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Swinging physics teacher problem!

  1. Jun 21, 2006 #1
    Hey all, I was doing a physics question in preparation for an exam, but I cant seem to figure it out!

    The question:

    "Your favourite physics teacher who is late for class attempts to swing from the roof of a 24 metre high building to the bottom of an idential building using a 24 metre rope as shown. She starts for rest with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the teacher. How high is the swinging physicist above level when the rope breaks?"

    [​IMG]

    What I tried was:
    Treat is as a centripetal force problem. Find the speed when T=2mg and W=mg. Once I find the speed, I can use conservation of energy to figure out what the height is. The problem is I dont know how to find speed! Any hints on how to figure out the Fnet when T=2mg and W=mg? Or am I doing this all wrong?

    Also, just to make sure, in the equation F=[m(v^2)]/r, F stands for Fnet, correct?
    If thats the case, then in the case of a rock attached to a string and being swung in a circle vertically:
    [​IMG]
    Then the net acceleration and force is not directed towards the centre anymore, correct?
     
    Last edited: Jun 21, 2006
  2. jcsd
  3. Jun 21, 2006 #2

    Hootenanny

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    Your well on your way, but don't forget to resolve the weight. So you have;

    [tex]F_{net} = T - mg\cos\theta[/tex]

    Where [itex]\theta[/itex] is the angle the rope makes with the vertical. So as an overall equation you have;

    [tex]T - mg\cos\theta = \frac{mv^2}{r}[/tex]

    Now, you can determine the velocity of the teacher at any point using conservation of energy.
     
  4. Jun 22, 2006 #3
    Obviously your favourite physics teacher isn't quite strong at physics.
    Tarzan would have done better.
     
  5. Mar 28, 2008 #4

    mld

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    hi.. i am working on the same problem.. however i can't figure it out, even with the help already given... any chance someone can post the complete solution?
    thanks for the help!
     
  6. Mar 28, 2008 #5

    Hootenanny

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    The forum rules prohibit the posting of complete solutions, however, we can help you through the problem if you show some effort. What have you attempted thus far? What are your thoughts/ideas?
     
    Last edited: Mar 28, 2008
  7. Mar 30, 2008 #6

    mld

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    Ok, well working with what you gave us, I have gotten the equation down to:
    T-mgcostheta=(mv^2)/r
    2mg-mgcostheta=(mv^2)/r
    2g-gcostheta=v^2/r
    2(9.8)-9.8costheta=v^2/24

    I'm not sure where to go from here. I know that the rope will break when the left side of the equation is equal to the right side, correct? I feel as if I'm a little out of my element here and any help would be great.
     
  8. Mar 31, 2008 #7

    Hootenanny

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    Your on the right lines. See the hint in one of my earlier posts,
     
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