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Centripetal merry-go-round problem

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.43 m/s and a centripetal acceleration of magnitude 1.87 m/s2. How far is the man from the center of the merry-go-round?


    2. Relevant equations
    Ac= v^2/r


    3. The attempt at a solution
    I understand I am looking for the radius in the equation, im just not sure about the other aspects of it. I get a radius of 6.29 the way i do it by: (3.43^2/1.87). Not confident though.
     
  2. jcsd
  3. Feb 26, 2008 #2

    Kurdt

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    Looks fine to me.
     
  4. Feb 26, 2008 #3
    alright, so the constant speed of 3.43 m/s can be plugged in as the velocity? and im confused on the wording "centripetal acceleration of magnitude... "
     
  5. Feb 26, 2008 #4
    "...centripetal acceleration of magnitude 1.87 m/s2" means that the magnitude of the centripetal acceleration vector is 1.87.
     
  6. Feb 26, 2008 #5

    Kurdt

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    Similarly the speed is the magnitude of the velocity vector, so yes you use the speed for v in the equation.
     
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