SQA SQA AH style-style practice: A motor delivers 2.5\ kW at 1500\ rev min^-1. Calculate the torque produced, given that power equals torque times angular velocity.

First convert the rotational speed to angular velocity in radians per second. ω = 2π N60 = 2π × 150060 = 157\ rad s^-1. The power relationship links power, torque and angular velocity. Relationship: P = Tω, so T = Pω. Substitution: T = 2500157 = 15.9\ N m. Markers reward converting rev per minute to 157\ rad s^-1, using P = Tω, and the torque of about 15.9\ N m. Forgetting the conversion to rad per second is the usual error.

ScotlandEngineering ScienceSyllabus dot point

How do gear trains and drive systems change speed, torque and direction, and how efficient are they?

Mechanisms transmitting motion: gear trains and compound gears, velocity ratio and mechanical advantage, the relationship between torque, speed and power, and drive-system efficiency.

An SQA Advanced Higher Engineering Science answer on mechanisms, covering simple and compound gear trains, velocity ratio and mechanical advantage, the relationship between torque, rotational speed and power, and the efficiency of a drive system.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-16

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

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Quick answer

A gear train changes speed, torque and direction. For a simple train the velocity ratio is the number of teeth on the driven gear divided by the number on the driver, $VR = \dfrac{T_{driven}}{T_{driver}}$ , and the output speed is the input speed divided by the $VR$ . Reducing the speed increases the torque in the same ratio (for an ideal train), because power is conserved: $P = T\omega$ , with $\omega = \dfrac{2\pi N}{60}$ converting rev per minute to rad s $^{-1}$ . A compound gear train multiplies the ratios of its stages, giving large reductions in a small space. No real drive is perfect: efficiency is the output power over the input power, $\eta = \dfrac{P_{out}}{P_{in}} \times 100\%$ , with the shortfall lost mainly to friction as heat.

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What this key area is asking
Gear trains and velocity ratio
Torque, speed and power
Compound gear trains
Drive-system efficiency
Examples in context
Try this

What this key area is asking

The SQA wants you to analyse mechanisms that transmit motion, in particular gear trains (simple and compound), to calculate velocity ratio and mechanical advantage, to use the relationship between torque, rotational speed and power, and to find the efficiency of a drive system. This is how a mechanism turns a motor's output into the speed and force a machine needs.

Gear trains and velocity ratio

Because meshing gears share the same tooth pitch, a larger gear has proportionally more teeth, so the tooth ratio equals the speed ratio. An idler gear placed between driver and driven reverses the direction of rotation but does not change the velocity ratio, since its teeth cancel. A larger driven gear turns more slowly than the driver, and a smaller one turns faster.

Torque, speed and power

Torque is the turning effect of a force, the rotational equivalent of force. Power links torque and speed: for a given power, a higher torque means a lower speed and the reverse. This is the key to gearing: an ideal gear train conserves power, so reducing the speed by a factor $VR$ multiplies the torque by the same factor. That is why low gears give a vehicle pulling power and high gears give speed.

Compound gear trains

A compound gear train has two or more stages, with two gears fixed on a common shaft so that the output of one stage drives the next. The overall velocity ratio is the product of the stage ratios:

VR_{total} = VR_1 \times VR_2 \times \dots

This lets a large overall reduction be achieved in a small, compact gearbox, where a single pair of gears would need impractically different sizes.

Drive-system efficiency

No drive is perfectly efficient: bearing and tooth friction, and any belt slip, turn some input power into heat. The efficiency is the useful output power as a percentage of the input. The lost power is exactly the friction that the dynamics area identifies, which is why a gearbox warms up in use.

Speed, torque and efficiency through a reduction gearbox

A motor supplies $1.8\ \text{kW}$ at a torque of $12\ \text{N m}$ to a gearbox with a velocity ratio of $4$ . The gearbox is $90\%$ efficient. Find the output torque (assuming ideal speed reduction) and the useful output power.

step 1 Find the ideal output torque

An ideal reduction multiplies torque by the velocity ratio: $T_{out} = T_{in} \times VR = 12 \times 4 = 48\ \text{N m}$ (before efficiency losses).

step 2 Find the useful output power

$P_{out} = \eta \times P_{in} = 0.90 \times 1.8 = 1.62\ \text{kW}$ .

step 3 Interpret the result

The output shaft runs at a quarter of the input speed with about four times the torque, but $10\%$ of the input power, $0.18\ \text{kW}$ , is lost to friction as heat, so the real output torque is a little below the ideal $48\ \text{N m}$ .

Examples in context

A car gearbox uses different velocity ratios so the engine can give pulling torque in low gear and high road speed in top gear. A clock uses a compound gear train to step the fast-running mainspring down to the slow hands. A winch uses a high-reduction gearbox to lift a heavy load with a small motor torque. A wind-turbine gearbox does the opposite, stepping the slow blades up to the fast speed a generator needs. In every case the velocity ratio sets the speed, the torque follows from constant power, and efficiency fixes how much output is actually delivered.

Try this

Q1. State the relationship for the velocity ratio of a simple gear train. [1 mark]

Cue. $VR = \dfrac{\text{teeth on driven}}{\text{teeth on driver}}$ .

Q2. A gearbox has stage ratios of $3$ and $5$ . Find the overall velocity ratio. [2 marks]

Cue. $VR_{total} = 3 \times 5 = 15$ .

Q3. State what happens to the output torque when a gear train reduces the speed (ideal case). [1 mark]

Cue. The torque increases by the same factor as the speed is reduced (constant power).

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 AH style4 marksA simple gear train has a driver of

20

teeth meshing with a driven gear of

60

teeth. The driver turns at

1500\ \text{rev min}^{-1}

. Calculate the velocity ratio and the output speed.

Show worked answer →

The velocity ratio of a gear train is the driven teeth over the driver teeth.

Relationship: $VR = \dfrac{\text{teeth on driven}}{\text{teeth on driver}}$ .

Substitution: $VR = \dfrac{60}{20} = 3$ .

The output speed is the input speed divided by the velocity ratio.

$N_{out} = \dfrac{1500}{3} = 500\ \text{rev min}^{-1}$ .

Markers reward $VR = 3$ from the tooth ratio, and the output speed of $500\ \text{rev min}^{-1}$ . A gear train that reduces speed increases torque in the same ratio (if ideal).

SQA AH style5 marksA motor delivers

2.5\ \text{kW}

1500\ \text{rev min}^{-1}

. Calculate the torque produced, given that power equals torque times angular velocity.

Show worked answer →

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

$\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 1500}{60} = 157\ \text{rad s}^{-1}$ .

The power relationship links power, torque and angular velocity.

Relationship: $P = T\omega$ , so $T = \dfrac{P}{\omega}$ .

Substitution: $T = \dfrac{2500}{157} = 15.9\ \text{N m}$ .

Markers reward converting rev per minute to $157\ \text{rad s}^{-1}$ , using $P = T\omega$ , and the torque of about $15.9\ \text{N m}$ . Forgetting the conversion to rad per second is the usual error.

Related dot points

Sources & how we know this

Advanced Higher Engineering Science Course Specification (C823 77) — SQA (2019)
Advanced Higher Engineering Science Data Booklet — SQA (2019)