When a running rotating mass is coupled to a stationary or slower one, a common speed is reached that follows from conservation of angular momentum: ω = (J₁ω₁ + J₂ω₂)/(J₁+J₂). This is the rotational counterpart of the perfectly plastic impact – angular momentum is conserved, and part of the rotational energy is lost as friction heat in the clutch.
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When a running rotating mass is coupled to a stationary or slower one, a common speed is reached that follows from conservation of angular momentum: ω = (J₁ω₁ + J₂ω₂)/(J₁+J₂). This is the rotational counterpart of the perfectly plastic impact – angular momentum is conserved, and part of the rotational energy is lost as friction heat in the clutch.
A motor rotor with J₁ = 0.08 kg·m² runs at n₁ = 1,500 rpm and is coupled to a stationary load with J₂ = 0.12 kg·m², i.e. n₂ = 0. The common speed is n = (0.08·1,500 + 0.12·0)/(0.08 + 0.12) = 120/0.20 = 600 rpm. The speed therefore drops to 40 per cent, because the added rotating mass is one and a half times the driving one. The energy balance shows the price: before coupling the motor side holds ½·0.08·157.08² = 987 J of rotational energy, afterwards both masses together only ½·0.20·62.83² = 395 J. The difference of 592 J – about 60 per cent – is converted into heat as friction work in the clutch and must be dissipated.
A slip-free final state is assumed: after coupling both masses run at the same speed. During the coupling process no external moments from drive or load act – only then is angular momentum conserved. This assumption holds well when the coupling process is much shorter than the time constants of drive and load; with slow engagement under load the drive torque shifts the result. Both inertias must refer to the same axis of rotation; an intervening gearbox must first be reduced through the square of the ratio. The time history of the coupling process, the clutch torques and forces, the slip time, the heating of the friction surfaces, the torsional flexibility of the shafts and the torsional vibrations arising from it, and bearing and seal losses are all excluded.
Engaging a running rotating mass onto a stationary one produces a common speed – often surprisingly far below the initial one. This calculator determines that synchronous speed from conservation of angular momentum and thereby also reveals how much energy the clutch must absorb as heat.
Rotation has a counterpart to the linear momentum m·v: angular momentum, L = J·ω, with the mass moment of inertia J and the angular velocity ω. The angular-momentum principle states that angular momentum changes only through an external moment – if none acts during coupling, it is conserved. From J₁ω₁ + J₂ω₂ = (J₁+J₂)·ω the common speed follows immediately. Rotational energy ½·J·ω², by contrast, is not conserved: it is smaller after coupling, and the difference is converted into heat as friction work in the clutch.
A carousel somebody jumps onto slows down – without anyone braking. It now shares its angular momentum with a larger rotating mass. A clutch behaves exactly the same way, and just as when jumping on, friction arises that turns energy into heat.
ω = (J₁·ω₁ + J₂·ω₂) / (J₁ + J₂)
Angular momentum of a rotating mass: L = J·ωConservation during coupling: J₁ω₁ + J₂ω₂ = (J₁+J₂)·ωCommon speed: ω = (J₁ω₁ + J₂ω₂)/(J₁+J₂)Stationary load (ω₂ = 0): ω = ω₁·J₁/(J₁+J₂)Rotational energy: Trot = ½·J·ω²Clutch friction work: W = ½J₁ω₁² + ½J₂ω₂² − ½(J₁+J₂)ω²Speed in radians per second: ω = 2π·n/60| Symbol / input | Meaning |
|---|---|
| Common speed n after coupling | The speed at which both masses run together after the clutch closes. It always lies between the two initial speeds and closer to the speed of the larger rotating mass. The value applies to the state immediately after synchronisation, i.e. before drive or load torques change the speed further. |
| Mass moment of inertia J₁ of the first rotating mass | The mass moment of inertia of the first rotating mass about the common axis – for example the motor rotor with its clutch disc. The mass moment of inertia describes how strongly a body resists angular acceleration. Source: the motor datasheet, a CAD mass calculation, or a formula collection such as J = ½·m·r² for a disc. Both inertias must refer to the same axis of rotation; with a gearbox in between, reduce them to a common shaft first. |
| Speed n₁ of the first rotating mass before coupling | The speed of the first rotating mass immediately before the clutch closes, usually in revolutions per minute. If the two masses turn in opposite directions, enter one of the speeds as negative – the common speed can then also become zero or change sign. |
| Mass moment of inertia J₂ of the second rotating mass | The mass moment of inertia of the second rotating mass about the same axis, for example the load side with gearbox and driven machine. The larger J₂ is relative to J₁, the more strongly the load side pulls the common speed towards its own initial value, and the more energy is converted into heat in the clutch. |
| Speed n₂ of the second rotating mass before coupling | The speed of the second rotating mass immediately before coupling, counted in the same direction as n₁. For the most common case – engaging a stationary load – enter zero here. A negative value means counter-rotation. |
J₁ and J₂ are the mass moments of inertia of the two rotating masses, referred to the same axis of rotation – for instance the motor rotor with its clutch disc on one side and the load side on the other. They come from datasheets, a CAD mass calculation, or a formula collection. n₁ and n₂ are the speeds immediately before coupling, usually in revolutions per minute; for a stationary load n₂ = 0. Counter-rotation is expressed by a negative sign.
First check that both inertias really refer to the same shaft – with a gearbox in between, the load side must first be reduced to the motor shaft through the square of the ratio. Then enter both inertias and initial speeds. The result is the speed immediately after synchronisation, still without the influence of drive and load torque. The energy balance is the interesting part afterwards: the rotational energy before coupling, ½·J₁·ω₁² + ½·J₂·ω₂², minus that afterwards, ½·(J₁+J₂)·ω², gives the friction work the clutch must absorb and dissipate.
A motor rotor with J₁ = 0.08 kg·m² runs at 1,500 rpm and is coupled to a stationary load with J₂ = 0.12 kg·m². Then n = (0.08·1,500 + 0.12·0)/0.20 = 120/0.20 = 600 rpm. The speed drops to 40 per cent, because the added rotating mass is one and a half times the driving one. The energy balance: 1,500 rpm equals 157.08 rad/s, so ½·0.08·157.08² = 987 J before coupling. Afterwards both run at 600 rpm = 62.83 rad/s, so ½·0.20·62.83² = 395 J. The difference of 592 J, about 60 per cent of the initial energy, is converted into heat in the clutch. With equal inertias it would be exactly half.
The common speed is the inertia-weighted mean of the initial speeds – always between them and closer to the speed of the larger rotating mass. The share of energy lost is revealing: it depends only on the ratio of the inertias, not on the absolute speed. With equal rotating masses and a stationary load, exactly half the rotational energy turns into heat – regardless of whether coupling happens at 100 or 10,000 revolutions per minute. This is why engaging large rotating masses is always thermally critical, and why plants with large flywheels use start-up clutches, speed matching or synchronising devices.
The calculation runs in kilograms, metres and radians per second internally. The inertias may be entered in kg·m² or kg·cm², the speeds independently in rpm or rad/s. Since the speeds enter linearly and only their ratio to the inertias matters, the result carries the same unit as the inputs – no conversion is needed for the speed calculation itself. For the energy balance it is: there ω must be inserted in radians per second.
Friction and start-up clutches in drive trains; engaging flywheels, presses and centrifuges; synchronisation in manual gearboxes; estimating the friction work and hence the thermal load on clutch linings; designing start-up aids for large inertias; the speed drop when adding load on test benches.
A slip-free final state is assumed, in which both masses run at the same speed, and that no external moments from drive or load act during the coupling process – only then is angular momentum conserved. This holds well as long as the coupling process is much shorter than the time constants of drive and load; with slow engagement under load the drive torque shifts the result. Both inertias must refer to the same axis of rotation. The time history of coupling, the clutch torques and forces, the slip time, the heating of the friction surfaces, the torsional flexibility of the shafts and the torsional vibrations arising from it, and bearing and seal losses are not covered.
Common mistake: The most consequential error is adding inertias from both sides of a gearbox without reduction – the load side must be converted to the shaft under consideration through the square of the ratio. Second, rotational energy is often expected to be conserved; only angular momentum is, and the energy loss is precisely the thermal load on the clutch. Third, the result is read as an operating speed although it describes only the state immediately after synchronisation – after that the drive accelerates the now larger rotating mass again. And fourth, the sign is forgotten for counter-rotation, which greatly underestimates the speed drop.
Friction and start-up clutches in drive trains; engaging flywheels, presses and centrifuges; synchronisation in manual gearboxes; estimating the friction work and hence the thermal load on clutch linings; designing start-up aids for large inertias; the speed drop when adding load on test benches.
J₁ and J₂ are the mass moments of inertia of the two rotating masses, referred to the same axis of rotation – for instance the motor rotor with its clutch disc on one side and the load side on the other. They come from datasheets, a CAD mass calculation, or a formula collection. n₁ and n₂ are the speeds immediately before coupling, usually in revolutions per minute; for a stationary load n₂ = 0. Counter-rotation is expressed by a negative sign.
A slip-free final state is assumed, in which both masses run at the same speed, and that no external moments from drive or load act during the coupling process – only then is angular momentum conserved. This holds well as long as the coupling process is much shorter than the time constants of drive and load; with slow engagement under load the drive torque shifts the result. Both inertias must refer to the same axis of rotation. The time history of coupling, the clutch torques and forces, the slip time, the heating of the friction surfaces, the torsional flexibility of the shafts and the torsional vibrations arising from it, and bearing and seal losses are not covered.