Angular Impulse (AQA A Level Physics): Revision Note

Angular impulse

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

$F=\frac{∆p}{∆t}$

• This leads to the definition of linear impulse:

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

$∆p=F∆t=∆\left(mv\right)$

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

$\tau =\frac{∆L}{∆t}$

• Where:

  • $\tau$ = 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 $\tau$ acting for a time $∆t$ produces a change in angular momentum $∆L$

$∆L=\tau ∆t=∆\left(I\omega \right)$

• 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, $∆L=\tau ∆t$

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