# Centre of gravity

1. Mar 4, 2006

### Benny

Hi, I would like some clarification on the following. Suppose I have a unit square A, symmetric about the xy plane. In addition to this, suppose that I cut a square B from A so that I am left with a lamina (A - B). Where B is a (1/4) * (1/4) square which is symmetric about the x-axis and lies in the half plane x >=0 with one of its sides along the y-axis.

If I know the centre of gravity of A and the centre of gravity of B, is it possible for me to find the COG of (A - G)? Any help would be good thanks.

2. Mar 4, 2006

### Hootenanny

Staff Emeritus
Hang on a minute. If B is symmetric about the x-axis, then how can $x\geq 0$?

3. Mar 4, 2006

### arildno

I assume you mean centre of mass rather than centre of gravity?

4. Mar 4, 2006

### Hootenanny

Staff Emeritus
[EDIT] Nevermind, I thought you mean the lamina B. You were describing the plane, my mistake.

5. Mar 4, 2006

### Benny

For the purposes of my question, it can be assumed that I am after the COM rather than the COG.

Does anyone have suggestions? I mean A by itself with constant density will have COM at the origin. The COM of B will be along the x-axis, half way across its length (ie. at x = (1/8)). However, I am unsure about how to find the COM if I remove B from A. That is, I want to find the COM of (A - B).

6. Mar 5, 2006

### arildno

Think of the four squares that A consists of (B being one of them).
Where are the COM's of the three other placed?
What will be the COMMON C.O.M. for these?

Last edited: Mar 5, 2006
7. Mar 5, 2006

### topsquark

Just butting in briefly. In this case the COM and COG will be the same point.

-Dan

8. Mar 5, 2006

### arildno

Nope. That is entirely dependent on whether you assume a constant or non-constant gravitational field.

9. Mar 5, 2006

### topsquark

As none such was mentioned in the statement of the problem...

Though I will say that COM is a safer concept to use in general.

-Dan

10. Mar 6, 2006

### Benny

I kind of agree with you guys about the COM and COG references. The text that I'm using never makes reference to COMs but rather, to COGs. However, the answer always turns out 'correct' (according to the given answers) if I consider them to be COM problems.

Anway, the question I posed consisted of a unit square A. If B is the (1/4) * (1/4) square as described, then the Area(A) = 16Area(B). As I see it, with B placed in A the way it is, it isn't possible to divide A into four squares - unless you mean something like taking the non-B parts of A and constructing squares from that.

My square is just a simple (the simplest I could think of) of what I am really trying to understand. This is, what effect does removing some fraction of some lamina, have on the COM of the resulting lamina. This question is probably difficult to answer in general. However, I believe that if anyone can give me some examples or even just some pointers on how to consider such problems, I would be able to work out the rest myself.

To draw an analogy. It's like sticking two cylinders on the ends of a rectangular block. Choose some point on the block say, its COM (denote this by G). If we want the moment of inertia of the composite object (cylidners + block) about an axis through G, all that is required is the computation of the individual moments of inertia of the cylinders and block about the axis through G. Then, the moment of inertia of the composite object about an axis through G is just the sum of the inidividual inertia.

I would like to know if there is a similar method for COMs.

11. Mar 6, 2006

### HallsofIvy

Staff Emeritus
Actually, I disagree with what is being said about "Center of Mass" and "Center of Gravity"- neither is appropriate here. A geometric figure such as a square does not have mass and so can have neither center of mass nor center of gravity. What you are seeking is the "centroid".

You can do this as a "weighted" average (oops! Now I'm talking about weight!). Think of A as being made of (A-B)+ B. Call the x coordinates of their respective centroids Ax, (A-B)x, Bx.
Then (Area of A)Ax= (Area of A-B) (A-B)x+ (Area of B)Bx. The same for the y coordinates.

12. Mar 6, 2006

### arildno

I agree. No density function has been given, so no mass can be assumed here, so centroid is the correct word for what you're after. My bad.

13. Mar 7, 2006

### Benny

Ok then, I was a little loose with the terms but it's good that my question was understood anyway. I was thinking about using ratios but I didn't have a firm idea in mind at the time. Thanks for explaining.