Is the distance that the centre of gravity moves when adding or subtracting the same shape from another shape the same?

My teacher told me to assume that circle is being added instead of subtracted and find the distance from the centre of the large circle (The centre of gravity before subtraction) to the new centre of gravity. However, I get 2 different answers when I consider that it is added vs subtracted (When I considered it to be subtracted, I used negative weight for the small circle.

Let mass per unit area = m
Let acceleration due to gravity = g
Let the centre of gravity = G
Let the distance from the centre of the large circle to G = d

Mass of large circle = 16πr²m
Mass of small circle = πr²m

Considering addition, $$ Moment\ around\ G = 16πr^2m·d - πr^2m·(3r-d) \\ 0 = 16πr^2m·d - πr^2m·(3r-d) \\ 16d = 3r-d \\ 17d = 3r \\ d = \frac{3}{17}r $$

Considering subtraction, $$ Moment\ around\ G = 16πr^2m·d + πr^2m·(3r-d) \\ 0 = 16πr^2m·d + πr^2m·(3r-d) \\ 16d = -3r+d \\ 15d = -3r \\ d = \frac{-3}{15}r \\ d = \frac{-1}{5}r $$

Or I made a stupid mistake... circle removed


1 Answer 1


You are right, of course. Subtraction means that the small circle is "empty", as appears on the drawing. The center of gravity moves to the left of the center of the large circle (negative $d$).

Addition, as you calculated it, would mean that the mass density of the small circle is twice that of the remainder, indeed moving the center of gravity to the right (positive $d$). But this is a totally different shape, not a circular hole but a heavier, "redder" circle.

To make sense of the question your teacher asked you, I think you should do the following calculation : assume you start from a shape with two identical "holes", placed symmetrically with respect to the center of the big circle. However weird this shape looks, by symmetry its center of gravity is still at the center of the large circle, and its mass is $14\pi m r^2$. And it is the only information you need, besides the mass of this weird shape.

Then you "fill" the left-side hole by adding a circle there. Now redo the entire calculation with the weird "two-holes" shape and addition of this small circle on the left. Then the result should be the same, because filling the left-hand side of the "two-holed" shape makes you recover the same shape.

Good luck for this calculation !


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