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UM Motion

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose the earth is a perfect sphere with R=6370 km. if a person weighs exactly 600.0 N at the north pole how much will the person weigh at the equator? (hint: the upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case) ans: 597.9 N

    2. Relevant equations


    3. The attempt at a solution

    I have no real clue on how to solve such a question, i plugged in for Fg but the answer i seem to get is higher than that of 597.9 N. Need help for a solution please!!!!!
  2. jcsd
  3. Jan 20, 2009 #2
    I will give you a hint.......A person standing on the equator experiences a centripetal force towards the centre of the earth .Work out what this is and draw a free body force diagram for the person.
  4. Jan 20, 2009 #3

    Doc Al

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    Apply Newton's 2nd law. What's different about being on the equator compared to being on the north pole?
  5. Jan 20, 2009 #4
    Think about the angular velocity at the north pole and the equator. Then calculate the centripetal acceleration at the equator and use this value to find the effective acceleration of gravity at the equator.
  6. Jan 20, 2009 #5
    how could i determine the angular velocity though? i am not given a time
  7. Jan 20, 2009 #6
    Last edited: Jan 20, 2009
  8. Jan 20, 2009 #7
    so i should use Fc=4pie^2Mr/ T^2 to solve for Force centripetal
  9. Jan 20, 2009 #8
    Yes, use this equation but it's a little easier to find the effective acceleration of gravity at the equator, g - acentripetal, then multiply by the mass.
  10. Jan 20, 2009 #9
    okay so it would be easier to just get the centripetal acceleration, then take gravity- acceleration centripetal then multiply by mass for F=ma right? but how would i get the acceleration centripetal when Ac=V^2/r and i don't have V?
  11. Jan 20, 2009 #10
    V= earth's angular velocity times the earth's radius.
  12. Jan 20, 2009 #11
    i am not familiar with earths angular velocity?
  13. Jan 20, 2009 #12


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    Really? You don't know how long a day is? Or you don't know how to calculate angular velocity from knowledge of the period of rotation?:wink:
  14. Jan 20, 2009 #13
    don't know how to calculate angular velocity from knowledge of the period of rotation? I've never heard of the term for one so maybe that's why im lost with what your saying
  15. Jan 20, 2009 #14


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    Angular frequency is what you really need for this problem; and it is defined by:

    [tex]\omega\equiv2\pi f=\frac{2\pi}{T}=\frac{|v|}{r}[/tex]

    Angular velocity is a vector quantity with magnitude [itex]\omega[/itex] and direct determined by a cross product.
  16. Jan 20, 2009 #15
    okay thank you i have gotten the correct answer, much appreciated
  17. Jan 20, 2009 #16
    It takes 24 hours(one day)for one rotation of 360 degrees or 2pi radians.The centripetal force is m vsquared/r or mr omega squared,use the equation you are happiest with.At the poles the upward push of the scale (R) will equal the persons weight(600N).At the equator there must be a resultant force(the centripetal force)because the earth is spinning and the person is accelerating towards its centre.It follows that R cannot be equal to the weight because the resultant force will then be zero and it further follows that weight minus R equals the centripetal force.From this you can calculate the recorded weight(R)
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