Picture a heavy wheel. It spins. It stores push. It keeps things smooth. Why does a flywheel help? Because it stores energy and gives it back. That push in a spin ties to torque. We call this flywheel torque.
Flywheel torque is the twist you need to speed up a flywheel. It is also the twist a flywheel can give when speed falls. The wheel has inertia.
It resists quick change. A Flywheel Torque Calculator helps you find that twist fast. You add mass data, size, and speed change. You get torque. You avoid guess work. You save time.
How to calculate flywheel torque
Let us work with a real case. I am an engineer in a small workshop. I guide a learner near a test bench. We have a steel flywheel. It is a solid disk. It has a mass of 12 kg. It has a radius of 0.15 m. We want the wheel to speed up from 200 rpm to 800 rpm in 4 seconds. We ask: what average torque do we need to make this happen? We also ask: how much energy does the flywheel store at 800 rpm?
Step-by-step calculation with formula
Step 1: Know key formulas.
- Moment of inertia for a solid disk: I = (1/2) m r^2
- Angular speed: ω = 2π × RPM / 60
- Angular acceleration: α = (ω2 − ω1) / Δt
- Torque for spin-up: T = I × α
- Kinetic energy in a flywheel: E = (1/2) I ω^2
Step 2: Gather numbers.
- m = 12 kg
- r = 0.15 m
- RPM1 = 200
- RPM2 = 800
- Δt = 4 s
Step 3: Compute moment of inertia.
- I = 0.5 × 12 × (0.15)^2
- (0.15)^2 = 0.0225
- I = 0.5 × 12 × 0.0225
- I = 6 × 0.0225
- I = 0.135 kg·m^2
Step 4: Convert rpm to rad/s.
- ω1 = 2π × 200 / 60
- ω1 = 2π × 3.333…
- ω1 ≈ 6.6667π ≈ 20.944 rad/s
- ω2 = 2π × 800 / 60
- ω2 = 2π × 13.333…
- ω2 ≈ 26.6667π ≈ 83.776 rad/s
Step 5: Compute angular acceleration.
- α = (ω2 − ω1) / Δt
- α = (83.776 − 20.944) / 4
- α = 62.832 / 4
- α = 15.708 rad/s^2
Step 6: Compute required average torque.
- T = I × α
- T = 0.135 × 15.708
- T ≈ 2.1206 N·m
Step 7: Compute stored energy at 800 rpm.
- E = 0.5 × I × ω2^2
- ω2^2 ≈ (83.776)^2 ≈ 7018.5
- E = 0.5 × 0.135 × 7018.5
- E = 0.0675 × 7018.5
- E ≈ 474.75 J
Step 8: Extra check for power during spin-up.
- Average power during ramp: P = T × ω_avg
- ω_avg = (ω1 + ω2)/2 = (20.944 + 83.776)/2 = 52.36 rad/s
- P ≈ 2.1206 × 52.36 ≈ 111.0 W
Notes that help:
- Bigger mass or radius increases I. Torque goes up.
- Faster ramp (shorter time) makes α larger. Torque goes up.
- Stored energy rises with ω^2. Speed is king.
Why we care here:
- Our motor must give at least 2.12 N·m on top of any load torque during the ramp.
- The system stores about 475 J at 800 rpm, so it can smooth dips in supply power.
FAQs
Does flywheel torque change at steady speed?
If speed stays constant and no losses act, net torque is near zero. In real life, friction and load ripple act, so the flywheel gives and takes small torque to keep speed steady.
Which formula should I pick for I?
Use the right shape. Solid disk: I = 0.5 m r^2. Thin rim: I = m r^2. For a custom shape, sum parts or use the parallel axis rule.
How do I size a flywheel for energy?
Pick E target first. Then choose I and speed: E = 0.5 I ω^2. Work within safe stress and max rpm limits.
Final words and a short trick:
- Manual trick: spin-up torque T ≈ I × Δω / Δt. Energy at speed E ≈ 0.5 I ω^2. For a solid disk, I ≈ 0.5 m r^2. Convert rpm to rad/s with ω ≈ 2π × rpm / 60.
- Why use a calculator: it keeps units clean, handles shape options, tests ramp times fast, and prevents under-sizing a motor or over-stressing a rim.