AQA A-Level Physics Paper 3 (3BC), June 2025: Question 1

8 marks · Medium difficulty · Short Answer

Calculate kinematic quantities, angular impulse, and compare angular deceleration from a disc brake system to electrical braking for an electric motor rotor.

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Question

Question 01 showing an electric motor rotor. Figure 1 shows an end view with driving torque clockwise. Figure 2 shows retarding torque anticlockwise. Given values: initial angular velocity 98.0 rad/s clockwise at t = 0, angular acceleration -303 rad/s^2. Part 01.1 asks to determine t1 when it stops (2 marks). Part 01.2 states it accelerates anticlockwise to -120 rad/s at t2 and asks for number of anticlockwise revolutions between t1 and t2 (2 marks). Part 01.3 gives moment of inertia 9.60 x 10^-2 kg m^2 and asks for angular impulse between t = 0 and t2 (1 mark). Part 01.4 introduces a disc brake in Figure 3 with two brake pads each applying 182 N at the edge of a disc of diameter 0.160 m and pad diameter 22 mm, with combined moment of inertia 0.101 kg m^2, asking to compare angular deceleration produced by the disc brake with that from reversing connections (3 marks).
Question text

01 The rotating part of an electric motor is called the rotor.

Figure 1 shows an end view of a rotor turning clockwise due to a driving torque from

the motor. In this question, the clockwise direction is treated as positive.

Figure 1 Figure 2

The rotor can be brought to rest rapidly by reversing the electrical supply connections

to the motor. Figure 2 shows the rotor at time t = 0 when the supply connections are

reversed.

The rotor then slows down due to a constant anticlockwise retarding torque so that it

stops at time t = t1.

The angular velocity of the rotor at t = 0 is 98.0 rad s−1 clockwise.

The applied torque on the rotor at t = 0 is anticlockwise.

The applied torque produces a constant angular acceleration of −303 rad s−2.

Friction torque is negligible.

01.1 Determine t1.

[2 marks]

3 t =

1 s

The electrical supply remains connected and the rotor now accelerates uniformly

anticlockwise with an acceleration of magnitude 303 rad s−2.

At a later time t = t , the angular velocity of the rotor is −120 rad s−1.

01.2 Determine the number of anticlockwise revolutions made by the rotor between

t1 and t2.

[2 marks]

number of revolutions =

01.3 The moment of inertia of the rotor about the axis of rotation is 9.60 × 10−2 kg m2.

Calculate the angular impulse on the rotor between t = 0 and t2.

[1 mark]

angular impulse4 = N m s

01.4 Another way of quickly stopping the motor is to use a disc brake. The motor is

brought to rest by switching off the power supply and then forcing two brake pads

against the disc.

Figure 3 shows the two brake pads each applying a constant retarding force of 182 N

to the disc.

Figure 3

moment of inertia of rotor and disc about the axis of rotation = 0.101 kg m2

diameter of disc = 0.160 m

diameter of pad = 22 mm

Compare the angular deceleration produced by the disc brake system with the

angular deceleration produced by reversing the electrical connections.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 01. 01.1: Uses omega_1 = 0 = omega_0 + alpha*t1 to give t1 = 0.32(3) s (2 marks). 01.2: Uses rotational kinematic equations to find angular displacement theta = 23.8 rad, then divides by 2pi to find number of revolutions = 3.78 (2 marks). 01.3: Angular impulse = delta(I*omega) = 9.6 x 10^-2 * (-120 - 98) = (-)21 N m s (1 mark). 01.4: Calculates effective radius r = 0.080 - 0.011 = 0.069 m, determines torque T = F*r using 182 or 364 N, uses alpha = T/I to find alpha = 249 rad/s^2, and concludes reversing connections provides greater deceleration (3 marks).

Question Answers Additional comments/Guidance Mark AO

Mark MP1 and MP2 independently

01.1 Two correct substitutions with consistent signs✓ 2 AO2

ω1 = 0 = ω0 + αt1 Allow their symbols.

ω1=0, ω0 = 98 , = -303

∆𝜔

Allow use of ∆𝑡 = 𝛼 provided signs consistent and

“0” is seen.

giving t1 = 0.32(3) (s) ✓ Calculator value: 0.32343234…

01.2 Two correct substitutions with consistent signs Expect to see 2 AO2

From t to t : (−120)2 = 2 (−303)θ

22 θ = (−)23.8 rad (or similar)

ω2 = ω1 + 2αθ

with

OR

ω2 = –120, ω1 = 0 α = –303

ω2 = ω1+ αt AND

Condone one substitution error. All substitutions

2 must have internally consistent signs.

θ = ω1t + ½ αt OR θ = ½(ω1+ ω2)t

OR

Divdes their θ by 2π✓ Condone “number of revolutions = 3” (1 sf only) on

answer line if correct answer seen in working

Divides correct θ by 2π to get number (of rev) = 3.78 ✓ Accept 2 or more sf answers

Condone final answer which has minus sign

– A-LEVEL PHYSICS – –

= Δ I ω = 9.6 × 10−2 × (-120 - 98

01.3 (Angular impulse ( ) )) Calculator value: 20.928 1 AO2

= (-) 21 (N m s) ✓

Penalise rounding errors

01.4 r = distance from axis to line of action of force 3 AO3

MAX 2 from ✓✓

• Calculates r = (0.080 − 0.011 =) 0.069 m NO POT penalty in bullet points.

• Determines T (= Fr) with their r and

F = 182 or 364 N

T Do not accept a force for their T

• Uses α = with their T

I Allow 250 or answers that round to 248 or 249

α = 249 rad s−2

AND Award 3 marks when correct α and correct

reversing connections provides greater deceleration than conclusion seen

disc brake ✓ Ignore minus sign

Total 8

How to answer it

Rotational Dynamics: Electric Motor & Braking Systems

AQA A-Level Physics • Turning Points / Engineering Physics • 8 Marks Total
📋 What This Question Tests
  • Equations of Rotational Kinematics: Applying rotational analogues of SUVAT equations ( ω = ω₀ + αt , ω² = ω₀² + 2αθ ) with strict adherence to sign conventions.
  • Unit Conversions: Converting angular displacement between radians and complete revolutions.
  • Angular Impulse: Calculating impulse from change in angular momentum ( Δ(Iω) ).
  • Torque & Moment of Inertia: Finding effective radius to centre of pressure, summing torque for multiple brake pads, and calculating angular deceleration using T = Iα .
Part 01.1 • 2 Marks

Determining Time to Come to Rest (t₁)

Kinematics of rotational motion under constant retarding torque

📐 Step-by-Step Calculation

  1. Identify initial and final states:
    ω₀ = +98.0 rad s⁻¹ (clockwise)
    ω₁ = 0 rad s⁻¹ (rest)
    α = -303 rad s⁻² (anticlockwise)
  2. Select formula:
    ω₁ = ω₀ + αt₁
  3. Substitute values:
    0 = 98.0 + (-303)t₁
    303t₁ = 98.0
  4. Solve for t₁ :
    t₁ = 98.0 / 303 = 0.3234... s
    t₁ = 0.32 s (or 0.323 s)

🧠 Exam Technique & Signing

  • Sign Convention: The question defines clockwise as positive (+). Since the retarding torque opposes rotation, the acceleration must be negative (-303 rad s⁻²).
  • Method Mark (MP1): Awarded for correct substitution with mutually consistent signs (e.g. 0 = 98 - 303t or 0 = -98 + 303t ).
  • Accuracy Mark (MP2): Awarded for 0.32 s (2 s.f.) or 0.323 s (3 s.f.).

❌ Common Errors

  • Forgetting that deceleration acts opposite to initial velocity, resulting in a negative time if signs aren't balanced properly.
  • Over-rounding intermediate values; always keep full calculator precision before writing down your final answer.
Mark Scheme: MP1 for ω₁ = 0 = ω₀ + αt₁ with consistent signs. MP2 for t₁ = 0.32(3) s .
Part 01.2 • 2 Marks

Revolutions Made During Anticlockwise Acceleration

Finding angular displacement between t₁ and t₂ and converting to revolutions

📐 Step-by-Step Calculation

  1. State conditions between t₁ and t₂ :
    ω₁ = 0 rad s⁻¹
    ω₂ = -120 rad s⁻¹
    α = -303 rad s⁻²
  2. Use the rotational kinematic equation:
    ω₂² = ω₁² + 2αθ
    (-120)² = 0² + 2(-303)θ
    14400 = -606θ ⇒ θ = -23.762 rad
    (Magnitude of angle θ = 23.762 rad)
  3. Convert radians into full revolutions:
    1 rev = 2π rad ≈ 6.2832 rad
    Number of revs = 23.762 / (2π)
    Number of revolutions = 3.78

💡 Alternative Valid Method

You can also determine the time interval Δt = t₂ - t₁ first:

  • -120 = 0 - 303(Δt) ⇒ Δt = 0.396 s
  • θ = ½(ω₁ + ω₂)Δt = ½(-120)(0.396) = -23.76 rad
  • Revolutions = 23.76 / 2π = 3.78

❌ Common Errors

  • Dividing by 360 instead of 2π: Remember that angular equations use radians, not degrees! There are 2π radians in 1 revolution.
  • Mixing time intervals: Using ω₀ = +98 rad s⁻¹ instead of ω₁ = 0 rad s⁻¹ as the initial speed for the interval between t₁ and t₂ .
Mark Scheme: MP1 for finding θ = 23.8 rad and dividing by 2π . MP2 for final answer of 3.78 (or rounding to 3.8).
Part 01.3 • 1 Mark

Angular Impulse Between t = 0 and t₂

Linking angular impulse to change in angular momentum

📐 Calculation

  1. Recall definition of Angular Impulse:
    Angular Impulse = Δ(Iω) = I(ω₂ - ω₀)
  2. Substitute values (note direction & signs!):
    I = 9.60 × 10⁻² kg m²
    ω₀ = +98.0 rad s⁻¹
    ω₂ = -120 rad s⁻¹
  3. Calculate:
    Δ(Iω) = 9.60 × 10⁻² × [(-120) - 98.0]
    Δ(Iω) = 9.60 × 10⁻² × (-218.0) = -20.928 N m s
  4. Angular Impulse = -21 N m s (or 21 N m s)

🧠 Examiner Insight

  • Direction changes sign: The rotor changes from +98.0 rad s⁻¹ (clockwise) to -120 rad s⁻¹ (anticlockwise).
  • Therefore, the change in angular velocity is (-120 - 98) = -218 rad s⁻¹ , not 120 - 98 = 22 rad s⁻¹ .
  • The mark scheme accepts both -21 N m s and 21 N m s (magnitude), but penalises incorrect arithmetic rounding (must round calculator value 20.928 to 21).
Mark Scheme: 1 mark for (-) 21 N m s . Rounding errors penalised.
Part 01.4 • 3 Marks

Braking System Comparison

Geometry of disc brake, retarding torque, and deceleration

📐 Step-by-Step Calculation

  1. Find effective radius (distance from axis to pad centre):
    Disc radius R = 0.160 / 2 = 0.080 m
    Pad radius r_pad = 0.022 / 2 = 0.011 m
    Line of action radius:
    r = 0.080 - 0.011 = 0.069 m
  2. Calculate total retarding torque:
    Two pads, each applying 182 N :
    Total retarding force = 2 × 182 N = 364 N
    T = Total Force × r = 364 × 0.069 = 25.116 N m
  3. Calculate angular deceleration:
    Combined moment of inertia I = 0.101 kg m²
    α = T / I = 25.116 / 0.101 = 248.67... rad s⁻² ≈ 249 rad s⁻²
  4. Make final comparative conclusion:
    Electrical reversal deceleration = 303 rad s⁻²
    Disc brake deceleration = 249 rad s⁻²
    Conclusion: Reversing electrical connections provides a greater angular deceleration.

🧠 Mark Scheme Breakdown

Awarding up to 2 marks from:

  • Calculating effective radius: r = 0.080 - 0.011 = 0.069 m .
  • Determining torque: T = 182 × r (single pad) or 364 × r (both pads).
  • Applying Newton's second rotational law: α = T / I with their calculated torque.

Final mark:

  • Obtaining α = 249 rad s⁻² (accepts 248–250) AND stating that reversing connections gives greater deceleration.

❌ Critical Pitfalls & Common Mistakes

  • Using the disc radius directly ( 0.080 m ): The brake pad has finite size ( 22 mm diameter). Its force acts at its geometric centre, which is 11 mm inward from the outer edge!
  • Forgetting there are TWO brake pads: The question states two pads act inwards, one on each side. Each creates friction at that radius, so either double the force ( 364 N ) or double the torque.
  • Using force instead of torque in α = T / I : Examiners will give zero marks for substituting 182 N or 364 N directly into the numerator without multiplying by radius r .
Mark Scheme: Max 2 marks for calculation steps ( r = 0.069 m , T = Fr , α = T/I ). 3rd mark for α = 249 rad s⁻² AND explicitly concluding that reversing electrical connections gives a greater deceleration.

Topics

Optional topics · 3.11 Engineering physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BC), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.