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 the rotational equivalent of momentum: L = Iω for a rigid body, or L = r × p for a particle. These worked answers correct the 20 problems of the AP Physics C worksheet on angular momentum, from rolling bicycle wheel to torque from calculus: each solution shows the formula, the substitution and the final answer with its unit.
Also, each answer shows the formula, the substitution and the result with its unit in bold. Where energy is lost, say where it goes. Compare your reasoning with the method, not only the number, because the free-response section awards marks for the physical argument.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 6
Problem 1 Rolling bicycle wheel
1. ω = v/R = 5.1 / 0.34 = 15 rad/s.2. f = ω/2π = 15 / 6.283 = 2.4 rev/s.
3. Thus, The top point moves at v + Rω = 2v = 10.2 m/s.
Problem 2 Cricket ball and the umpire
1. L = m v r⊥ = 0.16 × 30 × 1.5 = 7.2 kg m²/s.2. L = 0.16 × 30 × 0.50 = 2.4 kg m²/s.
3. Hence, L = r × p depends on the lever arm measured from the chosen point, so a point closer to the line gives a smaller value; L is zero for any point on the line itself.
Problem 3 Spinning disc
1. I = ½ M R² = ½ (3.0)(0.20)² = 0.060 kg m².2. L = Iω = 0.060 × 12 = 0.72 kg m²/s.
3. K = ½ I ω² = ½ (0.060)(144) = 4.3 J.
Problem 4 Skater pulling in her arms
1. I₁ω₁ = 0.40 I₁ ω₂, so ω₂ = 2.0 / 0.40 = 5.0 rad/s.2. Thus, K ∝ Iω², so K₂/K₁ = 0.40 × (5.0/2.0)² = 0.40 × 6.25 = 2.5. Her muscles supply the extra energy.
Problem 5 Order of the race
1. a = g sinθ / (1 + β) and β is 2/5, 1/2, 2/3, 1. The order is solid sphere, solid disc, thin shell, hoop.2. In a = g sinθ / (1 + β) both m and R cancel, because I = βmR² scales with both; only the mass distribution (β) matters.
Problem 6 Energy split
rotational fraction = β / (1 + β).1. 1/2 = 50 %.
2. Hence, 0.4 / 1.4 = 0.286 = 29 %.
Problem 7 Sphere down a ramp
1. v = √(2gh/(1 + β)) = √(2 × 9.81 × 0.35 / 1.4) = √4.905 = 2.2 m/s.2. ω = v/R = 2.215 / 0.040 = 55 rad/s.
3. K = mgh = 0.50 × 9.81 × 0.35 = 1.7 J.
Problem 8 Hoop on a 25° slope
1. a = g sinθ / 2 = 9.81 × 0.4226 / 2 = 2.07 m/s².2. Thus, t = √(2 × 3.0 / 2.073) = √2.894 = 1.7 s.
3. v = a t = 2.073 × 1.701 = 3.5 m/s (check: v² = 2aL = 12.4).
Problem 9 Friction on a cylinder
1. f = (β/(1 + β)) m g sinθ = (1/3)(4.0)(9.81)(0.500) = 6.5 N.2. Hence, N = mg cosθ = 4.0 × 9.81 × 0.866 = 34.0 N, so μs ≥ 6.54 / 34.0 = 0.19.
Problem 10 Clay on a turntable
1. I_clay = m r² = 0.50 × 0.16 = 0.080 kg m². L conserved: 0.80 × 6.0 = 0.88 ω, so ω = 5.5 rad/s.2. K₁ = ½(0.80)(36) = 14.4 J; K₂ = ½(0.88)(5.45)² = 13.1 J. In addition, lost: 1.3 J, turned into heat in the sticking process.
Problem 11 Child on a roundabout
1. I₁ = 250 + 40 × 2.0² = 410 kg m²; L = 410 × 0.80 = 328 kg m²/s.2. I₂ = 250 + 40 × 0.50² = 260 kg m²; ω₂ = 328 / 260 = 1.3 rad/s.
Problem 12 Bullet and rod
1. L = m v ℓ = 0.020 × 300 × 0.80 = 4.8 kg m²/s.2. Thus, I = ⅓(1.2)(0.80)² + 0.020(0.80)² = 0.256 + 0.0128 = 0.269 kg m².
3. ω = L/I = 4.8 / 0.2688 = 18 rad/s.
Problem 13 Reading an L–t graph
1. Στ = dL/dt = (14.0 − 2.0) / 4.0 = 3.0 N m.2. Hence, α = τ/I = 3.0 / 1.5 = 2.0 rad/s².
3. Then, the area under a torque–time graph is the change in angular momentum (the angular impulse).
Problem 14 Spot the error
1. No friction force is needed at constant speed, and static friction acts at a point at rest, so it does no work even when present.2. A ball rolling without slipping at constant speed keeps its kinetic energy; no net force acts on it, so it does not slow down (apart from small rolling resistance).
Problem 15 Angular momentum about the contact point
1. ω = 4.0/0.30 = 13.3 rad/s; I = ½(2.0)(0.09) = 0.090 kg m²; L_cm = 0.090 × 13.3 = 1.2 kg m²/s.2. Then add m v R = 2.0 × 4.0 × 0.30 = 2.4, so L = 1.2 + 2.4 = 3.6 kg m²/s (equal to (3/2) m v R).
Problem 16 Hoop against disc
1. Afterwards, hoop: v = √(2 × 9.81 × 2.0 / 2) = 4.4 m/s. disc: v = √(2 × 9.81 × 2.0 / 1.5) = 5.1 m/s.2. a_hoop / a_disc = (1/2)/(2/3) = 0.75, and both start at rest at the same moment, so the hoop has covered 0.75 × 5.0 = 3.75 m, which is 1.25 m behind.
Problem 17 Sphere on a loop-the-loop
1. At the top the weight alone supplies the centripetal force: m g = m v²/R, so v² = gR.2. Thus, Energy: m g h = m g (2R) + ½ m v²(1 + 0.4) = 2mgR + 0.7 m g R, so h = 2.7 R = 1.35 m.
Problem 18 Falling yo-yo
1. m g − T = m a and T R = I α = ½ m R² (a/R), so T = ½ m a. Then a = g/(1 + ½) = 6.5 m/s².2. T = ½ m a = ½ (0.20)(6.54) = 0.65 N.
3. Hence, v = √(2 a s) = √(2 × 6.54 × 1.5) = 4.4 m/s.
Problem 19 Child jumping on a roundabout
1. L = m v R = 35 × 4.0 × 2.5 = 350 kg m²/s. I = 300 + 35 × 2.5² = 518.75 kg m². ω = 350 / 518.75 = 0.67 rad/s.2. K₁ = ½(35)(16) = 280 J; K₂ = ½(518.75)(0.6747)² = 118 J. Also, lost: 162 J, dissipated as the child’s feet grip and slide.
Problem 20 Torque from calculus
1. L = ∫τ dt = ∫6.0t dt = 3.0t² kg m²/s.2. At t = 4.0 s, L = 3.0 × 16 = 48 kg m²/s, so ω = L/I = 48 / 2.0 = 24 rad/s. The torque is τ = 6.0 × 4.0 = 24 N m, so α = τ/I = 12 rad/s².
3. Thus, W = ΔK = L²/2I = 48² / 4.0 = 576 J. check: ∫τω dt = ∫9t³ dt = 9 × 256 / 4 = 576 J.
