35 thoughts on “HOMEWORK H32.B”

  1. will the distance from the y-centroid of the shape to the bottom circle be the y-centroid of the bottom circle plus the y-centroid or minus the y-centroid since it's in the negative direction?

    1. I think it would be subtracting because adding would lead to a centroid higher on the y axis while subtracting would yield a centroid closer to y = 0, which should be the case.

      1. Sean's physical feeling is right, that adding a bottom semi-circle brings y-centroid closer to zero. However, Ashley and Bowen's approach is more rigorous, and it allows you staying with the formula safely. I would recommend you to think in this way.

      1. Yes. Poisitive area for both and negative centroid for the smaller part of the circle. Then you weight the centroid with the areas and then solve for the whole figure (both semicircle) centroid, that end up being in the positive part of the plane since it has a larger area.

    2. I agree with previous reply. Both areas are positive, the centroid of larger semicircle is positive and that of the smaller semicircle should be negative. Like we add them together. Centroids of larger one + (-)centroids of smaller one

    1. I think your best bet to estimate that value would be to look at the tabulated values for the individual shapes. You know the second area moment has to be smaller than that of a full circle of radius R which is (pi*R^4)/4 but larger than that of a semicircle of radius R which is (pi*R^4)/8.

      As long as your answer was in that range, you'd be close.

      1. Zach's approach is easy to understand. Or you can assume the radius of the the bottom circle is d. Then put d=R/2 at the last step you can get what is asked in the problem. You can check it by setting d=R, then the y-centroid of the combined shape would be zero, and the area, the second area moment for a full circle as Zach mentioned.

    2. You can take the individual moments about their own centroids and then use the parallel axis theorem to find the combined second moment of the shaded portion about the overall centroid.

  2. I agree with ashley's method because if you want to find the contribution of areas on a composite areas centroid you use the displacement , the location to be exact ,not distance from a neutral point like the origin,for example , similar to H12 when negative positions were used.

    1. In that case you would be calculating the centroid in the z direction. Other than that, you follow a near identical process that you did for bending in the z direction.

  3. In that case you would be calculating the centroid in the z direction. Other than that, you follow a near identical process that you did for bending in the z direction.

    1. I dound the second moment of inertia for each one and weight it, as when finding the ycentroid for each. I am not sure if this is correct. Can someone afirm if this is correct path to solve the problem?

    2. There is only one centroid of the composite, so when finding the second moment, I think the distance is the separate part's distance from its individual centroid to the composite centroid, in this case in the y direction. At least that is my understanding.

  4. Can you do a table like the one we used to do in previous problems to find the centroids or is better just to follow equations and skip the table?

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