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Rotation Problem: Earth lengths/Apparent Weights

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    A friendly Brazilian has a mass of 150 kg. Being in Brazil, he rotates in a circle around the center of the earth once per day, The radius of this circle (which is essentially the radius of the Earth) is 6.40 x 10^6 m.

    I have found that the normal force is 1.4649 x 10^3.

    2. Relevant equations
    ac = v2/r , centripetal acceleration
    [itex]\Sigma[/itex]F=ma
    v = (2[itex]\pi[/itex]r)/T , T is period of one revolution


    3. The attempt at a solution
    I only got to mg - Fn = mac
    I know what the apparent weight is but I'm not sure about the normal force. If I can find this, then I can then find ac, v, and then T to find about the length of day.
     
  2. jcsd
  3. Oct 30, 2011 #2

    ehild

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    What is the question?

    ehild
     
  4. Oct 30, 2011 #3
    Sorry forgive me. The question is: How long would a day have to be for the Brazilian's apparent weight to be 1.46 x 10^3 N?

    Btw, the answer should be about 6.16 x 10^4 sec or 17.1 hours.
     
  5. Oct 30, 2011 #4

    ehild

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    The weight is equal to the force the man presses a horizontal support - scales, ground, chair ... So its is the same as the normal force.

    ehild
     
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