Rotational inertia (moment of inertia)

[mossman]

hi guys I am 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 don't know where I am going wrong here. the answers i entered showed up incorrect and I am wondering what I am doing wrong. thanks for your help guys!

 

[Doc Al]

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.)

 

[Moetasim]

Hello there,

It seems like you are on the right track with your thinking. The rotational inertia (also known as moment of inertia) of a rigid body is indeed the sum of the inertias of all its constituent mass points. In this case, the seesaw can be thought of as a rigid body, with each mass point (m_1, m_2, and m_bar) contributing to its overall rotational inertia.

The formula you used for the inertia of a thin rod rotating about its center (1/12mL^2) is correct. However, in this case, the seesaw is not rotating about its center, but about its pivot point. This means that the distance from the pivot point to each mass point will be different. For m_1 and m_2, this distance is L/2, but for m_bar, it will be different. You will need to use the parallel axis theorem to calculate the inertia of m_bar about the pivot point.

The parallel axis theorem states that the inertia of a body about any axis parallel to its center of mass is equal to the inertia about its center of mass plus the product of its mass and the square of the distance between the two axes. In this case, the center of mass of the seesaw is at its pivot point, and the distance between the two axes is L/2.

So, the inertia of m_bar about the pivot point will be 1/12m_barL^2 + m_bar(L/2)^2. Adding this to your previous answer of m_1(L/2)^2 + m_2(L/2)^2 will give you the total rotational inertia of the seesaw.

I hope this helps clarify your understanding of rotational inertia. Keep up the good work and keep practicing!