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I have no idea how to start this help

  1. Nov 29, 2005 #1
    A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path,(a) at the beginning of his motion, (b) at a height of 1.5 m above the bottom of the circular arc, and (c) at the bottom of his arc.

    I think on the first part you take the sum F=0 I think
     
  2. jcsd
  3. Nov 29, 2005 #2

    mezarashi

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    Use free-body diagrams (there will be angles involved at the 1.5m above bottom point). Use the principle of conservation of energy to find the energy at various heights. Assume he starts at v=0.

    The force summation in the direction of the length of the rope must be equal to F = v^2/r as he is undergoing circular motion. There are only two forces involved: rope tension and gravity.
     
  4. Nov 29, 2005 #3
    so will the angles made on the triangle be 45degrees? and I use them in my calculations?
     
  5. Nov 29, 2005 #4

    mezarashi

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    Where are you getting the 45 degrees from? Are you using trigonometry?
     
  6. Dec 5, 2006 #5
    Well the angle at 1.5m from the bottom of the circle would be arccos(2.5/4) = 0.89 rad --> pi/2 - 089 rad = 0.68 rad is the angle at that moment. Using the equation F = ma, you get F = m*V^2/r. However, I don't see how you can get velocity if you don't have time... Can someone help it out? I'm not getting it either. And I'm not sure when the angle I calculated comes into play
     
  7. Dec 5, 2006 #6
    b)ohhh, I see what u mean by using the conservation of energy. So, it is mgh = 0.5mv^2 + mgh,
    9.8 * 4 = 0.5 * v^2 + 9.8 * 1.5
    v = 7m/s.

    Plugging in the equation F = ma,
    F = m * v^2 / r

    F = 70kg * 7^2 / 4
    F = 857N
    So the answer is 857 N??? I think this number is too large...

    For a, the answer should be 0, because when you plug in F = m * v^2/r, since v is 0, F would also be 0
     
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