Flywheel Energy Storage Revision Notes | A Level AQA Physics

1. Why Flywheels Matter

Imagine a potter's wheel spinning steadily, or a car engine that would judder violently between each piston stroke without something to keep it turning smoothly. The device that solves both problems is the flywheel — one of the oldest and most elegant energy storage technologies in engineering.

A flywheel stores energy as rotational kinetic energy. When excess energy is available, the flywheel speeds up and absorbs it. When energy is needed, the flywheel slows down and releases it. This makes flywheels ideal for bridging the gap between uneven energy supply and smooth, continuous demand.

At A-level, the key question is: what determines how much energy a flywheel can store, and how do engineers exploit that in real systems? Those two questions map directly onto the two specification requirements for this topic.

2. Rotational Kinetic Energy and the Key Equation

You already know that a moving object has kinetic energy $E_k = \frac{1}{2}mv^2$. For a rotating body, the equivalent expression replaces mass with moment of inertia $I$ and linear speed $v$ with angular velocity $\omega$. The result is the rotational kinetic energy equation.

Notice that $\omega$ is squared. Doubling the angular velocity quadruples the stored energy — exactly the same relationship as linear kinetic energy and speed. This means spinning a flywheel faster is a very powerful way to increase its energy storage.

The moment of inertia $I$ depends on both the total mass of the flywheel and how that mass is distributed relative to the rotation axis. For a uniform solid disc of mass $M$ and radius $R$, the moment of inertia is $I = \frac{1}{2}MR^2$. For a thin ring (all mass at radius $R$), it is $I = MR^2$. The ring stores more energy for the same mass and radius because its mass is further from the axis.

3. Factors Affecting Energy Storage Capacity

The equation $E = \frac{1}{2}I\omega^2$ tells you everything you need to know about what controls how much energy a flywheel can store. There are two routes to increasing $E$: increase $I$, or increase $\omega$. Both have practical limits, and understanding those limits is what separates a good exam answer from a great one.

The moment of inertia $I$ increases with mass and with the square of the radius. Doubling the radius of a flywheel (while keeping mass the same) quadruples $I$, which quadruples the stored energy. This is why large-radius flywheels are preferred when space allows — concentrating mass at the rim (like a bicycle wheel rather than a solid disc) maximises $I$ for a given total mass.

• Greater mass → greater $I$ → greater stored energy at the same $\omega$
• Greater radius → greater $I$ (scales as $R^2$) → greater stored energy at the same $\omega$
• Mass concentrated at the rim → greater $I$ than a solid disc of the same mass and radius
• Greater angular velocity $\omega$ → much greater stored energy (scales as $\omega^2$)
• Material strength limits the maximum $\omega$: at high spin rates, tensile stress in the rim can cause the flywheel to fracture

Material choice is therefore a critical design factor. High-strength materials such as carbon-fibre composites allow much higher angular velocities before fracture, enabling compact, lightweight flywheels to store large amounts of energy. Steel flywheels are cheaper but must spin more slowly, so they tend to be larger and heavier for the same energy storage.

There is also a practical upper limit on radius set by the physical space available in the machine or vehicle. Engineers must balance all four factors — mass, radius, mass distribution, and angular velocity — against constraints of cost, space, and material strength.

4. Calculating Stored Energy: Worked Examples

Theory becomes exam-ready when you can apply it to numbers. The following worked example walks through a standard AQA-style calculation, showing every substitution and unit check explicitly.

Worked example: Calculating the rotational kinetic energy of a flywheel

A flywheel can be modelled as a uniform solid disc of mass 12 kg and radius 0.40 m. It rotates at 3000 revolutions per minute (rpm). Calculate the rotational kinetic energy stored in the flywheel. [4 marks]

1. $\omega = \frac{2\pi \times 3000}{60} = \frac{6000\pi}{60} = 100\pi \approx 314 \text{ rad s}^{-1}$ — [M1] Angular velocity must be in rad s⁻¹. Multiply revolutions per minute by $2\pi$ to get radians per minute, then divide by 60 to convert to per second. Never substitute rpm directly into the energy equation.
2. $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 12 \times (0.40)^2$ $(0.40)^2 = 0.40 \times 0.40 = 0.16$ $I = \frac{1}{2} \times 12 \times 0.16 = 0.96 \text{ kg m}^2$ — [M1] For a uniform solid disc, $I = \frac{1}{2}MR^2$. Expand the square explicitly: $0.40 \times 0.40 = 0.16$. Do not skip this step — squaring errors are the most common source of lost marks here.
3. $E = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.96 \times (100\pi)^2$ $(100\pi)^2 = 100\pi \times 100\pi = 10000\pi^2$ $E = \frac{1}{2} \times 0.96 \times 10000\pi^2$ $E = 0.48 \times 10000\pi^2 = 4800\pi^2$ — [M1] Expand $(100\pi)^2$ as $100\pi \times 100\pi = 10000\pi^2$ before multiplying. Keeping $\pi^2$ exact avoids rounding errors mid-calculation.
4. $E = 4800 \times \pi^2 \approx 4800 \times 9.870 \approx 47\,400 \text{ J}$ — [A1] $\pi^2 \approx 9.870$. The answer is approximately 47 kJ. Always state the unit. A sanity check: a 12 kg disc spinning at 3000 rpm storing ~47 kJ is physically reasonable for an industrial flywheel.

Answer: $E \approx 47\,400 \text{ J}$ (47.4 kJ to 3 significant figures)

Notice how the unit conversion in Step 1 is non-negotiable. The equation $E = \frac{1}{2}I\omega^2$ requires $\omega$ in rad s⁻¹. Substituting rpm directly gives an answer that is wrong by a factor of approximately $(2\pi/60)^2 \approx 1/91$, meaning the result would be about 91 times too small — and the examiner will not award the accuracy mark.

5. Smoothing Torque and Speed

Now that you can calculate stored energy, the next question is: why does that stored energy matter in a real machine? The answer lies in the relationship between torque, angular velocity, and the flywheel's resistance to change.

Many engines and machines deliver power in pulses rather than continuously. A single-cylinder petrol engine, for example, produces a torque spike once every two revolutions during the power stroke, and very little torque during the other strokes. Without intervention, the crankshaft would speed up and slow down dramatically with each cycle — producing vibration, noise, and mechanical stress.

A flywheel attached to the crankshaft resists these changes in angular velocity because of its large moment of inertia. During the power stroke, the flywheel absorbs the excess energy and speeds up slightly. During the other strokes, it releases that energy and slows down slightly. The net effect is a much smoother output speed and torque — the flywheel acts as a mechanical buffer.

The physics here connects directly back to Newton's second law for rotation: $\tau = I\alpha$, where $\tau$ is the net torque and $\alpha$ is the angular acceleration. For a given unbalanced torque $\tau$, a larger $I$ produces a smaller $\alpha$ — meaning the flywheel changes speed more slowly. A heavier, wider flywheel is therefore a more effective speed smoother.

6. Applications: Vehicles and Production Machines

With the physics of torque smoothing established, you can now analyse the two main application contexts the specification requires: energy storage in vehicles, and smoothing in industrial production machines.

In vehicles, flywheels are used in kinetic energy recovery systems (KERS). When a vehicle brakes, the kinetic energy that would otherwise be lost as heat in the brake pads is instead used to spin up a flywheel. When the vehicle accelerates again, the flywheel drives a generator or is mechanically coupled back to the drivetrain, releasing the stored energy. Formula 1 cars used mechanical KERS flywheels spinning at up to 60,000 rpm, storing enough energy to provide a significant power boost for several seconds per lap.

Hybrid buses in some cities use a similar principle. The flywheel spins up during braking at bus stops and releases energy to assist acceleration away from the stop — a cycle that repeats dozens of times per route. Because the energy is stored mechanically rather than chemically, the system responds faster than a battery and does not degrade with repeated charge-discharge cycles in the same way.

In industrial production machines, the challenge is different. Processes such as stamping, pressing, or punching require a large burst of energy over a very short time — far more than the motor driving the machine could deliver continuously. A flywheel stores energy steadily from the motor between strokes, then releases it rapidly during the working stroke. The motor runs at a near-constant load; the flywheel handles the peak demand. This protects the motor, reduces peak current draw, and allows a smaller, cheaper motor to drive a powerful press.

Comparing these applications reveals a design tension: vehicle flywheels favour high angular velocity and low mass (to save weight), while industrial flywheels favour high moment of inertia (to store large amounts of energy and smooth large torque spikes). The same physical principle — $E = \frac{1}{2}I\omega^2$ — underpins both, but the engineering priorities differ.

7. Exam Technique and Common Mistakes

Flywheel questions on AQA papers tend to test three things: correct use of the energy equation with unit conversion, qualitative explanation of how flywheels smooth torque and speed, and evaluation of design factors. Each of these has a characteristic trap.

For extended-response questions asking you to evaluate flywheel design, a strong answer will: open with the energy equation and identify both $I$ and $\omega$ as variables; explain how mass, radius, and mass distribution affect $I$; note that $\omega$ has a material-strength upper limit; and conclude by comparing the trade-offs for the specific context given (vehicle vs industrial).

Together, the two specification requirements for this topic — factors affecting energy storage capacity, and the applications of flywheels for smoothing and energy recovery — are both rooted in the single equation $E = \frac{1}{2}I\omega^2$ and the rotational form of Newton's second law $\tau = I\alpha$. Master those two relationships and you have the tools to answer any flywheel question the examiner sets.