The problems are on the rotational inertia worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Rotational inertia I measures how strongly an object resists changes in its rotation and depends on both mass and how far that mass is from the axis. These worked answers correct the 20 problems of the AP Physics 1 worksheet on rotational inertia, from units and symbols to child on a roundabout: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the law, the substitution and the final value in bold. When a problem has several steps, check that your signs and axis choice are consistent, and compare your units with the expected ones: kg m², N m or rad/s².
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 5
Problem 1 Units and symbols
1. kg m².
2. N m.
3. rad/s².
4. Rotational inertia measures how strongly an object resists a change in its angular velocity about a given axis, and it depends on mass and on how that mass is distributed relative to the axis.
Problem 2 A single point mass
1. I = mr² = 0.40 × (0.50)² = 0.10 kg m².
2. I = 0.40 × (1.0)² = 0.40 kg m², which is 4 times larger since r doubled.
Problem 3 Torque from a force
1. τ = rF = 0.40 × 25 = 10 N m.
2. τ = rF sinθ = 0.40 × 25 × sin 30° = 5.0 N m.
Problem 4 Angular acceleration
1. α = τ/I = 6.0 ÷ 0.30 = 20 rad/s².
2. t = ω/α = 40 ÷ 20 = 2.0 s.
Problem 5 Finding rotational inertia
1. I = τ/α = 12 ÷ 4.0 = 3.0 kg m².
2. With I = 6.0 kg m²: α = 12 ÷ 6.0 = 2.0 rad/s², half as large.
Problem 6 Hoop and disk
1. Hoop: I = MR² = 2.0 × (0.30)² = 0.18 kg m². Disk: I = ½MR² = 0.090 kg m².
2. The disk is easier to spin up: it has half the rotational inertia, because more of its mass is close to the axis, so the same torque gives twice the angular acceleration.
Problem 7 Dumbbell axes
1. Each mass is 0.50 m from the centre: I = 2 × 0.80 × (0.50)² = 0.40 kg m².
2. One mass is on the axis (r = 0) and the other is 1.0 m away: I = 0.80 × (1.0)² = 0.80 kg m².
3. The end axis gives twice the value, because the moving mass is twice as far out and r² makes it count four times, while only one mass contributes instead of two.
Problem 8 Masses at the corners of a square
1. The distance from the centre to a corner is half the diagonal: 0.60√2 ÷ 2 = 0.424 m, so r² = 0.18 m². I = 4 × 0.50 × 0.18 = 0.36 kg m².
2. From a corner: distances² are 0, 0.36, 0.36 and 0.72 m². I = 0.50 × (0 + 0.36 + 0.36 + 0.72) = 0.72 kg m².
3. Axis A (the centre), because its rotational inertia is half as large.
Problem 9 Pulling a rope on a wheel
1. τ = RF = 0.25 × 18 = 4.5 N m.
2. α = τ/I = 4.5 ÷ 0.50 = 9.0 rad/s².
3. ω = αt = 9.0 × 3.0 = 27 rad/s.
Problem 10 Slowing a flywheel
1. α = (0 − 60) ÷ 12 = −5.0 rad/s².
2. τ = Iα = 0.80 × (−5.0) = −4.0 N m.
3. F = τ/r = 4.0 ÷ 0.10 = 40 N.
Problem 11 Solid disk
1. I = ½MR² = ½ × 3.0 × (0.20)² = 0.060 kg m².
2. α = τ/I = 0.30 ÷ 0.060 = 5.0 rad/s².
Problem 12 Opening a door
1. τ = rF = 0.90 × 20 = 18 N m.
2. α = 18 ÷ 4.0 = 4.5 rad/s².
3. θ = ½αt², so t = √(2θ/α) = √(2 × 1.571 ÷ 4.5) = √0.698 = 0.84 s.
Problem 13 Choking up on a bat
1. Sliding the hands towards the middle puts the axis nearer the centre of mass, so less of the bat’s mass is far from the axis. Since I = Σmr², the rotational inertia drops sharply.
2. With τ = Iα, a smaller I gives a larger angular acceleration for the same torque from the hands.
Problem 14 Pulley and a hanging mass
1. I = ½MR² = ½ × 2.0 × (0.050)² = 2.5 × 10⁻³ kg m².
2. With T = ½Ma = 1.0a and mg − T = ma: 1.5 × 9.81 = (1.5 + 1.0)a, so a = 14.715 ÷ 2.5 = 5.9 m/s².
3. T = 1.0 × 5.886 = 5.9 N; α = a/R = 5.886 ÷ 0.050 = 1.2 × 10² rad/s² (118 rad/s²).
Problem 15 Two masses and a pulley
1. I = ½MR² = ½ × 2.0 × 0.010 = 0.010 kg m². The 3.0 kg mass moves down. Net force along the string: (m₁ − m₂)g = (m₁ + m₂ + ½M)a, so a = 2.0 × 9.81 ÷ 5.0 = 3.9 m/s² (3.924).
2. T₁ = m₁(g − a) = 3.0 × 5.886 = 18 N (17.7 N). T₂ = m₂(g + a) = 1.0 × 13.734 = 14 N (13.7 N).
3. (T₁ − T₂)R = (17.658 − 13.734) × 0.10 = 0.3924 N m. Iα = 0.010 × (3.924 ÷ 0.10) = 0.3924 N m. They agree.
Problem 16 Falling rod
1. τ = mg(L/2) = 1.2 × 9.81 × 0.40 = 4.7 N m.
2. I = (1/3)mL² = (1/3) × 1.2 × (0.80)² = 0.256 kg m². α = 4.709 ÷ 0.256 = 18 rad/s².
3. a = αL = 18.4 × 0.80 = 15 m/s², which is 1.5 times g. The free end accelerates faster than free fall because the rod’s other parts pull it round.
Problem 17 Unwinding from a cylinder
1. Mg − T = Ma.
2. Torque of the tension about the centre: TR = Iα = ½MR²(a/R), so T = ½Ma.
3. Mg − ½Ma = Ma, so a = (2/3)g = 6.5 m/s² (6.54 m/s²). T = ½ × 0.50 × 6.54 = 1.6 N.
Problem 18 Analysing an experiment
1. Gradient = (9.9 − 2.1) ÷ (0.5 − 0.1) = 7.8 ÷ 0.4 = 19.5 rad/s² per N m.
2. The gradient is 1/I, so I = 1 ÷ 19.5 = 0.051 kg m².
3. I = ½MR² = ½ × 0.80 × (0.36)² = 0.0518 kg m². Difference = (0.0518 − 0.0513) ÷ 0.0518 = 1 %.
Problem 19 Spot the error
1. More mass at the rim means more rotational inertia, and for the same torque a larger I gives a smaller angular acceleration, not a larger one.
2. Hoop: I = 4.0 × 0.25 = 1.0 kg m², so α = 2.0 ÷ 1.0 = 2.0 rad/s². Disk: I = 0.50 kg m², so α = 2.0 ÷ 0.50 = 4.0 rad/s².
3. αdisk : αhoop = 2 : 1.
Problem 20 Child on a roundabout
1. τ = RF = 2.0 × 80 = 160 N m.
2. I = 200 + 30 × (1.5)² = 267.5 kg m². α = 160 ÷ 267.5 = 0.60 rad/s².
3. I = 200 + 30 × (0.50)² = 207.5 kg m², so α = 160 ÷ 207.5 = 0.77 rad/s². Moving inwards reduces the child’s contribution mr² to the total, so the same torque gives a larger acceleration.
