How do you find the centroid of a circle with a missing square?

A

AlphaJohn

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Please see the image for the full question.
I know that you should divide the circle into different areas. For the left semicircle the centroid is at x=-4r/3pi, but I'm not sure how to tackle the right part to ultimately find the total centroid.

Feel free to also answer the questions after. Thank you very much in advance!


centroidquestion.JPG
 
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I won't touch the other questions, but for the centroid, I would use subtraction.

The centroid of the disk is the weighted average of the centroids of the shaded region and of the square, so subtract the square from the disk (each multiplied by its area).
 
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For the 2nd question - use parallel axis theorem after calculating the centroid.

Answer to 3rd and 4 th question depend on the level of mechanics being taught in the class. Are you supposed to use 2-D elastic conditions including stress concentrations?
 
Ok so if I am correct...

The left and right (full) semicircles will cancel out one another, leaving only the (negative) centroid of the square.
Using xbar = ΣXGi*Ai / ΣAi
xbar = (-a^2 * (a /2)) / (pi*r^2 - a^2)
=-a^3 / (2(pi*r^2 - a^2))
filling in a = 50mm and r=D/2=75 mm
xbar = -4.12

which means the coordinates of the centroid are (-4.12, 0) mm
 
AlphaJohn said:
Ok so if I am correct...

The left and right (full) semicircles will cancel out one another, leaving only the (negative) centroid of the square.
Using xbar = ΣXGi*Ai / ΣAi
xbar = (-a^2 * (a /2)) / (pi*r^2 - a^2)
=-a^3 / (2(pi*r^2 - a^2))
filling in a = 50mm and r=D/2=75 mm
xbar = -4.12

which means the coordinates of the centroid are (-4.12, 0) mm
Click to expand...
Numerically correct - where is your origin?

Now use parallel axis theorem to calculate the second moment of area.
 
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