50 thoughts on “HOMEWORK H32.A”

  1. I think so too. I am not sure if I am right but you could just find the y centroid of the shaded area and then use that directly to find the inertia.

    1. @Mohinish, you will need to find the inertia of the outer and the inner semi-circle about the overall centroid and then subtract inner from outer as Jason and Chalisa said

    1. I had the same question however I think that using this formula would give us the right value for the second moment of inertia of a circle

      1. I tried looking through the textbook and the examples and I thought that this might not be correct.

        The formula 1/8 pi R^4 would give the inertia about the origin but we need to find the inertia about the centroid

          1. I used 1/8piR^4 for the second moment! Which will help you find the first half of the equation.

        1. So if 1/8 pi R^4 is the inertia around the origin, could i take the distance in the parallel axis theorem to be the distance from the origin to the centroid of the whole shape we find in the first part of the question? (In which case the distance in the parallel axis theorem for both the bigger and smaller semi-circle will be the same)

          1. It's correct, Kushal! Whenever you are using second moment of inertia from table, you will always keep in mind, which axis that's around, and around which axis I am looking for? Then apply the parallel axis theorem.

        1. So then you would have two d's for each second moment of area and they would be the difference between the overall centroid and the respective section's centroid right? And then you subtract the cutout from the bigger semicircle's second moment of area? My answer came out pretty nasty but that's the approach I used so I just wanted to make sure.

  2. For a semicircle, the moment about its centroid is 1/8 pi r^4, correct? I found this equation online and just want to check that it is the correct one before I do all of the work.

    1. You can use the specific formula for inertia for a semicircle and with that and the parallel axis theorem you can get the inertia for the entire shaded region.

  3. I found the centroids of both semicircles but how to we find the overall centroid of the shaded region? Is it just outer centroid - inner centroid?

    1. You take a weighted average with the area of the larger semi-circle being positive and the area of the smaller semi-circle being negative as it is a cutout from the larger semi-circle.

  4. I keep receiving an answer with R^4. I reworked it and received a value with R. I am not sure which is correct. Did anyone else receive a final answer with R^4?

  5. When I use the parallel axis theorem, I'm confused on what to use for d. I saw a comment saying d is the distance to the centroid of the combined shape, but isn't it the distance from the neutral axis to the top or bottom of the area?

  6. In this type of questions are we gonna have the formulas on the exam? When solving with the formula given in class I get an answer with R^3 for the centroid. I did the y_bar(At) = y1_bar(area1) + y2_bar(area2). area2 since is the not shaded portion is negative. Then I solved for y_bar. I don't know if I am moving in the right direction and if finding and answer with R^3 is correct.

  7. Hi, I used Yc = (A1Y1 - A2Y2) / (A1 - A2) to find Yc, with Y = 4r / (3pi). I'm not entirely sure if that's the correct way of solving this. Could someone offer some insight? thanks a lot.

    1. For Yc, I think that is correct because you're essentially "averaging" the Y components of the two shapes with respect to their areas. Since the smaller semi-circle is cut out, you have to subtract instead of add the total area as you normally would with an average. And your "Y" is correct too! Just be careful with your "R".

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