por He gravity center of a body of measurable size is the point where its weight is considered to be applied. It is therefore one of the fundamental concepts of Statics.
The first approach in elementary physics problems consists in assuming that any object behaves like a point mass, that is, it lacks 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 particle model.
Naturally this is an approximation, which gives very good results for many applications. It is not an easy task to consider the individual behavior of the thousands and millions of particles that any object may contain.
However, the real dimensions of things must be taken into account if you want to obtain results that are closer to reality. Since we are generally in the immediate vicinity of the Earth, the force always present on any body is precisely the weight.
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Considerations to find the center of gravity
If body size is to be taken into account, where specifically is the weight going to be applied on the body? When you have a continuous object of arbitrary shape, its weight is a distributed force between each of its constituent particles.
Let these particles be m1, m2, m3… Each of them experiences its corresponding gravitational force m1g, m2g, m3g…, all of them parallel. This is so, since the Earth’s gravitational field is considered constant in the vast majority of cases, given that the objects are small compared to the size of the planet and are close to its surface.
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 Mass center. 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. Summation notation is useful for expressing the total mass of the body:
Symmetrical and homogeneous objects, which is equivalent to their density being uniform, have their center of gravity at the geometric center: cubes, parallelepipeds, rings, tires or spheres.
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 actual mass. Even so, if you want to analyze the forces that act on a ring, you have to apply the weight to this precise point.
In cases where the object has an arbitrary shape, if it is homogeneous, its center of mass can still be calculated by finding the centroid or centroid 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 the 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 distribution of masses are:
Naturally, the sum of all the masses is equivalent to the total mass of the system M, as indicated above.
The three equations are reduced to a compact form by considering the vector rcm or position vector of the center of mass:
And in the case of a continuous distribution of masses, where the particles are of differential size and cannot be distinguished in order 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 use has been made of the definition of mass density to express the differential of mass dm contained in a volume spread dV:
The density of the object can be constant, in which case it falls outside the integral, or it can be a function of the spatial coordinates and its dependence on (x,y,z) must be known to solve the integral.
Properties
Some important considerations about the center of mass are the following:
– 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 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. Tracking the movement of this point gives a global idea of the movement of the object and makes it easier to study its behavior.
-Finding the center of gravity of a body in static equilibrium
Suppose that you want to make the body in the previous figure be in static equilibrium, that is, it neither translates nor rotates about an arbitrary axis of rotation that can be O.
The net torsion moment of the weight with respect to O, according to figure 3 is: A force F applied vertically upwards in the center of gravity (or also above or below, on the axis that passes through it) would produce a opposite torque that would prevent rotation of the object and maintain rotational equilibrium. The magnitude of F is chosen so that the object does not move either and in this way we will have it in static equilibrium.
-Example worked out
A thin bar of uniform material has a length of 6 m and weighs 30 N. A 50-N weight is hung from its left end and a 20-N weight is hung from its right end. Find: a) The magnitude of the upward force necessary to maintain the balance 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.
In order for the bar + weight system to remain in translational equilibrium, the sum of the forces must be zero. The forces are vertical, if we consider upwards with a + sign and downwards with a – sign, then:
F- 50 – 20 – 30 N = 0
F = 100N
This force guarantees translational equilibrium. Taking the torque moments of all forces about an axis passing through the left end of the system and applying the definition:
t = r x F
The moments of all these forces about the selected point are perpendicular to the plane of the bar:
tF = xF = 100x
tW = -(l/2)mg =-3m . 30N = -90Nm
t1 = 0 (since the 50 N force passes through the selected axis of rotation and exerts no moment)
t2 = -lF2 = 6 m . 20N = -120Nm
Therefore:
100 x -90 -120 Nm = 0
x = 2.10m
The center of gravity of the bar + weights set is located 2.10 meters from the left end of the bar.
Difference with center of mass
The center of gravity coincides with the center of mass, as indicated, provided that the terrestrial gravitational field is constant for all points of the object to be considered. The gravitational field of the Earth is nothing other than the well-known and familiar value of g = 9.8 m/s2 directed vertically downwards.
Although the value of g varies with latitude and altitude, these usually do not affect the objects that are dealt with most of the time. It would be very different if a large body in the vicinity of the Earth is considered, for example an asteroid that came 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 it, since g it 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 how its mass is distributed in relation to its geometry.
The center of mass exists whether or not weight is applied. 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 the weight, as we have been able to appreciate 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 like 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).
Next, it is 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.
object balance
Let’s analyze the stability of a truck driving on the highway. When the center of gravity is above the bed of the truck, the truck will not tip over. The image on the left is the most stable position.
Even when the truck leans to the right, it will still be able to return to the stable equilibrium position, as in the middle drawing, since the vertical still passes through the base. However when this line passes outside the truck will tip over.
The diagram shows the forces at the fulcrum: the normal in yellow, the weight in green, and the static friction to the left in fuchsia. The normal and the friction are applied to the axis of rotation, so they do not exert torque. Therefore they will not contribute to tipping the truck.
The weight remains, which does exert a torque, luckily in an anticlockwise direction and 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 extreme right position, the torque of the weight changes to be clockwise. Unable to be countered for another time, the truck will tip over.
References
Bauer, W. 2011. Physics for Engineering and Science. Volume 1. Mc Graw Hill. 247-253.
Giancoli, D. 2006. Physics: Principles with Applications. 6th.. Ed Prentice Hall. 229-238.
Resnick, R. (1999). Physical. Vol.1.3ra Ed. in Spanish. Continental Publishing Company SA de CV 331-341.
Rex, A. 2011. Fundamentals of Physics. Pearson.146-155.
Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1.340-346.