Orientation · Module 9 of 10
Motion and Power Transmission
Machines rarely run at the speed a job needs, so gears and belts reshape motion. This module introduces rotational speed, the speed-torque trade of a gear ratio, and the power a spinning shaft carries.
Readiness check
This module is about spinning machines. Tick what you can do comfortably.
- Divide one speed by a ratio.
- Multiply by 2π and divide by 60.
- Recall that torque is a turning force.
- Multiply a torque by an angular speed.
- Recall that power is energy per unit time.
The core idea
Rotational speed is measured in revolutions per minute or radians per second, ω = 2πN/60. Gears and belts change speed by their ratio and, keeping power the same, change torque the opposite way. A rotating shaft transmits power P = Tω, torque times angular speed.
output speed = input speed / ratioω = 2πN / 60power P = TωMotors and engines produce power at a particular speed and torque, but the load usually needs a different pairing, and transmission elements make the conversion. Rotational speed is stated as revolutions per minute, N, or as an angular speed in radians per second, ω = 2πN/60; the factor of 2π converts turns to radians and the 60 converts minutes to seconds. A gear or belt drive changes speed by its ratio: gear down and the output turns slower. Because an ideal drive conserves power, and power is torque times angular speed, whatever speed is lost is gained back as torque, so a reduction multiplies torque by the same factor it divides speed. This speed-for-torque trade is the whole reason gearboxes exist. Finally, the power a shaft carries is P = Tω, torque times angular speed, which links this module back to energy: a shaft turning fast with modest torque can carry the same power as one turning slowly with large torque. These ideas open into the Machine Elements and Dynamics courses, where gears, shafts, and rotating systems are designed in detail.
The skills, taught in order
Five skills introduce rotating machines and power flow.
9.1 Rotational motion
Rotation is measured in revolutions per minute or radians per second, linked by ω = 2πN/60. Angular speed in rad/s is the form used in physics equations, so converting from rpm is a constant first step.
9.2 Gears and belts
Gears, belts, and chains transmit rotation between shafts and change its speed by their ratio, set by tooth counts or pulley sizes. They let one motor drive loads at many different speeds.
9.3 The speed-torque trade
Because an ideal drive conserves power, a reduction that divides speed by a ratio multiplies torque by the same ratio. You cannot gain both; you trade one for the other, which is why a low gear gives more pulling torque.
| Through a reduction | Speed | Torque |
|---|---|---|
| Ratio greater than 1 | decreases | increases |
| Power | unchanged (ideal) | |
Speed and torque move oppositely through a ratio; their product, the power, is preserved.
9.4 Power in rotation
A rotating shaft carries power P = Tω, torque times angular speed. This ties rotation back to energy: the same power can be delivered fast with low torque or slowly with high torque.
9.5 Machines and mechanisms
Real drivetrains combine gears, shafts, bearings, and linkages, each with efficiency below one. Seeing a machine as a chain of these elements is the systems view that Machine Elements and Dynamics build on.
Engineering connection: sizing a motor and gearbox for a conveyor uses exactly ω = 2πN/60 and P = Tω, the workflow that Machine Elements develops in full.
Worked example 1: a speed reduction
A motor runs at 1500 rpm and drives a load through a gear reduction with a ratio of 5. Find the output speed.
- ProblemFind the output speed for the reduction in Figure 1.
- Given / findInput 1500 rpm, ratio 5. Find the output speed.
- AssumptionsIdeal reduction; ratio is input speed over output speed.
- Modeloutput speed = input speed / ratio.
- EquationsNout = 1500 / 5
- SolveNout = 300 rpm.
- CheckA reduction slows the output, and 300 rpm is five times slower than 1500, as the ratio requires; the torque is five times larger.
- ConclusionThe load turns at 300 rpm with five times the motor's torque, the point of gearing down.
Worked example 2: power in a shaft
A shaft delivers a torque of 20 N·m while turning at 1500 rpm. Find the power transmitted.
- ProblemFind the power transmitted by the shaft in Figure 2.
- Given / findTorque 20 N·m, speed 1500 rpm. Find the power.
- AssumptionsSteady rotation; power is torque times angular speed.
- Modelω = 2πN/60; P = Tω.
- Equationsω = 2π(1500)/60 = 157.1 rad/sP = 20 × 157.1
- SolveP = 3141.6 W ≈ 3.14 kW.
- CheckUnits N·m × 1/s = N·m/s = W, correct; the value is a sensible small-motor power.
- ConclusionThe shaft carries about 3.14 kW; the same power could come from more torque at lower speed.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Dropping the 2π factor | Power off by about six times | "Is the speed in rad/s?" | Convert rpm with ω = 2πN/60. |
| Both speed and torque up | Power appears created | "Is power conserved?" | A ratio trades speed for torque, not both. |
| Ratio inverted | A reduction speeds the load up | "Is the output slower?" | A reduction divides output speed by the ratio. |
| Ignoring efficiency | Full torque expected at the load | "Any losses?" | Real drives deliver a little less than ideal. |
Practice ladder
Convert 3000 rpm to radians per second.
Show answer
ω = 2π(3000)/60 = 314.16 rad/s.
A motor at 1200 rpm drives a load through a ratio of 4. Find the output speed.
Show answer
Output = 1200 / 4 = 300 rpm.
A shaft carries 15 N·m at 600 rpm. Find the power transmitted.
Show answer
ω = 2π(600)/60 = 62.83 rad/s; P = 15 × 62.83 ≈ 942 W, about 0.94 kW.
A 3 kW motor at 1500 rpm drives a wheel through a 6:1 reduction. Find the wheel speed and estimate the ideal torque at the wheel.
What good work looks like
Wheel speed = 1500 / 6 = 250 rpm. Motor torque = P / ω = 3000 / (2π × 1500/60) ≈ 19.1 N·m; through 6:1 the wheel torque is about 6 × 19.1 ≈ 115 N·m ideal. A good answer uses P = Tω and the ratio, and notes real torque is a little lower.
Working with AI, and proving it yourself
Use AI as a guide, not an oracle
Portfolio task
Pick a geared device and work out its output speed and the power its shaft carries.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Convert rpm to rad/s.
ω = 2πN/60.
2. What does a reduction do to speed and torque?
Divides speed and multiplies torque by the ratio, at constant power.
3. Write rotational power.
P = Tω, torque times angular speed.
4. Why gear a motor down?
To gain torque for a slow, heavy load.
5. Can a ratio raise both speed and torque?
No; that would create power. It trades one for the other.
Textbook mapping
This module follows Wickert and Lewis, An Introduction to Mechanical Engineering, 3rd edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Rotational motion | Wickert and Lewis, Section 8.2, Rotational Motion |
| Gears and gear trains | Wickert and Lewis, Section 8.3, Gears and Gear Trains |
| Power transmission | Wickert and Lewis, Chapter 8, Motion and Power Transmission |
Section numbers refer to Wickert and Lewis, 3rd edition. Any edition with the same chapter titles is equivalent for study.