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Here's a tough one

  1. Oct 28, 2003 #1
    Well I've been working on this question for more than three days and I've come to the solution that I have no clue how to do it. It's the finally of six questions and being that it's homework I really don't expect you to just give me any answers, but some clear guidance would be really appreciated, being that I couldn't find a clue, in a field of clue, during clue mating season, while doing the clue mating ritual dance. Anyway here's the question:

    Your favourite physics teacher who is late for class attempts to swing from the roof of a 24m-tall building to the bottom of an identical building across from it, using a 24m-long rope. He starts from rest with the rope horizontal, but the rope will break if the tension force is twice the weight of the teacher. How high is the swinging physics teacher above level ground when the rope breaks.

    And that's the question, no mass, nothing of the sort. When we told our teacher we were lost, he told us to remember conservation of energy and that [mgh = mgh + 1/2m(v^2]

    To be quite honest that really didn't help me. As such any help you could provide would be greatly appreciated. Thanks.
  2. jcsd
  3. Oct 28, 2003 #2


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    The centripetal acceleration provided by the rope is v^2/r (circular motion).

    He'll be accelerating as he goes down the curve. If you work on re-arranging the equations, you'll see that the mass drops out.

    That help any?
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