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A puzzle

  1. Oct 14, 2004 #1
    Hi,
    I have tried to solve this puzzle:
    If we had a round meadow and a cow tied with a rope on the edge of that meadow. Hw long should be a rope if we wanted to let the cow eat a half of that meadow?
    I have tried to solve this puzzle time ago and now it attracted me again. Does anyone know how it could be solved?
    Regards,
    Niko
     
  2. jcsd
  3. Oct 14, 2004 #2

    Alkatran

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    What are the dimensions of the meadow?
    Meadow area = width*height
    Cow area = (rope length)^2 * pi
    cow can eat half:
    width*height/2 = (rope length)^2 * pi
    square root(width*height/(2*pi)) = rope length
     
  4. Oct 15, 2004 #3
    Meadow is rotund, not rectangular.
    The radius of the meadow is whatever size. We have to find a relation between rope length and radius length. Assume the radius length is 1.
     
  5. Oct 15, 2004 #4

    Alkatran

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    alright then

    m^2/2=r^2
    sqr(m^2/2) = r
    m/sqr(2) = r

    oooo hard :tongue2:
     
  6. Oct 15, 2004 #5

    HallsofIvy

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    You're still not reading the problem correctly. You are assuming the cow is tethered INSIDE the circle so that the area grazed is a complete circle. That's not true- the cow is tethered at the edge of the circle so the area she can graze is only a portion of a circle.
     
  7. Oct 17, 2004 #6

    uart

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    You need to get an expression for the area swept out by the tether rope as a function of it's lenght. Normalize the problem by taking the meadow to be unit radius and let the rope lenght be "r". You can get the following expression for the area "A" swept by the rope.

    A = r^2 acos(r/2) + acos( 1 - 0.5 r^2) - r sqrt( 1- 0.25 r^2)

    Now solve numerically to find the value or r which gives A=Pi/2, which turns out to be somewhere around 1.15 to 1.16 times the meadow radius.
     
  8. Oct 17, 2004 #7

    Gokul43201

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    Hasn't this come up before?
     
  9. Oct 18, 2004 #8

    uart

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    Not the I know of. Are you sure you're not thinking of the 0.999(repeated) != 1 thread. ;)
     
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