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Centrifical force of earth

  1. Oct 22, 2005 #1
    How long would a day be if the Earth were rotating so fast that objects at the equator had no weight?

    I do Fcnet = m(v^2/r) --> mg = m(v^2/r) --> g = (v^2/r)

    g = (v^2/r)
    but g = 0 right? so then how do i do this problem? that would make v = 0 and ya...
     
  2. jcsd
  3. Oct 22, 2005 #2

    Janus

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    Gold Member

    g is the acceleration due to gravity. Its value is independent of any spin of the Earth.
     
  4. Oct 22, 2005 #3
    ok thanks. unfortunately the answer isnt given so idk if im right until monday. i did:
    g = (v^2)/r
    9.8 = (v^2)/6.38e6
    v =7907

    40086722 (distance of Earth in m)/7907 = 5069 seconds
    = 1.4 hours per day.

    i have one more question and then im done.

    A train traveling at a constant speed rounds a curve of radius 275m. A pendulum suspende dfrom the ceiling swings out to an anglr of 17.5 throughot the turn. What is the speed of the train?

    Fcnet = m(v^2/r)
    Fk = m (v^2/r)
    coeff of friction * mg = m(v^2/r)

    i dont know the coefficient of friction or v. im supposed to solve for v. am i not supposed to use friction as the force? and how do i factor the 17.5 degree swing fo the pendulum into this?

    thanks a lot for the help
     
  5. Oct 23, 2005 #4

    Doc Al

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    Staff: Mentor

    train problem

    First figure out the acceleration of the train by analyzing the forces on the pendulum.
     
  6. Oct 23, 2005 #5
    would this be correct then?
    Fnet=ma
    mg(cos17.5)=ma
    9.8(cos17.5)=a
    9.34=a

    (v^2)/r = a
    (v^2)/275 = 9.34
    v^2 = 2568
    v = 50.7 m/s
     
  7. Oct 23, 2005 #6

    Doc Al

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    Staff: Mentor

    No. Start by identifying the forces acting on the pendulum mass. (There are two forces.) Then apply Newton's 2nd law to the vertical and horizontal components.
     
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