Inputs
Shape of the moving mass, its mass and governing diameter, the target speed at the motor shaft and the run-up time; optionally the gear ratio and the motor rotor inertia.
Drives · Drive sizing
Moment of inertia and acceleration torque
Calculate the moment of inertia of a moving mass, refer it to the motor shaft through the gear ratio, and determine the torque and power needed to bring it up to speed in the time you have. A linearly moving mass can be included through its drive radius.
Rigid-body model with constant angular acceleration. Only the acceleration torque is calculated – the load torque acting during the run-up from friction, gravity or process force adds to it. Torsional elasticity, backlash, slip, gearbox efficiency and the motor's torque-speed curve are not included.
JTa = J·α
02
Inertia and acceleration torque
Set the mass and the run-up.
Inputs and method
Moment of inertia and acceleration torque
Calculate the moment of inertia of a moving mass and the torque a drive needs to bring it up to speed in the time available.
Calculation
The moment of inertia follows from the geometry: ½·m·r² for a solid cylinder, m·r² for a thin ring, ⅖·m·r² for a solid sphere and ½·m·(ro²+ri²) for a hollow cylinder. A gearbox reduces it to J/i², which follows from conservation of kinetic energy; a linearly moving mass acts through the drive radius as m·r². From α = 2π·n/(60·t) follow Ta = J·α, the peak power as Ta·ω and the stored energy as ½·J·ω².
Example
A solid cylinder of 50 kg and 300 mm diameter has J = 0.5625 kg·m². To reach 1,500 min⁻¹ in 2 s, α = 78.54 rad/s² and therefore Ta = 44.18 N·m at 6.94 kW peak power. With a gear ratio of 5 the reduced inertia falls to 0.0225 kg·m².
Sources and limits: Classical rigid-body dynamics, quantities per ISO 80000-4. Every reduction is derived from conservation of kinetic energy. Acceleration torque only; load torque, torsional elasticity, backlash, slip, gearbox efficiency and the motor's torque-speed curve are not included.
Moment of inertia and acceleration torque in detail
Calculate the moment of inertia of a moving mass and the torque a drive needs to bring it up to speed in the time available.
What is the moment of inertia?
The moment of inertia describes how strongly a body resists a change of its rotational speed – the rotational counterpart of mass in linear motion. It depends not only on how heavy the body is but above all on how far its material sits from the axis of rotation: the distance enters quadratically. That is why reducing the diameter almost always achieves more than saving weight.
Formula and variables
Ta = J_total · α with α = 2π·n / (60·t)
Solid cylinder: J = ½·m·r² · Thin ring: J = m·r²Hollow cylinder: J = ½·m·(ro² + ri²) · Solid sphere: J = ⅖·m·r²Linearly moving mass via the drive radius: J = m·r²Gearbox: J_red = J / i² (from ½·J·ω² = ½·J_red·ω_motor²)P_max = Ta · ω · E = ½·J·ω²
| Symbol / input | Meaning |
|---|---|
| J, J_red | Inertia of the body and its value referred to the motor shaft. |
| i | Ratio of motor speed to load speed; it acts quadratically. |
| n, t, α | Target speed, run-up time and the resulting angular acceleration. |
| Ta, P, E | Acceleration torque, peak power and stored rotational energy. |
Choose the inputs correctly
Shape of the moving mass, its mass and governing diameter, the target speed at the motor shaft and the run-up time; optionally the gear ratio and the motor rotor inertia.
How to use the calculator
Choose the shape of the moving mass – the form then asks for exactly the dimensions that shape needs. Enter mass and diameter, then the target speed at the motor shaft and the run-up time you want. For a geared drive add the ratio, and for a servo sizing also the rotor inertia from the motor datasheet. The reported acceleration torque is the part you add to the load torque before selecting a motor.
Worked example
A solid cylinder of 50 kg and 300 mm diameter has J = 0.5625 kg·m². To reach 1,500 min⁻¹ in 2 s, α = 78.54 rad/s² and therefore Ta = 44.18 N·m at 6.94 kW peak power. With a gear ratio of 5 the reduced inertia falls to 0.0225 kg·m².
Which value decides what?
The acceleration torque says what the drive must provide on top of the load torque. The peak power at the end of the run-up is the value motor selection hangs on – not the mean power, which is only half of it. The ratio of reduced load inertia to motor inertia is the third important number: if it becomes very large, the drive can no longer be controlled cleanly regardless of whether the torque is sufficient.
Mass in kilograms, diameters in millimetres, speed in min⁻¹, time in seconds, inertia in kg·m², torque in N·m. Watch datasheets: some manufacturers quote inertia in kg·cm², a factor of 10,000.
Typical applications
Pre-sizing motors and gearboxes for flywheels, drums, rollers and machine-tool axes, estimating the run-up time of an existing machine, dimensioning braking resistors from the stored rotational energy, and checking the inertia ratio of servo drives.
Assumptions, limits and common mistakes
Classical rigid-body dynamics, quantities per ISO 80000-4. Every reduction is derived from conservation of kinetic energy. Acceleration torque only; load torque, torsional elasticity, backlash, slip, gearbox efficiency and the motor's torque-speed curve are not included.
Common mistake: For a linearly moving mass do not enter a dimension of the mass itself but the pitch diameter of the drive wheel – only that determines how strongly the mass acts at the shaft. Do not treat the gear ratio as linear: it enters quadratically. And do not reduce the motor inertia by the ratio; it already sits on the motor shaft.
Frequently asked questions
Why does the gear ratio enter quadratically?
Because kinetic energy must be conserved. If the load turns i times slower than the motor, its angular velocity is smaller by i and its energy ½·J·ω² smaller by i². For it to represent the same energy at the motor shaft, the inertia acting there must be J/i².
Why does a hollow cylinder have a larger inertia than a solid one?
At the same mass, yes. Inertia is mass times distance squared: material near the axis contributes little, material far out a great deal. If the bore is removed and the same mass is arranged further out, the inertia rises. If instead the outside diameter is fixed and the bore reduces the mass, it falls.
How do I include a linearly moving mass?
Through the radius of the driving wheel. If the mass moves at v = ω·r, equating energy ½·m·v² = ½·J·ω² gives J = m·r² directly. A 200 kg slide on a pinion of 100 mm pitch diameter therefore acts at the shaft as 200·0.05² = 0.5 kg·m².
Is the calculated torque enough for the motor?
No, it is only the acceleration part. During the run-up the load torque from friction, gravity or process force acts as well, and the motor must deliver the sum across the whole speed range – with induction motors the available torque near rated speed is considerably lower than at start-up.
What is the stored rotational energy for?
It is the energy that has to leave the system again when braking. With frequent start-stop cycles it governs the sizing of the braking resistor or the drive's regeneration capability, and it becomes the thermal load on the brake.