How Does Cutting a Circle from a Square Affect Its Center of Mass?

AI Thread Summary
Cutting a circle from a square affects the center of mass (COM) depending on the relative positions of the circle and square centers. If the circle and square share the same center, the COM remains unchanged. However, if they do not share a center, the COM shifts towards the center of the circle. The discussion highlights the importance of considering the mass distribution when calculating the new COM after the cut. Ultimately, the material's density does not affect the calculation as long as the dimensions remain constant.
mnandlall
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If I cut a perfect circle into a square piece of tin, how will the centre of mass be affected? I am making the assumption that I know the dimensions of the square as well as the diameter of the circle being cut. Let's say that I also know the position of the centre of the circle relative to the centre of the square.

Is there a specific relationship that exists here?
 
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If the circle and square share the same common center, then no- the COM will not be affected.
 
Alright, but what if they do not share the same common center?
 
Then it will move to the new center.
 
Do you mean to say that regardless of where I cut the circle, the new centre of mass will always be at the centre of the circle?
 
As long as the material is even- then yes.

Where else could the center of mass be, except at the center of the circle?
 
christianjb said:
As long as the material is even- then yes.

Where else could the center of mass be, except at the center of the circle?
What? No.

If you have a 6"x6" square of tin, and you cut a 1" circle at a spot 1" from the edge, the4 CoM will NOT be the centre of the circle!
 
Try turning the problem on its head. A 1" circle of tin 2" from a fulcrum has the same effect as a 1" circle missing from a piece of tin 2" from its fulcrum.
 
Okay, I will try to think about it that way, and see where I get. Thanks.
 
  • #10
Am I misunderstanding the question? I think I might be.

Are you asking where the COM of the foil minus the circle is? In that case, I agree, the COM will shift.
 
  • #11
Yes, this is what I was asking. Sorry if I was unclear about it.
 
  • #12
OK, one way of solving the problem is to consider how to combine two COM's to form their joint COM.

i.e. if the COM of A is at Ra, and the COM of B is at Rb, then the joint COM is at

Rab=(MaRa+MbRb)/(Ma+Mb), where Ma is the total mass of A.

If you cut out the circle without removing it- then the center of mass of the entire square (including the circle) has not changed. You can work out the COM of the circle and you can then balance an equation including the COM of the square with the circular cut.
 
  • #13
Do you need to know the mass of the object to figure out the answer? I mean, what if the object was made of a different material, perhaps gold. If all of the same dimensions are used, will there be a different answer?
 
  • #14
mnandlall said:
Do you need to know the mass of the object to figure out the answer? I mean, what if the object was made of a different material, perhaps gold. If all of the same dimensions are used, will there be a different answer?

I doubt it. Just use a mass 'M' for the total square, and it should drop out by the end.
 
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