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Rotational inertia (moment of inertia) 
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#1
Mar2308, 02:49 PM

P: 3

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! 


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