CENTER OF GRAVITY: PROPERTIES, CALCULATION, EXAMPLES - PHYSICAL - 2024

The center of gravity of a body of measurable size is the point where its weight is considered to be applied. It is therefore one of the main concepts of Statics.

The first approach in the problems of elementary physics consists of assuming that any object behaves like a point mass, that is, it has no dimensions and all the mass is concentrated in a single point. This is valid for a box, a car, a planet or a subatomic particle. This model is known as the particle model.

Figure 1. In the high jump the athlete manages so that his center of gravity is outside the body. Source: Pixabay

This is of course an approximation, which works very well for many applications. It is not an easy task to consider the individual behavior of the thousands and millions of particles that any object can contain.

However, the real dimensions of things must be taken into account if results that are closer to reality are to be obtained. Since we are generally in the vicinity of the Earth, the ever-present force on any body is precisely the weight.

Considerations for finding the center of gravity

If body size is to be taken into account, where specifically is weight to be applied? When you have an arbitrarily shaped continuous object, its weight is a force distributed between each of its constituent particles.

Let these particles be m ₁, m ₂, m ₃ … Each one of them experiences its corresponding gravitational force m ₁ g, m ₂ g, m ₃ g…, all of them parallel. This is so, since the gravitational field of the Earth is considered constant in the vast majority of cases, since the objects are small compared to the size of the planet and are close to its surface.

Figure 2. The weight of the object is a distributed mass. Source: self made.

The vector sum of these forces results in the weight of the object, applied to the point called the center of gravity denoted in the figure as CG, which then coincides with the center of mass. The center of mass in turn is the point where all the mass could be considered concentrated.

The resulting weight has magnitude Mg where M is the total mass of the object, and of course it is directed vertically towards the center of the Earth. The summation notation is useful for expressing the total mass of the body:

The center of gravity does not always coincide with a material point. For example, the CG of a ring is at its geometric center, where there is no mass itself. Even so, if you want to analyze the forces acting on a hoop, you have to apply the weight to this precise point.

In those cases in which the object has an arbitrary shape, if it is homogeneous, its center of mass can still be calculated by finding the centroid or center of gravity of the figure.

How is the center of gravity calculated?

In principle, if the center of gravity (CG) and the center of mass (cm) coincide as the gravitational field is uniform, then the cm can be calculated and weight applied to it.

Let us consider two cases: the first is one in which the mass distribution is discrete; that is, each mass that makes up the system can be counted and assigned a number i, as was done in the previous example.

The coordinates of the center of mass for a discrete mass distribution are:

Naturally the sum of all the masses equals the total mass of the system M, as indicated above.

The three equations are reduced to a compact form when considering the vector r _cm or position vector of the center of mass:

And in the case of a continuous mass distribution, where the particles are of differential size and cannot be distinguished to count them, the sum is replaced by an integral that is made over the volume occupied by the object in question:

Where r is the position vector of a differential mass dm and the definition of mass density has been used to express the mass differential dm contained in a volume differential dV:

Properties

Some important considerations about the center of mass are as follows:

- Although a reference system is required to establish the positions, the center of mass does not depend on the choice made of the system, since it is a property of the object.

- When the object has an axis or a plane of symmetry, the center of mass is on that axis or plane. Taking advantage of this circumstance saves calculation time.

- All external forces acting on the object can be applied to the center of mass. Keeping track of the motion of this point gives an overview of the object's motion and makes it easier to study its behavior.

-Finding the center of gravity of a body in static equilibrium

Suppose you want to make the body of the previous figure be in static equilibrium, that is, it does not translate or rotate about an arbitrary axis of rotation that can be O.

Figure 3. Scheme to calculate the torque of the weight with respect to point O.

-Resolved example

A thin bar of uniform material is 6 m long and weighs 30 N. A 50 N weight is hung at its left end and a 20 N weight is hung at its right end. Find: a) The magnitude of the upward force necessary to maintain the equilibrium of the bar, b) The center of gravity of the assembly.

Solution

The force diagram is shown in the following figure. The weight of the bar is applied at its center of gravity, which coincides with its geometric center. The only dimension of the bar taken into account is its length, since the statement reports that it is thin.

Figure 4. Diagram of forces for the bar.

For the bar + weights system to remain in translational equilibrium, the sum of the forces must be zero. The forces are vertical, if we consider up with sign + and down with sign - then:

F- 50 - 20 - 30 N = 0

F = 100 N

This force guarantees the translational balance. Taking the torsional moments of all the forces with respect to an axis that passes through the extreme left of the system and applying the definition:

t = rx F

The moments of all these forces about the selected point are perpendicular to the plane of the bar:

Thus:

The center of gravity of the bar + weights set is located 2.10 meters from the left end of the bar.

Difference from center of mass

The center of gravity coincides with the center of mass, as indicated, as long as the Earth's gravitational field is constant for all points of the object to be considered. The gravitational field of the Earth is nothing more than the well-known and familiar value of g = 9.8 m / s ² directed vertically downwards.

Although the value of g varies with latitude and altitude, these do not usually affect the objects that are most of the time discussed. It would be very different if you consider a large body in the vicinity of the Earth, for example an asteroid that is very close to the planet.

The asteroid has its own center of mass, but its center of gravity would no longer have to coincide with this, since g would probably experience substantial variations in magnitude, given the size of the asteroid and that the weights of each particle might not be parallel.

Another fundamental difference is that the center of mass is found regardless of whether or not there is a force called weight applied to the object. It is an intrinsic property of the object that reveals to us how its mass is distributed in relation to its geometry.

The center of mass exists whether there is weight applied or not. And it is located in the same position even if the object moves to another planet in which the gravitational field is different.

On the other hand, the center of gravity is clearly linked to the application of weight, as we have seen throughout the previous paragraphs.

Examples of center of gravity

Center of gravity of irregular objects

It is very easy to find out where the center of gravity of an irregular object such as a cup is. First, it is suspended from any point and from there a vertical line is drawn (in figure 5 it is the fuchsia line in the left image).

It is then suspended from another point and a new vertical is drawn (turquoise line in the right image). The intersection of both lines is the center of gravity of the cup.

Figure 5. CG location of a mug. Source: modified from Pixabay.

Balancing objects

Let's analyze the stability of a truck traveling on the road. When the center of gravity is above the base of the truck, the truck will not tip over. The image on the left is the most stable position.

Figure 6. Balancing the truck. Source: self made.

Even when the truck leans to the right, it will be able to return to a stable equilibrium position, as in the middle drawing, since the vertical still passes through the base. However when this line goes outside the truck will tip over.

The diagram shows the forces at the fulcrum: normal in yellow, weight in green, and static rub to the left in fuchsia. Normal and friction are applied to the axis of rotation, so they do not exert torque. Therefore they will not contribute to overturn the truck.

The weight remains, which does exert a torque, luckily counterclockwise and which tends to return the truck to its equilibrium position. Note that the vertical line passes through the support surface, which is the tire.

When the truck is in the far right position, the torque of the weight changes to clockwise. Unable to be countered for another time, the truck will overturn.

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 247-253.
2. Giancoli, D. 2006. Physics: Principles with Applications. 6th.. Ed Prentice Hall. 229-238.
3. Resnick, R. (1999). Physical. Vol. 1. 3rd Ed. In Spanish. Compañía Editorial Continental SA de CV 331-341.
4. Rex, A. 2011. Fundamentals of Physics. Pearson. 146-155.
5. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1,340-346.