Rotational Dynamics

τ = J × α

Quick answer

This calculator solves τ = J × α for Torque. With Moment of Inertia 0.5 kg·m², Angular Acceleration 1.1 rad/s², the result is 0.55 N·m. It solves in 3 directions, so you can enter the values you have and get the one you need.

Calculator

Torque (N·m)
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Formula

τ = J × α

Description

Newton's second law for rotation: torque equals moment of inertia times angular acceleration. This is the rotational analog of F = ma. The moment of inertia J represents how much torque is needed to achieve a given angular acceleration, analogous to how mass represents resistance to linear acceleration. Higher inertia loads require more torque (and therefore more motor current) to accelerate and decelerate. In servo systems, the motor must provide enough torque to overcome both the load torque and the inertial torque during acceleration phases.

Variables

  • τ — Net torque (N·m)
  • J — Moment of inertia (kg·m²)
  • α — Angular acceleration (rad/s²)

Practical Notes

For optimal servo performance, match the motor inertia to the load inertia (Jload/Jmotor ratio of 1:1 to 5:1). Through a gear reducer, the reflected inertia is Jreflected = Jload/N². Time to reach target speed: t = ω/α = J×ω/τ. Common inertia formulas: solid cylinder J = 0.5×m×r², thin ring J = m×r², solid sphere J = 0.4×m×r².

Symbols and units

Symbol Meaning Default unit Accepted units
tau Torque N·m N·m, mN·m
J Moment of Inertia — kg·m²
alpha Angular Acceleration — rad/s²

Step by step

  1. 1. What you know: Moment of Inertia 0.5 kg·m², Angular Acceleration 1.1 rad/s²
  2. 2. The relation: τ = J × α
  3. 3. Substitute: J = 0.5 kg·m², alpha = 1.1 rad/s²
  4. 4. Result: Torque = 0.55 N·m

Worked examples

Minimum range

Given: Moment of Inertia 0.025 kg·m², Angular Acceleration 0.055 rad/s²

Result: Torque = 1.375 mN·m

Low range

Given: Moment of Inertia 0.25 kg·m², Angular Acceleration 0.55 rad/s²

Result: Torque = 137.5 mN·m

Typical

Given: Moment of Inertia 0.5 kg·m², Angular Acceleration 1.1 rad/s²

Result: Torque = 0.55 N·m

High range

Given: Moment of Inertia 0.8 kg·m², Angular Acceleration 1.76 rad/s²

Result: Torque = 1.408 N·m

Maximum range

Given: Moment of Inertia 8 kg·m², Angular Acceleration 17.6 rad/s²

Result: Torque = 140.8 N·m

Solving for each variable

Solve for Torque

Moment of Inertia 0.025 kg·m², Angular Acceleration 0.055 rad/s² → 1.375 mN·m

Solve for Moment of Inertia

Torque 5 mN·m, Angular Acceleration 0.055 rad/s² → 0.09091 kg·m²

Solve for Angular Acceleration

Torque 5 mN·m, Moment of Inertia 0.025 kg·m² → 0.2 rad/s²

How Torque changes with Moment of Inertia

Moment of Inertia (input) Torque (result)
0.05 kg·m² 55 mN·m
0.125 kg·m² 137.5 mN·m
0.25 kg·m² 0.275 N·m
0.375 kg·m² 0.4125 N·m
0.5 kg·m² 0.55 N·m
0.75 kg·m² 0.825 N·m
1 kg·m² 1.1 N·m
2 kg·m² 2.2 N·m
3.5 kg·m² 3.85 N·m
5 kg·m² 5.5 N·m

The same result in other units

Unit Value
N·m 0.55
mN·m 550

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Frequently asked questions

How do you calculate Rotational Dynamics?

Use τ = J × α. With Moment of Inertia 0.5 kg·m², Angular Acceleration 1.1 rad/s², Torque works out to 0.55 N·m.

What is the formula for Rotational Dynamics?

The formula is τ = J × α. Newton's second law for rotation: torque equals moment of inertia times angular acceleration.

How do you find Moment of Inertia when you know the other values?

Rearrange for Moment of Inertia: with Torque 5 mN·m, Angular Acceleration 0.055 rad/s², Moment of Inertia is 0.09091 kg·m².

How do you find Angular Acceleration when you know the other values?

Rearrange for Angular Acceleration: with Torque 5 mN·m, Moment of Inertia 0.025 kg·m², Angular Acceleration is 0.2 rad/s².

What units does Rotational Dynamics use?

Torque is shown in N·m, mN·m. Values are held in SI base units internally, so you can mix prefixes without converting by hand.

Related Concepts

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