What is the process for finding the moment of inertia for a semicircle?

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The discussion focuses on the process of calculating the moment of inertia for a semicircle, outlining steps such as dividing the shape into portions, finding centroids, and applying specific formulas for inertia. Participants express confusion about the lack of calculus in the process and seek clarification on determining dimensions for the semicircle's non-circular sides. The importance of the parallel axis theorem is highlighted, along with the need for specific centroid formulas for different shapes. Overall, the conversation emphasizes the necessity of understanding geometric properties and dimensions to effectively calculate the moment of inertia. The thread concludes with a recognition of previously overlooked rules and formulas that aid in solving the problem.
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Homework Statement


statics7_zps3djyiigg.png

Find lx and ly.

The Attempt at a Solution


So after countless studying of this topic, and no notes from my professor to help, I thought I had the process figured out. But this initial image is throwing me off a bit.

Here's what I believe to be the right process for these problems (please correct me if I'm wrong)
1) First, divide the object into portions. Find the area of each portion.
2) Next, find the x centroid of each portion, which is done by halving the width of each portion.
3) Next, find the y centroid of each portion, which is done halving the height, and adding any additional height that is below the portion.
4) Next, you multiply the area of each portion by it's x and y centroid distances.
5) Next, sum up the total area, the total (x centroid * area), and total (y centroid * area)
6) Next, figure out the X and Y centroids for the total object by dividing the summed up (x centroid * area) and (y centroid * area) respectively by the total area.
7) Now you can figure out the moment of inertia x of each portion with the specific shape formula needed (for example a rectangle being [(1/12)*b*h^3]) and then adding to it (area)*(total Y centroid - the portions Y centroid). Or I guess you just subtract whichever is smaller between the total Y centroid and portion's Y centroid.
8) For the moment of inertia y, it's the same thing except the second part of the formula would be (area)*(total X centroid - the portions X centroid).
9) At this point, you can add up the inertia parts and get the total inertia for X and Y.

Is this process correct? I'm a bit suspicious that it's not since I didn't need to use any calculus (I could've sworn this chapter was all about calculus).

But anyways, back to my main point. In this particular problem I can't actually get this process started, since they don't give all of the dimensions or any angles to figure them out. I'm also not totally sure what geometric techniques I can use on a semicircle. What is the first step here? Thanks in advance for any help.
 
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javaistheman said:
2) Next, find the x centroid of each portion, which is done by halving the width of each portion.

How does that work for the quadrant?

There are sufficient dimensions on that drawing to work out the other dimensions you need.
 
CWatters said:
How does that work for the quadrant?

There are sufficient dimensions on that drawing to work out the other dimensions you need.

What do you mean? For the bottom left piece for example, the width is 20 so I divide that in half to get 10 for the distance.

How do I find the dimensions of the top left and bottom right pieces. For the semi circle I can do ([pi*30^2]/2). Then 40*20 to get the bottom left piece. Then I'm left with nothing else. Is it possible to figure out the non-circular sides of the semi-circle?

Edit: Just realized you're referring to the semi-circle when you say "quadrant". Yeah, that's what I'm trying to get clarity on. That shape is throwing me off.
 
javaistheman said:
How do I find the dimensions of the top left and bottom right pieces.

I'm sure you are going to kick yourself...see the mods I made to the drawing..

statics7_zps3djyiigg.png
 
CWatters said:
I'm sure you are going to kick yourself...see the mods I made to the drawing..

View attachment 80690

Ha! I knew there was a basic rule that I didn't know. Thanks a lot.

I also see now that the centroid of that quadrant has it's own formula too.
 
You make no mention of the parallel axis theorem, which you will need.
It seems to me you are given all the dimensions you need if you assume that everything which looks like a right angle is a right angle.
 
javaistheman said:
Ha! I knew there was a basic rule that I didn't know. Thanks a lot.

I also see now that the centroid of that quadrant has it's own formula too.

Here's a list of centroids for other geometrical figures:

http://en.wikipedia.org/wiki/List_of_centroids
 
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