Enter the acceleration (m/s^2), the mass (kg), and the radius (m) into the calculator to determine the Torque From Acceleration.
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Acceleration to Torque Formula
The calculator uses one of three formulas depending on the input set you choose.
Inertia + angular acceleration:
T = J * α
Speed ramp:
T = J * (ω_final - ω_start) / t
Linear acceleration at a radius:
T = m * a * r
- T = torque (N·m)
- J = mass moment of inertia (kg·m²)
- α = angular acceleration (rad/s²)
- ω_start, ω_final = initial and final angular speed (rad/s)
- t = acceleration time (s)
- m = mass being accelerated (kg)
- a = linear acceleration (m/s²)
- r = radius or lever arm where the force acts (m)
The first mode is the direct form. Enter the rotating inertia and the angular acceleration and you get the acceleration torque. Use it when α is already known.
The second mode computes α for you from a speed change over a time window, then multiplies by inertia. Use it for motor sizing where you have a target rpm and a ramp time.
The third mode handles linear loads driven through a pulley, drum, or pinion. The mass and linear acceleration give a tangential force, and that force times the radius gives the torque the shaft must deliver. This mode does not include the rotating inertia of the drum or pulley itself, so add that separately if it matters.
Reference Tables
Typical angular acceleration values you can compare against your result:
| Application | Typical α (rad/s²) | Notes |
|---|---|---|
| Conveyor start-up | 2 to 10 | Soft ramp, seconds-long |
| Industrial servo | 100 to 5,000 | Depends on inertia ratio |
| CNC spindle ramp | 50 to 500 | 0 to several thousand rpm |
| Vehicle wheel, hard launch | 20 to 80 | Tire radius limited |
| Hard disk, pickup speed | 200 to 1,000 | Small inertia, fast spin-up |
Unit conversions used by the calculator:
| From | To | Multiply by |
|---|---|---|
| lb·ft² | kg·m² | 0.04214 |
| oz·in² | kg·m² | 1.829e-5 |
| rpm | rad/s | 0.10472 |
| rev/s² | rad/s² | 6.2832 |
| N·m | lb-ft | 0.7376 |
| N·m | oz-in | 141.61 |
Examples and FAQ
Example 1. A motor must spin a load with J = 0.05 kg·m² from 0 to 1,500 rpm in 0.4 seconds. Convert 1,500 rpm to 157.08 rad/s. α = 157.08 / 0.4 = 392.7 rad/s². T = 0.05 × 392.7 = 19.6 N·m of acceleration torque. Add the running torque from friction or process load to size the motor.
Example 2. A 200 kg load is lifted by a drum of radius 0.1 m at 2 m/s². T = 200 × 2 × 0.1 = 40 N·m for the acceleration component. Gravity adds m × g × r = 200 × 9.81 × 0.1 = 196.2 N·m, so the total drum torque during the lift is about 236 N·m.
Does this give peak or continuous torque? It gives the torque required during the acceleration event. That value should be compared to the motor's peak torque rating, not its continuous rating, unless the acceleration is sustained.
Should I include the motor rotor inertia? Yes. Total inertia means the load inertia plus the rotor inertia plus any coupling, gearbox, and shaft inertia reflected to the motor shaft.
How do I reflect inertia through a gearbox? Divide the load-side inertia by the gear ratio squared to get the equivalent inertia at the motor shaft. Then use that value as J.
Why is my torque negative? The final speed is lower than the start speed, or the angular acceleration was entered as negative. The shaft is decelerating. Use the magnitude when sizing a brake or a motor that has to absorb the energy.
