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Centripetal Acceleration of washer spin cycle

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    On your first day at work for an appliance manufacturer, you are told to figure out what to do to the period of rotation during a washer spin cycle to triple the centripetal acceleration. You impress your boss by answering immediately.

    Express your answer in terms of T.

    2. Relevant equations

    a=(4pi^2R)/T^2

    3. The attempt at a solution

    I read the question, and I keep thinking they just want to triple the accleration...which is not the answer. I don't exactly know what they are asking for....
     
  2. jcsd
  3. Oct 7, 2007 #2

    Kurdt

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    Using that equation is a bit nasty since T is squared. You can tidy it up by remembering that [itex]R=v^2/a[/itex]. So once you have your equation You pretty much have your answer since you know how the two quantities are related. When one triples the other _______.
     
  4. Oct 7, 2007 #3
    This is part of an online homework and the program keeps telling me I am getting the wrong answer.

    It wants the answer in terms of T and I cannot have variables a, v, R or pi in there. I must be making an exteremely dumb mistake......
     
  5. Oct 7, 2007 #4

    Kurdt

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    If I said a was proportional to the inverse of T, that is:

    [tex] a \propto \frac{1}{T} [/tex]

    The you you want 3 x a, what will the above proportionality be? That is what would be done to T to make a 3 times larger?
     
  6. Jan 30, 2008 #5

    Kurdt

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    Why have you bumped this thread?
     
  7. Jan 30, 2008 #6
    Help with answer!!

    So does anyone have an idea what the answer is?

    I have this same question and for the answer they have it setup as

    T^3=_____________

    Thanks
     
  8. Jan 30, 2008 #7

    Kurdt

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    We don't provide the answers for you here. You have to show some of your own work.
     
  9. Jan 31, 2008 #8
    Try an algebraic approach if you can't get it by thinking about it.

    [tex]3a = 3\left[\frac{4\pi^2R}{T^2}\right][/tex]

    Now get rid of the 3 on the RHS by determining a coefficient for T within the square.
     
  10. Jan 31, 2008 #9
    Thank you BlindSide Youve been a great help.
     
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