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Rotational inertia (moment of inertia)

  1. Mar 23, 2008 #1
    hi guys im having a problem with a physics problem.

    the question is about a seesaw that has two masses, m_1 and m_2 at each end of the seesaw. the length of the seesaw is L, so each mass is a distance L/2 from the pivot (center). i answered the question for what the Inertia was for a massless seesaw, (massless rod), and the answer came out to be correct and was m_1(L/2)^2 + m_2(L/2)^2,

    now its asking me what it is if the swing does not have a negligible mass (m_bar), and what the new inertia would be. well the inertia of a thin rod rotating about its center is
    1/12mL^2.

    so now wouldn't the rotation of a rigid body be the sum of the inertias of the points with mass points?

    so the answer should be m_1(L/2)^2 + m_2(L/2)^2 + 1/12m_barL^2

    ive tried using hints but they didn't help, i dont know where im going wrong here. the answers i entered showed up incorrect and im wondering what im doing wrong. thanks for your help guys!
     
  2. jcsd
  3. Mar 23, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yep. Just add them.

    Yep.

    Looks right to me. (Those online systems can be picky, though.)
     
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