Questions about center of gravity

[Dame]

TL;DR Summary

A conversation about the center of gravity of a static object.

Hello,

I am very new to the concept of center of gravity and I have a question. I wanted to know if the center of gravity of an object is always in the same location in 3-D space. For example, if I was able to find the center of gravity for cylinder/rectangle when its lying flat on a horizontal surface. Would the center of gravity for the same cylinder/rectangle be in the same spot if I stood the object up or rotated the object about its length?

 

[Baluncore]

Welcome to PF.

The CofG is independent of orientation.
The centre of gravity of a 3D object is an identifiable point about which the mass is balanced in all three dimentions.

 

[Delta2]

The center of gravity does not change coordinates in the coordinate system that moves or rotates together with the body and provided that the body is rigid/solid and does not contain fluid/liquid/gas.

 

[ergospherical]

Although, it is worth to note that the centre of gravity of a body depends on external parameters, namely the gravitational field, and is permitted to vary in position even if the body is static. Specifically it is the point $\mathbf{r}_0(t)$ which satisfies $\int_B \rho(\mathbf{r},t) (\mathbf{r} - \mathbf{r}_0(t)) \times \mathbf{g}(\mathbf{r},t) dV = \mathbf{0}$ where the integral is taken over a set $B \subseteq \mathbb{R}^3$ enclosing the entire body.

 

[Ibix]

To expand slightly on the previous comment, center of gravity and center of mass are slightly different concepts under some circumstances. In a uniform gravitational field they are the same thing, but in a non-uniform gravitational field they are not.

Replies #2 and #3 are either talking about center of mass or assuming a uniform gravitational field. Reply #4 is not. This thread is tagged as I level, so non-uniform fields may be relevant.

Perhaps @Dame could say if they understand the distinction between the centers of mass and gravity, which would help to pitch the level of the conversation.

 

[Delta2]

To make a long story short: The center of gravity does not change as long as the body is rigid, and moves within a uniform gravitational field.

 

[Richard R Richard]

Expanding

Centroid, center of mass, and center of gravity are not exactly the same
they only match if

• the gravitational field is uniform

• the density of the object is uniform

for large objects in which the gravitational field varies with height, as is the case on Earth, the center of gravity will differ from the center of mass. in small objects of everyday life the difference is negligible,
Example: The center of mass of an object of negligible base S and height h and constant density would be calculated$$ \displaystyle \mathbf r _{\text {cm}} = \dfrac {\int \limits_V \rho \mathbf r dV} {\int \rho dV} = \dfrac {\int \limits_V \mathbf r dV} {V} = \dfrac {\int \limits_0 ^ H \mathbf hS dh} {SH} = \dfrac {\int \limits_0 ^ H \mathbf h dh} {H} = \dfrac H2 $$
and the center of gravity
$$\displaystyle \mathbf r _{\text {cg}} = \dfrac {\int \limits_V \rho \mathbf rg (r) dV} {\int_V \rho g (r) dV} = \dfrac { \int \limits_V \mathbf rg (r) dV} {\int \limits_V g (r) dV} =$$ $$ \dfrac {\int \limits_0 ^ H \mathbf hS g (h) dh} {S \int \limits_0 ^ H g (h) dh} = \dfrac {\int \limits_0 ^ H \mathbf hg (h) dh} {\int \limits_0 ^ H g (h) dh} $$as you can see if $g (h) = constant$ both expressions matchbut generally $g (h) = \dfrac {GM} {(R_{Earth} + h) ^ 2}$ for the variation of gravity as a function of the mass and the terrestrial radius, it is not worth it consider the difference when $H << R_{Earth}$

Nor is it the same, the centroid only takes into account the geometry of the object, if the object has a constant density then the centroid and the center of mass coincide, and if the gravity is also constant it also coincides with the center of gravity.

Following the example,

$$ \displaystyle \mathbf r _{\text {ce}} = \dfrac {\int \limits_V \mathbf r dV} {\int \limits_V dV} = \dfrac {\int \limits_V \mathbf r dV } {V} = \dfrac {\int \limits_0 ^ H \mathbf hS dh} {SH} = \dfrac {\int \limits_0 ^ H \mathbf h dh} {H} = \dfrac H2 $$

Centroid	Center of mass	Center of gravity
$\displaystyle\mathbf r_{\text{ce}} = \dfrac{ \int \limits_V \mathbf r dV}{ \int \limits_VdV}$	$\displaystyle\mathbf r_{\text{cm}} = \dfrac{ \int \limits_V \mathbf r\rho(V) dV}{ \int \limits_V\rho(V)dV}$	$\displaystyle\mathbf r_{\text{cg}} = \dfrac{ \int \limits_V \mathbf r\rho(V) g(V)dV}{ \int \limits_V\rho(V)g(V)dV}$

 

[anorlunda]

@Richard R Richard , none of your Tex math worked. Click on the LateX Guide link at the bottom of the post editor to see how it works here on PF.

 

[Delta2]

Interesting if you put the commands inside

[TEX]...[/TEX]

it doesn't work. Instead if you put the command with lower case letters like

[tex]...[/tex]

it works. You should also to enclose the tex commands within double # or double $ like this for example

##\int f(x)dx ##

or

$$\int f(x)dx$$

 

[Richard R Richard]

Sorry I accidentally posted when I was modifying and editing in preview

 

[Dame]

Thank you guys for the warm welcome. I think all the answers have definitely shed some light. But, I think @Ibix had the best response so far.

I do have a few follow up questions though that's a little involved. The scenario I am about describe is taking place in a uniform gravitational field. Would it be possible to drop an object(lets say something mostly homogeneous like a brick) in free fall from a small height of 1 m and have it impact the ground at close to or similar to the orientation you released it from?

Is it safe to assume that I will need to hold the object from, or close to, its CofG in order to have a chance of achieving this?

 

[Baluncore]

All the particles of the brick, including the CofG, will accelerate towards the Earth at the same rate. If you release it cleanly, then I see no reason why it should rotate. If you impart a rotation on release, it will continue to rotate during the fall. You do not have to release it from it's CofG.

 

[Richard R Richard]

I am not so sure what he says, his conclusions are only true if the gravitational field is constant, and the Earth does not have it, the particles of the brick furthest from the center of the Earth accelerate with less intensity than the closest ones, not We note that the acceleration difference is very small compared to what the chemical bonds of the brick resist.
I reiterate my appreciations about certain answers that are not correct throughout this thread, proposing a couple of examples for your analysis. If you hang an object by its center of gravity, the object will not rotate, since it has each moment created by the weight force of each of its particles balanced with respect to the vertical. But if you hang it by its center of mass the object can rotate, the same if you hang it from the centroid point.
The first example two spheres of mass m joined by a fine thread that separates their centers a distance h. If they are arranged in a radial direction to the ground, the wire will take tension due to the acceleration difference between the different heights.
The second example is that you try to hang a letter "C" by its center of mass, with a thread of negligible mass, it will become a letter "U" in a few moments.

 

[Richard R Richard]

Just to provide some numerical data, since it seems that it is writing alone, that we imagine an equilateral triangle of uniform density of side equal to the radius of the Earth, located its base or vertex closest to the Earth at 2 radius of the Earth with respect to the surface of the earth, for simplicity suppose we only take into account the vertical component of gravity and not the entire distance to the center. Thus, the triangle with vertex down has the CM at $2 + \dfrac {1} {\sqrt3} = 2.577Re$ radii from the surface or at $3 + \dfrac {1} {\sqrt3}$ radii with respect to the center of the earth, the calculation of the CG places it below the CM, of course, at $2.56 Re$. And if it is inverted vertex up, the CM is at $2 + \dfrac {1} {2 \sqrt3} = 2.28Re$ from the surface and the CG at $2.22 Re$ from the surface. So if the vertex triangle is released downwards, it will be in an unstable equilibrium, any disturbance will make it rotate, to try to have its CG as close to the ground and as far away from its CM, the stable equilibrium position is then it will be vertex up. This answers the main question of the thread.

Would the center of gravity for the same cylinder/rectangle be in the same spot if I stood the object up or rotated the object about its length?

The variable distance calculated between CM and CG in the cases presented shows that the CG changes its position relative to the CM, by orienting the object in a different way with respect to a vertical line that passes through its CM and the center of the earth.

Pd, if you need the counts or the formulas, to analyze them, please give me time to write them in LATEX, but any page that calculates double integrals will allow you to obtain the same result quickly.

Greetings

 

[Delta2]

@Richard R Richard in all of your examples at #13 and #14 the size of the system must be tens (or more) of kilometers long, otherwise the difference in $g=G\frac{M_{Earth}}{(R_{Earth}+x)^2}$ through out the system will be negligible.

For ordinary sized systems/objects we can say that gravitational field of Earth is (almost perfectly) uniform throughout the system/object and thus the center of mass of system (almost perfectly) coincide with the center of gravity.

 

[Richard R Richard]

Totally agree, in the insignificance about for objects of daily life, the OP will tell us if this was what he was asking.