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Rotational motion of a turntable

  1. Jul 15, 2010 #1
    1. The problem statement, all variables and given/known data
    The turntable of a record player rotates at 33.33rev/min and takes 20.0s to reach this speed from rest. Calculate: a) its angular acceleration, b) the number of revolutions it makes before reaching its final speed.

    2. Relevant equations
    angular acceleration = angular speed/ time
    1 rev = 2pie rad
    angular displacement = angular speed x time

    3. The attempt at a solution
    part a:
    I converted the angular speed to rad/s. I solved for angular acceleration, which is delta angular speed divided by time.

    3.49rad/s / 20.0s = 0.175 rad/s^2 Is that correct?

    part b:
    I used the equation, angular speed = angular displacement/time.

    therefore, angular displacement = angular speed x time
    3.49rad/s x 20.0s = 69.8rad

    I then converted rad into revolutions --> 11.1 revolutions. Is that correct??

  2. jcsd
  3. Jul 15, 2010 #2
    Part A is correct.

    For part B you're on the right track, but remember that would only work if the record were spinning at a constant speed of 3.49rad/s. But it is always changing as it goes from 0rad/s up to 3.49rad/s.

    You want to do basically the same thing you did, except in place of 3.49rad/s you want to use the average angular velocity.
  4. Jul 15, 2010 #3

    So for part b, average angular velocity will be 3.49/2 = 1.75 rad/s.

    therefore, angular displacement = 1.75 x 20.0 = 35 rad --> 5.57 revolutions.
  5. Jul 15, 2010 #4
    Yup, that's it :D

    Although, technically if you round here it would be 1.74 rad/s because of the round even rule, that many people are unaware of.
    But don't mind that lol, the answer is right.
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