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Centripital Acceleration

  1. Sep 12, 2007 #1
    You plan to throw stones by using a sling of length 1.5 m which you whirl over your head. Suppose you wish to throw a stone a distance of 24 m. What must be the centripetal acceleration of the stone just before its release if it is to reach this distance? Assume that the release height is 3.0 m.

    So from this question, I was able to seperate the motion of the stone into x and y.

    Xf= X0+Vt

    Yf=Yi+Vt-gt^2

    so putting them together I get
    t= (xf-x0)/vt

    yf=yi+V((xf-x0)/vt)-g((xf-x0)/vt)^2

    Am I on the right track? I got an intial velocity of 10.2242. I'm not sure. I was told by my teacher to not worry about angles and other things like that.
     
  2. jcsd
  3. Sep 12, 2007 #2
    My first question would be, are you whirling it in a horizontal circle above your head? Or are you whirling it in a vertical plane. If it's the horizontal case, then the amount of time it's in the air is going to be independent of the initial velocity; it'll simply be the amount of time it would be in the air due to free fall from the height you're whirling it at.

    In the case that you're whirling it in a vertical plane, then you would have to worry about the angle (you'd want something around 45 degrees to achieve the maximum distance at the lowest velocity.) But, since you were told not to worry about degrees, I'm guessing that it's the first case.
     
  4. Sep 12, 2007 #3
    It's a horizontal circle. What i posted is something he put on the board to help us. I'm just wondering if anyone else got a calculation close to mine
     
  5. Sep 12, 2007 #4
    Am I on the right track?
     
  6. Sep 12, 2007 #5

    learningphysics

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    Yes, you are on the right track... find the initial velocity first... but that's not what I'm getting for initial velocity:

    you need (1/2)gt^2 here...
     
  7. Sep 13, 2007 #6
    Hmm....how far off am I
     
  8. Sep 13, 2007 #7
    I'm still getting 10.2242
     
  9. Sep 13, 2007 #8

    learningphysics

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    I'm not getting that... can you show the calculations in more detail?
     
  10. Sep 13, 2007 #9
    0=3+V((24)/v)-4.9((24)/v)^2
     
  11. Sep 13, 2007 #10
    and that yielded 10.2242
     
  12. Sep 13, 2007 #11

    learningphysics

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    "Yf=Yi+Vt-gt^2" is wrong.

    this should be

    Yf=Yi-(1/2)gt^2

    The initial vertical velocity is 0.
     
  13. Sep 13, 2007 #12
    Okay, so would I solve for T in the X equation and plug it into the Y one.

    I got 78.0047, and that was also a wrong answer.
     
  14. Sep 13, 2007 #13

    learningphysics

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    can you show your calculation?
     
  15. Sep 13, 2007 #14

    learningphysics

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    What do you get when you solve the above?
     
  16. Sep 13, 2007 #15
    0=24-(1/2)(9.8)(24/v)^2
     
  17. Sep 13, 2007 #16

    learningphysics

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    wrong equation... see my previous post. The 24 should be a '3'.
     
  18. Sep 13, 2007 #17
    I kept plugging x positions into the y-coordinates

    could you explain to me why the intial velocity is zero
     
  19. Sep 13, 2007 #18

    learningphysics

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    initial vertical velocity is 0... not total velocity... what number did you get?

    Also note... the question asks for the centripetal acceleration, not the initial velocity. You need the initial velocity... then from that get the centripetal acceleration.
     
  20. Sep 13, 2007 #19
    i got 30.672, the answer was right

    I'm just not sure why the intial velocity is zero. I mean the rock is spinning.
     
  21. Sep 13, 2007 #20

    learningphysics

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    initial vertical velocity = 0... initial horizontal velocity = 30.672... it is spinning horizontally, not vertically.
     
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