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Rotational Kinematics

  1. Oct 17, 2004 #1
    Hello,

    I have a problem where I honestly have no idea how to get started. I have a slight idea what might be invloved but other than that, I am stumped.

    Here's the problem: At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.6m above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.4 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

    What's throwing me is that there is a height involved and none of the examples provided in my book or from my professor for that matter included a problem with height. I think I read somewhere however that d = s but again, I don't understand how to work this. I know I will have to use one of the Rotational Kinematic equations, but like I said this is about all I know.

    Thanks for any help provided.
     
  2. jcsd
  3. Oct 17, 2004 #2

    arildno

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    1. Find the time it takes to fall 8.6 meters.
    The fact that you are rotating, will not affect that time.
    2. You may now determine the number of cycles the diver has made.
     
  4. Oct 17, 2004 #3
    Ok, So what I know is:

    Theta: -
    Omega initial: 1.4 rev/s
    Omega final: -
    alpha: -
    time: ?

    Again I'm still stuck on the distance portion of that. Do I use one of the regular kinematic equations to find time or do I have to convert the 8.6m to something else to use one of the rotational kinematic equations?
     
  5. Oct 17, 2004 #4

    arildno

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    You've been given the AVERAGE angular speed. Use that as a constant...
    "Do I use one of the regular kinematic equations to find time"
    Yes!
    Setting your water surface at y=0, you have vertical position at time "t":
    [tex]y(t)=8.6-\frac{gt^{2}}{2}[/tex]
    Find the "t"-value T so that y(T)=0
     
  6. Oct 17, 2004 #5
    Hmm, I don't think I did this correctly. My answer is 1.32.
     
  7. Oct 17, 2004 #6

    arildno

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    1.32 what? seconds or revolutions??
     
  8. Oct 17, 2004 #7
    Oh seconds. Sorry about that.
     
  9. Oct 17, 2004 #8

    arildno

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    Seems about right..
     
  10. Oct 17, 2004 #9
    Oh, ok. Well, I finished out the problem the way I thought I should and have come to get an answer of 11.93. Does that sound right? I'm thinking that 11 revolutions is a bit much in 8 meters.
     
  11. Oct 17, 2004 #10

    arildno

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    However did you manage that??????
    1.4(rev/s)*1.32s=11.93(revs)??????????????????????
     
  12. Oct 17, 2004 #11
    I used what I thought I was supposed to do. LOL I used a kinematic equation. Obviously that was not the right thing to do. I am a complete novice when it comes to physics. If my book doesn't have clear and explicit examples to follow, then I am sunk when it comes to doing a problem.

    Anyway, the correct answer would be 1.85 revolutions.

    Thank you for your help. It is much appreciated. :smile:
     
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