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Homework Help: Velocity in uniform circular motion

  1. Mar 14, 2005 #1
    A 300-m tall 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?
  2. jcsd
  3. Mar 14, 2005 #2


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    {Period of Earth's Rotation} = T = (86400 sec)
    {Radius of Earth} = R
    {Tower Bottom Speed} = 2*Pi*R/T
    {Tower Top Speed} = 2*Pi*(R + 300 meters)/T

    {Tower Top Speed} - {Tower Bottom Speed} = {2*Pi*(R + 300 meters)/T} - {2*Pi*R/T} =
    = 2*Pi*(300)/T
    = 2*Pi*(300)/(86400)
    = (0.02182 m/sec)

  4. Mar 14, 2005 #3
    hey bro how did you get the T=86400 sec? thanx for your help, it was the correct answer!
  5. Mar 14, 2005 #4


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    Gold Member

    Mugzieee, how long is a day?
  6. Mar 14, 2005 #5
    sorry, that was a thoughtless question which i realized afterwards. thank you :blushing:
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