What is step 3. geared output?

The output speed is 290\,rev/min, a 5:1 reduction. The output angular speed is:

QLDEngineeringSyllabus dot point

How are turning force, rotational speed and power related in a machine?

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

A QCE Engineering Unit 4 answer on torque and power. Covers torque as force times radius, the relationship between torque, angular speed and power, conversion of rev/min to rad/s, and how power is conserved through a drive, with worked arithmetic.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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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.