AQA A Level Physics

Revision Notes

11.1.2 Angular Acceleration Equations

Equations for Uniform Angular Acceleration

  • The kinematic equations of motion for uniform linear acceleration can also be re-written for rotational motion
  • The four kinematic equations for uniform linear acceleration are

v space equals space u space plus space a t

s space equals space u t space plus space 1 half a t squared

v squared space equals space u squared space plus space 2 a s

s space equals space fraction numerator open parentheses u space plus space v close parentheses t over denominator 2 end fraction

  • This leads to the four kinematic equations for uniform rotational acceleration

omega subscript 2 space equals space omega subscript 1 space plus space alpha t

increment theta space equals space omega subscript 1 t space plus space 1 half alpha t squared

omega subscript 2 squared space equals space omega subscript 1 squared space plus space 2 alpha increment theta

increment theta space equals space fraction numerator open parentheses omega subscript 1 space plus space omega subscript 2 close parentheses t over denominator 2 end fraction

  • The five linear variables have been swapped for the rotational equivalents, as shown in the table below
Variable Linear Rotational
displacement s θ
initial velocity u ω1
final velocity v ω2
acceleration a α
time t t

Worked example

The turntable of a record player is spinning at an angular velocity of 45 RPM just before it is turned off. It then decelerates at a constant rate of 0.8 rad s−2.

Determine the number of rotations the turntable completes before coming to a stop.

Answer:

Step 1: List the known quantities

  • Initial angular velocity, omega subscript 1 = 45 RPM
  • Final angular velocity, omega subscript 2 = 0
  • Angular acceleration, alpha = −0.8 rad s−2
  • Angular displacement, increment theta = ?

Step 2: Convert the angular velocity from RPM to rad s−1

  • One revolution corresponds to 2π radians, and RPM = revolutions per minute, so

omega space equals space 2 straight pi f and f space equals space fraction numerator R P M over denominator 60 end fraction(to convert to seconds)

omega subscript 1 space equals space fraction numerator 2 straight pi cross times RPM over denominator 60 end fraction space equals space fraction numerator 2 straight pi cross times 45 over denominator 60 end fraction space equals space fraction numerator 3 straight pi over denominator 2 end fraction space rad space straight s to the power of negative 1 end exponent

Step 3: Select the most appropriate kinematic equation

  • We know the values of omega subscript 1omega subscript 2 and alpha, and we are looking for angular displacement theta, so the best equation to use would be

omega subscript 2 squared space equals space omega subscript 1 squared space plus space 2 alpha increment theta

Step 4: Rearrange and calculate the angular displacement increment theta

0 space equals space omega subscript 1 squared space plus space 2 alpha increment theta

negative 2 alpha increment theta space equals space omega subscript 1 squared space

increment theta space equals space fraction numerator omega subscript 1 squared over denominator negative 2 alpha end fraction space equals space fraction numerator open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses squared over denominator negative 2 cross times negative 0.8 end fraction

Angular displacement, increment theta = 13.88 rad

Step 5: Determine the number of rotations in increment theta

  • There are 2π radians in 1 rotation
  • Therefore, the number of rotations = fraction numerator 13.88 over denominator 2 straight pi end fraction = 2.2 
  • This means the turntable spins 2.2 times before coming to a stop

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Ashika

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