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Centripetal Force of merry-go-round

  1. Nov 2, 2003 #1
    a) A 22 kg child is riding a merry-go-round that is rotating at 40 rpm. What centripetal Force must she exert to stay on if she is 1.25 m from its center???

    40 rpm = 4.19 rad/s
    r = 1.25 m

    Ac = (4.19 rad/s)^2 * (1.25 m) = 21.95 m/s^2

    F = (22 kg) * ( 21.95 m/s^2) = 482.9 N


    b) What cent Force does she need to stay on one that rotates at 3 rpm if she is 8 m from its center

    3 rpm = .314 rad/s

    Ac = (.314 rad/s)^2 (8 m) = .789 m/s^2

    Force = (22 kg) (.789 m/s^2) = 17.36 N

    c) compare each force with her weight.

    Force 1 is 21.95 x's her weight
    Force 2 is .789 x's her weight

    how does this look
    thank
    nautica
     
  2. jcsd
  3. Nov 3, 2003 #2

    HallsofIvy

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    Looks good to me.
     
  4. Nov 3, 2003 #3
    Cool, thanks

    Nautica
     
  5. Nov 6, 2003 #4
    This is the wrong forum. This should be posted in K12 section not College.
     
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