The problems are on the moment of inertia worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Moment of inertia is I = ∫r² dm, the sum of each mass element times the square of its distance from the rotation axis. These worked answers correct the 20 problems of the AP Physics C worksheet on moment of inertia, from dumbbell inertia to clay hits a disc: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the principle, the substitution and the final result with units. The derivations show every integration step, so check your own working against them line by line. Use g = 9.81 m/s² unless told otherwise, and keep two or three significant figures.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 5
Problem 2 Units
1. kg·m².2. N·m. A joule is energy, a scalar; torque is a vector quantity with a different meaning, even though the dimensions match.
Problem 3 Torque from a spanner
1. τ = rF = 0.25 × 30 = 7.5 N·m.2. τ = rF sin 30° = 7.5 × 0.50 = 3.8 N·m.
Problem 4 Angular acceleration
α = τ/I = 6.0/0.40 = 15 rad/s².Problem 5 Using the table
1. I = (2/5)MR² = 0.40 × 5.0 × 0.010 = 0.020 kg·m².2. I = MR² = 5.0 × 0.010 = 0.050 kg·m².
Problem 6 Ring against disc
1. The ring (MR² against ½MR²), since all its mass is at the largest radius.2. The disc has the larger α, because α = τ/I and its I is smaller.
Problem 7 Rod about two axes
1. Icm = ML²/12 = 0.60/12 = 0.050 kg·m².2. I = Icm + M(L/2)² = 0.050 + 0.60 × 0.25 = 0.20 kg·m², equal to ML²/3.
Problem 8 Disc about the rim
1. ½MR² = ½ × 2.0 × 0.040 = 0.040 kg·m².2. I = 0.040 + 2.0 × (0.20)² = 0.12 kg·m².
Problem 9 Rod with varying density
1. M = ∫₀^1.5 2.0x² dx = (2.0/3)(1.5)³ = 2.25 kg.2. I = ∫x² dm = ∫₀^1.5 2.0x⁴ dx = (2.0/5)(1.5)⁵ = 0.40 × 7.594 = 3.0 kg·m².
Problem 10 Flywheel on a rope
1. τ = RF = 0.30 × 12 = 3.6 N·m.2. I = ½MR² = ½ × 6.0 × 0.090 = 0.27 kg·m².
3. α = 3.6/0.27 = 13 rad/s². ω = αt = 13.3 × 2.5 = 33 rad/s.
Problem 11 Skater spinning
1. L = Iω is conserved: ω₂ = 3.0 × 4.0/1.2 = 10 rad/s.2. K₁ = ½ × 3.0 × 16 = 24 J; K₂ = ½ × 1.2 × 100 = 60 J. The skater does 36 J of work pulling her arms in.
Problem 12 Reading ω against t
1. α = 10/4.0 = 2.5 rad/s².2. τ = Iα = 0.50 × 2.5 = 1.25 N·m.
3. θ = area = ½ × 4.0 × 10 = 20 rad.
Problem 13 Plank pivot point
1. The centre is at 1.2 m, so d = 1.2 − 0.40 = 0.80 m.2. I = ML²/12 + Md² = 3.84 + 8.0 × 0.64 = 3.84 + 5.12 = 9.0 kg·m².
3. The theorem applies only when one axis passes through the centre of mass; ML²/3 is the value about an end, so that sum is invalid (it gives 16.6 kg·m²).
Problem 14 Swinging signboard bar
1. I = ML²/3 = 2.5 × 2.56/3 = 2.1 kg·m².2. τ = Mg(L/2) = 2.5 × 9.81 × 0.80 = 19.6 N·m, so α = 19.6/2.13 = 9.2 rad/s².
3. The centre accelerates downward at a = α(L/2) = 7.36 m/s². Then Mg − F = Ma gives F = 2.5 × (9.81 − 7.36) = 6.1 N (one quarter of the weight).
Problem 15 Disc by integration
1. σ = M/(πR²), dm = σ2πr dr = (2M/R²) r dr.2. I = ∫₀^R r²(2M/R²) r dr = (2M/R²)(R⁴/4) = ½MR².
3. I = ½ × 0.80 × 0.0225 = 9.0 × 10⁻³ kg·m².
Problem 16 Annular disc
1. σ = M/[π(R₂² − R₁²)]. I = ∫R₁^R₂ r² σ2πr dr = (σπ/2)(R₂⁴ − R₁⁴) = (σπ/2)(R₂² − R₁²)(R₂² + R₁²) = ½M(R₁² + R₂²).2. I = ½ × 3.0 × (0.010 + 0.040) = 0.075 kg·m².
Problem 17 Massive pulley
1. m₁g − T₁ = m₁a; T₂ − m₂g = m₂a; (T₁ − T₂)R = Iα = ½MR² × a/R, so T₁ − T₂ = ½Ma. Adding: (m₁ − m₂)g = (m₁ + m₂ + M/2)a, so a = 9.81/5.5 = 1.8 m/s².2. T₁ = 3.0(9.81 − 1.78) = 24 N; T₂ = 2.0(9.81 + 1.78) = 23 N.
Problem 18 Rolling cylinder
1. Along the slope: Mg sin θ − f = Ma; torque: fR = Iα = Ia/R. Eliminating f: a = g sin θ/(1 + I/MR²) = 9.81 × 0.50/1.5 = 3.3 m/s².2. v = √(2as) = √(2 × 3.27 × 2.0) = 3.6 m/s.
Problem 19 Compound pendulum arm
1. I = ML²/3 + mL² = 0.20 + 0.30 = 0.50 kg·m².2. τ = (M × 0.50 + m × 1.0)g = (0.30 + 0.30) × 9.81 = 5.89 N·m, so α = 5.89/0.50 = 12 rad/s².
Problem 20 Clay hits a disc
1. L = mvR = 0.050 × 6.0 × 0.30 = 0.090 kg·m²/s.2. I = ½ × 1.0 × 0.090 + 0.050 × 0.090 = 0.0495 kg·m². ω = L/I = 1.8 rad/s.
3. K₁ = ½ × 0.050 × 36 = 0.90 J; K₂ = ½ × 0.0495 × 3.31 = 0.082 J; loss = 0.82 J.
