Flywheel Energy Storage Calculator

Compute stored rotational energy, required RPM, tip speed, and specific energy for any flywheel geometry.

⚙️ Flywheel Energy Storage Calculator
Flywheel Shape
Flywheel Mass10.0 kg
kg
1 kg1000 kg
Outer Radius0.200 m
m
0.05 m1.0 m
Spin Speed3,000 RPM
RPM
100 RPM60,000 RPM
Flywheel Shape
Flywheel Mass20.0 kg
kg
1 kg1000 kg
Outer Radius0.250 m
m
0.05 m1.0 m
Target Energy0.0050 kWh
kWh
0.001 kWh10 kWh
Stored Energy
Formula Used
Energy (J)
Energy (kJ)
Energy (kWh)
Moment of Inertia
Angular Velocity
Spin Speed (RPM)
Tip Speed
Specific Energy

⚙️ What is a Flywheel Energy Storage Calculator?

Flywheel energy storage is a mechanical method of storing kinetic energy by spinning a rotor at high speed. A flywheel energy storage calculator computes how much energy is locked in a rotating mass based on the rotor geometry, mass, and spin speed using the formula E = half I omega squared, where I is the moment of inertia and omega is the angular velocity in radians per second.

Flywheels are used in a wide range of real-world applications. In data centers and hospitals, flywheel uninterruptible power supplies (UPS) replace lead-acid battery banks, providing bridge power for 10 to 30 seconds until a diesel generator starts. In rail and bus transit systems, flywheels capture regenerative braking energy and release it during acceleration, cutting energy consumption by 20 to 30 percent. Grid operators deploy large flywheel arrays for frequency regulation, absorbing surplus energy within milliseconds. Industrial presses and stamping machines use flywheels to smooth out cyclic power demand spikes.

A common misconception is that heavier flywheels always store more energy. In fact, energy scales with the square of angular velocity, so doubling RPM at the same mass quadruples stored energy. This is why modern composite flywheels spin at 20,000 to 60,000 RPM rather than the 1,000 to 3,000 RPM of traditional cast-iron industrial flywheels. The shape factor k also matters: a thin-ring design concentrates mass at the rim and achieves k = 1.0, storing twice as much energy per kilogram of rotor material compared to a solid disk at k = 0.5.

This calculator handles both modes engineers actually use: computing stored energy from a known geometry and spin speed, and working backwards to find the RPM required to store a target energy. It also reports tip speed (the tangential velocity at the rim), which is the key stress parameter that limits flywheel design, and specific energy in both J/kg and Wh/kg for comparing flywheel performance against batteries and other storage technologies.

📐 Formula

E  =  ½ × I × ω²  =  ½ × k × m × r² × ω²
E = stored rotational kinetic energy (Joules)
I = moment of inertia (kg·m²) = k × m × r²
k = shape factor: 0.5 solid disk, 1.0 thin ring, 0.625 thick ring (r_i = 0.5 r_o), 0.4 solid sphere
m = flywheel mass (kg)
r = outer radius (m)
ω = angular velocity (rad/s) = RPM × 2π ÷ 60
v_tip = tip speed (m/s) = ω × r
Example: Solid disk, 10 kg, 0.2 m radius, 3,000 RPM: I = 0.5 × 10 × 0.04 = 0.2 kg·m²; ω = 314.16 rad/s; E = 0.5 × 0.2 × 314.16² = 9,870 J

📖 How to Use This Calculator

Steps

1
Select the flywheel shape from the dropdown: solid disk or cylinder (most common, k = 0.5), thin ring or hoop (k = 1.0), thick ring with inner radius equal to half the outer radius (k = 0.625), or solid sphere (k = 0.4).
2
Enter flywheel mass and outer radius using the sliders or type directly into the fields. Use the actual rotor mass (excluding housing and bearings) and the maximum outer radius of the spinning part.
3
Choose your mode: in Energy Storage mode, enter the operating RPM to find stored energy. In Find RPM mode, switch tabs and enter your target stored energy in kWh to get the required spin speed.
4
Click Calculate to see stored energy in J, kJ, and kWh, moment of inertia, angular velocity in rad/s, RPM, tip speed in m/s, and specific energy in J/kg and Wh/kg.

💡 Example Calculations

Example 1 — Small Solid Disk at 3,000 RPM

Solid disk flywheel, 10 kg, 0.20 m radius, 3,000 RPM

1
Shape factor for solid disk: k = 0.5. Moment of inertia: I = 0.5 × 10 kg × (0.20 m)² = 0.5 × 10 × 0.04 = 0.200 kg·m².
2
Convert RPM to angular velocity: ω = 3,000 × 2π ÷ 60 = 314.16 rad/s. Tip speed: v = 314.16 × 0.20 = 62.83 m/s.
3
Stored energy: E = 0.5 × 0.200 × 314.16² = 0.1 × 98,696 = 9,870 J = 9.870 kJ = 0.002742 kWh.
Stored Energy = 9,870 J (9.87 kJ) | Tip Speed = 62.8 m/s | Specific Energy = 987 J/kg (0.274 Wh/kg)
Try this example →

Example 2 — Industrial Thin-Ring Flywheel at 6,000 RPM

Thin-ring flywheel, 50 kg, 0.40 m radius, 6,000 RPM

1
Shape factor for thin ring: k = 1.0. Moment of inertia: I = 1.0 × 50 × (0.40)² = 50 × 0.16 = 8.00 kg·m².
2
Angular velocity: ω = 6,000 × 2π ÷ 60 = 628.32 rad/s. Tip speed: v = 628.32 × 0.40 = 251.3 m/s (within steel design limits).
3
Stored energy: E = 0.5 × 8.00 × 628.32² = 4.0 × 394,786 = 1,579,144 J = 1.579 MJ = 0.4387 kWh.
Stored Energy = 1.579 MJ (0.439 kWh) | Tip Speed = 251 m/s | Specific Energy = 31,583 J/kg (8.77 Wh/kg)
Try this example →

Example 3 — Find RPM for a Target Energy

Solid disk flywheel, 20 kg, 0.25 m radius, target 0.005 kWh

1
Switch to Find RPM mode. Target energy: E = 0.005 kWh = 0.005 × 3,600,000 = 18,000 J. Moment of inertia: I = 0.5 × 20 × (0.25)² = 0.5 × 20 × 0.0625 = 0.625 kg·m².
2
Required angular velocity: ω = √(2 × 18,000 ÷ 0.625) = √(57,600) = 240.0 rad/s.
3
Convert to RPM: RPM = 240.0 × 60 ÷ (2π) = 2,291.8 RPM. Tip speed: v = 240.0 × 0.25 = 60.0 m/s.
Required RPM = 2,292 RPM | Angular Velocity = 240.0 rad/s | Tip Speed = 60.0 m/s
Try this example →

❓ Frequently Asked Questions

What is the formula for flywheel energy storage?+
The stored rotational kinetic energy is E = half times I times omega squared. Expanding the moment of inertia: E = half times k times m times r squared times omega squared. Here k is the shape factor (0.5 for a solid disk), m is mass in kg, r is outer radius in m, and omega is angular velocity in rad/s, calculated as RPM times 2pi divided by 60.
What shape factor k should I use for my flywheel?+
Use k = 0.5 for a solid disk or cylinder (most industrial flywheels), k = 1.0 for a thin-walled hoop or ring (mass concentrated at the rim), k = 0.625 for a thick ring with inner radius equal to half the outer radius, and k = 0.4 for a solid sphere. Thin-ring designs maximize k, storing the most energy per unit mass when tip speed is the design limit.
How do I convert RPM to rad/s for the flywheel formula?+
Multiply RPM by 2pi and divide by 60: omega (rad/s) = RPM times 2pi divided by 60. For example, 3,000 RPM = 3,000 times 6.2832 divided by 60 = 314.16 rad/s. The calculator does this automatically when you enter RPM.
What is flywheel tip speed and what is the safe limit?+
Tip speed is the tangential velocity at the outermost rim: v = omega times r. It determines centrifugal hoop stress. For structural steel, the practical limit is roughly 200 to 500 m/s before tensile failure risk becomes significant. High-strength carbon-fiber composite rotors can sustain tip speeds above 1,000 m/s, which is why composite designs achieve 10 to 20 times more specific energy than steel ones.
How does flywheel energy compare to a lithium-ion battery?+
A typical steel flywheel delivers 10 to 50 Wh/kg specific energy, while advanced composite flywheels reach 100 to 500 Wh/kg, comparable to lithium-ion batteries at 100 to 265 Wh/kg. However, flywheels offer much higher power density (discharge rate), near-unlimited cycle life, no chemical degradation, and sub-millisecond response time. Batteries have lower self-discharge and lower cost per kWh for long-duration storage.
Why does doubling RPM quadruple the stored flywheel energy?+
Because energy is proportional to omega squared. E = half times I times omega squared, so if omega doubles (2x RPM), omega squared increases by a factor of 4, and stored energy quadruples. Similarly, doubling the radius also quadruples energy because I = k times m times r squared and r squared appears in the formula. This makes high-speed rotors far more efficient than low-speed, heavy ones.
What is the moment of inertia of a flywheel?+
Moment of inertia (I) measures the resistance of a rotating body to changes in angular velocity. For a flywheel it is I = k times m times r squared, where k depends on geometry. A 10 kg solid disk with 0.2 m radius has I = 0.5 times 10 times 0.04 = 0.2 kg times m squared. Larger I means more energy stored per unit of angular velocity squared, but also requires more torque to spin up or slow down.
What are common uses of flywheel energy storage systems?+
Flywheel energy storage is used in: data center UPS systems (bridge power for 10-30 s), regenerative braking on trains and hybrid buses, grid-frequency regulation (rapid response to demand fluctuations), hospital and airport backup power, industrial stamping presses and punching machines (smoothing power demand spikes), and formula racing KERS (kinetic energy recovery systems).
Why do modern flywheels spin in a vacuum with magnetic bearings?+
At high RPM, aerodynamic drag losses become significant and generate heat. Evacuating the flywheel housing to a few millibars eliminates air drag almost entirely. Magnetic bearings (active or passive) eliminate mechanical contact, removing lubrication needs and friction losses. Together, these measures allow round-trip energy efficiencies of 85 to 95 percent and maintenance-free operational lifetimes of 20 years or more.
How long can a flywheel store energy before it runs down?+
Self-discharge depends on bearing friction and aerodynamic drag. A conventional ball-bearing flywheel in air loses significant energy within minutes to hours. A vacuum-housed magnetic-bearing flywheel loses roughly 1 to 5 percent of stored energy per hour. This makes flywheels best suited for short-duration, high-cycle storage (seconds to minutes) rather than long-duration overnight or multi-day storage, where batteries or pumped hydro are preferred.
How do I size a flywheel to store a specific number of kWh?+
Use the Find RPM mode: enter your flywheel mass, outer radius, and shape, then enter the target energy in kWh. The calculator returns the required RPM and angular velocity. Check that the resulting tip speed is within the material limit for your rotor (typically under 500 m/s for steel, up to 1,000 m/s for composite). If tip speed is too high, increase mass or radius and recalculate.