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Homework Help: Centripetal Acceleration HELP. Is it challenging? Well come find out!

  1. Oct 22, 2013 #1
    1. An athlete whirls a 7.48 kg hammer tied to
    the end of a 1.4 m chain in a horizontal circle.
    The hammer moves at the rate of 1.76 rev/s.
    What is the centripetal acceleration of the
    hammer? Assume his arm length is included
    in the length given for the chain.
    Answer in units of m/s

    2. Equations I am pretty sure I need are
    v = 2∏r/T
    ac= v2/r
    not sure if I need any others.

    3. I tried finding v first, and I assumed the 1.76 rev/s was T, so I did 2*∏*1.4(about 8.80) and divided it by 1.76, getting about 4.99 m/s
    and I plugged it in to the ac equation.
    So I squared the 4.99, getting about 24.98, and dividng it by 1.4, which gets me about 17.843.
    But that answer is wrong.
    I don't know what the answer is.

    I feel like what I am doing wrong is the 1.76 rev/s is not T, or it is, but I have to change it so it's in like m/s or something like that. If that's the case, then I need help finding that because I have no idea how to do that.

    Thank you for any help!
  2. jcsd
  3. Oct 23, 2013 #2
    in your equation for v, T is the time to complete one revolution (make sure you see why)

    1.76 rev/s

    is that what you want? or is it seconds per rev?
  4. Oct 23, 2013 #3
    I think I see where you are getting at.
    For every second that passes, it makes 1.76 revolutions.
    But to find the velocity, I need to put T as a time per revolution. So would I just do the inverse of T?
    So if i do the inverse for T, wouldn't that just be multiplying T with the 2∏r?
    But wouldn't I need to the the inverse of everything else as well? What you do to one side of the equation, you do to the other.

    So 1/v = T/2∏r
    And then do the inverse again to get the answer. But isn't that the same is the answer I would have gotten before? Am I even going in the right direction?
  5. Oct 23, 2013 #4
    well 1.76 is not T. Your messing around with thinking that it is. 1.76 is the angular speed, completely different from T.

    the equation v = 2∏r/T is correct. Find T.
  6. Oct 23, 2013 #5
    If I want to turn the rev/s into common terms, would I multiply 1,76 by 2∏r?
  7. Oct 23, 2013 #6
    I have yet to learn about angular speed in my class, so I know practically nothing on it.
  8. Oct 23, 2013 #7

    With what you have now, you have 1.76 revolutions over one second. Just do the inverse of that: (1/1.76) to find your period, T

    This is the relationship between frequency and period.

    You are given a frequency of 1.76 rev/sec.

    Frequency is (1/T) T is the period
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