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Forces on a Ferris Wheel

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that you're on a circular ferris wheel with a radius of 22 m. You know that your mass is 64 kg on the very bottom part of the ferris wheel, even though your actual mass is 58 kg. How long does it take for the ferris wheel to go one revolution?

    2. Relevant equations
    I believe centripetal force? F = mv^2/r
    Gravitational force: F = mg
    Circumference: C = 2*pi*r


    3. The attempt at a solution
    So, what I did was set the centripetal force equal to the gravitational force:
    mv^2/r = mg

    which implies:

    v = Sqrt[g*r] which gave me the value 14.6 m/s. I then found the circumference of the ferris wheel to be 2*pi*22 = 138.23 m, and used the relation t = displacement/velocity to obtain 9.41 s.

    This, however, came out to be wrong... is my problem setup incorrectly?
     
  2. jcsd
  3. Nov 4, 2011 #2

    PhanthomJay

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    The centripetal force is not in general equal to the gravitational force. It is equal to the net force in the centripetal direction (inward toward the center of the circle). You have to identify both forces acting on you. One is the weight. What's the other? And what therefore is the net force? Incidentally, the problem incorrectly states that there is a change in mass. I hope you catch this error.
     
  4. Nov 5, 2011 #3
    I made a mistake, the problem stated that you measured yourself on a scale on the very bottom of the circle to be 64 kg. I don't see the other force...
     
  5. Nov 5, 2011 #4

    PhanthomJay

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    If the scale reads 64 kg, you must first convert it to force units in Newtons. That force, sometimes called your 'apparent' weight, is the normal contact force of the scale acting upward on you. The other force acting on you is the gravity force (your actual weight (in Newtons!) acting always downward on you). The vector sum of these 2 forces is the net force acting on you.
     
  6. Nov 5, 2011 #5
    Oh... I think I see it. So, the normal force would be (64kg)(g) and the gravitational force would be (59kg)(g), thus the net force is 5*g upwards, which implies that this is what the centripetal force is equal to?

    Is my reasoning correct?
     
  7. Nov 5, 2011 #6

    BruceW

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    That sounds right to me. (Actually, it is the centripetal acceleration, if you check the units).
     
  8. Nov 5, 2011 #7

    PhanthomJay

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    Yes, that is the net force upward (toward trhe center of the wheel) in Newtons, where g = 9.8 m/sec^2, although note you mentioned m = 58 kg in your original post, not 59 kg.. Now you must solve for the time it takes the ferris wheel to go one revolution at the constant speed you will first calculate.
     
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