# Centre of Gravity and Centre of Mass

1. Jan 16, 2013

### sorax123

Hey folks,
I was doing some moments problems the other day, and I started considering centre of mass and centre of gravity and have gotten myself rather unsure. My textbook defines centre of mass as "the point through which a single force has no turning effect", which I have an intuitive grasp of. However, it then goes on to talk about how something will topple if the centre of MASS lies outside the base. But in previous studies, the reason for toppling was given as the line of action of weight lying outside the base. But surely that corresponds to the centre of GRAVITY lying outside the base... This has led to me thinking about when a body lies in a non uniform gravitational field and hence the centre of gravity is not in the same place as the centre of mass. I understand the centre of gravity as the point from which the weight can be thought to act. So If you have a body in the aforementioned situation, surely it should rotate as the CoG would cause a moment about the Centre of Mass. Would this body then rotate about the CoM?
I was wondering if someone could define centre of gravity and explain the difference between the two so that i can be sure, and also allowing me to answer some of my puzzling issues.
Many thanks,
Dom

2. Jan 16, 2013

### Staff: Mentor

I don't think there is a difference between the two. Your center of mass should be the same as your center of gravity.

3. Jan 16, 2013

### Staff: Mentor

That's only true in a uniform gravitational field.

4. Jan 16, 2013

### Staff: Mentor

Perhaps I'm misunderstanding what "center of gravity" is then. What's the difference between CoM and CoG?

5. Jan 16, 2013

### DrewD

If something is toppled, it must have been standing to begin with (at least that's how I interpret the english) and therefore is being supported by a normal force that may not be acting on the CM.

note: I know this doesn't address difference between CG and CM, but I don't think that is an integral part of your confusion. My apologies if I'm wrong.

Last edited: Jan 16, 2013
6. Jan 16, 2013

### sorax123

Yeah, my confusion lies in the effect of each centre. Imagine a tractor which is placed on a hydraulic machine which can tilt it. It topples when the line of action of weight is outside the wheel base. If CoG is what I understand it to be- the point at which the resultant moment due to weight is 0- then surely it is when the CoG is outside the wheel base when it topples, not when the CoM is outside the base? Am I misunderstanding the two terms or is the book being clumsy and using CoM and CoG interchangabely hence causing my confusion?

7. Jan 16, 2013

### Staff: Mentor

In a uniform gravitational field, the center of gravity and center of mass are the same. Most books--since they assume a uniform gravitational field--use the terms interchangeably. No big deal.

8. Jan 16, 2013

### sorax123

Fair enough, thanks for clearing that up. :)

9. Jan 16, 2013

### Staff: Mentor

CoM is just the average location of the mass of a body. Nothing to do with gravity, explicity.

The CoG is often defined as a point about which there is no torque due to gravity, which as long as the gravitational field is uniform is the same as the center of mass.

10. Jan 16, 2013

### Studiot

Just to reinforce what Doc Al has said,

Imagine a body floating free in space (no gravity).

If you apply a force (eg a rocket thrust) on a line through the centre of mass the body will be driven along that line without spinning.

If you apply the same thrust off this line it will not only drive the mass forwards it will also cause it to start spinning.

11. Jan 16, 2013

### AlephZero

The COM is defined from the mass distribution of the body, i.e.$$\int_V \rho(\mathbf r) \mathbf r\,dv\; \Big/\! \int_V \rho(\mathbf r) \,dv$$

The COG is defined from the force distribution caused by gravity, i.e. $$\int_V \rho(\mathbf r)g(\mathbf r) \mathbf r\,dv\; \Big/\! \int_V \rho(\mathbf r)g(\mathbf r) \,dv$$

If $g(\mathbf r)$ is independent of the position $\mathbf r$, the two are equal.

The most common "practical" case where the two definitions are not equal is when working in a rotating (non-inertial) coordinate system, though I won't complain if you object to using the word "gravity" in that situation.

12. Jan 16, 2013

### Staff: Mentor

I see. Apparently I just had no idea what I was talking about then lol. I guess I misunderstood what a "non uniform gravitational field" is as well.

13. Jan 16, 2013

### haruspex

14. Jan 16, 2013

### Staff: Mentor

That's right.