit −0.60/+0.60 transfers with units.
  • Credit force-pair → equal opposite torque → same-interval angular impulse derivation.
  • Credit disk-alone external torque explanation.

Part c

The pair ends at +0.80 kg·m²/s; brake plus Earth gains +0.20 kg·m²/s. Relative to the surroundings’ initial value, enlarged total is unchanged: 0.80+0.20=1.0. The brake interaction transfers angular momentum across the pair’s boundary; enlarging the system makes that transfer internal, provided other external torque is negligible.

  1. Use pair ΔL = −0.20 for the braking interval.
  2. Assign the equal opposite transfer to the surroundings about the same axis.
  3. Add the changes or include the surroundings’ initial L consistently.
Scoring criteria
  • Credit pair +0.80, surroundings gain +0.20 and unchanged combined total.
  • Credit system-boundary and common-axis conditions.

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Connect rolling acceleration and contact work

A rigid 3.0 kg solid disk of radius 0.20 m rolls from rest down a stationary rigid incline with sinβ = 0.50. Its provided inertia is I_cm = ½MR² = 0.060 kg·m². Use g = 10 m/s². There is enough static friction to prevent slipping, and deformation losses are negligible. Downhill translation and the corresponding clockwise rotation are positive.

Part a

For constant R and continuing no slip: s_cm=RΔθ, v_cm=Rω and a_cm=Rα in the compatible sign convention. Balances: Mg sinβ−f=Ma_cm; fR=I_cmα. Hence a_cm = Mg sinβ/(M+I_cm/R²)=15/(3+1.5)=3.33 m/s²; α=16.7 rad/s²; f=5.0 N uphill. The acceleration relation requires the no-slip constraint throughout the interval, not merely a matching instantaneous speed.

  1. A rim arc RΔθ equals the center travel along the stationary surface when no slip persists.
  2. Take corresponding time rates to obtain v and a constraints for fixed radius.
  3. Eliminate α in the torque balance to get f=I a/R²; substitute into force balance.
Scoring criteria
  • Credit displacement, speed and acceleration constraints with no-slip/fixed-R scope.
  • Credit both balances and correct uphill friction direction.
  • Credit a 3.33 m/s², α16.7 rad/s², f5.0 N.

Part b

s_cm=0.60 m; Δθ=3.0 rad; v_cm=2.0 m/s; ω=10 rad/s. Thus 0.60=0.20×3.0, 2.0=0.20×10, and 3.33≈0.20×16.7. K_trans=6.0 J; K_rot=3.0 J. The height drop is s sinβ=0.30 m, giving MgΔh=9.0 J, equal to the sum.

  1. Use constant acceleration from rest: s=½at² and Δθ=½αt².
  2. Use v=at and ω=αt.
  3. Use center translation plus CM rotation once, and compare with the actual vertical height drop.
Scoring criteria
  • Credit all four motion values and three numerical constraint checks.
  • Credit 6.0 and 3.0 J energies, 0.30 m height and 9.0 J sum.

Part c

The contact point is instantaneously at rest in the stationary-ramp frame, so actual friction power F·v_contact is zero. Center-of-mass contribution is −fv_cm=−10 W; rotation contribution is (fR)ω=+10 W; sum zero. Static friction redistributes the kinetic-energy shares. With slipping, the contact has relative motion and kinetic friction dissipates mechanical energy into internal forms; the no-slip constraints need not hold.

  1. Identify the velocity where the force actually acts, not only v_cm.
  2. Calculate the opposing translation contribution and aiding spin contribution in the defined senses.
  3. Do not extend the ideal rigid/stationary/no-slip zero-work result to moving surfaces or slipping.
Scoring criteria
  • Credit zero contact velocity/power with the stationary ideal-surface assumptions.
  • Credit −10/+10 W cancellation.
  • Credit qualitative dissipation and constraint failure for slipping, without an unsupported lossless claim.

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Justify the satellite approximation and derive escape

A gravity-only, collision-free satellite–central-body model has GM = 3.0 × 10¹⁴ m³/s², G = 6.67 × 10⁻¹¹ N·m²/kg² and center separation r₀ = 6.0 × 10⁶ m. The satellite mass is 200 kg. Use an inertial center-of-mass frame, and U=0 at infinite separation. Treat central-body motion as negligible only after justifying it.

Part a

M = (GM)/G ≈ 4.50 × 10²⁴ kg. The center-of-mass condition Mr_M=mr_m gives r_M/r_m=m/M ≈ 4.45 × 10⁻²³. Since r_M+r_m=r₀, r_M≈(m/M)r₀ ≈ 2.67 × 10⁻¹⁶ m and r_m≈6.0 × 10⁶ m. Central motion is negligible compared with the satellite’s orbit. Comparable masses give comparable orbital radii, so neither body can be held fixed in the center-of-mass frame.

  1. Divide the supplied GM parameter by G to infer a mass consistent with the gravitational data.
  2. Use the same common center of mass for both radii: Mr_M=mr_m.
  3. Apply the calculated mass ratio before approximating total separation as the satellite radius.
  4. State the relative size of the approximation rather than claiming the central body is exactly stationary.
Scoring criteria
  • Credit M≈4.50×10²⁴ kg and r_M/r_m≈4.45×10⁻²³ from the supplied data.
  • Credit central radius about 2.67×10⁻¹⁶ m and negligible relative motion.
  • Credit the comparable-mass failure of a fixed-central-body approximation.

Part b

v_c=√(GM/r₀)≈7.07×10³ m/s. K=5.0×10⁹ J; U=−1.0×10¹⁰ J; E=−5.0×10⁹ J; L=m r₀ v_c≈8.49×10¹² kg·m²/s. All K,U,E,L stay constant along this gravity-only circular orbit. Fixed radius fixes U; gravity perpendicular to velocity holds speed and K fixed; conservative gravity keeps E fixed; radial gravity has zero torque about the central body and keeps L fixed.

  1. Set GMm/r₀²=mv_c²/r₀, yielding v_c²=GM/r₀=5.0×10⁷.
  2. Compute K and U with the stated reference, then add them for E.
  3. Use θ=90° for circular radius/velocity to get L=mr₀v_c.
Scoring criteria
  • Credit circular-force derivation and all four values with units/signs.
  • Credit explicitly naming all four as constant and giving appropriate physical reasons.

Part c

At threshold r→∞, U→0 and v→0, so E=0. Conservation gives ½mv_esc²−GMm/r₀=0, yielding v_esc=√(2GM/r₀)=1.00×10⁴ m/s. It assumes gravity from this central body is the only force, central recoil is negligible, and the outward trajectory avoids collision; the formula is a speed magnitude.

  1. Choose the infinity reference before setting final energy.
  2. Equate initial mechanical energy to its threshold asymptotic value0.
  3. Cancel m and take the positive square root.
Scoring criteria
  • Credit E=0 derived from both limiting terms.
  • Credit full algebra to √(2GM/r₀) and 1.00×10⁴ m/s.
  • Credit gravity-only and collision-free/outward assumptions.

Part d

From ½mv²−GMm/r=0, v=√(2GM/r). At 4r₀, v=v_esc/2=5.00×10³ m/s. It remains positive for every finite r and tends to zero only as r→∞; there is no finite stopping radius in this model. Above threshold E>0 and v at infinity tends to √(2E/m)>0.

  1. Use the conserved zero energy at every later radius.
  2. Increasing r by4 reduces v² by4 and v by2.
  3. At infinity U tends to0, so the remaining K is the total E; zero or positive E sets the limit.
Scoring criteria
  • Credit v(r) derivation and5.00×10³ m/s at4r₀.
  • Credit positive finite-distance speed and asymptotic zero at infinity.
  • Credit above-threshold positive limiting speed.

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