The problems are on the angular momentum worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Angular momentum is L = Iω, and it stays constant when no net external torque acts, so a spinner who pulls mass inward spins faster. These worked answers correct the 20 problems of the AP Physics 1 worksheet on angular momentum, from spinning flywheel energy to testing conservation of angular momentum: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the law used, the substitution and the final result in bold. For conservation problems, check L before and after, and then compare kinetic energies to see whether energy was added or lost.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 6
Problem 1 Spinning flywheel energy
1. K = ½Iω² = ½ × 0.50 × (40)² = 4.0 × 10² J.
2. At 80 rad/s: K = ½ × 0.50 × (80)² = 1.6 × 10³ J, four times larger because K ∝ ω².
Problem 2 Revolutions per minute
1. ω = 120 × 2π ÷ 60 = 12.6 rad/s.
2. K = ½ × 0.20 × (12.57)² = 16 J (15.8 J).
Problem 3 Angular momentum of a turntable
1. L = Iω = 0.40 × 15 = 6.0 kg m²/s.
2. L = 0.40 × 10 = 4.0 kg m²/s.
Problem 4 Rolling wheel
1. ω = v/R = 7.0 ÷ 0.35 = 20 rad/s.
2. 20 ÷ 2π = 3.2 revolutions per second.
Problem 5 A skater pulls in her arms
1. Iiωi = Ifωf: ωf = 3.0 × 2.0 ÷ 1.5 = 4.0 rad/s.
2. Angular momentum is conserved, because there is no net external torque on the skater.
Problem 6 Work done by a torque
1. W = τΔθ = 4.0 × 5.0 = 20 J.
2. W = ½Iω²: ω = √(2W/I) = √(2 × 20 ÷ 0.40) = √100 = 10 rad/s.
Problem 7 Skater’s energy
1. ωf = 2.4 × 3.0 ÷ 0.80 = 9.0 rad/s.
2. Ki = ½ × 2.4 × 9.0 = 10.8 J; Kf = ½ × 0.80 × 81 = 32.4 J.
3. The skater’s muscles do 21.6 J of work pulling her arms inwards, against the outward (centripetal) requirement of the circular motion.
Problem 8 Rolling ball
1. Ktrans = ½mv² = ½ × 0.50 × 9.0 = 2.25 J.
2. Krot = ½Iω² = ½ × (2/5)mR² × (v/R)² = (1/5)mv² = 0.20 × 0.50 × 9.0 = 0.90 J.
3. Fraction = 0.90 ÷ 3.15 = 29 % (2/7).
Problem 9 Rolling down a ramp
1. Disk: v = √(4gh/3) = √(4 × 9.81 × 1.5 ÷ 3) = √19.62 = 4.4 m/s.
2. Hoop (I/mR² = 1): mgh = mv², v = √(gh) = √14.7 = 3.8 m/s.
3. The disk arrives first because it reaches a higher speed.
Problem 10 Motor and a disk
1. α = τ/I = 2.5 ÷ 0.25 = 10 rad/s², so ω = αt = 80 rad/s.
2. K = ½ × 0.25 × (80)² = 800 J; Pavg = 800 ÷ 8.0 = 100 W.
3. P = τω = 2.5 × 80 = 200 W, twice the average because power grows linearly with ω.
Problem 11 Clay on a turntable
1. Idisk = ½ × 2.0 × (0.30)² = 0.090 kg m²; Iclay = 0.20 × (0.25)² = 0.0125 kg m².
2. ωf = 0.090 × 5.0 ÷ 0.1025 = 4.4 rad/s.
3. Ki = ½ × 0.090 × 25 = 1.125 J; Kf = ½ × 0.1025 × (4.39)² = 0.988 J. Lost = 0.14 J (about 12 %).
Problem 12 Angular momentum of a moving ball
1. L = mvr = 0.20 × 5.0 × 0.40 = 0.40 kg m²/s (r is the perpendicular distance from O to the line of motion).
2. No net torque acts about O (no force, ignoring gravity and friction), and the perpendicular distance stays 0.40 m, so L is constant.
Problem 13 Storing energy in a flywheel
1. ω = 3000 × 2π ÷ 60 = 314 rad/s.
2. K = ½ × 2.0 × (314.2)² = 9.9 × 10⁴ J (98.7 kJ).
3. t = E/P = 98 700 ÷ 100 = 9.9 × 10² s, about 16 minutes.
Problem 14 Ranking rolling objects
1. v = √(2gh/(1 + I/mR²)): hoop √(gh); disk √(4gh/3); sphere √(10gh/7).
2. gh = 4.905 m²/s²: hoop = 2.2 m/s; disk = √6.54 = 2.6 m/s; sphere = √7.007 = 2.6 m/s (2.65 m/s).
3. Sphere first, then disk, then hoop, since a smaller I/mR² means less energy in spin.
Problem 15 Putty hits a rod
1. L = mvr = 0.20 × 3.0 × 0.60 = 0.36 kg m²/s.
2. I = (1/3)ML² + mr² = (1/3) × 1.0 × 0.36 + 0.20 × 0.36 = 0.12 + 0.072 = 0.19 kg m².
3. ω = L/I = 0.36 ÷ 0.192 = 1.9 rad/s. Ki = ½ × 0.20 × 9.0 = 0.90 J; Kf = ½ × 0.192 × (1.875)² = 0.34 J; lost = 0.56 J.
Problem 16 Acceleration on a slope
1. mg sinθ − f = ma.
2. Torque of friction about the centre: fR = Iα = ½mR² (a/R), so f = ½ma.
3. mg sinθ = (3/2)ma, so a = (2/3)g sinθ = (2/3) × 9.81 × 0.50 = 3.3 m/s² (3.27). f = ½ × 2.0 × 3.27 = 3.3 N, directed up the slope.
Problem 17 Rolling hoop up a hill
1. K = ½mv² + ½Iω² = ½mv² + ½mR²(v/R)² = mv² = 1.5 × 16 = 24 J.
2. mgh = 24 J, h = 24 ÷ (1.5 × 9.81) = 1.6 m.
3. Sliding block: h = v²/(2g) = 16 ÷ 19.62 = 0.82 m, only half as high, because the rolling hoop must keep half of its energy in rotation.
Problem 18 Jumping onto a roundabout
1. Li = Iω + mvR = 200 × 1.2 + 30 × 4.0 × 2.0 = 240 + 240 = 480 kg m²/s.
2. If = 200 + 30 × (2.0)² = 320 kg m²; ωf = 480 ÷ 320 = 1.5 rad/s.
3. Ki = ½ × 200 × (1.2)² + ½ × 30 × (4.0)² = 144 + 240 = 384 J; Kf = ½ × 320 × (1.5)² = 360 J; lost = 24 J.
Problem 19 Spot the error
1. ωf = Iiωi/If = 3.0 × 2.0 ÷ 1.0 = 6.0 rad/s.
2. Ki = ½ × 3.0 × 4.0 = 6.0 J; Kf = ½ × 1.0 × 36 = 18 J.
3. Constant L does not imply constant K. K = L²/2I, so reducing I raises K; here the skater’s muscles supply the extra 12 J.
Problem 20 Testing conservation of angular momentum
1. Gradient = (11.9 − 3.1) ÷ (8.0 − 2.0) = 8.8 ÷ 6.0 = 1.47.
2. ω = L × (1/I), so the gradient is the angular momentum: 1.47 kg m²/s.
3. Difference = (1.50 − 1.47) ÷ 1.50 = 2 %, which supports conservation of L within the scatter of the data.
