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Angular Acceleration

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A 26.0 in diameter wheel is rotation initially at 210.0 revolutions per minute. It slows down uniformly and comes to rest in 15.0 seconds.

    a.) what is the angular acceleration?
    b.) through how many revolutions did it turn in those 15.0s?



    2. Relevant equations



    3. The attempt at a solution

    a.)

    f= 210.0 / 60s = 3.5 Hz

    ang. accel= 2(pi)f= 2(pi) x 3.5
    = 21.99 rad/s

    21.99 / 15.0 = 1.47 rad/s^2

    b.) 1/2(210 + 0)(15.0)
    = 1/2(3150)
    = 1575
    = 1.60 x 10^3 rev

    Can anyone confirm?????
     
  2. jcsd
  3. Jul 12, 2010 #2
    There is a problem here your wheel is slowing down but makes more revolutions in 15s then in a min at the initial velocity.What equation are you using there ?
     
  4. Jul 12, 2010 #3
    for b.) i was using theta = 1/2(initial velocity + final velocity) x time
     
  5. Jul 12, 2010 #4
    Check your units. You should not use t=15s. Your units must be consistent.
     
  6. Jul 12, 2010 #5


    so 1/2(3.5 + 0) (15)
    = 1/2 (52.5)
    = 26.25 rev
     
  7. Jul 12, 2010 #6
    Now use another equation to check your answer.
     
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