Find the moment of inertia of a thin wire of mass

[airbauer33]

[SOLVED] Moment of inertia

Homework Statement

Find the moment of inertia of a thin wire of mass, m, bent into a square without a top, with each side being of length, b, about its axis of symmetry. (looks like field goal uprights)

Homework Equations

I = I1 + I2 + I3

The Attempt at a Solution

I know the two uprights I are (m/3)b^2. I found a solution on the net that shows the moment of inertia of the horizontal part being 0 because its center of mass is at the origin. I think that there should still be a 1/12(m/3)b^2 contribution from the horizontal part.

 

[Mindscrape]

Hmm, I don't know if I necessarily agree with your statement for the uprights. Did you try to superpose two rods? You have to remember that the mass given to you is for the whole square thing. The rod equations you are using are not accounting for that. You should also consider that the equation you are using uses the bottom of the rod as the rotation axis. Hint: you may want to consider the parallel axis theorem.

Sorry I can't be much more help, but I don't know what you mean by solution on the net. Where is your origin defined?

 

[ogb p]

I think the correct answer should be (m/3)b^2 + 1/12(m/3)b^2 = (m/3)(4/3)b^2.
Your attempt at a solution is correct. The moment of inertia of the thin wire bent into a square without a top can be calculated by dividing it into three parts: the two uprights and the horizontal part. Each upright has a moment of inertia of (m/3)b^2, as you correctly stated. However, the horizontal part also contributes to the moment of inertia, and its value is indeed 1/12(m/3)b^2. Therefore, the total moment of inertia would be (m/3)b^2 + 1/12(m/3)b^2 = (m/3)(4/3)b^2. Great job on your attempt at solving this problem!