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Angular acceleration of a seesaw

  1. Nov 6, 2008 #1

    There is a seesaw. The pivot point is closer to the left side than the right side. Let`s call the left length of the beam L1 and the mass on the left side m1. The right sideof the beam is L2 and the mass is m2. Information is L1 = 0.990 m, L2 = 1.78m, m1 = 4.40 kg, and m2 = 2.35 kg. You need to find angular acceleration.

    2. Relevant equations

    torque=I(alpha) alpha is angular acceleration and I have calculated the torque below.

    3. The attempt at a solution

    So I`ve calculated torque doing L1m1g-L2m2g and found it to be roughly 1.6954 N*m.
    My problem is that I have no idea how to find I in this situation. I`ve read that section of my textbook several times and it just makes no sense to me.
    I appreciate and hints or help. Thank you in advance! :)
  2. jcsd
  3. Nov 6, 2008 #2

    George Jones

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    How is the mass distributed? Is mass 1 concentrated at the left end of the beam, and mass 2 distributed at the right end? If so, then just use the basic definition of I for point masses.