Center of gravity (it's not center of mass )

[dilasluis]

Center of gravity (it's not center of mass!)

Hi, I'm a very desperate person right now.
For two days (I'm totally dumb) I've been trying to calculate the center of gravity of two rigid bodys... Look, it's not the center of mass (I'm not that dumb!).
OK, here's the problem...(at least one of them)

description of the body:
It's a body made of 3 bars: 1st bar is at y=a (0<x<a), 2nd bar is at x=a (-a<y<a) and 3rd bar is at y=-a (0<x<a).
The center of the axis xOy (x=0, y=0) it's the atractive body m who may be seen as a pontual mass (it does not matter).
I've calculated the center of mass and it's $$x_C = \frac{3}{4} a$$
but it seems impossible to calculate the center of gravity (I know it's not because it's value is $$x_G = \sqrt{2} a$$).

Please, someone, could you explain me how did Rimrott got to that result (I've tried all the ways you could imagine - in two days or you go mad or you even try making integrals literaly by point)

Help me!

 

[AKG]

The center of the axis xOy (x=0, y=0) it's the atractive body m who may be seen as a pontual mass (it does not matter).

This sentence makes no sense. For one, I'm pretty sure "pontual" is not a word. Also, I'm pretty sure the center of mass is the center of gravity. Unless you're using some non-conventional definitions, I don't see why the two would be different (check this page).

 

[Doc Al]

Also, I'm pretty sure the center of mass is the center of gravity.

In a uniform gravitational field, the center of gravity is the same as the center of mass. That is not the case here.

 

[dilasluis]

Sometimes we're so sure of our own mistakes...

This sentence makes no sense. For one, I'm pretty sure "pontual" is not a word.

You are rong, pontual is inded a word. It's portuguese for punctual (mass of infinitesimal size). I hope infinitesimal does exist in english... But you're not forced to know portuguese...

Unless you're using some non-conventional definitions, I don't see why the two would be different (check this page).

Like you, my friend, a lot of people think that the center of mass is the same as the center of gravity... YOU ARE TOTALLY RONG. They are only the same for a sphere (because you can actually replace the sphere for a punctual mass) and for constant gravity fields. For example, suppose that the gravity field of the Earth did not vary with $$\frac{1}{r^2}$$ and it was constant then every body in the victinity of Earth would have they're center of mass coincident with they're center of gravity. BUT THAT'S NOT THRUTH. So only spheres on Earth have it's center of mass coincident with it's center of gravity.

DEFINITIONS: The center of gravity is the place where you would put an infinitesimal small mass (punctual) and it would be atracted by the same force as a bigger body with the same mass.

So, if there is not a body to atract you, you would not have a center of gravity but you would still have a center of mass.

By the way, I found out that Rimrott's book has a rong deduction (that is for sure because I can prove it) and that probably (and this is probably) the solutions he has about the question I posted is rong. If I can prove that, that would mean 2 days of doing the same stuff all over again because I've trusted one other book.

 

[AKG]

like you, my friend, a lot of people think that the center of mass is the same as the center of gravity... YOU ARE TOTALLY RONG. They are only the same for a sphere (because you can actually replace the sphere for a punctual mass) and for constant gravity fields.

That's not correct. Yes, I was not considering constant gravitational fields, however, objects other than spheres can have their center of mass located in the same place as their center of gravity. A thin ring around the Earth would be one example I could think of. I thought I remembered having learned that center of mass and gravity were different, but couldn't remember the reason why. When I went to look it up, mathworld.com erroneously told me they were the same...

 

[pmb_phy]

In a uniform gravitational field, the center of gravity is the same as the center of mass. That is not the case here.

That makes no sense. The topic is the center of mass/gravity. What does the gravitational field have to do with this? There are at least two mass distributions which generate a uniform gravitational field. The center of mass makes sense for only one of them and its well defined in that case. The terms "Center of mass" and "Center of gravity" are usually taken to mean identically the same thing. If you disagree then please state the definition of each.

Pete

 

[darkar]

Center of mass is the point at which all of the mas of an object or system may be considered to be concentrated. Whereas the center of gravity is the point where all of the weight of an object may be considered to be concentrated in representing the object as a particle. In most cases we meet the center of mass coincide with the center of gravity, but in a non uniform field, they might not coincide. so the center of mass will not be the same as center of gravity.

You may think that a long thin bar, one part is inside the effective gravitational field, then the c.o.m. will not be the same as c.o.g.

Hope is right.

 

[Doc Al]

That makes no sense. The topic is the center of mass/gravity. What does the gravitational field have to do with this?
There are at least two mass distributions which generate a uniform gravitational field. The center of mass makes sense for only one of them and its well defined in that case. The terms "Center of mass" and "Center of gravity" are usually taken to mean identically the same thing. If you disagree then please state the definition of each.

The center of mass is the average location of the mass of an object. This of course has nothing to do with the gravitational field.

The center of gravity is the average location of the weight of an object. In a uniform gravitational field, each mass element would weigh the same so the center of gravity is identical to the center of mass. In a non-uniform gravitational field--like in dilasluis's problem--the center of gravity is not the same as the center of mass.

I'll try to dig up a precise definition. Like you, I rarely see problems in which center of mass and center of gravity are not synonymous.

 

[arivero]

First of all, let me to say that this definition of "relative center of gravity" (and I say relative because it is not intrinsic to the body) seems not very useful, because the dynamics decomposes naturally around the center of mass, not about this relative center.

Secondly, for an spherical hull attracted by a punctual mass in its centre (ie a Dyson's sphere), the center of mass is at the center of the sphere, but the relative center of gravity should be at infinity, as total force is zero.

Thirdly, the problem seems too easy. One needs to calculate the total force between the U shaped object and the punctual mass, simply by integrating along the object. Then one asks for which distance R will G mM/R^2 be equal to the calculated force.

 

[dilasluis]

A thin ring around the Earth would be one example I could think of.

In the reallity it's not. Earth is oblate and that means that is gravity field is not the same in every direction. So, if the Earth were a sphere the center of mass of the ring would be coincident with the center of mass of the Earth (i corrected this part because I was rong) and it's center of gravity would be at infinit because the kepler force is zero (forgot this one!). But, because Earth is not a sphere, the center of gravity of the ring would be somewhere inside earth. (I think this cases are the least important for the matter I am studying - satellite's attitude and dynamic...)
It's the same thing as the hollow sphere around the punctual mass...

You inded posted a concept i never though of. Everything would be different if you had the ring not around Earth but in it's vicinity. It's center of gravity would depend on it's attitude because, although it has 2D symmetry if you had the ring in any plan parallel to x0y or xOz or yOz (where O is the center of the earth) plane the mass center and the gravity center would be the same. But if your ring was in a plane that made an angle $$\alpha$$ with any of the Earth axis planes (even better, if each of the princial axis of the ring would made a different angle with each of the axis of earth) the center of mass would not be coincident with the center of gravity.

 

[arivero]

$$2\int_0^{\pi/4} \cos\theta {G m \lambda (d l_1(\theta) ) \over (a/\cos \theta)^2 }+ 2 \int_{\pi/4}^{\pi/2} ...d l_2(\theta) = G {4 a \lambda m \over x_G^2}$$
IE for each segment $$l_1, l_2$$ one calculates the projection of the force over the x axis, and then one equals it to the point gravity formula. It can be noticed above that the result depends on the geometry of the system, but G, m and lambda (the density of the bar) drop away.

 

[dilasluis]

$$2\int_0^{\pi/4} \cos\theta {G m \lambda (d l_1(\theta) ) \over (a/\cos \theta)^2 }+ 2 \int_{\pi/4}^{\pi/2} ...d l_2(\theta) = G {4 a \lambda m \over x_G^2}$$
IE for each segments l1 l2 one calculates the projection of the force over the x axis, then one equals it to the point gravity formula.

First, you are using polar coordinates for something that is as cartesian as it can be... and I dind't get your formulation at all. I understood what you meant by it but I think it's not right and I can't prove that isn't because I don't got it...

Simplier is doing like this:

$$\frac{1}{\rho_G^2} = \frac{1}{m} \int_m \frac{\rho_c}{\rho} \frac{1}{\rho^2} dm$$

Because you know that the center of gravity $$\rho_G$$ is somewhere parallel to the center of mass $$\rho_C$$ because the attitude of the body is tangential and it it's symmetryc around the 0x axis (this are distances from the punctual mass).

so the problem was not finding a formulation for the problem (at leat a new one) because I've accomplished that, look (and I think this more coherent than your's)

$$\frac{1}{\rho_G^2} = \frac{\lambda \rho_C}{m} \{ \int_{-\rho_C}^{a - \rho_C} \frac{1}{((\rho_c + x)^2 + a^2)^{\frac{3}{2}}} dx + ... \}$$

But if you tell me that your formulation (like I've said I can't use it because I don't understand it...) gives a result of $$\sqrt(2)$$ I will be very pleased if you would explain it to me...

 

[arivero]

I mangled a bit the formulation, yep, because I was not sure it you were actually asking for a classroom exercise. But OK, but going cartesian we should get something similar to yours. Let's see

$$2\int_0^a {a \over \sqrt{a^2 +l_1^2 }} {G m \lambda d l_1 \over a^2 +l_1^2 }+ 2 \int_0^a ... d l_2 = G {4 a \lambda m \over x_G^2}$$
Thus
$$\int_0^a {d l_1 \over (a^2 +l_1^2)^{3/2} }+ \int_0^a ... d l_2 = {2 \over x_G^2}$$
At a first glance it seems we do not agree about the integration limits but it is just because you insist on using the Xc point, which is irrelevant for the problem. Hmm, let's add the other integral explicitly
$$\int_0^a {d l_1 \over (a^2 +l_1^2)^{3/2} }+ \int_0^a {l_2 \over a} {d l_2 \over (a^2 +l_2^2)^{3/2} } = {2 \over x_G^2}$$

Does it has some sense to you?

 

[arivero]

a bit algebra goes to

$$\int_0^1 {dx \over a^2 (1 +x^2)^{3/2} }+ \int_0^1 {x d x \over a^2 (1 +x^2)^{3/2} } = {2 \over x_G^2}$$

and then
$$x_G={\sqrt{2} a\over \sqrt{ \int_0^1 {1+x \over (1 +x^2)^{3/2} }dx }}$$

If this is not the result, and I have not done any trivial error in the algebra, then it is that I have not understood the (your) definition of "center of gravity". Still, I can non see how this concept can be useful for anything, except for, er, intuitive, visualization perhaps.


PS: looking for instance http://www.sosmath.com/tables/integral/integ11/integ11.html
one sees that the first integral is $$1/\sqrt{2}$$, the second one is $$-1/\sqrt{2} + 1$$, so the sum is $$1$$ and the total result is the one you asked for. QED and all that.

 

[pmb_phy]

The center of gravity is the average location of the weight of an object.

Ah. Okay. I had forgotten that. However it's more correct to describe it as that point through which a single point, equal to the sum of the gravitational forces on the separate object, effectively acts.

The term "weight" refers only to situations in which an object is being supported whereas the center of gravity can refer to objects are not supported.

However in a non-uniform gravitational field there usually isn't any point which can move as a sinlge point, i.e. in GR terms, extended objects don't always move on geodesics.

For example: Consider a circular loop which is parallel to the XY plane and whose center is on the z-axis. Let there be a point particle at R = (0, 0, 0) and let the mass of the point be much greater thatn the mass of the loop. Then there is no point on the loop which moves as a point particle. In fact the motion will be oscillatory. It will be unstable but it will oscillate. There can be no point which can be attached to a body which would behave like a single point. Hence to think of the center of gravity as being different than the center of mass is not meaningful defined as such.

 

[KingNothing]

The center of gravity is the average location of the weight of an object. In a uniform gravitational field, each mass element would weigh the same so the center of gravity is identical to the center of mass. In a non-uniform gravitational field--like in dilasluis's problem--the center of gravity is not the same as the center of mass.

Thanks Doc Al, that clarified things a lot for me.

 

[Olias]

Is it not so that a high-jumping athelete's centre of mass can be external to his/her body close to the the high-bar?

The athelete's centre of Mass can go under the bar while he moves over the bar, thus a moving bodies centre of mass, does not actually need to be 'central' or internal to moving bodies, the parabolic motion is the path where centre of mass tends to equlibriate, but not in all situations?

 

[arivero]

A good example, Olias. Still, the centre of mass is inside the convex hull of the body, for sure. But it is interesting to remark that it is just such kind of problems, having external plus internal forces plus angular momenta, the kind of thing that does centre of mass to be an useful concept.

 

[Olias]

I have a come across a better example, but its way to late in UK now, but interesting things to ponder in this thread, thanks.

 

[dilasluis]

It still doens't make any sense to me because, in my honest opinion, I can't figure where does some terms on your integral come from (and that was the problem with your earlier demonstartion).

Let me tell you something, when you, in dynamics, make calculations on objects behavior your are always making mistakes (most of them are negligible) when you consider the center of mass. When you talk, for example, on equilibrating (I don't know if this is the term...) a fork on a glass you are trying to find it's center of gravity, not it's center of mass! Telling me that is only a detail, i can accept it but do you know, for instance, that if you could improve the third decimal case (after it all is error) on the coefficient of Drag of an airplane you would make millions!

But that is not the subject...

The point is that you manage to find the right result but I doubt your means... So, because I am in serious doubt you have to explain me some things and then I will accept your deduction...

a bit algebra goes to

$$\int_0^1 {dx \over a^2 (1 +x^2)^{3/2} }+ \int_0^1 {x d x \over a^2 (1 +x^2)^{3/2} } = {2 \over x_G^2}$$

and then
$$x_G={\sqrt{2} a\over \sqrt{ \int_0^1 {1+x \over (1 +x^2)^{3/2} }dx }}$$

1st
Where does the $$x$$ standing besides dx present on your second integral comes from?

2nd How can you integrate the second part over x if x is a constant there? (ok, maybe you made a mistake and it's y you are implying but that just send's me back to 1st).

$$2\int_0^a {a \over \sqrt{a^2 +l_1^2 }} {G m \lambda d l_1 \over a^2 +l_1^2 }+ 2 \int_0^a ... d l_2 = G {4 a \lambda m \over x_G^2}$$
Thus
$$\int_0^a {d l_1 \over (a^2 +l_1^2)^{3/2} }+ \int_0^a ... d l_2 = {2 \over x_G^2}$$
$$\int_0^a {d l_1 \over (a^2 +l_1^2)^{3/2} }+ \int_0^a {l_2 \over a} {d l_2 \over (a^2 +l_2^2)^{3/2} } = {2 \over x_G^2}$$

3rd You must have a lot of imagination (not I, because I didn't invent the concept of the center of gravity - it's not mine - it really exist and you should read about it before you say such not so smart afirmations... but that's not what we are here for...)
why did you divide (hope this is the correct term) the Gravitation term by $$a^2 +l_1^2$$? I could only put it there because of the "meaningless" $$\rho_C$$.

4th You are not doing algebra there (or I am really mistaken) you are doing magic!

When you take $$a$$ from the brackets (i think this is the term for () ) you have to divide x by it...
And besides that, when you took that same $$a$$ you wouldn't have $$a^2$$ you would have $$a^4$$ because it was inside something that had 3/2 as exponent so, because it was elevated to 2, 2 cut's with 2 and you have it elevated by 3 but because you had the other a that came from the right side you would have 4...

For all this reasons I'm really surprised that you got the results right...

But ok, explain me what you did... I'm going to put here my integral and you are going to tell me what do you think is wrong! OK?

$$\frac{1}{x_G^2} = \frac{x_c \lambda}{m} \{ 2 \int_{-x_C}^{a-x_C} \frac{ dx}{((x_C + x)^2 + a^2)^{\frac{3}{2}}} + 2 \int_{0}^{a} \frac{dy}{(a^2 + y^2)^{\frac{3}{2}}} \}$$

 

[dilasluis]

Is it not so that a high-jumping athelete's centre of mass can be external to his/her body close to the the high-bar?

The athelete's centre of Mass can go under the bar while he moves over the bar, thus a moving bodies centre of mass, does not actually need to be 'central' or internal to moving bodies, the parabolic motion is the path where centre of mass tends to equlibriate, but not in all situations?

Sorry, this is going really bad... Only if the athlete we're hollow it's center of mass could be outside his/her body... But if you tell me that is center of gravity could be outside his body that would be right - one thing, please, study before you tell me that this is not true... They aren't the same, because they are not, they have different definitions.
It's like most of you are stuck with you're own preconceved idea that were taugh in high school... it's like saying that the density of air is constant and we all know it depend's on a lot of things (we we're taugh in high school the definition of density by ICAO ISA at sea level) and if you insist on calculating a flow over a body with the same density that you have on ISA sea level you are going to make mistakes. What I'm trying to say is just... STUDY!


Definitions: Center of mass- property of a body, never changes it's location
Center of gravity - property of a sistem of 2 or more bodies where one of the body creates a gravitational field. It depend's on the object's attitudes (orientation) in respect with the gravitational field created by that body.

 

[arildno]

dilasius:
First of all, I'd like to commend you for being clear about the conceptual difference between C.M. and C.G. (too few are..)
However, your integral is completely wrong :

$$\frac{1}{x_G^2} = \frac{x_c \lambda}{m} \{ 2 \int_{-x_C}^{a-x_C} \frac{ dx}{((x_C + x)^2 + a^2)^{\frac{3}{2}}} + 2 \int_{0}^{a} \frac{dy}{(a^2 + y^2)^{\frac{3}{2}}} \}$$

1.
The horizontal gravitational force acted upon a line segment of length "dl" centered at (x,y) from a particle with mass m at the origin is:
$$dF=-\frac{Gm\lambda{dl}*x}{(x^{2}+y^{2})^{\frac{3}{2}}}$$

No introduction of "Xc" is needed or recommended here!
2.
The total mass of the rods are: $$M=4a\lambda$$
3.
We gain the following equation for Xg:
$$-\frac{1}{x_{G}^{2}}=-\frac{1}{4a}(2\int_{0}^{a}\frac{xdx}{(x^{2}+a^{2})^{\frac{3}{2}}}+\int_{-a}^{a}\frac{ady}{(y^{2}+a^{2})^{\frac{3}{2}}})$$

The first integral represent the contributions from the horizontal rods, whereas the second is from the vertical rod.

This gives arivero's answer!

 

[arivero]

When you talk, for example, on equilibrating (I don't know if this is the term...) a fork on a glass you are trying to find it's center of gravity, not it's center of mass!

We use center of mass because all the external forces over a body can be decomposed naturally as if it were a single force applied to the center of mass plus a torque applied to the body. This decomposition does not need any specifics about the nature of the external force. On the contrary, the definition of center of gravity, a point such that the whole mass concentrated there should give an equivalent gravity force, does need to know Newton's formula, or it seems to me.

So my resolution works in this way. I first calculate the intensity of the total force, then I look for the distance needed to get the same intensity for the attraction of two masses m, M, the second M being the total mass of the U shaped object.

As the object is symmetric, I only need the force along the x axis, because the force along y cancels. The integrals are done for half of the object, and a factor 2 compensates this. You have not objected to it, so I guess this is clear.

1st
Where does the $$x$$ standing besides dx present on your second integral comes from?
2nd How can you integrate the second part over x if x is a constant there? (ok, maybe you made a mistake and it's y you are implying but that just send's me back to 1st).

Sorry, sloppy notation here. I used "x" as a dumb variable when algebraic substituting from l, instead of the customary "u". To go over your notation, one must substitute $$l_1\to y, l_2\to x$$. As you see, the first integral is for the vertical x=a bar, while the second one is for the horizontal y=a.

3rd You must have a lot of imagination (not I, because I didn't invent the concept of the center of gravity - it's not mine -)

I apologise here too... please interpret plurally the "your", as if it were opposite to "our". I am of the oppinion that the ownership, if any, of a scientific concept, pertains to the whole community using it. IMO, this is also the reason to use "our" when speaking in a classroom.

why did you divide (hope this is the correct term) the Gravitation term by $$a^2 +l_1^2$$?.

Because in my notation $$l_1$$ is equal to the the vertical coordinate; then $$a^2 +l_1^2$$ is the square distance from the infinitesimal $$dl_1$$ segment of the bar to the central mass "m". I need to divide by square distance because this is gravity formula, in this case $$\delta F=G {m \delta M_1 \over r^2_{m,dl_1}}$$.
We project this force onto the horizontal axis using the other factor in the integral.

4th You are not doing algebra there (or I am really mistaken) you are doing magic!

When you take $$a$$ from the brackets (i think this is the term for () ) you have to divide x by it...

Yes, yes... all of this has been carried into the change of variable, $$x=l_i/a$$; please note the corresponding change in the limits of the integral. As I said above, it is unfortunate I called x to the dumb variable instead of traditional u,v. It sees this has been cause of most of the confusion interpreting my answer.

 

[KingNothing]

I understand mass, gravity, and what a center is, but that's about all so far.

 

[dilasluis]

Finally

dilasius:
First of all, I'd like to commend you for being clear about the conceptual difference between C.M. and C.G. (too few are..)
However, your integral is completely wrong :


1.
The horizontal gravitational force acted upon a line segment of length "dl" centered at (x,y) from a particle with mass m at the origin is:
$$dF=-\frac{Gm\lambda{dl}*x}{(x^{2}+y^{2})^{\frac{3}{2}}}$$

No introduction of "Xc" is needed or recommended here!
2.
The total mass of the rods are: $$M=4a\lambda$$
3.
We gain the following equation for Xg:
$$-\frac{1}{x_{G}^{2}}=-\frac{1}{4a}(2\int_{0}^{a}\frac{xdx}{(x^{2}+a^{2})^{\frac{3}{2}}}+\int_{-a}^{a}\frac{ady}{(y^{2}+a^{2})^{\frac{3}{2}}})$$

The first integral represent the contributions from the horizontal rods, whereas the second is from the vertical rod.

This gives arivero's answer!

I finally understood it! My error was, when I substitute:
$$\hat{r}$$ (the versor of r) by $$\frac{x_c}{r}$$ I already knew that the solution was on the xx axis but I couldn't do that because I was using x and y and $$x_C$$ was a constant. What I should have done was to substitute $$\hat{r}$$ by $$\frac{\vec{r}}{r}$$ that is it's definiton. And then I should assume $$\vec{r} = x \vec{e_x}$$ it in the xx axis becuase this were where the solution should be, and it is. And that would gave the same result...
Thank you all a lot, especially you because I undestood what you meant by it and that made real clear to me what I was doing wrong.
I only saw your post today...

 

[KingNothing]

Where would a 'totally' un-uniform gravitational field exist, and where can a totally uniform one exist (in real life, i can think of some theoretically)

 

[kuenmao]

Where would a 'totally' un-uniform gravitational field exist, and where can a totally uniform one exist (in real life, i can think of some theoretically)

Er...totally un-uniform? I mean...what's the difference between "not uniform" and "totally not uniform"? Whatever the case, just think of the Earth. At every different distance from the Earth's centre of mass, we can find different gravitational acceleration. That's not uniform.

When the body is very small compared to the source of the field, we can assume that the field is uniform. For example, we are all more or less 6400 km from the centre of mass of the Earth, and that the ground is more or less flat in a small area, so we can assume that the gravitational field around us is uniform, and it points downward with a strength of 9.80665 N/kg (this unit shows the idea of "field strength" better than m/s2)

 

[KingNothing]

Er...totally un-uniform? I mean...what's the difference between "not uniform" and "totally not uniform"? Whatever the case, just think of the Earth. At every different distance from the Earth's centre of mass, we can find different gravitational acceleration. That's not uniform.

When the body is very small compared to the source of the field, we can assume that the field is uniform. For example, we are all more or less 6400 km from the centre of mass of the Earth, and that the ground is more or less flat in a small area, so we can assume that the gravitational field around us is uniform, and it points downward with a strength of 9.80665 N/kg (this unit shows the idea of "field strength" better than m/s2)

That is precisely the case I wanted to avoid by using the word `totally`...although perhaps i don't know the correct terminology...what I mean is, a field un-uniform (like Earth's) but way way way more extreme. as in, a huge difference from one meter to the next.

I hope you know what I mean, by not trying to simply say 'uniform' or 'ununiform', but that there may be a 'more uniform' or 'less uniform' field.

 

[muhammadusman]

In a uniform gravitational field, the center of gravity is the same as the center of mass. That is not the case here.

but if the field is not uniform then what will be the case. do u have an appropriate answer