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A child stands on the edge of a merry-go-round

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A 15 kg child stands on the edge of a small merry-go-round of radius 2.0 m. If the merry-go-round turns at a steady rate of 11 rpm, calculate the magnitude and direction of the net force
    on the child.


    2. Relevant equations
    I'm thinking FG = G(m/r2)


    3. The attempt at a solution
    FG = 11(15/2^2)
    = 41.25

    The direction would be clockwise?
     
  2. jcsd
  3. Oct 18, 2012 #2
    For circular motion, F = mrω^2

    ω = 2π/T where T is the time period for one rotation.

    Also, the force in circular motion always points in a particular direction. Do you know it?
     
  4. Oct 18, 2012 #3
    So it would be,

    F= (15)(2)(2π/11)^2
    F= 9.7N

    The force in circular motion always points towards the center or rotation.
     
  5. Oct 18, 2012 #4
    Almost, but you have to convert your value for T into the correct units first. It might be easier to use ω = 2πf, where frequency is revolutions per second.
     
  6. Oct 18, 2012 #5
    f= (15)(2)(2π(11/60))^2
    f= 30 * (2π * .183)^2
    f= 34n
     
  7. Oct 18, 2012 #6
    The equation is correct but I think you forgot to square the bracket.
     
  8. Oct 18, 2012 #7
    Oops.

    correct answer is 39N.


    Thanks for your help!!!
     
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