Angular Impulse (AQA A Level Physics): Revision Note

Exam code: 7408

Ashika

Written by: Ashika

Reviewed by: Caroline Carroll

Updated on

Angular Impulse

  • In linear motion, the resultant force on a body can be defined as the rate of change of linear momentum:

F = pt

  • This leads to the definition of linear impulse:

An average resultant force F acting for a time t produces a change in linear momentum p

p = Ft = (mv)

  • Similarly, the resultant torque on a body can be defined as the rate of change of angular momentum:

τ = Lt

  • Where:

    • τ = resultant torque on a body (N m)

    • L = change in angular momentum (kg m2 s−1)

    • t = time interval (s)

  • This leads to the definition of angular impulse:

An average resultant torque τ acting for a time t produces a change in angular momentum L

L = τt = (Iω)

  • Angular impulse is measured in kg m2 s−1, or N m s

  • This equation requires the use of a constant resultant torque

    • If the resultant torque changes, then an average of the values must be used

  • Angular impulse describes the effect of a torque acting over a time interval

    • This means a small torque acting over a long time has the same effect as a large torque acting over a short time

Angular Impulse on a Torque-Time Graph

  • The area under a torque-time graph is equal to the angular impulse or the change in angular momentum

    • This is because the area, angular impulse, is a fraction of the base × height, torque × time, L = τt

    • The fraction of torque × time depends upon the shape of the area under the graph

1-4-8--angular-impulse-on-a-torque-time-graph-ib-2025-physics

When the torque is not constant, the angular impulse is the area under a torque–time graph

Worked Example

The graph shows the variation of time t with the net torque τ on an object which has a moment of inertia of 6.0 kg m2.

1-4-8-angular-impulse-graph-worked-example-ib-2025-physics

At t = 0, the object rotates with an angular velocity of 2.0 rad s−1 clockwise.

Determine the magnitude and direction of rotation of the angular velocity at t = 5 s.

In this question, take anticlockwise as the positive direction.

Answer:

  • The area under a torque-time graph is equal to angular impulse, or the change in angular momentum

L = τ×t

1-4-8-angular-impulse-graph-worked-example-ma-ib-2025-physics
  • The area under the positive curve (triangle) = 12×10×3 = 15 N m s

  • The area under the negative curve (rectangle) = 5×2 = 10 N m s

  • Therefore, the overall change in angular momentum is

L = 15  10 = 5 N m s

  • The change in angular momentum is equal to

L = (Iω) = I(ωf  ωi)

  • Where

    • Moment of inertia, I = 6.0 kg m2

    • Initial angular velocity, ωi = −2.0 rad s−1 (clockwise is the negative direction)

  • Therefore, when t = 5 s, the angular velocity is

6×(ωf  (2)) = 5

6ωf +(2×6) = 5

6ωf = 5  12

ωf = 76 = 1.17

Anti-clockwise is positive, so ωf = 1.17 rad s−1 in the clockwise direction

Examiner Tips and Tricks

Many applications for angular impulse will be related to sports. These are similar to the linear impulse topic.

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Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.