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Angular Acceleration - What am I doing wrong?

  1. Sep 24, 2009 #1
    I am trying to answer the following question:

    A cyclist starts from rest and pedals such that the wheels of the bike have a constant angular acceleration. After 12.0 s, the wheels have made 26 revolutions. What is the angular acceleration of the wheels?

    My attempt at a solution:

    This problem seems extremely simple, but for some reason I'm not getting the answer.

    From the question I can determine that the initial angular velocity of the cyclist is equal to zero at t = 0.

    When t = 12.0s:

    I worked out the 26 revolutions in 12 seconds to 13.6 rad/s by

    [tex]\frac{26}{12.0}[/tex] = 2.17 rev/s = [tex]\frac{130 rev}{1 min}[/tex]

    Therefore, [tex]\frac{130 rev}{1 min}[/tex] x [tex]\frac{2\pi rad}{1 rev}[/tex] x [tex]\frac{1 min}{60 s}[/tex] = 13.6 rad/s

    [tex]\omega[/tex]f= 13.6 rad/s

    So, [tex]\alpha[/tex] = [tex]\frac{\Delta\omega}{\Delta(t)}[/tex]

    [tex]\alpha[/tex] = [tex]\frac{13.6 rad/s - 0 rad/s}{12.0 s}[/tex]

    [tex]\alpha[/tex] = 1.13 rad/s2

    Also, I'm assumining that the wheels are moving in a clockwise direction therefore making the velocity and the acceleration negative; giving me a final answer of -1.13 rad/s2

    Now I know this answer is wrong, I just don't know why.

    Can someone give me a hand? Thanks in advance.
  2. jcsd
  3. Sep 24, 2009 #2


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    Hold on a minute! Angle divided by time is angular velocity only if the acceleration is zero (constant angular speed). You have to use the rotational kinematics equations.
  4. Sep 24, 2009 #3
    Oh wow! Thank you so much.
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