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Homework Help: Swinging on a rope minimum horizontal speed

  1. Dec 13, 2007 #1
    A hiker plans to swing on a rope across a ravine in the mountains, as illustrated in the figure, where L = 4.9 m and x = 2.7 m, and to drop when she is just above the far edge. At what minimum horizontal speed should she be moving when she starts to swing(in m/s)?


    I tried using the equation for range (x=(vo^2sin2(angle))/g but it didnt work, and im guessing im on the complete wrong track.
  2. jcsd
  3. Dec 13, 2007 #2

    Shooting Star

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    Homework Helper

    Why don't you do it the other way round? Suppose the hiker starts from the right at zero speed. Like a pendulum, she has to swing to the left side, and you can find the speed on the left side. Reversing the situation will give you that required speed.

    If you don't like this approach, I will of course give you the solution.
  4. Dec 13, 2007 #3
    You can also think about it this way, when the hiker is above the second edge, he has gained potential energy but lost kinetic energy. Equate the two and you get your minimum velocity.
  5. Dec 13, 2007 #4

    Doc Al

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    That range equation is for finding the horizontal distance traveled by a projectile--where the only force acting on the object is gravity. In this problem, the hiker is not a projectile--she's hanging on to a rope!

    Use conservation of mechanical energy.
  6. Dec 13, 2007 #5

    Shooting Star

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    Draw a proper diagram. The two positions of the rope make up an isosceles triangle. Drop a perpendicular from the right end-point (where she’s holding the rope), to the vertical rope on the left. The length of that perpendicular is x. Let the increase in vertical height from left position to the right position be h. Now you have a right angled triangle relating x, L and h.

    The KE at the start on the left is becoming PE on the right, because she has gained a height of h. Now can you say what should have been the initial KE to rise through a height h?
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