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How do belt and chain drives transmit rotation between shafts, and how do you calculate torque and the power transmitted by a rotating shaft?

Belt, chain and pulley drive systems and their speed ratio, the concept of torque, and the power transmitted by a rotating shaft.

An SQA Higher Engineering Science answer on belt, chain and pulley drive systems and their speed ratio, the concept of torque as a turning moment, and how to calculate the power transmitted by a rotating shaft.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A belt or chain drive transmits rotation between shafts that are too far apart for gears. The speed ratio depends on the pulley (or sprocket) sizes: the smaller pulley turns faster, in inverse proportion to diameter, $N_{\text{driven}} = N_{\text{driver}} \times \dfrac{D_{\text{driver}}}{D_{\text{driven}}}$ . A belt drive keeps the same direction of rotation; a crossed belt reverses it; a chain (like gears) avoids slip. Torque is the turning effect of a force, $T = Fr$ , in newton metres. The power transmitted by a rotating shaft is $P = T\omega$ , where $\omega$ is the angular velocity in rad/s, found from rev/min by $\omega = \dfrac{2\pi N}{60}$ . A reduction (larger driven pulley) lowers speed and raises torque.

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What this key area is asking
Belt, chain and pulley drives
Torque
Power transmitted by a rotating shaft
Putting it together
Examples in context
Try this

What this key area is asking

The SQA wants you to analyse belt, chain and pulley drive systems and their speed ratio, understand torque as a turning effect, and calculate the power transmitted by a rotating shaft. These extend the mechanisms ideas from gears to flexible drives and to the rotational power that flows through a machine.

Belt, chain and pulley drives

A belt drive uses a flexible belt around two pulleys to carry rotation from one shaft to another. A chain drive uses a chain around toothed sprockets. Both connect shafts that are too far apart for gears to mesh.

A belt drive can slip (useful as overload protection, but it loses exact timing). A chain does not slip, so it keeps a fixed speed ratio, like gears but over a distance.
An ordinary (open) belt keeps the same direction of rotation on both shafts; a crossed belt reverses the driven shaft.

The speed ratio of a belt or chain drive is set by the pulley diameters (or sprocket teeth): the smaller wheel turns faster.

A larger driven pulley turns slower, a reduction, which increases the available torque, exactly as for a reduction gear.

Torque

Torque is what a shaft delivers to turn a load: a motor is rated by the torque it can produce. The bigger the force, or the longer the radius at which it acts, the greater the torque, which is why a longer spanner loosens a stiff bolt more easily.

Power transmitted by a rotating shaft

A rotating shaft carries power. The power transmitted is the product of the torque and the angular velocity.

The single most important step is converting the rotational speed from rev/min to rad/s before using $P = T\omega$ . Forgetting this gives an answer 60 (and $2\pi$ ) times wrong.

Power from a drive system

A motor produces 8 N m of torque at 1800 rev/min. It drives, through a belt, a shaft that should run at 600 rev/min. Find the motor's output power and the driven shaft speed ratio.

step 1 Convert the motor speed to rad/s

$\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 1800}{60} = 188.5\ \text{rad/s}$ .

step 2 Motor output power

$P = T\omega = 8 \times 188.5 = 1508\ \text{W} \approx 1.5\ \text{kW}$ .

step 3 Required speed ratio

The shaft must run at 600 rev/min from a 1800 rev/min motor, a ratio of $\dfrac{1800}{600} = 3$ , so a 3:1 reduction belt drive is needed (driven pulley three times the driver diameter). Ignoring losses, the same 1.5 kW is transmitted, but at lower speed and higher torque on the driven shaft.

Putting it together

A drive system links a power source to a load: the motor supplies torque at a speed, the belt or chain (or gears) changes the speed ratio, and the power $P = T\omega$ flows through (less friction losses). Because power is roughly conserved across an ideal drive, lowering the speed with a reduction must raise the torque in the same proportion. This is the rotational version of the force-distance trade-off seen with levers and gears.

Examples in context

A washing machine uses a belt from a small motor pulley to a large drum pulley: a big reduction so the high-speed motor turns the heavy drum slowly with plenty of torque for a full load. A bicycle is a chain drive; changing to a larger rear sprocket increases the reduction, raising torque for hills at the cost of speed. Car engine accessories (alternator, water pump) are driven by a belt from the crankshaft, with pulley sizes chosen so each runs at its correct speed from the one engine. In each, the speed ratio and the transmitted power $P = T\omega$ describe how the drive behaves.

Try this

Q1. A 40 mm driver pulley at 2000 rev/min drives an 80 mm pulley. Find the driven speed. [2 marks]

Cue. $N = 2000 \times \frac{40}{80} = 1000\ \text{rev/min}$ .

Q2. A force of 50 N acts at a radius of 0.2 m. Find the torque. [1 mark]

Cue. $T = Fr = 50 \times 0.2 = 10\ \text{N m}$ .

Q3. A shaft delivers 10 N m at an angular velocity of 100 rad/s. Find the power. [2 marks]

Cue. $P = T\omega = 10 \times 100 = 1000\ \text{W} = 1\ \text{kW}$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher (specimen)3 marksA driver pulley of diameter 50 mm runs at 1500 rev/min and drives a pulley of diameter 150 mm by a belt. Calculate the speed of the driven pulley.

Show worked answer →

For a belt drive, the smaller pulley turns faster, in inverse proportion to the diameters.

Relationship: $N_{\text{driven}} = N_{\text{driver}} \times \dfrac{D_{\text{driver}}}{D_{\text{driven}}}$ .

Substitution: $N_{\text{driven}} = 1500 \times \dfrac{50}{150}$ .

Answer: $N_{\text{driven}} = 1500 \times \dfrac{1}{3} = 500\ \text{rev/min}$ .

Markers reward the inverse diameter ratio (driver diameter on top), a clean substitution, and the answer of 500 rev/min. The larger driven pulley turns slower, a 3:1 reduction, which increases the torque available.

SQA Higher (specimen)4 marksA motor shaft transmits a torque of 12 N m at a rotational speed of 1500 rev/min. Calculate the angular velocity in rad/s and the power transmitted by the shaft.

Show worked answer →

First convert the rotational speed to an angular velocity in radians per second.

$\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 1500}{60} = 157\ \text{rad/s}$ .

The power transmitted by a rotating shaft is torque times angular velocity.

$P = T\omega = 12 \times 157 = 1885\ \text{W} \approx 1.9\ \text{kW}$ .

Markers reward the conversion of rev/min to rad/s (dividing by 60 and multiplying by two pi), then using $P = T\omega$ with omega in rad/s to get about 1.9 kW. Using rev/min directly in $P = T\omega$ is the usual error.

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Sources & how we know this

SQA Higher Engineering Science Course Specification — SQA (2019)
Higher Engineering Science Course Specification (PDF) — SQA (2019)