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Pivotion of plank

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    a thin plank of mass m and length l is pivoted at 1 end.The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.

    2. Relevant equations

    conservation of energy

    3. The attempt at a solution
    when horizontal 2 forces mg and force of pivot. mg we know .force of pivot to find.PE=mgl/2cos theta where theta is angle with vertical.conserving energy i get mgl/2cos theta=1/2*I omega^2.then i found net force m^2 g^2+m*omega^2*l/2 whole underroot.but not getting correct answer
  2. jcsd
  3. Oct 16, 2012 #2

    Simon Bridge

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    Homework Helper

    What makes you think it is not the correct answer?

    reason it out - change in gravitational PE is mgh ... what did you use for h and why?
    (note: cos(30)=1/2)

    Where would the force on the pivot come from?
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