RPM to rad/s Converter for Shaft Power and Flywheels

Shaft Speed in the Unit That P = T·ω Actually Needs

Two of the most-used expressions in drivetrain work are unforgiving about units. Shaft power is P = T·ω, and the kinetic energy stored in a rotating mass is KE = ½Iω². Both give an answer in SI units only when ω is supplied in radians per second. Nameplates, tachometers, gearbox catalogues and torque-speed curves are all marked in revolutions per minute, so the first arithmetic step in almost every shaft calculation is turning that RPM figure into an angular velocity. Skip it and the power figure comes out 9.55 times too large — an error big enough to specify a motor an entire frame size wrong.

Conversion factor: ω = RPM × 2π/60 = RPM × π/30 ≈ 0.104719755. A 1 450 RPM motor turns at 1 450 × 0.104719755 = 151.8436 rad/s. Delivering 250 N·m at that speed is 250 × 151.8436 = 37 960.9 W, or 37.96 kW.

Why the factor cannot be skipped

Radians Are Dimensionless, Revolutions Are Not

A radian is arc length divided by radius, so it cancels away and N·m × rad/s collapses cleanly to N·m/s, which is the watt. A revolution is a count of turns with no such geometric definition, which is why it cannot be dropped into the same product.

Where the 9 549 Shortcut Hides

Rearranging P(kW) = T·RPM·π/30 ÷ 1000 puts the whole conversion into a single divisor of 30 000/π = 9 549.2966. That is the same 2π/60 in disguise, packaged so a torque and a speed can be combined without ever writing ω down.

Energy Follows the Square

In ½Iω² an error in ω is squared before it reaches the answer. Using RPM in place of rad/s inflates flywheel energy by more than ninety times, which is why energy-storage sizing is where the missing factor usually gets caught.

Working a Shaft Duty from a Nameplate Speed

A typical duty check runs from nameplate to angular velocity to power, then back again once the gearbox ratio changes the picture.

Enter the rated or measured shaft speed

Type the RPM from the motor plate, the tacho or the torque-speed curve into the left field. The rad/s value appears as you type, so a whole speed range can be walked through in one sitting. A comma decimal separator works as well as a point.

Multiply by torque to get watts

The angular velocity in rad/s is already the power in watts per newton-metre of torque, so multiplying it by the shaft torque gives watts directly. Divide by 1 000 for kilowatts and cross-check against the T × RPM ÷ 9 549.30 shortcut.

Reverse it for a gearbox output

Simulation and dynamics work hand back angular velocities in rad/s that have to be quoted as a catalogue speed. Press the swap button (↔), or just type in the rad/s field, and the RPM equivalent comes back on the other side.

Copy ω into the energy calculation

The copy button on each field hands over the plain number with no unit attached, ready to be squared in a ½Iω² flywheel sum or dropped into a τ = I·α acceleration check without any tidying.

Shaft Speeds, Angular Velocity and Power per Newton-Metre

Because power per newton-metre of torque in watts is numerically identical to ω, one table serves both purposes. Read the middle column for dynamics work and the right-hand columns to judge, at a glance, how much torque a given duty is going to demand.

Shaft speed	ω (rad/s)	Watts per N·m	Power at 100 N·m
100 RPM — geared output shaft	10.4720	10.47 W	1.05 kW
500 RPM — slow mixer drive	52.3599	52.36 W	5.24 kW
750 RPM — pump gearbox output shaft	78.5398	78.54 W	7.85 kW
1 000 RPM — conveyor head shaft	104.7198	104.72 W	10.47 kW
1 450 RPM — typical 4-pole motor shaft	151.8436	151.84 W	15.18 kW
1 750 RPM — fan drive shaft	183.2596	183.26 W	18.33 kW
2 900 RPM — centrifugal pump shaft	303.6873	303.69 W	30.37 kW
3 600 RPM — compressor input shaft	376.9911	376.99 W	37.70 kW

Walking a Torque-Speed Curve

Points along a curve are converted one after another, each ω copied out as a bare number, so a power curve can be built from a speed-and-torque table without a spreadsheet formula in sight.

Both Sides of a Reduction

Input and output speeds of a gearbox go through the same two fields, so the ratio can be checked in rad/s where the dynamics live and in RPM where the catalogue lives.

Rotational and Cyclic Units Side by Side

The searchable unit lists reach beyond RPM and rad/s, so a shaft speed can be lined up against degrees per second or revolutions per second when a sensor or a datasheet insists on those.

Torque, Power and Flywheel Questions

Why does P = T·ω refuse to work with a speed in RPM?

Power is work per unit time, and work done by a torque is the torque multiplied by the angle swept in radians. Because the radian is a ratio of two lengths it carries no dimension, so N·m × rad/s reduces to N·m/s, which is exactly the watt. A revolution per minute is a count on a different time base, and multiplying torque by it produces a number 9.5493 times larger than the true wattage. That is the arithmetic behind the classic mistake of quoting a 4 kW duty as roughly 38 kW.

Where does the 9 549 in kW = N·m × RPM ÷ 9 549 come from?

It is the whole unit conversion folded into one divisor. Starting from P = T·ω and substituting ω = RPM·π/30 gives P in watts as T·RPM·π/30; dividing by 1 000 for kilowatts leaves the constant 30 000/π = 9 549.2966. Checking it against the long route: 250 N·m at 1 450 RPM is 250 × 151.8436 = 37 960.9 W, and 250 × 1 450 ÷ 9 549.30 = 37.96 kW. Reference books that round the divisor to 9 550 introduce an error of about 0.007%, which is far below the tolerance of any nameplate figure.

Why does doubling flywheel speed quadruple the stored energy?

Because ω is squared in KE = ½Iω². A rotor with I = 0.5 kg·m² spun to 3 000 RPM reaches 314.1593 rad/s and holds 24.7 kJ; at 6 000 RPM it reaches 628.3185 rad/s and holds 98.7 kJ, precisely four times as much for the same mass. This is why speed rather than mass is the lever in energy-storage design, and why burst containment becomes the governing constraint before the inertia figure does. A heavier example: a 25 kg·m² press flywheel at 1 500 RPM stores about 0.31 MJ.

How does a gear ratio change angular velocity and torque together?

A reduction divides ω by the ratio and multiplies torque by it, leaving the product — the power — unchanged apart from losses. Take a 1 450 RPM motor at 151.8436 rad/s through a 10:1 box: the output turns at 145 RPM, or 15.1844 rad/s, and an input torque of 25 N·m arrives as about 240 N·m once a 96% efficiency is allowed for. One extra effect catches people out: inertia reflected back to the motor scales with the square of the ratio, so a modest load inertia can dominate the acceleration torque once it is seen through a large reduction.

What units does angular acceleration take in τ = I·α?

Radians per second squared, never RPM per second, for the same reason the power expression needs rad/s. Bringing a rotor of I = 0.5 kg·m² from rest to 3 000 RPM in 4 seconds means a change of 314.1593 rad/s, so α = 314.1593 ÷ 4 = 78.5398 rad/s² and the accelerating torque is 0.5 × 78.5398 = 39.27 N·m. That figure sits on top of whatever load torque the machine already demands, so a drive sized only for steady running will hit its current limit and simply take longer to run up.