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Conservation of Energy, Simple Harmonic Motion

  1. Dec 3, 2009 #1

    jacksonpeeble

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    1. The problem statement, all variables and given/known data
    A 0.200-m uniform bar has a mass of 0.790 kg and is released from rest in the vertical position, as the drawing indicates. The spring is initially unstrained and has a spring constant of k = 20.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.
    10_60.gif

    2. Relevant equations
    Eup=mgh=Edown=.5Iw2+.5ky2

    [tex]\sqrt{3(mgL-ky^{2}}[/tex]

    3. The attempt at a solution
    [tex]\sqrt{.1^{2}+.2^{2}}-.1=.124[/tex]

    [tex]\sqrt{\frac{3(.79 kg)(9.8 m/s/s)(.2 m)-(20 N/m)(.124 m)^{2}}{.790}}=2.343 m/s[/tex]

    I think that I went about this the right way and simply made an arithmetic error or plugged in something incorrect. Any help will be greatly appreciated, as I must have this assignment completed by midnight.
     

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    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2

    jacksonpeeble

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    Any suggestions?
     
  4. Dec 4, 2009 #3

    Redbelly98

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    Just saw this, looks like it's too late.

    In the energy equation, you omitted the kinetic energy of the bar's center-of-mass, (1/2)mv2. (v is the velocity of the center of the bar.)
     
  5. Dec 4, 2009 #4

    jacksonpeeble

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    Thank you, I appreciate the help anyway so that I can do the problems in the future. I was able to turn in that problem later for partial credit - a friend helped me.

    :]
     
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