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cmrgator
Oct23-03, 07:01 PM
Imagine a square with side length "a". Now, divide the square into 4 equal squares with side length "a/2". In the top righthand corner of the large square, a circle with radius "a/4" is cut out, which also removes the top corner piece. What is the new center of mass? (The origin is at the center of the large square). Thanks for your help!

On Radioactive Waves
Oct25-03, 02:16 AM
Here is an easy way to think about it.

Cut the circle out of the remaing 3 quarters of the large square, now you have a symmetrical shape about both x and y axis. The center of mass is the center, so we can now ignore this part.

Now what is the center of mass of the 3 circles?

Tom D
Oct25-03, 06:43 PM
Sorry, I don't see how that helps because of the cut out corner piece.
Anyway, even with the corner piece, I wouldn't know what to do.[g)]

On Radioactive Waves
Oct26-03, 01:51 PM
Oops, I missed the part about the corner cut out...


I guess its back to x_bar y_bar in the first place then....

Well, it could still be done the way I was suggesting, don't know if its any easier though....but the point was to take out all area symmetrical about both axis, to simplify the problem.