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Cow Eats Grass!

  1. Mar 15, 2008 #1
    There is a big field full of green grass. At its centre is a circular fence, of radius R. a cow is tied to the circumference of the fence using a rope whose length is exactly 2[tex]\Pi[/tex]R. Now, the cow starts eating the grass. we have to find the maximum area of the grass it can eat. the cow can not enter into the circular region bounded by the fence, which is solid.

    MY ATTEMPT AT THE SOLUTION:

    the cow can eat all the grass located in a semicircular region of radius equal to the rope length, lying on the opposite side of the tangent drawn at the point of tying...
    so that adds [tex]\Pi[/tex]^3 x R^2 sq units of grass to the cow's account.

    once the cow finishes this it crosses the tangent and the moment it does so the rope starts winding at the fence. let us assume that a length of rope subtending an angle [tex]\theta[/tex] is now wound. then L-R[tex]\theta[/tex] is the temporary radius of the semicircle where the cow can graze. but it is not possible to find out the area beacuse we will inevitably count a patch of area several times..... i have tried hard and am unable to find my way... help will be thankfully received....
     
  2. jcsd
  3. Mar 15, 2008 #2

    Ouabache

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    Since the problem states the cow is tied to the circumference of the fence (and not a point along the fence), another reasonable interpretation, is that the cow can move freely around the entire circumference of the circle and graze out perpendicular to the fence at distance 2[itex]\pi[/itex]R. (where R is radius of the inner circle). So the maximum area the cow will cover can be the shape of an annular ring. (shown in black on that reference).
     
    Last edited: Mar 15, 2008
  4. Mar 15, 2008 #3
    No, the problem means that the rope winds up.... because the point where the cow is tied up is fixed, and the i think we need to evaluate a tricky integral to get the answer... which i am unable to do...
    the the rope can wind fully around the fence as its length equals the circumference. if we need to find out the area which it can graze on one half.... the other half would be a reflection... i can't provide a figure.. but it is quite easy to visualize... but i am unable to find even this area....
     
  5. Mar 15, 2008 #4

    HallsofIvy

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    Yes, but that makes it much too easy! The problem normally is stated as "A cow is tied to a point on the circular fence.

    If I had given this problem to a class and a student gave your answer, I would say, "Yes, you are completely correct- I stated the problem incorrectly." And then whack him really hard!
     
  6. Mar 15, 2008 #5
    i surely agree that i have made an error in stating the problem.... well i wrote it up in my own words and may be some parts of it were not clear... but here are some essential points which i stress again-
    1. the cow may not enter into the fence.
    2. the cow is tied to a fixed point on the fence so that it may not move to a distance greater than the rope length from that point
    3. the cow will try to eat all the grass that is in its reach ( it has been starving!)
    4.the fence is solid (we may safely assume it to be a smooth circular wall as well) so that if the cow crosses the tangent line at that point, the rope begins to wind up the fence.
    5. the cow may go round the fence completely and reach the point wehre it is tied- as the rope length equals the circumference of the fence.
    6. of course the rope is inextensible unbreakable flexible etc.

    inconvinience caused is regretted; this explanation may present a more clear picture...
     
  7. Mar 15, 2008 #6

    tiny-tim

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    … its coordinates are … ?

    Hi puneeth! :smile:

    The first step is to write the equation of this curve.

    For every length xR, the rope goes an angle x round the fence, and then goes off at a right-angle for a distance (2π - x)R.

    So its coordinates are … ? :smile:
     
  8. Mar 15, 2008 #7

    Ouabache

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    "<HallsofIvy>.... And then whack him really hard! " :rofl:

    I believe what HallsofIvy means is that if this were an exam question, it is a good idea to go up to the professor (TA, proctor..), during the exam and clarify any ambiguities in the wording. But for a homework problem, it is valid to solve a problem based on the given information and be sure to state your assumptions.
     
    Last edited: Mar 15, 2008
  9. Apr 5, 2011 #8
    Unfortunately I don't have access to the scanner so I am unable to upload a drawing of my explanation. Hopefully, I'll be able to do it sufficiently well in words. All points below are in polar coordinates.

    Assume that the center of the circle is at [tex](r,\theta) = (0,0)[/tex] and the cow is attached to point (R,0). There are two regions to consider.

    1. The semicircular region on the right of (R,0), whose area can be trivially calculated. Call this A.

    2. For any point to the left (and above) of (R,0), the rope will wrap around part of the circumference. Let the angular region of wrapping be [tex]\phi[/tex]. The remaining length of the rope (of length [tex]R(2\pi - \phi)[/tex]) forms a tangent to the circle at point [tex](R,\phi)[/tex]. This fact can easily be used to compute the coordinates of the end point of the stretched out rope. This will give the locus of all such points [tex]\equiv r(\phi)[/tex].

    The area to be calculated is
    [tex]\int\frac{1}{2}r(\phi)^2 d\theta = B[/tex]

    where [tex]\theta[/tex] is of course the polar angle for the point [tex](r,\theta)[/tex] and is different from [tex]\phi[/tex].

    The limits of the integral is [tex]0\leq\phi\leq\pi[/tex]. Explain these limits!

    The maximum area the cow can graze on is then A+2B. (If you were able to explain the limits, then this should be clear).
     
    Last edited: Apr 5, 2011
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