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Circular Motion

  1. May 1, 2007 #1

    danago

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    Gold Member

    A 40kg gymnast is swinging on a horizontal bar. Her center of mass is 1.2m from the bar, and right at the top of the circle she is travelling in, her body has a speed of 1m/s.

    What force must she hold onto the bar with right at the bottom of the swing if she is to continue swinging?


    Well first thing i did was calculate the mechanical energy in the system right at the top of the swing.

    [tex]
    \begin{array}{c}
    E_M = E_k + E_p \\
    = \frac{{mv^2 }}{2} + mgh \\
    = 980J \\
    \end{array}
    [/tex]

    Since energy is conserved, i used this to calculate the tangental velocity at the bottom of the swing.

    [tex]
    \begin{array}{l}
    980 = 20v^2 \\
    v = 7ms^{ - 1} \\
    \end{array}
    [/tex]

    Using this, i can calculate the centripetal force (net force).

    [tex]
    \begin{array}{c}
    F_c = \frac{{mv^2 }}{r} \\
    = 1633.\overline {33} \\
    \end{array}
    [/tex]

    At the bottom of the swing, the force she holds on with and the weight force act in opposite directions, and i can say that:

    [tex]
    \begin{array}{c}
    \sum F = F - mg \\
    F = \sum F + mg \\
    = 2033.\overline {33} \\
    \end{array}
    [/tex]

    The answer the book gives is different however. Im not really sure what i have done wrong. Any help?

    Thanks,
    Dan.
     
  2. jcsd
  3. May 1, 2007 #2
    The total energy is off by 20J.
     
  4. May 1, 2007 #3

    danago

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    Gold Member

    If i use g=9.8ms-2, i get a total energy of 960.8J, but the book is using g=10. Is that the 20J difference youre talking about? Or have i missed something else?
     
  5. May 2, 2007 #4

    danago

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    Gold Member

    anyone have any input?
     
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