rotational inertia (moment of inertia)

mossman

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!

Doc Al

Yep. Just add them.

Yep.

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