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Rotational Motion and Equilibrium

  1. Apr 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A 2.5 kg pulley of radius 0.15m is pivoted about an axis through its center. What constant torque is required for the pulley to reach an angular speed 25rad/s after rotating 3.0 revolutions, starting from rest?

    2. Relevant equations

    torque = (mr^2)(angular acceleration)

    3. The attempt at a solution

    First I solved for time using t= omega/angular speed = 6(pie) rad / 25 rad/s = 0.75s.
    Then, I solved for angular acceleration = 33 rad/s

    Solving for torque, using the above equation, I got 1.9 m-N

    The textbook I have says this is the wrong answer. What have I done?

    The book says the answer is 0.47 m-N
  2. jcsd
  3. Apr 9, 2007 #2
    That seems wrong. Why is omega [itex]6\pi[/itex] radians? Isn't omega the angular speed? What you're calculating is the amount of time it would take for the pulley to rotate [itex]6\pi[/itex] radians at 25 rad/s.
  4. Apr 9, 2007 #3


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    In using t=omega/angular speed you are assuming that the angular speed is constant. As it starts from rest, this can't be right.
  5. Apr 10, 2007 #4
    Torque/I = angular acceleration. w=0+at, where a=angular accn. 6pie=0.5at^2, using equations of rotational motion. Solve to get torque.
  6. Apr 10, 2007 #5
    Oh, instead of omega above, I meant theta.
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