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Frictionless physics problem

  1. Feb 13, 2008 #1
    1. A 1.5 kg rock is being twirled in a circle on a frictionless surface using a horizontal rope. The radius of the circle is 2.00m and the rope makes 100 revolutions in 1.00 minutes. What is the tension in the rope?

    This problem has totally confused me. I don't even know what equation to use that will relate tension to the other givens in the problem!

    2. What is the apparent weight of a 75.0 kg person traveling at 100 km/hour:
    a) over the peak of a hill with a radius of curvature = to 500m
    and
    b) at the bottom of a hollow of the same radius

    I'm totally lost!!! HELP!!!!!!!!!!!!!:cry:
     
  2. jcsd
  3. Feb 13, 2008 #2
    1) Here are the equations you need:
    Circumference = 2*pi*Radius
    Speed = Distance / Time
    Centripetal Force = Mass * (Speed^2) / Radius
    So, if you know the Centripetal Force, what do you think the tension on the rope would be?

    2) You should be able to use the equation for centripetal force I gave you here to find the force needed to keep the person on his path. It will be the same magnitude for both at the bottom of the hill and the top, just opposite in direction. When at the top of the hill, some of the force of gravity will be that force keeping the person on the path, and the remaining part will be the apparent weight. At the bottom of the hill, the force of gravity is pushing the person off the path, meaning the normal force pushing the person back on the path (the person's apparent weight) is greater.

    Force of Gravity = Mass * 9.8m/s^2 down
     
  4. Feb 13, 2008 #3
    I still don't know what equation to use to relate tension to the other values.....
     
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