Why do we need a separate center of gravity idea when we have CoM?

[Ryder Rude]

TL;DR Summary

The derivation of Center of Mass doesn't assume anything about the nature of external forces involved. Then why do we need a separate point called 'center of gravity' when working with non uniform gravitational fields?

While deriving the formula for the position vector of the center of mass of a system of $n$ particles, we assume some external force $\vec{F_i}$ on each particle and internal forces $\vec{f_{ij}}$ between any two particles.

In the derivation, we come to the conclusion that the position vector $\vec{r}$ that we computed obeys $\frac{d^2\vec{r}}{dt^2}=\frac{1}{M}\sum \vec{F_i}$, where $M$ is the total mass of the system.

This result means that the point behaves as if it was a point particle of mass $M$ all the external forces were directly acting on it.

Since the derivation didn't assume anything about the nature of the external forces $\vec{F_i}$, the above result should be true even if the system of particles is in a non-uniform gravitational field.

So why do we need a separate center of gravity idea when we are working with non-uniform gravitational fields?

 

[A.T.]

Since the derivation didn't assume anything about the nature of the external forces $\vec{F_i}$, the above result should be true even if the system of particles is in a non-uniform gravitational field.

It is still true.

So why do we need a separate center of gravity idea when we are working with non-uniform gravitational fields?

Because we sometimes also care about the rotation of the body, not just the center of mass motion.

 

[vanhees71]

The motivation to introduce the center of mass (in Newtonian mechanics) is, as you describe, that for a closed system it moves always with constant velocity, i.e., as a free body (as observed in an intertial reference frame of course).

If there are external forces like the description of the motion of a system of particles in the gravitational field of the Earth, in general there's not a simple equation of motion for the center of mass. Only in the special case, where you can approximate the external gravitational field as homogeneous, i.e.,
$$m_i \ddot{\vec{x}}_i=\sum_{j \neq i} \vec{F}_{ij}+m_i \vec{g},$$
you get a simple equation of motion, i.e.,
$$M \ddot{\vec{R}}=M \vec{g},$$
i.e., the center of mass behaves like a free-falling body, because in the sum over $i$ and $j$ the interaction forces all cancel due to Newton's 3rd Law. This also doesn't change for more general cases, but the sum over the external force is usually not anymore simply related to the center of mass.

Take, e.g., the motion of the body in the Gravitational field of the Earth, when it's no longer appropriate to treat it as homogeneous. Simplify the Earth to a "point" sitting in the origin of the coordinate system you get
$$M \ddot{\vec{R}}=-\sum_i \frac{G m_i M_{\text{Earth}} \vec{r}_i}{|\vec{r}_i|^3},$$
which is not simply a function of $\vec{R}$.

 

[Ryder Rude]

It is still true.Because we sometimes also care about the rotation of the body, not just the center of mass motion.

So, say, there is another representative point of a system of particles, which gives us information about the system which is different from the information the center of mass gives. We call this representative point 'center of gravity'.

Though what information (related to rotations) does this point give? (Similar to how the momentum of center of mass is the vector sum of the momenta of the particles of the system)

And why is the idea of this new point restricted to the problems of gravity? Can we define this for general external forces?

 

[Ryder Rude]

The motivation to introduce the center of mass (in Newtonian mechanics) is, as you describe, that for a closed system it moves always with constant velocity, i.e., as a free body (as observed in an intertial reference frame of course).

If there are external forces like the description of the motion of a system of particles in the gravitational field of the Earth, in general there's not a simple equation of motion for the center of mass. Only in the special case, where you can approximate the external gravitational field as homogeneous, i.e.,
$$m_i \ddot{\vec{x}}_i=\sum_{j \neq i} \vec{F}_{ij}+m_i \vec{g},$$
you get a simple equation of motion, i.e.,
$$M \ddot{\vec{R}}=M \vec{g},$$
i.e., the center of mass behaves like a free-falling body, because in the sum over $i$ and $j$ the interaction forces all cancel due to Newton's 3rd Law. This also doesn't change for more general cases, but the sum over the external force is usually not anymore simply related to the center of mass.

Take, e.g., the motion of the body in the Gravitational field of the Earth, when it's no longer appropriate to treat it as homogeneous. Simplify the Earth to a "point" sitting in the origin of the coordinate system you get
$$M \ddot{\vec{R}}=-\sum_i \frac{G m_i M_{\text{Earth}} \vec{r}_i}{|\vec{r}_i|^3},$$
which is not simply a function of $\vec{R}$.

So the center of mass is not convenient to work with because of the complications of adding up non-uniform gravitation forces.

Instead, we work with a new point called 'center of gravity', which is easy to calculate, and it gives us some other information about the system than what the center of mass gives. Is this correct?

 

[A.T.]

So, say, there is another representative point of a system of particles, which gives us information about the system which is different from the information the center of mass gives. We call this representative point 'center of gravity'.

Center of gravity is not so much a representative point of the particles (like center of mass is), but more a representative point of application for the external force (gravity).

Though what information (related to rotations) does this point give?

Gravity can create a torque around the center of mass, if the center of gravity is not identical to it.

And why is the idea of this new point restricted to the problems of gravity? Can we define this for general external forces?

It is comparable to center of pressure or center of buoyancy. To simplify calculations we often model forces as applied at a single representative point.

 

[Ryder Rude]

Center of gravity is not so much a representative point of the particles (like center of mass is), but more a representative point of application for the external force (gravity).

But that is also the whole idea of center of mass. It is the point on which the external forces can be thought of as acting directly

How can center of gravity be describing the same idea and also be different from center of mass?

It is is comparable to center of pressure or center of buoyancy. To simplify calculations we often model forces as applied at a single representative point.

What if we have a situation where, say, some countable number of forces are acting on a countable number of points on a rigid body? Is there any such thing as 'center of external forces' for this problem?

 

[A.T.]

But that is also the whole idea of center of mass. It is the point on which the external forces can be thought of as acting directly

Only for calculating the linear acceleration of the center of mass. Not for angular acceleration.

How can center of gravity be describing the same idea and also be different from center of mass?

Compare the precise derivations of the points, not informal descriptions of the "idea".

What if we have a situation where, say, some countable number of forces are acting on a countable number of points on a rigid body? Is there any such thing as 'center of external forces' for this problem?

You can define it, if you find it useful to lump them all together.

 

[Ryder Rude]

You can define it, if you find it useful to lump them all together.

And if we do define it, its purpose will be related to measure the angular acceleration of the system? Here I'm making an analogy with center of mass:

If we track the location of the center of mass over time, and find it to be accelerating with acceleration $\vec{a}$, then we conclude that the sum of external forces on the system is $M\vec{a}$

Similarly, if we track the location of 'center of external forces' over time and find it as having angular acceleration $\vec{\alpha}$, then we conclude that the sum of external torques on the object is $I\vec{\alpha}$.

Is the above fact correct, or is a similar fact correct? I'm just guessing here. If it's indeed correct, why don't we ever study 'center of external forces' in school courses? It seems like a natural thing to complement center of mass.

 

[sophiecentaur]

How can center of gravity be describing the same idea and also be different from center of mass?

Centre of mass relates to the object itself. Centre of Gravity puts the object in some sort of gravitational field. But the two terms are used interchangeably in most contexts (School Physics context, for instance) and often give the same 'answer'.

 

[Dale]

But that is also the whole idea of center of mass. It is the point on which the external forces can be thought of as acting directly

This is not correct. The center of mass summarizes an object’s inertia. It is sort of a summary of $m$ in $F=ma$ for an extended object.

The center of gravity or the center of buoyancy or the center of whatever force is the quantity that summarizes the $F$ in $F=ma$.

You need two different quantities because they are different things. If it helps then remember that mass quantifies inertia. So the center of mass is the center of inertia. That may make it more clear why it is important to have it as a distinct concept from the center of gravity.

 

[A.T.]

If it's indeed correct, why don't we ever study 'center of external forces' in school courses? It seems like a natural thing to complement center of mass.

People use what is practical in solving problems, not what "seems natural".

 

[Ryder Rude]

This is not correct.

Why wouldn't it be correct? The center of mass point literally follows the path that a particle of mass $M$ would've followed if the forces on the system were thought of as directly acting on that point.

Also, how do we actually summarise the external forces? What's the formula? I can't seem to find anything on google. When I search 'center of force', I get 'central forces', so i guess the term 'center of force' isn't even used.

I'm expecting something similar to :

$$M\frac{d^2 (\frac{1}{M}\sum m_i\vec{r_i})}{dt^2}=\sum \vec{F_i}$$

Is there a similar identitiy for 'center of force'?

 

[Ryder Rude]

People use what is practical in solving problems, not what "seems natural".

It's worth knowing as a concept though. We don't use Cramer's rule practically, but we still know it. 'Center of force' should be taught in general instead of just 'center of gravity'.

You didn't say if my analogy was correct in the post you replied to

 

[A.T.]

i guess the term 'center of force' isn't even used.

It's called "point of application".

 

[sophiecentaur]

The center of mass point literally follows the path that a particle of mass M would've followed if the forces on the system were thought of as directly acting on that point.

Why are you insisting on this? CoM = CoG works as you say when each elemental part of the object's mass is subject to the same 'force per kg'. But with your insisted definition, you could push an object at any point and it would behave the same, which is obviously not the case. Your definition works only where it works.
I can't see why you continue to argue your case.

 

[Ryder Rude]

It's called "point of application".

From what I found, I think we're trying to find the point on which applying the net force would produce the same linear and angular acceleration of the system, as the individual forces together produce.

So in this sense, the individual forces can be thought of as acting together on this point.

 

[A.T.]

From what I found, I think we're trying to find the point on which applying the net force would produce the same linear and angular acceleration of the system, as the individual forces together produce.

But that point isn't unique.

 

[Ryder Rude]

But that point isn't unique.

Did I miss anything? There are not many links about 'Point of application'. It's not on even wikipedia

 

[sophiecentaur]

From what I found, I think we're trying to find the point on which applying the net force would produce the same linear and angular acceleration of the system, as the individual forces together produce.

So in this sense, the individual forces can be thought of as acting together on this point.

"In this sense" says it all. You are taking one specific instance but your idea doesn't satisfy a general requirement. Is there any point, at which you could apply a single force to a non uniform ladder in space that would produce the same effect as applying a single, normal force to rung no. 9 (except rung no. 9, of course)?

You need some exceptional high grade Classical Physics if you want to argue against what Classical Physics already tells you. Try looking for a reason that you may not have got this quite right instead of hanging onto your untenable position. There is never any shame in seeing the light.

 

[A.T.]

Did I miss anything?

The conditions you stated do not uniquely define a single point:

...the point on which applying the net force would produce the same linear and angular acceleration of the system, as the individual forces together produce.

You can always move a point that satisfies the above parallel to the net force, and still get the same linear and angular acceleration.

 

[hutchphd]

Summary:: The derivation of Center of Mass doesn't assume anything about the nature of external forces involved. Then why do we need a separate point called 'center of gravity' when working with non uniform gravitational fields?

So why do we need a separate center of gravity idea when we are working with non-uniform gravitational fields?

Because in a nonuniform graitational field the two points do not generally coincide.

Are we finished now?

 

[Ryder Rude]

"In this sense" says it all. You are taking one specific instance but your idea doesn't satisfy a general requirement. Is there any point, at which you could apply a single force to a non uniform ladder in space that would produce the same effect as applying a single, normal force to rung no. 9 (except rung no. 9, of course)?

Not exactly the same effect (as in, the particles won't move in the exact same way) But maybe the same net effect? (as in, if you add up the add up the angular and linear momenta of the prticles, you get the same result in both cases)

You need some exceptional high grade Classical Physics if you want to argue against what Classical Physics already tells you. Try looking for a reason that you may not have got this quite right instead of hanging onto your untenable position. There is never any shame in seeing the light.

Not sure what you're saying. I'm not arguing against anything that Classical physics tells me. I'm only trying to understand one Classical physics idea.

Back when I said that the forces could be thought of as acting directly on the CoM, I only meant that the CoM point moves as if all the forces were acting on it. I was only talking about the behavior of the CoM point (and not of the entire system). I was myself confused there. Now I understand that the CoG point represents a very different idea (as something which talks about what would happen to the whole system if all the forces were thought of as acting at CoG)

 

[Ryder Rude]

The conditions you stated do not uniquely define a single point:

You can always move a point that satisfies the above parallel to the net force, and still get the same linear and angular acceleration.

Yeah, I got that from your previous reply. But what additional conditions are missing? The formula was like taking the weighted average of the distances by using forces at the points as weights.

 

[A.T.]

Yeah, I got that from your previous reply. But what additional conditions are missing? The formula was like taking the weighted average of the distances by using forces at the points as weights.

That approach will give you a specific point. But you can move that point parallel to the force vector, and still get the same linear and angular accelerations. So it's not that "special", which might be one reason why it is not a key concept like center of mass.

 

[Ryder Rude]

That approach will give you a specific point. But you can move that point parallel to the force vector, and still get the same linear and angular accelerations. So it's not that "special", which might be one reason why it is not a key concept like center of mass.

We could still talk about a point of application in 1D, line of application in 2D, and plane of application in 3D. I think the fact that it's a property of the forces acting instead of a property of the system makes it way less fundamental than CoM.

This point is a good thing to talk about only when the problem is to balance both linear and angular accelerations of an object. This property is fundamental to force fields, similar to how CoM is fundamental to objects. This property is a function of subsets of points in the force fields. (Like, when a body moves through a force field, it occupies some subset of points in space at a particular time, which uniquely determines this property)

 

[kuruman]

A simple sample calculation might help. Let's find the CoM and CoG of a vertical pole of length $L$ on a planet of radius $R$ and mass $M$. Let $\lambda$ be the mass per unit length of the pole.
First find the distance of the CoM from the center of the planet.
$$R_{m}=\frac{\int_R^{R+L}r~dm}{\int_R^{R+L}dm}=\frac{\int_R^{R+L}r~\lambda dr}{\int_R^{R+L}\lambda dr}=R+\frac{L}{2}.$$Clearly the CoM is at the midpoint of the pole.

Next find the distance of the CoG from the center of the planet.
$$R_{g}=\frac{\int_R^{R+L}r~dF_g}{\int_R^{R+L}dF_g}= \frac{\int_R^{R+L}r~\lambda dr~g(r)}{\int_R^{R+L}\lambda dr~g(r)} =\frac{\int_R^{R+L}r~\lambda dr~\frac{GM}{r^2}}{\int_R^{R+L}\lambda dr~\frac{GM}{r^2}} =\frac{R}{L}(L+R)\ln\left(\frac{R+L}{R}\right).$$Clearly, the CoG is not at the midpoint of the pole. For $R>L$, a series expansion gives $$R_g=R+\frac{L}{2}-\frac{L^2}{6R}+\mathcal{O}(n)$$Note that the CoG is closer to the planet's center than the CoM. For a 10-m pole on the surface of the Earth, the difference is about 3 microns.

 

[Ryder Rude]

A simple sample calculation might help. Let's find the CoM and CoG of a vertical pole of length $L$ on a planet of radius $R$ and mass $M$. Let $\lambda$ be the mass per unit length of the pole.
First find the distance of the CoM from the center of the planet.
$$R_{m}=\frac{\int_R^{R+L}r~dm}{\int_R^{R+L}dm}=\frac{\int_R^{R+L}r~\lambda dr}{\int_R^{R+L}\lambda dr}=R+\frac{L}{2}.$$Clearly the CoM is at the midpoint of the pole.

Next find the distance of the CoG from the center of the planet.
$$R_{g}=\frac{\int_R^{R+L}r~dF_g}{\int_R^{R+L}dF_g}= \frac{\int_R^{R+L}r~\lambda dr~g(r)}{\int_R^{R+L}\lambda dr~g(r)} =\frac{\int_R^{R+L}r~\lambda dr~\frac{GM}{r^2}}{\int_R^{R+L}\lambda dr~\frac{GM}{r^2}} =\frac{R}{L}(L+R)\ln\left(\frac{R+L}{R}\right).$$Clearly, the CoG is not at the midpoint of the pole. For $R>L$, a series expansion gives $$R_g=R+\frac{L}{2}-\frac{L^2}{6R}+\mathcal{O}(n)$$Note that the CoG is closer to the planet's center than the CoM. For a 10-m pole on the surface of the Earth, the difference is about 3 microns.

It should be $R + \frac{L}{2}$ for the CoM, right?

 

[kuruman]

It should be $R + \frac{L}{2}$ for the CoM, right?

Isn't that what I have?

 

[Ryder Rude]

Isn't that what I have?

You accidentally wrote just $R$

 

[Ryder Rude]

Isn't that what I have?

Sorry I was viewing it on mobile. I had to slide it to see it full

 

[Dale]

The center of mass point literally follows the path that a particle of mass M would've followed if the forces on the system were thought of as directly acting on that point.

If the center of gravity (or buoyancy or any other distributed force) is different from the center of mass then the force acts through the center of gravity (etc) and therefore produces a torque about the center of mass. The motion of the body then cannot be described simply by the path of the center of mass. The object will rotate about the center of mass. Modeling this torque is the reason for distinguishing the center of mass from the center of gravity (etc).