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Circular motion problem (A2 level)

  1. Mar 26, 2009 #1
    Hey, I am having a problem understanding something to do with circular motion, its the last part of a question.

    1. The problem statement, all variables and given/known data
    part 1 - Calculate the angular speed of a girl standing at the equator, i used angular speed = 2pi/(60x60x24) = 7.3x10^-5 rad/s

    part 2 - The radius of the earth is 6400km and the girl has a mass of 60kg, calculate the resultant force on the girl necessary for this circular motion. I used
    F=mass x radius x angular speed ^2 and got 2.03N

    Part 3 - (The problem part) If the girl was standing on weighing scales calibrated in Newtons, what reading would she get?
    I used reading = mg + resultant Force which got me an answer of 590.6N (taking g = 9.81) However the mark scheme stated that i had to take away the resultant force rather than add it, I was just wondering why this was the case? I thought that the force of the girl would be towards the centre of the earth and therefore must be added rather than taken away :confused:

    Thanks in advance
     
  2. jcsd
  3. Mar 26, 2009 #2

    Doc Al

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    Staff: Mentor

    The net force on the girl is towards the center of the earth. What individual forces act on the girl?

    Hint: If the earth didn't rotate, the forces on the girl would add to zero: The upward normal force (which is what the scale reads) and the downward gravitational force (mg) would be equal. How does that change when you consider the earth's rotation?
     
  4. Mar 26, 2009 #3
    well the forces acting downwards are her weight and the centripetal force? and upwards is the contact force between her and the scales?
     
  5. Mar 26, 2009 #4

    Doc Al

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    Right.
    The "centripetal force" is just the name we give to the net force, not a separate force in itself. The net force is downward, though.
    Right.

    So the two actual forces on her are her weight and the contact force. They must add up to equal the net centripetal force. (Set up that equation.)
     
  6. Mar 26, 2009 #5
    so is it Force = weight (mg) - contact force?
     
  7. Mar 26, 2009 #6

    Doc Al

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    Right! (By "Force" I assume you mean "centripetal force".)

    So solve for the contact force. Is it more or less than her weight?
     
  8. Mar 27, 2009 #7
    so its (9.81 x 60)-2.03 = 586.57N
     
  9. Mar 27, 2009 #8

    Doc Al

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    Good!
     
  10. Mar 27, 2009 #9
    :) oh i get it now, thanks for all the help :)
     
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