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Rotational inertia (moment of inertia)

  • Thread starter mossman
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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!
 

Answers and Replies

Doc Al
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so now wouldn't the rotation of a rigid body be the sum of the inertias of the points with mass points?
Yep. Just add them.

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

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

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