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Tangential velocity of the earth

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the speed with which the earth would have to turn to rotate on its axis so that a person on the equator would weigh 3/4 as much


    2. Relevant equations
    VT=r*ω ; Vi=469 m/s is tangential velocity of earth

    ƩF=M*ac=m*Vt^2/r



    3. The attempt at a solution

    The positive direction is toward the center of the earth.

    From ƩF=m*ac

    Initial: m*g-Ni=m*Vi^2/r

    Final: 3/4m*g-Nf=m*Vf^2/r

    Since m*g is the same for initial and final state I assume that Ni=Nf

    Therefore:3/4m*g-[mg-m*Vi^2/r]=m*Vf^2/r or

    Vf=Sqrt(Vi^2-r*g/4)

    I have a sign error. I end up taking the square root of a negative number but the physics looks OK. Suggestions?
     
  2. jcsd
  3. Jan 1, 2013 #2

    haruspex

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    Nf is to be 3/4 of what? And what equation tells you what it will actually be? (Your 'Final' equation is completely wrong.)
     
  4. Jan 1, 2013 #3
    OK, Final:m*g-3/4*N=m*Vf^2/r
     
  5. Jan 1, 2013 #4

    haruspex

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    Looks right, if N is what you wrote as Ni previously,
     
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