Once upon a time there was a planet called Pluto. Well, this gelid body still orbits the Sun, but is not considered to be a planet anymore. Pluto, a dwarf planet, has a moon almost half its size, called Charon. Usually, moons orbit around their respective planets. For example, our moon orbits the Earth drawing an almost perfect circle around it. You can think of our planet as the center around which this gravitational effect takes place.
Pluto’s case is special, though. Since Charon is so large compared to it, Pluto is no longer the center around which it revolves. Actually, these two bodies form a system which orbits around a new center —yes, Pluto orbits it too! In this case, that point lies outside of Pluto. We call this the center of mass of the Pluto-Charon system or its barycenter. Take a look at the following image. Can you see how Pluto (the bigger mass) also rotates around a point in the middle?
The center of mass is a useful concept to describe how planets and moons move in outer space, but also to understand the specific effect of forces acting on any body. Have you ever wondered why, when you apply a force to an object, it can sometimes move linearly and sometimes rotate, all depending on where you exert the force? If so, then this article is for you!
How to find the center of mass
The center of mass of a single body of homogeneous density coincides with its centroid, which depends on its symmetry. For regular shapes, it is located at the intersection of symmetry planes or axes.
For a system of two or more bodies, follow these steps:
- Determine the masses of the individual objects
- Locate a coordinate system with its origin at the center of one of the bodies
- Determine the distance along each of the axes from the origin to the center of each body.
- Use the following equations to find the location of the center of mass along each axis:
| Cx=m1x1+m2x2+…m1+m2+… | (1) |
| Cy=m1y1+m2y2+…m1+m2+… | (2) |
| Cz=m1z1+m2z2+…m1+m2+… | (3) |
Were m1, m2, … are the masses, and x1, x2,…, y1, y2,… and z1, z2,… the positions of the bodies along each axis of the selected coordinate system.
What is the center of mass
Imagine you want to redecorate your living room and need to move a rolling couch —one with extremely well-lubricated wheels— from its current position. You decide to push it on its right side as shown in the next image. If you try this, and friction with the floor can be ignored, you will probably notice your couch will rotate instead of just displacing. The resulting movement is represented by the dotted arrows. So, why is the couch’s final movement not in a straight line along the direction of the applied force?
Now, imagine you push the couch again but this time right in the center. In this case, there will be no rotation at all, and the couch will move linearly in the same direction of the force, as the following image shows. So what is the difference between this scenario and the first one?
In a lot of physics exercises objects are usually treated as points. We often assume the mass that forms a body to be concentrated in a single point of space, on which all forces act. Now, if the net force applied to that point is not zero, it generates an acceleration in the same direction of said force. This is what Newton’s second law of motion indicates:
| Fnet=ma | (4) |
Where m is the object’s total mass, and a its resultant acceleration. Nevertheless, when entire objects are considered instead of imaginary points, the result is different. This is because the mass is not concentrated but distributed over the body’s volume. Let’s consider this in detail.
Think about a rod with two small masses attached to its ends. If you throw it in the air so that a friend of yours catches it some meters away, it will both translate and rotate. Translation is a linear movement of a body, while rotation is a circular motion around an axis. The next image shows how the rod rotates and, at the same time, translates, drawing a parabola in the air.
Adapted from: Hugh D. Young, Roger A. Freedman, University Physics, vol. 1. (2009)
If we could isolate both types of movement, we could observe the rod rotating about an axis located in its middle. On the other hand, that same axis flies through the air along a parabolic path. This translation can be explained based on equation 4, considering both the initial impulse and the effect of gravity. Nevertheless, Newton’s second law of motion does not say anything about the rotation we observe.
Adapted from: Hugh D. Young, Roger A. Freedman, University Physics, vol. 1. (2009)
To explain the rotation we need to understand torque first. Everytime an object rotates there exists an axis around which this rotation takes place. For example, car tires spin around the point in which they are fixed to the shaft transmitting the movement from the engine. The amount of rotation or angular momentum of a body can be altered by torque, which is the result of a force applied a certain distance away from the rotation axis, similar to a lever.
The increase in the amount of rotation of a body is due to the existence of an angular acceleration. So, an applied torque causes the object’s rotational movement to accelerate. This is equivalent to the amount of linear movement of any body being altered by a force, as described by equation 4. The rotational equivalent of this effect can be written as:
| Tnet=I | (5) |
Where Tnet is the net torque applied to the body, I the body’s moment of inertia, and the resultant angular acceleration. You have probably noticed that equations 4 and 5 look very similar. This is because they represent the same law of motion, but in two different frames of reference: the first one linear, and the second one angular.
If you throw the rod by its middle you can generate translation and no rotation, of course. So, similar to our couch experiment, the effect of applying a force to a rigid body depends on where the force acts. So far, it seems there is a point which displaces and around which rotation takes place. Let’s try and find where it is located exactly.
In our flying rod it is reasonable to think this point sits right in its middle. There, masses are equally separated and, if rotation takes place around it, both will have the same linear velocity. If that was not the rotation axis, the rod would wiggle in the air, which is not the observed behavior.
If we consider the rod connecting both masses to be massless, we can calculate the position from which these are equally located in space. If we set an x-axis parallel to the rod with its origin in the mass to the left, as the following image shows, we can express the distances of both masses as x1=0 and x2=l, where l is the length of the rod.
Then, the average location of the masses that compose this 2-mass system, can be calculated as:
| Cx=m1x1+m2x2m1+m2=m1(0)+m2lm1+m2=m2lm1+m2 | (6) |
This is simply the weighted average of the masses’ locations. If you are not sure about the meaning of this, go ahead and read the section called “What is a weighted average?”. Since both masses are equal, then m1+m2=2m2. So the average location of the masses that compose this system is given by:
| Cx=m2l2m2=l2 | (7) |
As we expected, the average location of masses lies in the middle of the rod. This is easy to understand since both masses are equal. Remember Cx is the only point on the rod that exhibits pure translation when a force is applied directly to it. This is because all masses are distributed equally around the line of the force, so no net torque is generated. If a force acts through a different point on the rod it will cause rotation around Cx, instead of linear movement. Since at this point all masses are equally distributed, we call it the center of mass.
Objects always tend to rotate around their center of mass when a force is applied away from it. On the other hand, any net force applied directly to the center of mass causes a linear movement. This explains why, in our rolling couch experiment, it rotates or displaces depending on where you push it.
Why do objects tend to rotate around their center of mass?
Let’s go back to our flying rod example. If the rotation of a free body would not occur around its center of mass, then this point would not only translate but also rotate around the new rotation axis, as shown in the next image. Now, a rotation of any point implies it is constantly accelerating, since its velocity is always changing direction. Remember acceleration is the change in time of either the magnitude or the direction of a body’s velocity vector.
Now, for an acceleration to exist, there has to be a force causing it. This force would have to be constantly applied to the center of mass to change its direction of movement. Nevertheless, this is not possible in our flying rod example, since no forces other than gravity are acting on it once it leaves your hand. You might ask yourself, can gravity cause the change in direction of the center of mass? The answer is no, since it is a force with a constant direction: towards the center of The Earth.
Keep in mind the rotation of the rod around its center of mass, once it has left your hand, does not require a net external force to redirect the masses at its ends. As long as friction with air can be ignored, the rod will continue rotating non stop. This is because both masses accelerate in opposite directions, so the net acceleration of the system is null and the necessary external net force is therefore zero, as shown in the following image. Furthermore, since no torque needs to be applied, no angular acceleration is caused, as described by equation 5, thus keeping the angular velocity constant. This is not the case if the rod would rotate around an axis different from its center of mass, since torques on each side would be different, given the distance from the rotation axis to each mass is no longer the same.
In summary, the rotation of a free body around an axis different from its center of mass is simply not possible, because it would require the existence of an additional external force that causes an acceleration of the center of mass.
How to calculate the position of the center of mass
The center of mass is defined as the point of space in which the weighted distances to all of the masses that form a body or a system of bodies adds to zero. This point of space, C(x,y,z), can be determined for any collection of masses by calculating its different components along the x-, y- and z-axes.
For a single body of homogeneous density, the center of mass lies on the same point as the centroid. This is the geometrical center of the body. For simple regular shapes, like spheres, squares, rings, cylinders, etc., the centroid is located where two or more axes or planes of symmetry cross each other. The following image shows some examples of the centroids of some regular shapes, with some of their symmetry axes shown as dotted lines:
Notice the centroid, and therefore the center of mass, of a ring would lie in its middle, where there is no material. This is because the center of mass is a concept and not necessarily a physical point inside a body. The effect of applying a force to the edge of the ring will be a rotational movement around its center of mass. If you want to displace the ring, you will have to exert a force that passes through the center of mass.
The concept of center of mass also applies to groups of bodies, like planets and moons. Though not physically connected, the force of gravity acts as the rod in our previous example: it connects one body to all others surrounding it. So, how do you determine the center of mass of a system composed of a planet and its moon?
Let’s consider our initial example: Pluto and Charon. The following image summarizes the situation. Pluto has an approximate mass of 131021kg, and Charon one of 1,591021kg. Both bodies are separated by a distance of 19570 km.
In order to calculate the average location of both masses, we need to establish a coordinate system. We conveniently place the origin in the center of Pluto so that its location is xP=0. Then, for Charon, xC=19570 km. Now, since we know the value of both masses, we can calculate the weighted average of the their positions:
| Cx=m1x1+m2x2m1+m2=131021kg (0)+1,591021kg (19570 km)131021kg+1,591021=2132,7 km | (8) |
Since Pluto’s radius is 1188,3 km, the center of mass of the system that it forms with its moon, Charon, lies outside of it, almost 1000 km away from its surface. Both Pluto and Charon orbit around this point thanks to the effect of the force of gravity they exert on one another.
The same procedure can be used to determine the center of mass of any 2- or 3-dimensional system. If needed, the weighted averages of mass locations along the y- and z-axes can be found through the use of equations 2 and 3, respectively. For this, the distances from the origin of the coordinate system to each of the masses along each of the axes (x, y, and z) need to be determined separately.
Example 1: calculate the coordinates (x, y) of the center of mass of the following 3-body system using the proposed coordinate system. Masses and distances are given.
Note: do not be confused about the 3 kg mass being smaller than the 1 kg mass. Remember all bodies can have different densities, and that the center of mass depends on the bodies’s masses, not their volumes!
Answer: Cx=1,11 m and Cy=1,36 m.
What is a weighted average
Imagine you have already passed all of your physics examinations and want to calculate your final grade. Let’s say not all exams have the same importance. For example, imagine you took two minor tests, sent a laboratory report and presented the final exam. Probably, the last one will have a greater influence on your final grade.
To assign importance to an examination, schools often assign a percentage of the final grade to each of them. The higher this percentage is, the more weight the test will have on your overall performance. In our example, it could be something like this:
Table1: Example of physics grade report
| Examination | Percentage (weight) | Your result(0 to 10 scale) |
| Test 1 | 15% | 7 |
| Test 2 | 15% | 10 |
| Laboratory report | 30% | 6 |
| Final exam | 40% | 9 |
As you can see, the greater the percentage assigned to a test, the more important it is, and therefore the greater the impact it should have on your final grade. So, how do you calculate it? If you calculate a simple average, you are not reflecting the importance of each or your grades in the final result, since you will only use the actual grades and the total number of them, like this:
| Average=7+10+6+94=324=8 | (9) |
One way to consider the importance of each value involved in an average is to calculate a weighted average. To do so, you multiply each value with its weight, then add the results together and divide them by the sum of all the weights. In our grading example, we would get:
| Weighted average=(70,15)+(100,15)+(60,3)+(90,4)0,15+0,15+0,3+0,4=7,951=7,95 | (10) |
Congratulations! You have passed your physics course.