Torque and rotational power for QCE Engineering Unit 4

Syllabus dot point

“Calculate torque as the turning effect of a force, and relate torque and rotational speed to mechanical power transmitted by a rotating shaft”

QCEEngineeringUnit 4: Machines and mechanisms6 min readUpdated 2 Jun 2026

Quick answer

Torque is the turning effect of a force about an axis, $\tau = F \times r$ , where $r$ is the perpendicular distance from the axis to the line of the force; its unit is the newton metre (N m). The mechanical power transmitted by a rotating shaft is the product of torque and angular speed, $P = \tau \omega$ , where $\omega$ is in radians per second and $P$ is in watts. Rotational speed in rev/min converts to rad/s by multiplying by $2\pi/60$ . Because an ideal drive conserves power, $\tau \omega$ is constant through it, so reducing speed raises torque and raising speed lowers torque. This is why a motor's high-speed, low-torque output is geared down to drive a heavy load.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to calculate torque, the turning effect of a force on a shaft, and link it to rotational speed to find the mechanical power a machine transmits. Torque and power are the quantities that describe what a motor or engine actually delivers, and they are the key to understanding why gears and drives trade speed against force. This is the rotational counterpart of the force-and-work content.

The answer

Torque: the turning effect

Torque measures how strongly a force twists a shaft about its axis:

\tau = F \times r

where $F$ is the applied force and $r$ is the perpendicular distance from the axis of rotation to the line of action of the force (the radius at which the force acts). The unit is the newton metre (N m). A larger radius gives more torque for the same force, which is why a long spanner loosens a tight bolt that a short one cannot.

Power in rotating systems

Power is the rate of doing work. For a rotating shaft, the power transmitted is the product of the torque and the angular speed:

P = \tau \omega

Here $\omega$ is the angular speed in radians per second and $P$ is in watts (W). Angular speed in revolutions per minute converts to rad/s with:

\omega = N \times \frac{2\pi}{60}

So a motor rated at a certain power can deliver high torque only at low speed, or high speed only at low torque, because the product is fixed by its power.

Power is conserved through a drive

An ideal gear train, belt or chain transmits power without loss, so the power in equals the power out:

\tau_{in}\,\omega_{in} = \tau_{out}\,\omega_{out}

This single equation explains the speed-torque trade-off of every mechanism in Unit 4. If a gearbox halves the output speed, it doubles the output torque, because their product must stay equal to the input power. In a real drive some power is lost to friction, so the output power is the input power multiplied by the efficiency.

Torque and power of a motor

Find the torque a motor delivers

An electric motor is rated at $P = 1.5\,\text{kW}$ and runs at $N = 1450\,\text{rev/min}$ . Find its output torque. Then find the torque available if a gearbox reduces the speed to $290\,\text{rev/min}$ at $95\%$ efficiency.

Step 1. angular speed of the motor:

\omega = 1450 \times \frac{2\pi}{60} = 151.8\,\text{rad/s}

Step 2. motor torque (from $P = \tau\omega$ ):

\tau = \frac{P}{\omega} = \frac{1500}{151.8} = 9.88\,\text{N m}

Step 3. geared output. The output speed is $290\,\text{rev/min}$ , a $5:1$ reduction. The output angular speed is:

\omega_{out} = 290 \times \frac{2\pi}{60} = 30.4\,\text{rad/s}

Output power after losses: $P_{out} = 0.95 \times 1500 = 1425\,\text{W}$ . So the output torque is:

\tau_{out} = \frac{1425}{30.4} = 46.9\,\text{N m}

The torque rose from $9.88\,\text{N m}$ to $46.9\,\text{N m}$ , close to five times, as expected for a $5:1$ reduction, slightly less because of the $5\%$ loss. This shows how gearing converts a motor's fast, weak output into a slow, strong one.

Why this matters for machines and mechanisms

Torque and power are how engineers specify and match machine components. A motor is chosen by its power and speed; the gearbox then converts that into the torque the load needs. Every gear-ratio, belt-drive and chain-drive calculation in Unit 4 rests on the fact that power is conserved, so $\tau\omega$ stays constant. Getting the torque-power-speed triangle right is essential for the Unit 4 engineered solution.

Common errors

Leaving speed in rev/min in the power formula: $P = \tau\omega$ needs $\omega$ in rad/s. Forgetting the $2\pi/60$ conversion gives a power wrong by a large factor.
Confusing torque with power: Torque is a turning force (N m); power is the rate of doing work (W). A high-torque motor is not automatically a high-power one if it runs slowly.
Forgetting efficiency in a real drive: Output power equals input power only for an ideal drive; with losses, multiply by the efficiency before finding the output torque.
Using the wrong radius for torque: Torque uses the perpendicular distance from the axis to the force's line of action, not the length of the lever measured along its slant.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks

A motor shaft delivers a torque of

25\ \text{N m}

while rotating at

1200\ \text{rev/min}

. Determine the angular speed in radians per second and the mechanical power transmitted by the shaft.

Show worked answer →

A 7 mark determine question rewards the unit conversion then the power.

Convert rotational speed to angular velocity: $\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 1200}{60} = 125.7\ \text{rad/s}$ .

Power transmitted: $P = T\omega = 25 \times 125.7 = 3142\ \text{W} \approx 3.14\ \text{kW}$ .

Markers reward the conversion from rev/min to rad/s, applying $P = T\omega$ , and the power of about $3.14\ \text{kW}$ with correct units.

Practice4 marks

Explain why, ignoring losses, a gear reduction that lowers output speed must increase output torque, with reference to power.

Show worked answer →

A 4 mark explain answer needs the conservation-of-power argument.

Ignoring friction, power in equals power out, and rotational power is $P = T\omega$ . If a reduction gear lowers the output angular speed $\omega$ , then to keep $P$ constant the output torque $T$ must rise in the same proportion. So a speed reduction is a torque increase, which is why low gears give more turning effort.

Markers reward using $P = T\omega$ with power conserved to show torque rises as speed falls.

Practise this

Sources & how we know this

Engineering General senior syllabus 2025 — QCAA