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Rotational Dynamics - moment of inertia finding

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    2 children, masses 35kg and 40kg sit at opposite ends of a 3.4m seesaw with mass 25kg with the fulcrum at midpoint. with the seesaw horizontal find
    a) the net torque
    b)angular acceleration

    2. Relevant equations

    tau=rF=Ialpha

    3. The attempt at a solution
    i got the right answer for a ----83.385Nm which i know is equal to Ialpha. my problem is finding the moment of inertial i have tried mr^2 which i knew wouldnt work (tried it just to make sure) anyways i need some help
     
  2. jcsd
  3. May 24, 2010 #2

    Doc Al

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    Staff: Mentor

    Actually mr^2 will work for part of the answer. To find the total moment of inertia, add up the contributions of each part: The two children plus the seesaw itself. Hint: Treat the children as point masses; treat the seesaw as a thin rod.
     
  4. May 24, 2010 #3
    so for the the total moment of inertia: 40*1.7^2+-35*1.7^2+Irod?
     
  5. May 24, 2010 #4

    Doc Al

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    Staff: Mentor

    Yes. (But get rid of that minus sign--I suspect it's just a typo.)
     
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