Center of gravity of a portion of cylinder

AI Thread Summary
The discussion focuses on determining the center of gravity (CG) of a portion of a cylinder filled with mass. Initial calculations using geometric methods for rectangles and triangles did not align with results from 3D modeling software, which indicated a CG of 41.89 from one end. It was suggested that the circular cross-sections of the cylinder complicate the calculation, requiring numerical integration techniques like Simpson's rule for accurate results. Participants emphasized the importance of considering the varying depth of contents at the cylinder's ends and recommended using a spreadsheet for calculations. Accurate CG determination involves creating a table of cross-section areas and distances, particularly near the cylinder's ends.
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  1. Figure # 1
  2. Figure # 2 reproduced from http://www.lmnoeng.com/Volume/InclinedCyl.htmhttp://www.lmnoeng.com/Volume/InclinedCyl.htm

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The blue is the mass inside a cylinder. In steady condition, I would want to know where its Center of gravity point shall be from either of the ends. Technically its like one mentioned in figure # 2 .It can be regarded as a portion of cylinder

Trials :
1) Divided them as a rectangle and a triangle and tried to combine the CG . (answer did not match with 3D models)
Workout :

Combined Center of gravity = area of Rectangle*Center of gravity of Rectangle + Area of triangle* Center of gravity of triangle/(Area of Rectangle+ Area of Triangle)
In this case, (taking all from left end : CG=(500*50+250*33.33)/(500+250)

No clue why it doesn’t match with the 3D modeling software’s answers. Per software, Center of gravity = 41.89 from left Any ideas ?

figure 1.jpg


figure 2.jpg
 
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I don't understand how your first picture combines with the second.
 
It's just to give an idea. However, if figure 2 is viewed from the side it translates to what I have in figure 1 except for the depth on right hand side of the filled portion
 
If you look a Fig. 2, you'll see that cross sections cut thru the cylinder perpendicular to the centerline axis form circular segments. You can't simply use the dimensions from the side projection in Fig. 1 and calculate a meaningful result for a circular cylinder. If you had a rectangular tank, you could get away with it.

You can calculate an approximate value of the longitudinal c.g. of the contents by using numerical integration (e.g., Simpson's first rule) by figuring the area at a series of equally spaced cross sections and then calculating moments from one end of the cylinder. The final result should give you the volume of the contents and the first moment about one end, and you can calculate the c.g. by dividing the moment by the volume.
 
You can calculate an approximate value of the longitudinal c.g. of the contents by using numerical integration (e.g. said:
Is there any example I can look at ?
 
subbby said:
Is there any example I can look at ?

Yes, in this attachment:

http://www.pomorci.com/Skole/Stabilitet/STABILITY 10,11,12.pdf

You'll also want some info about calculating the areas of the various circular segments:

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

In your calculation, the distance of the cross section from one end of the cylinder will be x, and the area of the cross section of the contents will be y, which is also called the ordinate. You'll want to create a table of these x and y values and then apply the Simpson's multipliers and the ordinate values. This type of calculation works well using a spreadsheet program like Excel.

Pay close attention to the end of the cylinder where the depth of the contents gradually vanishes. You may want to use more closely spaced cross sections in this region to improve accuracy.
 
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