Center of mass of a disk with a hole

[Tomas Carvalho]

Homework Statement

We have a circle of radius 4 with center at the origin of a referential, and a circle hole in it of radius 1 and center at (-2,0). We're supposed to calculate the center of mass.

2.MY QUESTION.
I know, by the usual formulas for calculating center of mass I get 2/15 as the x of the center of mass, however I don't know why this other approach doesn't seem to work:
I'm assuming that any line passing through the center of mass divides the body into two equal mass parts. As such, if we draw a vertical line through the center of mass we're looking for, the area inside the cricle between that line and the one passing through the center has an area of pi/2 (simple calculations so that area to the left is equal to ares to the right of line), so calling b the x of the center of mass:
The indefinite integral from 0 to b of sqrt(16-x^2) should be equal to (pi/2)/2)=pi/4, but once I input this on wolfram alpha I get an approximate value of 0.196 which doesn't seem correct.
How so?

 

[Tomas Carvalho]

I wouldn't say any line. It has to be the line going through the center of the hole and the circle. That's the line of symmetry that divides the object into two equal parts.

I know it doesn't divide it into two equal parts, but I thought it divided it into two parts of equal area as long as it passed through the center of mass. Maybe that's not always true...

 

[Orodruin]

I'm assuming that any line passing through the center of mass divides the body into two equal mass parts.

This is generally not correct.

Consider a configuration with just two point masses ##M## and ##m \ll M##. Clearly, the CoM lies between the point masses (much closer to ##M##, but still between them) and the side with ##M## in it has much more mass than the one with ##m##.

 

[kuruman]

Sorry, I deleted my original message because it didn't convey what I meant. I meant that you should look for the cm on the line joining the center of the hole and the big circle. This kind of problem is usually done by considering the hole as a circle of negative mass (or density) then using the standard center of mass equation.

 

[Tomas Carvalho]

This is generally not correct.

Consider a configuration with just two point masses ##M## and ##m \ll M##. Clearly, the CoM lies between the point masses (much closer to ##M##, but still between them) and the side with ##M## in it has much more mass than the one with ##m##.

I get what you mean, but it seems different. I we have a homogenous body, shouldn't this hold? I get your example, but if in the case you described, the two bodies had the same density (which is the case for a homogenous body), then wouldn't the center of mass lie inside the larger body (despite being between the two centers of mass)?

 

[Orodruin]

Sorry, I deleted my original message because it didn't convey what I meant. I meant that you should look for the cm on the line joining the center of the hole and the big circle. This kind of problem is usually done by considering the hole as a circle of negative mass (or density) then using the standard center of mass equation.

I believe this is what he has done to find 2/15 and the real question was "why can't I do it this other way?"

 

[Orodruin]

I we have a homogenous body, shouldn't this hold?

No. There is no relevant difference in the case of a homogeneous body. For example, consider a T-shaped body with a very long horizontal line, the basic idea is the same.

 

[Orodruin]

if in the case you described, the two bodies had the same density (which is the case for a homogenous body), then wouldn't the center of mass lie inside the larger body (despite being between the two centers of mass)?

Also, just to answer this question: No. Imagine that you have two spheres with the same density but one is much larger than the other. If you separate the spheres by a large enough distance, then the centre of mass will lie outside of both spheres.

 

[kuruman]

I believe this is what he has done to find 2/15 and the real question was "why can't I do it this other way?"

Yes, you're right.
To @Tomas Carvalho: another way of seeing why your method does not work is this. If you choose your reference at the vertical line that passes through the CM, then the proper equation to write for the position of the CM is $$X_{CM} = 0 =\frac{M_{right}~x_{right}+M_{left}~x_{left}}{M_{right}+M_{left}},$$where ##x_{right}## and ##x_{left}## are the CM positions of the right and left pieces. This gives the relation ##M_{right}~x_{right}=-M_{left}~x_{left}##. You get the result that the masses on either side of the CM line are equal only if their CM's are equidistant from the line (on either side of it). In other words, it's the absolute values of the products that must be equal, not the just the masses. You over-constrained the problem when you asserted that the masses are equal.