What Is Acceleration to Torque?
Torque and acceleration are two sides of the same coin. Torque is the rotational twist a shaft, motor or wheel delivers; acceleration is how fast an object's velocity changes. Connect them through Newton's second law and you can answer a very practical question: how much torque do I need to accelerate this thing?
That question shows up everywhere — choosing an EV motor, checking whether a winch can drag a load uphill, or working out what wheel torque a car needs to hit its claimed 0–60 time. One compact formula does all of it.
The Formula
Start from two relationships every physics student meets early:
[
F = m \cdot a \quad\text{(Newton's second law)}
]
[
T = F \cdot r \quad\text{(definition of torque)}
]
Substitute one into the other:
[
T = m \cdot a \cdot r
]
Where:
- T: torque (N·m)
- m: mass being accelerated (kg)
- a: linear acceleration (m/s²)
- r: radius from the axis of rotation to where the force acts (m)
Because the units multiply cleanly — kg × m/s² × m = N·m — no conversion factors are needed as long as you work in kilograms, meters and seconds. The result converts easily afterwards: 1 N·m = 0.7376 lb·ft = 0.1020 kgf·m.
Worked Example: Wheel Torque for a Car
A 1,200 kg car accelerates at 3 m/s², and its driven tires have a rolling radius of 0.32 m. What wheel torque does that take?
First find the force:
[
F = m \cdot a = 1200 \times 3 = 3600 \text{ N}
]
Then convert force to torque at the tire contact patch:
[
T = F \cdot r = 3600 \times 0.32 = 1152 \text{ N·m}
]
So roughly 1,150 N·m (≈ 850 lb·ft) of torque must reach the driven wheels. Note that the engine itself can make far less than this, because low gears multiply torque — first gear alone often multiplies by 3–4× before the final drive adds another ~3–4×.
Interpreting the Result
The number T = m·a·r is the torque needed to sustain the acceleration — a demand figure, not a measurement of what an engine produces. Three practical readings follow from it:
| If you want… | Then… |
|---|---|
| Faster acceleration | Torque must rise proportionally — double the acceleration means double the wheel torque. |
| The same acceleration in a heavier vehicle | Torque scales directly with mass; 400 kg extra at 3 m/s² costs another ~384 N·m at a 0.32 m radius. |
| Less torque demand | A larger effective radius lowers the torque needed for the same force — but usually raises the speed the shaft must turn. |
Quick self-check: run the example above through the calculator (mass 1200, acceleration 3, radius 0.32). You should get 1152.00 N·m — if you see that, everything is wired correctly.
Quick Recap
- T = m·a·r comes from combining F = ma with T = F·r.
- Work in kg, m/s² and m and the answer drops straight out in N·m; multiply by 0.7376 for lb·ft.
- The result is the torque required at the rotating component — gear ratios bridge the gap to engine torque.
- Use the calculator above to size motors or sanity-check performance targets.
If you want to move in the other direction — from rotation speed to torque — try the RPM to torque calculator.