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Homework Help: Angular velocity of a tower

  1. Feb 19, 2008 #1
    A 300m tower is built on the equator. How much faster does a point at the top of the tower move than a point at the bottom?

    The earth is 6400km
    thus c=2pi * 6400000m
    and c of the top of the tower is c= 2pi *6400300m

    Now what???


  2. jcsd
  3. Feb 19, 2008 #2
    Do you use the period of 24hr for both the earth angular velocity and the tower angular velocity???
    I need the period for the tower to solve the problem
  4. Feb 20, 2008 #3
    angular velocity= 2pi r/ period

    how do I find the period of the tower so I can find the angular velocity of the tower and subtract it from the angular velocity of the earth???
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