Specification: AQA A-level Physics (7408) 3.11.1.2 Engineering physics option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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Rotational kinetic energy: the kinetic energy of a body rotating about an axis, Eₖ = ½Iω², where I is the moment of inertia about that axis and ω is the angular velocity in rad s⁻¹. It is the rotational analogue of ½mv².
Flywheel: a heavy wheel used to store rotational kinetic energy. Uses include smoothing the torque and speed of engines and presses, storing energy from regenerative braking, and short-term energy storage for the grid.
Energy storage capacity: a flywheel stores more energy if it has a larger moment of inertia (more mass, concentrated in the rim, at a larger radius) and a higher angular velocity. The maximum ω is limited by the strength of the material: too fast and the flywheel breaks apart.
The equation
rotational kinetic energy = ½ × moment of inertia × (angular velocity)² Eₖ = ½Iω²
| Quantity | Symbol | Unit |
|---|---|---|
| rotational kinetic energy | Eₖ | joules (J) |
| moment of inertia about the axis | I | kilogram metres squared (kg m²) |
| angular velocity | ω | radians per second (rad s⁻¹) |
Rearranged: I = 2Eₖ ÷ ω² | ω = √(2Eₖ ÷ I) | Convert rev min⁻¹ to rad s⁻¹ first: ω = (rev min⁻¹) × 2π ÷ 60. Eₖ ∝ ω², so doubling the angular velocity stores four times the energy.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A wheel with a moment of inertia of 0.40 kg m² rotates at 50 rad s⁻¹. Calculate its rotational kinetic energy.
Show solution
F: Eₖ = ½Iω²
I: Eₖ = ½ × 0.40 × 50²
F: Eₖ is already the subject
A: 500 J
Q2 (Core) A flywheel with a moment of inertia of 2.5 kg m² rotates at 120 rad s⁻¹. Calculate its rotational kinetic energy in kJ.
Show solution
F: Eₖ = ½Iω²
I: Eₖ = ½ × 2.5 × 120²
F: Eₖ is already the subject
A: 18 000 J = 18 kJ
Q3 (Core) A flywheel stores 45 kJ of rotational kinetic energy when it rotates at 2400 rev min⁻¹. Calculate its moment of inertia, to 2 significant figures.
Show solution
Convert first: 45 kJ = 45 000 J; ω = 2400 × 2π ÷ 60 = 251.3… rad s⁻¹
F: Eₖ = ½Iω²
I: 45 000 = ½ × I × 251.3…²
F: I = 2 × 45 000 ÷ 251.3…² = 90 000 ÷ 63 165…
A: I = 1.424… kg m² = 1.4 kg m² (2 s.f.)
Watch out: substituting 2400 rev min⁻¹ directly gives an answer about 91 times too small. Always convert to rad s⁻¹.
Q4 (Core) A small grinding wheel has a moment of inertia of 2.0 × 10−3 kg m². Calculate the angular velocity at which it stores 160 J, in rad s⁻¹ and in rev min⁻¹ (to 2 significant figures).
Show solution
F: Eₖ = ½Iω²
I: 160 = ½ × 2.0 × 10−3 × ω²
F: ω = √(2 × 160 ÷ 2.0 × 10−3) = √160 000
A: ω = 400 rad s⁻¹
in rev min⁻¹: 400 × 60 ÷ 2π = 3819.7… = 3800 rev min⁻¹ (2 s.f.)
Q5 (Exam) A flywheel is a uniform disc of mass 40 kg and radius 0.30 m (I = ½MR²). Calculate the energy it stores when rotating at 3600 rev min⁻¹, to 2 significant figures.
Show solution
Step 1 – moment of inertia
I = ½ × 40 × 0.30² = 1.8 kg m²
Convert first: ω = 3600 × 2π ÷ 60 = 120π = 376.9… rad s⁻¹
Step 2 – energy
F: Eₖ = ½Iω²
I: Eₖ = ½ × 1.8 × 376.9…²
F: Eₖ is already the subject
A: 127 910… J = 1.3 × 105 J (2 s.f.)
Q6 (Exam) A flywheel with a moment of inertia of 0.60 kg m² slows from 4500 rev min⁻¹ to 3000 rev min⁻¹ as it supplies energy to a machine. Calculate the energy released, to 2 significant figures.
Show solution
Convert first: ω₁ = 4500 × 2π ÷ 60 = 471.2… rad s⁻¹; ω₂ = 3000 × 2π ÷ 60 = 314.1… rad s⁻¹
F: ΔEₖ = ½Iω₁² − ½Iω₂² = ½I(ω₁² − ω₂²)
I: ΔEₖ = ½ × 0.60 × (471.2…² − 314.1…²)
F: ΔEₖ = 0.30 × (222 066… − 98 696…)
A: 37 011… J = 3.7 × 104 J (2 s.f.)
Watch out: ½I(ω₁ − ω₂)² is not the energy change. It gives 7402 J, five times too small. Find each energy, then subtract.
Q7 (Exam) An electric motor with a useful power output of 1.5 kW brings a flywheel of moment of inertia 3.5 kg m² from rest to 2000 rev min⁻¹. Assuming all the useful output is stored by the flywheel, calculate the time taken, to 2 significant figures.
Show solution
Convert first: ω = 2000 × 2π ÷ 60 = 209.4… rad s⁻¹; 1.5 kW = 1500 W
Step 1 – energy stored
F: Eₖ = ½Iω²
I: Eₖ = ½ × 3.5 × 209.4…²
A: Eₖ = 76 763… J
Step 2 – time
t = E ÷ P = 76 763… ÷ 1500 = 51.17… s
A: 51 s (2 s.f.)
In reality friction in the bearings means it takes longer.
Q8 (Exam) A 1200 kg car fitted with a flywheel energy recovery system brakes from 20 m s⁻¹ to 10 m s⁻¹. Half of the car’s lost kinetic energy is transferred to a flywheel of moment of inertia 0.020 kg m², which was already rotating at 20 000 rev min⁻¹. Calculate the new angular velocity of the flywheel in rev min⁻¹, to 2 significant figures.
Show solution
Step 1 – kinetic energy lost by the car
ΔE = ½ × 1200 × (20² − 10²) = 180 000 J, so the flywheel gains 90 000 J
Step 2 – flywheel’s initial energy
ω₁ = 20 000 × 2π ÷ 60 = 2094.3… rad s⁻¹; Eₖ = ½ × 0.020 × 2094.3…² = 43 864… J
Step 3 – new angular velocity
new Eₖ = 43 864… + 90 000 = 133 864… J
F: Eₖ = ½Iω²
I: 133 864… = ½ × 0.020 × ω²
F: ω = √(2 × 133 864… ÷ 0.020) = 3658.7… rad s⁻¹
A: 3658.7… × 60 ÷ 2π = 34 938… = 3.5 × 104 rev min⁻¹ (2 s.f.)
Watch out: add the energies, not the angular velocities. The flywheel already had energy before braking began.
Q9 (Stretch) A uniform solid cylinder of mass 2.0 kg and radius 0.050 m (I = ½mr²) is released from rest and rolls without slipping down a slope, descending a vertical height of 0.80 m. Ignore energy losses. (a) Show that I = 2.5 × 10−3 kg m². (b) Calculate its speed at the bottom, to 2 significant figures. (c) Calculate its angular velocity at the bottom, to 2 significant figures. (d) Explain why it is slower than a block sliding down a frictionless slope of the same height. (g = 9.81 m s⁻²)
Show solution
Step 1 – (a) moment of inertia
I = ½ × 2.0 × 0.050² = 2.5 × 10−3 kg m²
Step 2 – energy equation
mgh = ½mv² + ½Iω², and rolling without slipping means v = ωr, so ½Iω² = ½(½mr²)(v ÷ r)² = ¼mv²
mgh = ½mv² + ¼mv² = ¾mv², so v² = 4gh ÷ 3
Step 3 – (b) speed
I: v² = 4 × 9.81 × 0.80 ÷ 3 = 10.464
A: v = 3.234… m s⁻¹ = 3.2 m s⁻¹ (2 s.f.)
Step 4 – (c) angular velocity
ω = v ÷ r = 3.234… ÷ 0.050 = 64.69… rad s⁻¹
A: 65 rad s⁻¹ (2 s.f.)
Step 5 – (d)
the same 15.696 J of gravitational potential energy is shared: one third becomes rotational kinetic energy, so less is left as translational kinetic energy. A sliding block would reach √(2gh) = 3.96… m s⁻¹.
Watch out: the mass cancels, so every uniform solid cylinder reaches the same speed, whatever its mass or radius.
Q10 (Stretch) A grid energy storage flywheel is a carbon-fibre ring of mass 120 kg and radius 0.30 m, modelled as a thin ring (I = MR²). Its maximum safe speed is 16 000 rev min⁻¹, and it is never allowed to fall below half this speed. (a) Calculate the maximum energy stored, to 2 significant figures. (b) Calculate the usable energy, to 2 significant figures. (c) Calculate how long the flywheel could supply a steady 25 kW, to 2 significant figures. (d) Suggest why the maximum speed is limited.
Show solution
Step 1 – moment of inertia and ω
I = 120 × 0.30² = 10.8 kg m²; ωmax = 16 000 × 2π ÷ 60 = 1675.5… rad s⁻¹
Step 2 – (a) maximum energy
F: Eₖ = ½Iω²
I: Eₖ = ½ × 10.8 × 1675.5…²
A: 1.515… × 107 J = 1.5 × 107 J (2 s.f.)
Step 3 – (b) usable energy
at half speed Eₖ ∝ ω² gives one quarter of the maximum, so usable energy = ¾ × 1.515… × 107 = 1.136… × 107 J
A: 1.1 × 107 J (2 s.f.)
Step 4 – (c) time
t = E ÷ P = 1.136… × 107 ÷ 25 000 = 454.7… s
A: 450 s (2 s.f.), about 7.6 minutes
Step 5 – (d)
the rim moves at v = ωr = 1675.5… × 0.30 = 502.6… m s⁻¹. The stress in the material increases with speed, so above a certain ω the flywheel would break apart; strong, low-density materials such as carbon fibre allow higher speeds.
Watch out: half the speed does not mean half the energy. Because Eₖ ∝ ω², 75% of the stored energy is released between full and half speed.
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