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

  1. Jul 3, 2004 #1
    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 [tex] x_C = \frac{3}{4} a [/tex]
    but it seems impossible to calculate the center of gravity (I know it's not because it's value is [tex] x_G = \sqrt{2} a [/tex]).

    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!!!
  2. jcsd
  3. Jul 3, 2004 #2


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    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).
  4. Jul 3, 2004 #3

    Doc Al

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    In a uniform gravitational field, the center of gravity is the same as the center of mass. That is not the case here.
  5. Jul 3, 2004 #4
    Sometimes we're so sure of our own mistakes...

    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...

    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 [tex] \frac{1}{r^2} [/tex] 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.
    Last edited: Jul 3, 2004
  6. Jul 4, 2004 #5


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    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 learnt that center of mass and gravity were different, but couldn't remember the reason why. When I went to look it up, erroneously told me they were the same... :frown:
  7. Jul 4, 2004 #6
    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.

  8. Jul 4, 2004 #7
    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.
  9. Jul 4, 2004 #8

    Doc Al

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    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.
  10. Jul 4, 2004 #9


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    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.
  11. Jul 4, 2004 #10
    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 [tex] \alpha [/tex] 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.
    Last edited: Jul 4, 2004
  12. Jul 4, 2004 #11


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    [tex]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}[/tex]
    IE for each segment [tex]l_1, l_2[/tex] 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.
    Last edited: Jul 4, 2004
  13. Jul 4, 2004 #12
    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:

    [tex] \frac{1}{\rho_G^2} = \frac{1}{m} \int_m \frac{\rho_c}{\rho} \frac{1}{\rho^2} dm [/tex]

    Because you know that the center of gravity [tex] \rho_G [/tex] is somewhere parallel to the center of mass [tex] \rho_C [/tex] 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)

    [tex] \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 + .... \} [/tex]

    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 [tex] \sqrt(2) [/tex] I will be very pleased if you would explain it to me...
  14. Jul 4, 2004 #13


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    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. Lets see

    [tex]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}[/tex]
    [tex]\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}[/tex]
    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, lets add the other integral explicitly
    [tex]\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}[/tex]

    Does it has some sense to you?
    Last edited: Jul 4, 2004
  15. Jul 4, 2004 #14


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    a bit algebra goes to

    [tex]\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}[/tex]

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

    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
    one sees that the first integral is [tex]1/\sqrt{2}[/tex], the second one is [tex]-1/\sqrt{2} + 1[/tex], so the sum is [tex]1[/tex] and the total result is the one you asked for. QED and all that.
    Last edited: Jul 4, 2004
  16. Jul 4, 2004 #15
    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 seperate 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.
  17. Jul 4, 2004 #16
    Thanks Doc Al, that clarified things a lot for me.
  18. Jul 4, 2004 #17
    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?
  19. Jul 4, 2004 #18


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    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.
  20. Jul 4, 2004 #19
    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.
  21. Jul 5, 2004 #20
    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...

    Where does the [tex] x[/tex] 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).

    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 [tex] a^2 +l_1^2 [/tex]? I could only put it there because of the "meaningless" [tex] \rho_C [/tex].

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

    When you take [tex] a [/tex] 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 [tex] a [/tex] you wouldn't have [tex] a^2 [/tex] you would have [tex] a^4 [/tex] 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?

    [tex] \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}}} \} [/tex]
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