CORE LINKS · DESIGN GUIDES
Rotate a load with a motor
Equal masses can need different torque because their axes and mass distributions differ. Separate steady resistance from the torque needed to change speed.
What you will understand
- Mass alone does not determine rotational inertia
- Separate steady and acceleration torque
- Compare calculated torque with motor limits
Blue: defined conditions · Red: calculated quantities. Diagrams are schematic.
- m, D: disk mass and outside diameter
- n: rotational speed / T: drive torque
- J: mass moment of inertia about the chosen axis
The examples below are illustrative. Each example button loads its fixed values, units and mode. Ordinary calculator links do not load an example.
Define the axis and rotating parts
Include the table, fixture and payload in J. Mass farther from the axis is harder to accelerate. This differs in units and purpose from a beam’s second moment of area I.
For practical review
Identify the shaft to which each inertia belongs before reflecting it to the motor. The parallel-axis theorem does not change axis orientation. Eccentric gravity torque and imbalance are not described by J alone.
Uniform cylinder about its centre axis: J = mD²/8 / parallel shifted axis: J = Jc + me²
Define speed and acceleration time
Define the target rpm and time to accelerate from rest. Rpm counts revolutions per minute; rad/s measures angle per second. Use consistent units in equations.
- Do not include acceleration torque again in steady resistance TL
- Also define the complete cycle, including reversals and dwell
ω = 2πn/60, α = ω/t
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Add resistance and acceleration torque
For a direct, lossless drive, T = TL + (Jload + Jm)α includes motor-side inertia. With gearing, account for shaft speeds, reflected inertia and efficiency.
For practical review
Here i is motor speed divided by load speed. Reflected load inertia is Jload/i², but motor speed increases. A larger ratio is not universally better; check permissible speed, efficiency and backlash.
Example: accelerate a 10 kg disk to 60 rpm in 0.5 s
Inputs
- Uniform disk m = 10 kg, D = 200 mm, centre axis
- Jload = 0.05 kg·m²; Jm = 0.001 kg·m² (assumed)
- Direct drive i = 1, efficiency 100%, TL = 0.5 N·m (assumed)
- Target speed 60 rpm; acceleration time 0.5 s
Results
- Angular acceleration α ≈ 12.566 rad/s²
- Acceleration torque ≈ 0.641 N·m; steady torque = 0.500 N·m
- Total during acceleration ≈ 1.141 N·m
This is a calculated demand, not a motor rating recommendation. Replace Jm with actual rotor/input-side inertia and TL with actual resistance. Deceleration, back-driving and heating are excluded.
Load the fixed values, units and mode for this illustration. Then edit them for your conditions.
Centre-axis solid-disk mode; inner diameter, width, height and offset are unused.
Check peak, continuous and stopping requirements
Peak capability and duration depend on the motor and drive. Check available torque at the operating speed, RMS demand over the complete cycle and stopping energy.
- Torque-speed envelope, peak duration, power supply and drive
- RMS torque, duty, cooling and ambient temperature
- Deceleration/regeneration, brake holding and emergency stopping
- Shaft, coupling and gearbox strength and permissible inertia
Keep a review checklist
This checklist is a personal reminder, not a safety score or an automatic approval. It is not sent or saved.
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Reference information and illustrative calculations. This content does not guarantee suitability or safety of a design. Confirm the actual materials, component specifications and operating conditions.