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Length of the rope around two circles

  1. Jun 14, 2006 #1

    To determine the shortest rope that can be wrapped around the two circles

    any ideas? i have forgotten everything about geometry. My nephew asked me this
  2. jcsd
  3. Jun 14, 2006 #2


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    Well you basically have four segments: two circular arcs and two straight lines. The problem is that you have to find where exactly the rope is tangent to the circles (aka where the straight lines turn into circular arcs) To find that point it looks like you can strategically place some right triangles and use trigonometry to solve it.
  4. Jun 15, 2006 #3
    It's been a while but I am almost positive that you just combine the diameters or radii and just use the circumference formula c = pi*d = 2pi*r

    So, the shortest length of rope that can we wound around two circles of diameters 6ft and 18ft is pi*(6+18) = 24pi ft.
  5. Jun 15, 2006 #4


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    You do understand that "a straight line is the shortes distance between two points" don't you? Your solution, wrapping the rope around the two circles, has the rope going from the point of tangency ln one circle, around the circle to the point of contact, then around the other circle to the point of tangency on the other circle. That is much longer than just the straight line from one point of tangency to the other!
    In this case "almost positive" isn't good enough.
  6. Jun 16, 2006 #5
    You are absolutley right HallsofIvy. I apologize. I was more than a little dazed and confused when I made my original reply to the thread.
  7. Jun 16, 2006 #6
    if i do use triangles, what length do i use, the radii and what else, i am almost there.

    and my nephew hasnt taken trig yet, not sure if i can be any of help if i explain to him.
    Last edited: Jun 16, 2006
  8. Jun 16, 2006 #7


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    Well, you might use an algebraic approach; I'll help you along a bit:

    Let the radii be r and R, respectively.

    Let the r-circle have centre at (0,0), the R-circle at (R+r,0)
    Thus, the two circles have the representations:
    [tex]x^{2}+y^{2}=r^{2}, (x-(r+R))^{2}+y^{2}=R^{2}[/tex]
    Let a point of tangency at the r-circle be denoted as [itex](x_{0},y_{0})[/itex]
    (Note that the x-value of the tangency points on the r-circle will be unique, whereas there must be two y-values).

    The unit tangent at [itex](x_{0},y_{0})[/itex] in the direction of the other circle is [itex]\vec{T}=(\frac{y_{0}}{r},-\frac{x_{0}}{r})[/itex], and we may parametrize its tangent line as:

    Now, put this expression into the equation for the R-circle:
    Or, we may rewrite to:
    which is a second degree equation in t, as expressed with the other variables!

    Now, note that if [itex]\vec{L}[/itex] is also tangent to some point on the R-circle, such a point will be the ONLY point on the R-circle coinciding with [itex]\vec{L}[/itex]
    If [itex]\vec{L}[/itex] is NOT a tangent line to the R-circle, it will either intersect this circle twice, or not at all.

    Thus, if we require that the discriminant of the second-degree equation is zero, then we know that the t-value thus gained gets us the point of tangency on the R-circle.
    The requirement that the discriminant be zero, furnish us with the last equation in order to determine [itex](x_{0},y_{0})[/itex] in terms of r and R (the other being the trivial requirement that the point lies on the r-circle.

    Hope this helps. :smile:

    In the above, I have of course ignored that the two circles have a common tangent line also for their point of contact. You'll find this solution as well when going through the calculations.
    Last edited: Jun 16, 2006
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