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You are not logged in. #2 20121210 19:04:47
Re: centroidshi Maiya, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #3 20121210 23:55:19
Re: centroidsHello; #9 20121211 01:09:52
Re: centroidsHere's the one for a quadrant. Divide the quadrant into infinitesimally thin strips, y tall and dx wide Assume the mass is proportional to area so I'll just work with areas. If the centroid is X units from the origin then moments about y axis for whole shape is and for the thin strips (each at distance x from the y axis) and added up using integration Equating these two expressions for the total moment: Check in Wiki. tick. You can regard a semi circle as two quadrants fitted together, so the position is the same. So now I have the required Cs for the components. Got to do some chores so I'll come back to finish the job this evening ≈ 8pm GMT. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #10 20121211 06:39:13
Re: centroidshi Maiya, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #11 20121211 09:08:31
Re: centroidsProgress report. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #13 20121211 23:47:41
Re: centroidshi Maiya, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #15 20121212 01:28:53
Re: centroidsHere's what I've got: curve joining C to B curve joining O to B substitution into CB gives B is the point (100√3,100) Area of half boomerang (CDBA) is found by integrating CD minus CA from 0 to 100, and adding integral DB less rectangle below red line. Here's one of these (the rest are similar) Substitute x = 200 sinu, dx/du = 200 cosu, x = 0 => u = 0, x = 100 => u = pi/6 Phew! that took some typing, so it would be great if you were able to do the rest of this stage. If you post back when you've got an answer I'll show you how to finish it off. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #19 20121214 03:02:50
Re: centroidshi Maiya, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #23 20121221 07:16:28
Re: centroidsFirst a correction: Check estimate by square counting = about 46 squares at 20 x 20 each = 18400 so far so good. This is the area between the curve CD and the x axis. Now for the area between the curve CA and the x axis. I could use integration, but as this area is a quarter circle plus a rectangle I might as well do it that way. Just a little less than the previous answer seems good. Now for the area between DB and the line of symmetry. Note. the curve is the same as the first so I can write the integral straight away but with new limits. Check estimate by square counting = 8 x 400 = 3200 So the area of half a boomerang is first answer minus second answer plus third answer = 4429.7153521131 Back at post 11 I estimated this answer would be between 4000 and 5000 so I'm happy with that. That's step 2 complete. I'll start step 3 in a new post. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #24 20121221 10:01:18
Re: centroidsThe centroid is found from this formula: For a uniform shape the weight can be replaced by the area: In step 2 I calculated the area of the whole shape, so now I have to work out the RHS. first component under CD: second component quarter circle CA (I'll use the formula I did way back in post 9 third component 100 x 100 square under this fourth component under DB fifth component rectangle under this Now the required total is 1st  2nd  3rd + 4th  5th = 500000 Finally for step 4, So let's say 113 from the y axis on the line of symmetry. That looks reasonable so it may be right (meaning not obvious that it is wrong). I'll check it at the weekend. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei 