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Torque for Angular Acceleration

This is the same as the rotational torque formula: torque equals moment of inertia times angular acceleration. It tells you how much torque a motor must provide to spin something up. For example, if a robot arm has an inertia of 0.1 kg·m² and you want it to accelerate at 10 rad/s², you need 1 N·m of torque. It is used to size motors and gearboxes for robot joints, conveyors, and rotating machinery.

Robotics Dynamics Calculator
VORZEP · ROTATIONAL DYNAMICS TOOLS

Torque for Angular Acceleration Calculator

Solve the rotational dynamics relation T = I · α for any of its three unknowns — torque, moment of inertia, or angular acceleration — with step-by-step math, a live rotational schematic, and companion dynamics metrics.

T = I · α
Solve for
Torque (T) Rotational force that causes angular acceleration. Typical: 0.1–1 N·m for small servo motors, 1–10 N·m for robot joints, 50–300 N·m for car wheel motors, 1,000+ N·m for industrial machinery.
N·m
Torque must be greater than zero.
= 20.00 N·m = 14.75 ft·lbf = 2.039 kgf·m
Moment of Inertia (I) Resistance of a body to angular acceleration; depends on mass distribution about the axis. Typical: 0.001–0.01 kg·m² for small robot links, 0.1–1 kg·m² for human limbs, 10–50 kg·m² for car wheels, 1,000+ kg·m² for industrial flywheels.
kg·m²
Moment of inertia must be greater than zero.
= 2.000 kg·m² = 20,000 kg·cm² = 47.45 lb·ft²
Angular Acceleration (α) Rate of change of angular velocity. Typical: 0.1–1 rad/s² for gentle motion, 1–10 rad/s² for responsive robot joints, 10–100 rad/s² for fast servo axes, 1,000+ rad/s² for hard disk drive actuators.
rad/s²
Angular acceleration must be greater than zero.
= 10.00 rad/s² = 572.96 °/s² = 1.592 rev/s²
TORQUE (T)
20.00 N·m
20.00 N·m · 14.75 ft·lbf · 2.039 kgf·m
MODERATE — ROBOT JOINT / SMALL MOTOR
T = I · α
Step-by-step solution
Torque vs. angular acceleration
Current I I × 0.5 I × 2.0
Torque vs. moment of inertia
Current α α × 0.5 α × 2.0
Sensitivity
Companion rotational metrics
Rotational dynamics — live schematic
ROTATIONAL DYNAMICS · T = I · α axis I = 2.000 kg·m² T = 20.00 N·m α = 10.00 rad/s² ROTATION STATE T = 20.00 N·m I = 2.000 kg·m² α = 10.00 rad/s² T = I · α = 2.000 × 10.00 = 20.00 N·m
Applied torque Angular acceleration Mass distribution (I)
What the rotational dynamics relation really measures
Ttorque
=
Imoment of inertia
×
αangular acceleration

Torque is the rotational analog of force

Just as force causes linear acceleration, torque causes angular acceleration. The relation T = Iα is the rotational form of Newton's second law, F = ma. Torque is the product of force and lever arm (T = F · r), so a small force applied far from the axis produces the same torque as a large force applied close in. This is why a long wrench makes it easier to loosen a bolt, and why robot arm joints are placed at the base to minimize the torque required.

Moment of inertia is rotational mass

I plays the role of mass in rotational motion: it measures how much torque is needed to produce a given angular acceleration. Unlike mass, which depends only on how much matter there is, I depends on how that matter is distributed about the rotation axis. Moving mass farther from the axis increases I by the square of the distance (I = Σmᵢrᵢ²). This is why a hollow cylinder has a larger I than a solid one of the same mass and radius, and why a figure skater spins faster when pulling her arms in.

Angular acceleration is rate of change of spin

α = dω/dt is how quickly the angular velocity changes. It is measured in rad/s² and is related to the linear tangential acceleration by a_t = α · r. A robot arm that must reach its target in 0.5 s from rest needs a large α at the start and a large negative α to stop — both require torque. In practice, α is limited by the available torque and the inertia, and by structural and thermal limits on the motor and gearbox.

Gearboxes trade torque for speed

A gearbox with ratio N multiplies torque by N and divides angular speed by N (ignoring losses). This is why small, fast motors can drive heavy, slow robot joints: the motor spins quickly with modest torque, and the gearbox converts that into slow rotation with large torque. The inertia seen by the motor is divided by N², so a high ratio also makes the load easier to accelerate. But gearboxes add friction, backlash, and inertia of their own, so the choice of ratio is a trade-off between torque, speed, precision, and efficiency.

How to Use This Calculator

The rotational dynamics relation links three quantities: T, I, and α. Pick which one you want the tool to solve, then fill in the other two.

1. Choose what to solve for

Pick T, I, or α from the toggle. The calculator rearranges T = I · α automatically for whichever you choose.

2. Use realistic presets

Presets cover typical values: torque from 0.5 N·m (small servo) to 500 N·m (industrial), inertia from 0.01 kg·m² (small link) to 100 kg·m² (flywheel), and angular acceleration from 0.5 rad/s² to 200 rad/s².

3. Match units carefully

Use SI units throughout: T in N·m, I in kg·m², α in rad/s². If you have I in kg·cm², divide by 10,000. If you have I in lb·ft², multiply by 0.04214. The hint under each field shows the value in alternative units.

4. Read the result and verdict

The result card shows the solved value plus equivalent units. A colour-coded verdict classifies the torque from very low (tiny servo) through moderate (robot joint) to very high (industrial machinery).

5. Check the schematic and charts

The schematic shows the rotating disk, the applied torque, the angular acceleration, and the mass distribution that defines I. The chart plots torque against α for three inertia values.

6. Switch to Advanced for depth

Advanced mode adds a chart of T vs. I, a sensitivity check on a +1% input change, and companion metrics including angular momentum, rotational kinetic energy, and spin-up time.

What This Tool Offers

Built around the rotational dynamics relation T = I · α, the fundamental link between torque, moment of inertia, and angular acceleration.

Three-way solving

Solve for torque, moment of inertia, or angular acceleration. The calculator rearranges the formula for whichever variable you choose.

Clean rotational schematic

A rotating disk with the applied torque, angular acceleration, and mass distribution that defines the moment of inertia.

Torque-class verdict

Each result is classified from very low (tiny servo) through moderate (robot joint) to very high (industrial), with a coloured badge.

Multi-unit hints

Each field shows the value converted to related units — ft·lbf and kgf·m for torque; kg·cm² and lb·ft² for inertia; °/s² and rev/s² for α.

Dual comparison charts

T vs. α for three inertia values, plus an advanced T vs. I chart for three α values.

Reference tables

Typical torque by application, moments of inertia for common shapes, and a full variable reference table.

Reference Tables & Conversions

A torque unit converter, a T = Iα lookup grid, and typical values.

Torque unit converter

N·m
ft·lbf
kgf·m
in·lbf
dyn·cm
= 20.00 N·m = 14.75 ft·lbf = 2.039 kgf·m = 177.0 in·lbf = 2.000×10⁸ dyn·cm

Torque Lookup at α = 10 rad/s²

Computed from T = I · α. Values in N·m.

I (kg·m²)α = 1 rad/s²α = 5 rad/s²α = 10 rad/s²α = 50 rad/s²α = 200 rad/s²

Typical Torques by Application

Representative values for motors, joints, and machinery.

ApplicationTypical T (N·m)Typical I (kg·m²)Typical α (rad/s²)Notes
Small hobby servo0.1 – 0.50.0001 – 0.00150 – 200RC models, small robots
Robot arm joint (small)1 – 50.001 – 0.0120 – 100Desktop robot arms
Robot arm joint (industrial)50 – 5000.1 – 52 – 206-axis industrial robots
Drone propeller motor0.5 – 30.0001 – 0.001500 – 2000High-speed brushless
Electric car wheel motor200 – 40010 – 2020 – 50Direct-drive hub motor
Car engine crankshaft100 – 4000.1 – 0.5100 – 500Combustion pulses
Wind turbine rotor10⁵ – 10⁶10⁵ – 10⁷0.001 – 0.01Very large, slow
Industrial flywheel10³ – 10⁵10² – 10⁴0.1 – 10Energy storage
Hard disk drive actuator0.001 – 0.0110⁻⁷ – 10⁻⁶10³ – 10⁴Extremely fast, tiny

Moments of Inertia for Common Shapes

ShapeMoment of inertia INotes
Point mass at radius rm · r²Simplest case
Solid cylinder (radius r)½ · m · r²Axis through center
Hollow cylinder (radius r)m · r²Thin-walled, axis through center
Solid sphere (radius r)⅖ · m · r²Axis through center
Hollow sphere (radius r)⅔ · m · r²Thin shell
Thin rod (length L, center axis)1/12 · m · L²Perpendicular to rod
Thin rod (length L, end axis)⅓ · m · L²About one end
Rectangular plate (a × b, center)1/12 · m · (a² + b²)Perpendicular to plate
Solid disk (thickness t)½ · m · r²Same as solid cylinder
Annular disk (inner r₁, outer r₂)½ · m · (r₁² + r₂²)Ring with thickness

Variable Reference

SymbolMeaningTypical rangeUnits
TTorque10⁻³ – 10⁶N·m
IMoment of inertia10⁻⁷ – 10⁷kg·m²
αAngular acceleration10⁻³ – 10⁴rad/s²
ωAngular velocity—rad/s
LAngular momentum (L = Iω)—kg·m²/s
KErotRotational kinetic energy (½Iω²)—J
PRotational power (P = Tω)—W

What Does T = I · α Measure?

The rotational dynamics relation gives the torque required to produce a given angular acceleration of a body with a given moment of inertia. It is the rotational form of Newton's second law, F = ma, and the starting point for analyzing any rotating system: robot joints, wheels, turbines, flywheels, and gyroscopes. The relation is exact for rigid bodies rotating about a fixed axis and is the basis for motor sizing, gearbox selection, and motion control design.

Torque Formula

The defining relationship is:

T = I · α

Rearranged to solve for the other two unknowns:

I = T / α
α = T / I

The torque can also be written in terms of force and lever arm:

T = F · r · sin(θ)

Companion quantities in rotational dynamics:

L = I · ω (angular momentum)
KErot = ½ · I · ω² (rotational kinetic energy)
P = T · ω (rotational power)

The equation of motion for a rotating body with damping and an external load is:

T − Tload − b·ω = I · α

Rotational Dynamics Variables

SymbolMeaningNotes
TTorqueRotational force; product of force and lever arm
IMoment of inertiaRotational mass; depends on mass distribution about the axis
αAngular accelerationRate of change of angular velocity; dω/dt
ωAngular velocityRate of rotation; rad/s
LAngular momentumL = Iω; conserved when net torque is zero
KErotRotational kinetic energy½Iω²; energy stored in rotation
PRotational powerP = Tω; rate of work done by torque

How Each Variable Affects Torque

Torque is directly proportional to both moment of inertia and angular acceleration. Doubling the inertia doubles the torque required for the same α; doubling α doubles the torque for the same inertia. In motor sizing, this is the fundamental constraint: the motor must supply enough torque to accelerate the load at the desired rate, plus enough to overcome friction, gravity, and any external load. The peak torque requirement usually occurs during acceleration, not during steady-state operation.

The moment of inertia depends on both the mass and its distribution about the axis. Moving mass farther from the axis increases I by the square of the distance (I = Σmᵢrᵢ²). This is why a robot arm with the payload at the end has a much larger I than one with the payload near the joint, and why it takes much more torque to accelerate a long arm than a short one. In practice, designers minimize I by placing heavy components (motors, batteries) near the rotation axis and by using lightweight materials for the outer parts of the arm. Gearboxes help by dividing the reflected load inertia by N², making the load easier for the motor to accelerate.

Torque Calculation Examples

Given: I = 2 kg·m², α = 10 rad/s² — solve for T
T = 2 × 10 = 20 N·m
Given: T = 20 N·m, α = 10 rad/s² — solve for I
I = 20 / 10 = 2 kg·m²
Given: T = 20 N·m, I = 2 kg·m² — solve for α
α = 20 / 2 = 10 rad/s²
Given: solid cylinder, m = 5 kg, r = 0.2 m — solve for I
I = ½mr² = ½ × 5 × 0.2² = 0.1 kg·m²
Given: I = 0.1 kg·m², α = 50 rad/s² — solve for T
T = 0.1 × 50 = 5 N·m
Given: I = 2 kg·m², ω = 30 rad/s — solve for rotational KE
KErot = ½ × 2 × 30² = 900 J

Frequently Asked Questions

Force is a push or pull that tends to cause linear motion; torque is a twist that tends to cause rotation. Torque is the product of force and the perpendicular distance from the axis (T = F · r · sinθ). A force applied directly through the axis produces no torque; the same force applied at a distance r produces torque T = F · r. This is why door handles are placed far from the hinges and why a long wrench is easier to turn than a short one. Both force and torque are vectors, but torque has units of N·m (not J), because it is a vector quantity and not energy.

Because I = Σmᵢrᵢ², where rᵢ is the perpendicular distance from each mass element to the axis. If you change the axis, the distances change, so I changes. A thin rod has a small I about its long axis (most mass is close to the axis) but a much larger I about an axis through its center perpendicular to its length (most mass is far from the axis). The parallel-axis theorem relates the two: I = I_cm + m·d², where I_cm is the moment about the center of mass and d is the distance between the axes. This is why the choice of axis must always be specified when quoting I.

For a composite body, sum the moments of inertia of its parts about the same axis. Use the parallel-axis theorem to shift each part's I from its own center of mass to the common axis: I = Σ(I_cm,i + mᵢ · dᵢ²). For CAD models, most software can compute I directly from the geometry and density. For hand calculations, break the body into simple shapes (cylinders, plates, rods), look up their formulas, shift to the common axis, and add. If the body has a complex shape that cannot be decomposed easily, use a physical pendulum test or a torsion pendulum to measure I experimentally.

Rotational power is P = T · ω, the product of torque and angular velocity. A motor that produces 10 N·m at 100 rad/s delivers 1,000 W (1 kW) of mechanical power. A motor that produces 100 N·m at 10 rad/s delivers the same power. This is why gearboxes are so useful: they trade speed for torque without changing power (except for losses). The peak power of a motor occurs at the point where T · ω is maximized, which is usually at about half the no-load speed. For a given power rating, you can have high torque and low speed, or low torque and high speed, but not both.

Motor sizing involves four steps: (1) estimate the load inertia I_load about the joint axis, including the link, payload, and any gearing reflected to the motor side; (2) decide the required angular acceleration α for the motion profile (often from a trapezoidal velocity profile and the time available); (3) compute the acceleration torque T_acc = I · α, then add friction torque T_fric, gravity torque T_grav (which depends on the arm angle), and any external load torque; (4) add a safety factor of 1.5–2.0 for peak torque, and check that the motor's continuous torque rating exceeds the RMS torque over the duty cycle to avoid overheating. The required power is then P = T · ω at the operating point.

Torque is the cause of change in rotation; angular momentum is the quantity of rotation. The relationship is τ = dL/dt, the rotational analog of F = dp/dt. Angular momentum is L = I · ω and is conserved when the net external torque is zero. This conservation law explains why a spinning figure skater speeds up when pulling her arms in (I decreases, so ω increases to keep L constant), why a gyroscope maintains its orientation, and why a satellite can change its attitude by spinning internal flywheels. Torque, in contrast, is what you apply to change L — it is a force-like quantity, not a conserved one.

Yes. Torque is a vector, and its sign indicates the direction of rotation (usually positive for counterclockwise, negative for clockwise, by the right-hand rule). A negative torque produces a negative angular acceleration, which either slows down a positive rotation or speeds up a negative one. In motor sizing, both signs matter: a motor must produce positive torque to accelerate in one direction and negative torque (or a brake) to decelerate. The magnitude |T| determines how quickly the motion changes; the sign determines which way. In this calculator, we use magnitudes throughout for simplicity, but in a full dynamic analysis the signs must be tracked.

Calculating...

Variables

Symbol Quantity Unit
T Required torque N.m
I Moment of inertia of the rotating system kg.m2
alpha Desired angular acceleration rad/s2

What it means

This is the rotational equivalent of F = ma. It relates the torque (τ) applied to a body to its moment of inertia (I) and desired angular acceleration (α). In robotics, this is used in inverse dynamics to compute the torque required at each joint to achieve a planned motion. It is also used to size motors and to simulate dynamic behaviour. Understanding this equation is essential for designing robotic systems that can accelerate and move as intended without stalling.

Worked example

Torque for Angular Acceleration – Two Detailed Examples

Real‑World
Scenario: A robotic joint has a moment of inertia I = 2 kg·m² and requires an angular acceleration α = 3 rad/s². The torque required T = I × α = 2 × 3 = 6 N·m. The motor must be able to deliver this torque to achieve the desired acceleration. The robot designer uses this to select a motor and gearbox that can provide sufficient torque for the joint's dynamic requirements.
ParameterValue
I (kg·m²)2
α (rad/s²)3
1T = 2 × 3 = 6 N·m
Result 6 N·m ✓ Required torque
Scenario: A motor drives a load with I = 10 kg·m² and requires α = 0.5 rad/s². The torque is T = 10 × 0.5 = 5 N·m. This lower torque requirement allows a smaller, more efficient motor to be used. The engineer uses this calculation to optimise the system design, balancing performance with cost and energy consumption.
ParameterValue
I10
α0.5
1T = 10 × 0.5 = 5 N·m
Result 5 N·m ✓ Lower torque
Insight: Torque required for angular acceleration is the product of moment of inertia and angular acceleration. This is the rotational analogue of Newton's second law (F = ma).

Common mistakes

  • Same as ID 88 – see that entry.
  • Note: This duplicate uses T instead of τ – ensure you use the same symbol consistently.

Applications

Torque for angular acceleration, T = I·α, relates the torque required to achieve a given angular acceleration for a rotating body. This is used to size motors and actuators for rotational motion. In robotics, it helps select motors for joints and to design gearboxes. By calculating the required torque, engineers can ensure that the actuator can provide sufficient acceleration to meet dynamic performance specifications. This formula is also applied in balancing systems and in vibration analysis. Understanding this relation is critical for designing responsive and robust mechatronic systems.

  • Actuator sizing for robotic joints and turntables
  • Design of gearboxes and transmission systems
  • Dynamic analysis of rotating machinery
  • Control system design for angular motion
  • Simulation of mechanical systems with rotational components

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  • Published July 13, 2026
  • Last updated September 23, 2026