Unit 6 Practice Set: Energy and Momentum of Rotating Systems

How to use this set

This is a full-length, AP-style set for all of Unit 6 (topics 6.1 to 6.6): 20 multiple choice questions and 4 free response questions. Each free response question is one of the newer AP types: mathematical routines, translation between representations, experimental design, and qualitative quantitative translation.

Need a refresher first? Go back to the Unit 6 overview or warm up with the Unit 6 games.

Part A: Multiple choice

1. A solid sphere of mass 2.0 kg and radius 0.10 m spins about an axis through its center at 30 rad/s. (I = ⅖MR².) What is its rotational kinetic energy?

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B. I = ⅖ × 2.0 × 0.10² = 0.0080 kg·m². K = ½Iω² = ½ × 0.0080 × 900 = 3.6 J. Choice C forgets the ½; choice D uses I = MR², the formula for a hoop.

2. A hoop, a solid disk and a solid sphere have the same mass and radius. Each is given the same rotational kinetic energy about its center. Which spins fastest?

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C. ω = √(2K/I). Equal K, so the smallest I spins fastest. The sphere (⅖MR²) has less I than the disk (½MR²) and the hoop (MR²), because more of its mass is near the axis.

3. A constant torque of 12 N·m turns a grinding wheel through 5.0 complete revolutions. How much work does the torque do?

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C. W = τΔθ with Δθ in radians: 5.0 rev × 2π = 31.4 rad. W = 12 × 31.4 = 377 J ≈ 380 J. Choice A forgets to convert revolutions to radians.

4. The graph shows the torque on a wheel (I = 0.50 kg·m²) as a function of its angle of rotation. The wheel starts from rest. What is its angular speed at θ = 5.0 rad?

8 3 5 0 θ (rad) τ (N·m)
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C. Work = area under the τ-θ graph: rectangle 8 × 3 = 24 J plus triangle ½ × 2 × 8 = 8 J, total 32 J. All of it becomes rotational KE: ½ × 0.50 × ω² = 32, so ω² = 128 and ω = 11.3 rad/s. Choice B uses only the rectangle (24 J).

5. A drill motor keeps the bit turning at a steady 40 rad/s while it exerts a torque of 15 N·m. What power does the motor deliver?

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C. P = τω = 15 × 40 = 600 W. (W = τΔθ, so the rate of doing work is τ × Δθ/Δt = τω.)

6. A disk with I = 2.0 kg·m² spins at 10 rad/s. A brake pad exerts a constant friction torque of 4.0 N·m. Through what angle does the disk turn before it stops?

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C. Rotational work-energy: the brake's negative work removes all the KE. K = ½ × 2.0 × 10² = 100 J. τΔθ = 100 J, so Δθ = 100 / 4.0 = 25 rad (about 4 turns).

7. A 0.20 kg ball moves at 5.0 m/s in a straight line. The line passes 0.40 m from a point P (that is the perpendicular distance). What is the ball's angular momentum about P?

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B. L = m v r⊥ = 0.20 × 5.0 × 0.40 = 0.40 kg·m²/s. An object moving in a straight line does have angular momentum about any point that is not on its line, and it stays constant if no force acts.

8. A wheel (I = 0.25 kg·m²) starts at rest. The graph shows the torque on it during a short push. What is the wheel's angular speed after the push?

6 0.25 0.50 t (s) τ (N·m)
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C. Angular impulse = area under τ-t = ½ × 0.50 × 6 = 1.5 N·m·s = ΔL. Δω = ΔL / I = 1.5 / 0.25 = 6.0 rad/s. Choice D uses the full rectangle instead of the triangle.

9. Wheels A and B start at rest. Wheel B has twice the rotational inertia of A. Each receives the same angular impulse. Compare their final rotational kinetic energies.

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C. Same angular impulse means the same L. K = ½Iω² = L²/(2I). Same L and double I gives half the K. (B spins at half the rate: ½ × 2I × (ω/2)² = ½ × ½Iω².)

10. A skater spins at 1.5 rad/s with I = 4.0 kg·m², then pulls in her arms so I = 1.6 kg·m². Ignore friction with the ice. Which statement is correct?

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B. No external torque, so L is conserved: 4.0 × 1.5 = 1.6 × ω, ω = 3.75 rad/s. K goes from ½ × 4.0 × 1.5² = 4.5 J to ½ × 1.6 × 3.75² = 11.25 J. The extra energy comes from work her muscles do pulling her arms inward. "No external torque" keeps L constant, not ω and not K.

11. A turntable (I = 0.030 kg·m²) spins freely at 4.0 rad/s. A 0.20 kg lump of clay is dropped straight down and sticks 0.15 m from the axis. What is the new angular speed?

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B. The clay falls straight down, so it brings no angular momentum about the vertical axis. Iclay = mr² = 0.20 × 0.15² = 0.0045 kg·m². L conserved: 0.030 × 4.0 = (0.030 + 0.0045)ω, ω = 0.12 / 0.0345 = 3.5 rad/s.

12. A 30 kg child runs at 4.0 m/s along a line tangent to the edge of a merry-go-round (radius 2.0 m, I = 480 kg·m²) that is at rest, and jumps on at the edge. What is the angular speed just after?

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B. Child's L about the axis = mvr = 30 × 4.0 × 2.0 = 240 kg·m²/s. After: I = 480 + 30 × 2.0² = 600 kg·m². ω = 240 / 600 = 0.40 rad/s. Choice C forgets to add the child's own rotational inertia.

13. For the child and merry-go-round in question 12, what happens to the total kinetic energy of the child-merry-go-round system when the child jumps on?

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C. Before: ½ × 30 × 4.0² = 240 J. After: ½ × 600 × 0.40² = 48 J. This is a rotational "perfectly inelastic collision": angular momentum is conserved but kinetic energy is not.

14. Four objects start from rest at the top of the same incline: a hoop, a solid cylinder and a solid sphere that roll without slipping, and a block that slides with no friction. In what order do they reach the bottom?

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C. For rolling, a = g sinθ / (1 + β), where I = βMR². The block has β = 0 (no spin), the sphere 0.4, the cylinder 0.5, the hoop 1. Less of the energy goes into spinning, so more goes into moving down the ramp.

15. A solid cylinder (I = ½MR²) starts from rest and rolls without slipping down a ramp, dropping a height of 0.90 m. What is its speed at the bottom?

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B. Mgh = ½Mv² + ½(½MR²)(v/R)² = ¾Mv². v = √(4gh/3) = √(4 × 9.8 × 0.90 / 3) = √11.76 = 3.43 m/s. Choice D is √(2gh), a sliding block with no spin.

16. A thin hoop rolls without slipping across the floor. What fraction of its total kinetic energy is rotational?

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B. With v = ωR, Krot = ½(MR²)(v/R)² = ½Mv², exactly equal to Ktrans. So half of the total is rotational.

17. A solid ball rolls without slipping at 3.0 m/s on a level floor, then rolls up a ramp without slipping. How high does it rise before stopping?

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C. Total K = ½Mv² + ⅕Mv² = 0.7Mv². Static friction does no work on an object that rolls without slipping, so all of it becomes Mgh: h = 0.7v²/g = 0.7 × 9.0 / 9.8 = 0.64 m. Choice B (v²/2g) forgets the spin energy, which also turns into height.

18. A bowling ball is thrown so it slides along the lane without spinning at first. Kinetic friction from the lane acts on it until it begins to roll without slipping. During the sliding phase, which statement is true?

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B. Kinetic friction points backward, so it slows the center (v drops). About the center it makes a torque that speeds up the spin (ω rises). While the ball slips, kinetic friction turns some kinetic energy into thermal energy, so total K drops. Once v = ωR it rolls and the sliding friction stops.

19. A moon orbits planet X in a circle of radius r. Planet Y has twice the mass of X. A moon orbits Y in a circle of the same radius r. Compared with the moon of X, the moon of Y has

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C. v = √(GM/r): double M gives √2 times v. T = 2πr/v, so T is divided by √2.

20. A satellite is on an elliptical orbit around Earth. At which point is the kinetic energy of the satellite largest, and why?

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B. E = K + U is constant for the satellite-Earth system. U is most negative (lowest) at the smallest r, so K is largest there. (Angular momentum is the same everywhere, so D is wrong; the speed is largest at the closest point so that mvr stays constant.)

Part B: Free response

FR 1. Mathematical routines: the falling block and the pulley AP

A light string is wrapped around a pulley that is a uniform solid disk (mass M = 2.0 kg, radius R = 0.10 m, I = ½MR²) that turns on a frictionless axle. A block of mass m = 0.50 kg hangs from the string and is released from rest. The string does not slip.

  1. Use energy conservation for the block-pulley-Earth system to find the block's speed after it has fallen 1.2 m.
  2. Find the pulley's angular speed at that moment.
  3. Find the work done on the pulley by the string, and use it to find the string tension.
  4. The disk pulley is replaced by a thin ring of the same mass and radius. Without a full calculation, state whether the block's speed after falling 1.2 m is larger, smaller or the same, and explain.
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(a) No slipping, so the pulley's rim speed equals the block's speed: ω = v/R. Energy: mgh = ½mv² + ½Iω² = ½mv² + ½(½MR²)(v/R)² = ½(m + M/2)v². v = √(2mgh / (m + M/2)) = √(2 × 0.50 × 9.8 × 1.2 / 1.5) = √7.84 = 2.8 m/s. Points: 1 for including the pulley's rotational KE, 1 for using v = ωR, 1 for the correct answer with units.

(b) ω = v/R = 2.8 / 0.10 = 28 rad/s. 1 point.

(c) The string's torque is the only torque on the pulley, so its work equals the pulley's K: W = ½ × (½ × 2.0 × 0.10²) × 28² = ½ × 0.010 × 784 = 3.9 J. The string unwinds 1.2 m, so the pulley turns Δθ = 1.2 / 0.10 = 12 rad. W = τΔθ = (TR)Δθ = T × 1.2 m, so T = 3.92 / 1.2 = 3.3 N. Check with Newton's second law: a = mg/(m + M/2) = 3.27 m/s², T = m(g − a) = 0.50 × 6.53 = 3.3 N. Matches, and T < mg = 4.9 N as it must for a block speeding up downward. Points: 1 for W = ΔKrot, 1 for W = τΔθ = T·(distance), 1 for T.

(d) Smaller. A ring has I = MR², twice the disk's. The same released energy mgh now has to give more energy to the spinning pulley, so less is left for the block: ½(m + M)v² = mgh gives v = √(11.76 / 2.5) = 2.2 m/s. 1 point for "smaller" with reasoning about the larger share of energy going into rotation.

FR 2. Translation between representations: the spinning skater AP

A skater spins on ice with negligible friction. The graph shows her angular velocity. From 0 to 2.0 s her arms are out and her rotational inertia is 3.0 kg·m². Between 2.0 s and 3.0 s she pulls her arms in. After 3.0 s she holds them in.

2 5 2 3 5 t (s) ω (rad/s)
  1. Describe the graph of her angular momentum L against time from 0 to 5 s. Give the value(s).
  2. Find her rotational inertia after 3.0 s. Describe how I changes between 2.0 s and 3.0 s.
  3. Give the values for an energy bar chart of her rotational kinetic energy at t = 1 s and t = 4 s. Where did the difference come from?
  4. A classmate says: "Her angular velocity went up, so there must have been a torque on her." Explain what is right or wrong with this claim.
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(a) The ice exerts no torque about her spin axis, so L is constant: a horizontal line at L = Iω = 3.0 × 2.0 = 6.0 kg·m²/s for the whole 5 s, including while she pulls her arms in. 1 point for horizontal, 1 point for the value.

(b) I = L/ω = 6.0 / 5.0 = 1.2 kg·m². Between 2 and 3 s, I drops from 3.0 to 1.2 kg·m² while ω rises, always with Iω = 6.0 (so I = 6.0/ω: a falling curve, not a straight line). 1 point for 1.2 kg·m², 1 point for linking the drop in I to the rise in ω.

(c) K at 1 s = ½ × 3.0 × 2.0² = 6.0 J. K at 4 s = ½ × 1.2 × 5.0² = 15 J. The bars show K growing by 9.0 J. That energy came from work her muscles did pulling her arms inward (chemical energy in her body turned into rotational KE). 1 point each for the two values, 1 point for the source.

(d) Wrong. A torque would change her angular momentum, but L stayed 6.0 kg·m²/s. Her ω increased because her rotational inertia decreased, with no external torque at all. (Her arms do push on her body, but those are internal forces of the skater system.) 1 point.

FR 3. Experimental design: what shape is the can? AP

A student has a closed metal can and wants to find the number β in I = βMR² for the can without opening it. She has a long ramp, a meterstick, and a photogate that measures speed at the bottom of the ramp.

  1. Describe a procedure she could use, including what she varies, what she measures, and how she keeps the can rolling without slipping.
  2. Show that for a can released from rest that rolls without slipping, v² = 2gh / (1 + β), where h is the vertical drop.
  3. Her data are below. Say what she should plot to get a straight line, find the slope, and use it to find β. Which kind of object behaves like this can?
Drop h (m)0.100.200.300.400.50
Speed at bottom v (m/s)1.131.631.972.302.55
  1. A second student says the can's mass must be measured too. Is she right? Explain.
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(a) Release the can from rest at several different starting heights h (measured vertically from the photogate's level with the meterstick). Record the speed at the bottom with the photogate. Repeat each height 3 times and average. Use a gentle slope and a rough surface (or rubber strip) so the can rolls without slipping, and check by eye that it does not skid. Points: 1 for the independent variable (h) and how it is measured, 1 for the measured quantity (v) and repeats, 1 for a way to ensure rolling without slipping.

(b) Energy of the can-Earth system: Mgh = ½Mv² + ½(βMR²)(v/R)² = ½(1 + β)Mv², so v² = 2gh / (1 + β). 1 point.

(c) Plot v² against h. The v² values are 1.28, 2.66, 3.88, 5.29, 6.50 m²/s². A best-fit line through them has slope about 13.1 m/s² (and passes very close to the origin). The slope equals 2g / (1 + β), so 1 + β = 19.6 / 13.1 = 1.50 and β ≈ 0.50. The can behaves like a solid cylinder (I = ½MR²), for example a can full of frozen soup. Points: 1 for linearizing (v² vs h), 1 for the slope, 1 for β and the shape.

(d) No. M cancels in Mgh = ½(1 + β)Mv², and R cancels too. Only the way the mass is spread (β) matters. 1 point.

FR 4. Qualitative quantitative translation: the soup can race AP

Two cans have the same mass and the same radius. Can S is full of solid frozen soup, so it rolls like a solid cylinder (I = ½MR²). Can H is empty with very thin walls and light ends, so it rolls like a hoop (I ≈ MR²). Both are released together from rest at the top of the same ramp and roll without slipping.

  1. Without using equations, explain which can reaches the bottom first.
  2. Derive an expression for the speed of each can at the bottom of a ramp of vertical height h, in terms of g and h.
  3. Find the ratio vS / vH at the bottom. Explain how your result in (b) supports your reasoning in (a).
  4. The ramp's height is doubled. Does the ratio from (c) change? Explain.
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(a) Both cans lose the same gravitational potential energy (same mass, same drop). That energy is shared between moving along the ramp and spinning. Can H has its mass farther from the axis, so it needs more energy to spin at the same rate. More of its energy goes into spinning and less into moving forward, so at every point it is slower. Can S wins. Points: 1 for equal energy available, 1 for the larger share of rotational energy for H, 1 for the conclusion.

(b) Mgh = ½Mv² + ½Iω² with ω = v/R. Can S: Mgh = ½Mv² + ¼Mv² = ¾Mv², so vS = √(4gh/3). Can H: Mgh = ½Mv² + ½Mv² = Mv², so vH = √(gh). 1 point each.

(c) vS / vH = √(4/3) = 1.15. The equations show the same reason as (a): the coefficient in front of Mv² is larger for H (1 instead of ¾) because of its larger I, so for the same Mgh its v is smaller. Both cans have constant acceleration, so a larger final speed also means a shorter time. 1 point for the ratio, 1 point for connecting the larger coefficient to the larger I.

(d) No. Both speeds are proportional to √h, so doubling h multiplies both by √2 and the ratio stays √(4/3) = 1.15. 1 point.