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Supplementary Prob 1 Conservation of Energy.

  1. Mar 18, 2004 #1
    A girl of mass m ties a rope of length R to a tree branch over a creek and starts to swing from rest at a point that is a distance R/2 lower than the branch. What is the minimum breaking tension for the rope if it is not to break and drop the girl into the creek?
     
  2. jcsd
  3. Mar 18, 2004 #2

    ShawnD

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    Her potential energy will change into kinetic energy.

    [tex]\frac{1}{2}mv^2 = mgh[/tex]

    [tex]v = \sqrt{2gh}[/tex]

    substitute values into that

    [tex]v = \sqrt{19.6(\frac{R}{2})}[/tex]

    The tension in the rope is based on the girl's centripetal acceleration AND the force of gravity (am I right now?)

    [tex]F = ma (gravity) + ma (centripetal)[/tex]

    [tex]F = m(9.8) + m(\frac{v^2}{R})[/tex]

    [tex]F = 9.8m + m\frac{(\sqrt{19.6\frac{R}{2}})^2}{R}[/tex]

    [tex]F = 9.8m + m\frac{19.6\frac{R}{2}}{R}[/tex]

    try to simplify from there
     
    Last edited: Mar 18, 2004
  4. Mar 18, 2004 #3
    May you please explain more on why The tension in the rope would be based on the girl's centripetal acceleration?
     
  5. Mar 18, 2004 #4

    HallsofIvy

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    Because force is mass times acceleration!
     
  6. Mar 18, 2004 #5

    ShawnD

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    I think I made a mistake, I did not factor in gravity. Re-read my previous post now that I have fixed it.
     
    Last edited: Mar 19, 2004
  7. Mar 18, 2004 #6
    Ok true, I start getting the concept!
    thank you very much
     
  8. Mar 18, 2004 #7
    I am lost now, where you didn't factor gravity? the minimum tension would be 19.6m.
     
    Last edited: Mar 19, 2004
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