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Centripetal Acceleration with distance

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data
    You plan to throw stones by using a sling of length 0.7 m which you whirl over your head. Suppose you wish to throw a stone a distance of 36 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.


    2. Relevant equations
    ac=v2/r


    3. The attempt at a solution
    I tried to find the velocity in the x-direction but I didn't have an angle to work with. I then solved x=v0cosθ(t) for t and plugged that into y=y0+v0sinθ(t)-g/2(t)2. This gave me -yo=xtanθ-((gx2)/(2v02cos2θ). That didn't cancel anything out.
    I'm not sure where to go from there.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2012 #2

    PeterO

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    Homework Helper

    For this sling, it is the speed of the projectile which determines how far it goes. You will either have to be given or assume an angle of release to find that speed.
    Once you know the speed you can calculate the centripetal acceleration.
     
  4. Feb 12, 2012 #3
    I found time t it takes for the stone to fall 3.0m.

    t=√(yo2/g)

    then plugged that t in to find the velocity it would take the stone to go 36m.

    v0,x=x/t

    I plugged that v, and my r, in v2/r to get my centripetal acceleration although it gave me a too large of number. Where am i going wrong?
     
  5. Feb 12, 2012 #4

    PeterO

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    Homework Helper

    Show me your numbers. I could work it the way you describe.
     
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