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Specification: AQA A-level Physics (7408) 3.11.1.6 Engineering physics option  |  Paper: 3 Section B  |  Data and formulae booklet: given in the exam

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The equation

work done = torque × angular displacement    W = Tθ    and    power = torque × angular velocity    P = Tω

Quantity	Symbol	Unit
work done by the torque	W	joules (J)
power	P	watts (W)
torque	T	newton metres (N m)
angular displacement	θ	radians (rad)
angular velocity	ω	radians per second (rad s⁻¹)

Rearranged: T = W ÷ θ  |  θ = W ÷ T  |  T = P ÷ ω  |  ω = P ÷ T  |  Convert revolutions to radians (× 2π) and rev min⁻¹ to rad s⁻¹ (× 2π ÷ 60) first. The N m in W = Tθ is a joule only because the radian has no unit; torque itself is always written in N m, never J.

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Core: Q1–4.   Exam standard: Q5–8.   Stretch: Q9–10.

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Questions

Q1 (Core) A constant torque of 15 N m turns a shaft through 40 rad. Calculate the work done.

Show solution

F: W = Tθ
I: W = 15 × 40
F: W is already the subject
A: 600 J

Q2 (Core) A turbine shaft delivers a torque of 250 N m while rotating at 300 rad s⁻¹. Calculate the power transmitted in kW.

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F: P = Tω
I: P = 250 × 300
F: P is already the subject
A: 75 000 W = 75 kW

Q3 (Core) A constant torque of 8.0 N m is applied to a winch drum, which turns through 50 revolutions. Calculate the work done, to 2 significant figures.

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Convert first: θ = 50 × 2π = 100π = 314.1… rad
F: W = Tθ
I: W = 8.0 × 314.1…
F: W is already the subject
A: 2513.2… J = 2.5 × 10³ J (2 s.f.)
Watch out: 8.0 × 50 = 400 J is wrong. θ must be in radians.

Q4 (Core) A car engine has a maximum output power of 110 kW, which it produces at 5500 rev min⁻¹. Calculate the output torque at this engine speed, to 2 significant figures.

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Convert first: 110 kW = 110 000 W; ω = 5500 × 2π ÷ 60 = 575.9… rad s⁻¹
F: P = Tω
I: 110 000 = T × 575.9…
F: T = 110 000 ÷ 575.9…
A: 190.9… = 190 N m (2 s.f.)

Q5 (Exam) A flywheel of moment of inertia 0.20 kg m² is rotating at 2400 rev min⁻¹ when its drive is disconnected. A constant frictional torque of 0.50 N m brings it to rest. Calculate (a) the work done against friction and (b) the number of revolutions it makes before stopping, both to 2 significant figures.

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Convert first: ω = 2400 × 2π ÷ 60 = 251.3… rad s⁻¹
Step 1 – (a) work done = kinetic energy lost
Eₖ = ½Iω² = ½ × 0.20 × 251.3…² = 6316.5… J
A: 6.3 × 10³ J (2 s.f.)
Step 2 – (b) angle turned
F: W = Tθ
I: 6316.5… = 0.50 × θ
F: θ = 6316.5… ÷ 0.50 = 12 633… rad
12 633… ÷ 2π = 2010.6…
A: 2.0 × 10³ revolutions (2 s.f.)

Q6 (Exam) An electric motor takes an input power of 2.0 kW. Its output shaft delivers a torque of 9.5 N m at 1450 rev min⁻¹. Calculate (a) the output power and (b) the efficiency of the motor, both to 2 significant figures.

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Convert first: ω = 1450 × 2π ÷ 60 = 151.8… rad s⁻¹; 2.0 kW = 2000 W
Step 1 – (a) output power
F: P = Tω
I: P = 9.5 × 151.8…
F: P is already the subject
A: 1442.5… W = 1.4 kW (2 s.f.)
Step 2 – (b) efficiency
efficiency = output power ÷ input power = 1442.5… ÷ 2000 = 0.7212…
A: 72% (2 s.f.)

Q7 (Exam) A motor drives a fan at a constant 3000 rev min⁻¹. The total frictional torque on the fan (bearings and air resistance) is 0.15 N m. Calculate (a) the power the motor must supply to keep the fan turning, to 2 significant figures, and (b) the energy transferred in 1.0 hour, to 2 significant figures.

Show solution

Convert first: ω = 3000 × 2π ÷ 60 = 314.1… rad s⁻¹; 1.0 hour = 3600 s
Step 1 – (a) power
at constant ω, the driving torque equals the frictional torque, 0.15 N m
F: P = Tω
I: P = 0.15 × 314.1…
F: P is already the subject
A: 47.12… W = 47 W (2 s.f.)
Step 2 – (b) energy
W = Pt = 47.12… × 3600 = 169 646… J
A: 1.7 × 10⁵ J (2 s.f.)

Q8 (Exam) A winch drum of radius 0.25 m lifts a 400 kg load vertically at a constant 0.60 m s⁻¹. Calculate (a) the torque on the drum, (b) its angular velocity and (c) the output power of the winch, to 2 significant figures. (g = 9.81 m s⁻²)

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Step 1 – (a) torque
at constant speed the tension equals the weight: F = mg = 400 × 9.81 = 3924 N
T = Fr = 3924 × 0.25 = 981 N m
Step 2 – (b) angular velocity
ω = v ÷ r = 0.60 ÷ 0.25 = 2.4 rad s⁻¹
Step 3 – (c) power
F: P = Tω
I: P = 981 × 2.4
F: P is already the subject
A: 2354.4 W = 2.4 kW (2 s.f.)
Check with P = Fv: 3924 × 0.60 = 2354.4 W

Q9 (Stretch) A car engine produces an output torque of 160 N m at 4200 rev min⁻¹. (a) Calculate the output power of the engine, to 2 significant figures. (b) The driver is accelerating hard in this gear, and at one instant the car is travelling at 30 m s⁻¹. 90% of the engine’s output power reaches the wheels, which have a radius of 0.31 m. Calculate the driving force at the wheels and the torque on the wheels, both to 2 significant figures.

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Convert first: ω = 4200 × 2π ÷ 60 = 439.8… rad s⁻¹
Step 1 – (a) engine output power
F: P = Tω
I: P = 160 × 439.8…
A: 70 371… W = 70 kW (2 s.f.)
Step 2 – power at the wheels
0.90 × 70 371… = 63 334… W
Step 3 – (b) driving force
F = P ÷ v = 63 334… ÷ 30 = 2111.1… N
A: 2.1 × 10³ N (2 s.f.)
This is much more than the resistive forces at 30 m s⁻¹ (typically about 500 N for a car), which is why the car accelerates: at a steady cruising speed the engine would deliver far less power.
Step 4 – wheel torque
ω_wheel = v ÷ r = 30 ÷ 0.31 = 96.77… rad s⁻¹; T = P ÷ ω = 63 334… ÷ 96.77… = 654.4… N m
A: 650 N m (2 s.f.)
Check: T = Fr = 2111.1… × 0.31 = 654.4… N m. The gearbox multiplies the torque because the wheels turn more slowly than the engine.
Watch out: the engine’s ω and the wheels’ ω are different. Use the wheels’ ω with the power that reaches them.

Q10 (Stretch) A motor applies a constant torque of 12 N m to a flywheel of moment of inertia 0.80 kg m², initially at rest. A constant frictional torque of 2.0 N m acts. The flywheel makes 25 revolutions. Calculate (a) the work done by the motor, (b) the work done against friction, (c) the final angular velocity of the flywheel and (d) the power output of the motor at that instant. Give all answers to 2 significant figures.

Show solution

Convert first: θ = 25 × 2π = 50π = 157.0… rad
Step 1 – (a) work done by the motor
F: W = Tθ
I: W = 12 × 157.0…
A: 1884.9… J = 1.9 × 10³ J (2 s.f.)
Step 2 – (b) work done against friction
W_f = 2.0 × 157.0… = 314.1… J
A: 310 J (2 s.f.)
Step 3 – (c) final angular velocity
Eₖ = 1884.9… − 314.1… = 1570.7… J
F: Eₖ = ½Iω²
I: 1570.7… = ½ × 0.80 × ω²
F: ω = √(2 × 1570.7… ÷ 0.80) = √3926.9…
A: 62.66… = 63 rad s⁻¹ (2 s.f.)
Check with T = Iα: α = (12 − 2.0) ÷ 0.80 = 12.5 rad s⁻²; ω² = 2 × 12.5 × 157.0… = 3926.9…
Step 4 – (d) power
F: P = Tω
I: P = 12 × 62.66…
A: 751.9… W = 750 W (2 s.f.)
Watch out: in (d) use the motor’s torque, 12 N m, not the resultant torque. P = Tω gives the power of whichever torque you put in.

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