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Rotational Kinematics

  1. Nov 3, 2003 #1
    A phonograph record slows from an initial 45 rpm at a rate of .05 rad/s^2.

    a) how long does it take to come to rest???

    orig rotational velocity = 45 rpm = 4.71 rad/sec
    acc = -.05 rad/s^2

    0 = 4.71 rad/s + (-.050 rad/s^2) t
    t = 94.2 s

    b) How many revolutions does it make before stopping???

    (w + w)/2 = (4.71 rad/sec)/2 = 2.355 rad/s * 94.2 s = 221.84 rads =
    35.31 rev

    Last one for the week. How does this look???

    Thanks agaiin
    Nautica
     
  2. jcsd
  3. Nov 3, 2003 #2
    I think it's correct.

    Here's a shorter and more precise way to do (b):

    Without slowing down, you get
    45*94.2/60 = 70.65 revolutions.
    When slowing down, you know the answer is half of that:
    35.325 revolutions.
     
  4. Nov 3, 2003 #3
    Thanks, but for some reason the longer version makes more sense to me.

    Nautica
     
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