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Centripetal generated by rotation of the Earth

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data
    An object weighs 100N at the South Pole. How much does it weigh at the equator?

    Given: Earth's rotational spin speed 465m/s
    Diameter of the Earth is 1.274 x 10^7 m

    2. Relevant equations
    Fc = m x v^2 / r

    3. The attempt at a solution
    Well I have this general understanding that the object should weigh less since it's experience uniform circular motion, and that it "wants" to fly off tangentially (newton's first law) but is constrained by the force of friction, right? It seems I have calculated the answer of the centripetal force to be 0.3 N, but what I don't understand is that why I subtract it from 100N. The confusion lies in where the normal force and centripetal force vectors are located. It would be greatly appreciated if anyone could give a lucid answer to this question. The answer by the way, is 99.7N
  2. jcsd
  3. Nov 10, 2009 #2


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    The object weighs 100N so there is a force of 100N due to gravity pulling it down

    At the equator there is a 0.3N force pushing it up, so if you put it on a pair of scales there would be 100N down and 0.3N up so the scales would read 99.7N
  4. Nov 10, 2009 #3
    Yes, but how do you get 0.3 N pushing up? The centripetal force towards the center of earth is 0.3 N, so the normal force is also 0.3 N. If the object is at the south pole. The only two forces would be Fg pulling it towards the center of the Earth and Fn, pushing the object upwards. But now we introduce two more forces and if you were to draw (or describe) how these force vectors are used on a FBD, what would it look like?
  5. Nov 10, 2009 #4


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    No - the centrepetal force is outwards.
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