Lesson 18
System of Particles
I.System of Particles
A.Background
If we have a collection of particles, each of the particles will have different position vectors and may have different masses, be traveling with different velocities and different accelerations. This leads to the following question:
If we consider all the particles as a single system, which particle or location is the one described by Newton’s 2nd Law?
The motion of a system of particles or extended object is equivalent to that of a single particle whose mass is the total mass of the system and is located at a point in space called the center of mass.
B.Definition of the Center of Mass
The center-of-mass is by definition the average location of the system:
C.Important Facts
1.The center-of-mass of a system will often reside at a location in which no particle exists!!
Example: Where is the center-of-mass (average location) of the following donut?
2.Analysis of compound objects or systems is often easier using the concept of the center of mass.
Example: The center-of-mass of a player trying to dunk a basketball must always stay on the parabolic path for projectile motion that we worked out in chapter 3.
The center-of-mass of the player-ball system must stay on the parabolic path determined by the player’s initial velocity. When the player kicks their feet downward, the player’s body and the ball must move upward to keep the center-of-mass on the same path.
PROBLEM 1: Determine the center-of-mass for the following set of point particles.
D.Continuous Bodies
Although all bodies are actually composed of small discrete particles (atoms and molecules), it is often easier to make the calculations by treating the body as a continuous mass distribution. In this case, the summations can be replaced by integrals.
In order to evaluate these integrals, you must first be able to write the mass differential dm. This is done using the concept of density.
a)Linear Mass Density -
Linear mass density for a differential length dx and mass dm of an object is defined by the equation
Rearranging this equation gives us the mass differential in terms of a linear object’s density:
You will either have to be given the functional dependence of the mass density or the object will have to be an object of uniform density where you can find the density by taking the entire length of the object:
PROBLEM 2: Prove that the center of mass of a uniform density steel thin rod is at the center of the rod.
PROBLEM 3: Find the center of mass of the thin concrete strip shown below given that its linear density is.
b)Surface (Areal) Mass Density -
Surface mass density for a differential area dA and mass dm of an object is defined by the equation
Rearranging this equation gives us the mass differential in terms of an object’s areal mass density:
This converts our mass integral into a surface integral.
c)Volume Mass Density -
Volume mass density for a differential volume dV and mass dm of an object is defined by the equation
Rearranging this equation gives us the mass differential in terms of an object’s volume density:
This converts our mass integral into a volume integral.
Since skill in evaluating multiple integrals is not required for this course, only the simplest types of area or volume density problems are done in your textbook. However, it is important for most students to practice the formal steps in setting up even these simpler problems since they will see more complicated problems in other classes which require math skills involving either vector integration or the evaluation of multiple scalar integrals. These harder problems include working with non-uniform charge distributions in PHYS2424 and non-uniform mass distributions in the Sophomore Engineering Principles I class. It is common for engineers to mix materials (i.e. add steel rebar to concrete) to produce non-uniform materials with improved properties or to change the physical dimensions of a system like thickness in order to handle non-uniform loading of forces (i.e. water pushes hardest at the bottom of the dam).
E.Tricks of the Trade
1.You can often break a problem a complicated non-uniform object into several uniform shapes whose individual center-of-masses are either known or can be easily calculated.
2.You can treat any holes drilled into an object’s by first calculating a solid object and then accounting for the holes by treating them as extra particles with “negative” mass.
PROBLEM 4: Find the center-of-mass of the following plate of mass M and uniform density.
3.The center of mass of many common homogenous objects has already been calculated. The results of these calculations can be found in tables in engineering and physics text books.
4.You can also find the center of mass of irregular objects experimentally. For everyday objects on the Earth, the pull of gravity is approximately constant so the center of gravity (the effective point where the Earth pulls on an extended object) is the same point in space as the center of mass. By holding an object at different points and allowing it to rotate under the influence of gravity, it is possible to find the center of gravity.
This is also the point at which you could place a support to balance the object!! You may have seen acrobats who were balancing in strange formations or small toy birds balancing on their beaks. Now, you should be able to understand how toy works.
II.Newton II and Center-of-Mass
A.Velocity of the Center-of-Mass
Using the definition of velocity, we have that the velocity of the center-of-mass is
If we now insert the definition of the center-of-mass into our equation above, we have
Thus, we see that the velocity of the center-of-mass is just the average velocity of the system of particles.
B.Deriving Newton’s 2nd Law for A System of Particles
We will start by finding the total linear momentum of the system of particles.
The time-rate-of-change of the total linear momentum of the system is then given by
Newton’s 2nd Law for a particle says that the time-rate-of-change of momentum for a particular particle is equal to the net external force applied to that particle so we can apply this to each particle in the system
We can now break the forces applied to each particle into two parts: (those due to agents external to the system and those due to interactions with other particles inthe system).
Newton’s 3rd Law tells us that so the second summation of internal forces is zero. Thus, we have our final result that
This is Newton’s 2nd Law for a system of particles with the total linear momentum of the system only changed by the forces outside the system even though the particles are allowed to move with respect to each other like in spilled water or an exploding bomb!!
We now consider some examples where the center-of-mass concept helps us to analyze complicated problems.
EXAMPLE 1: A uniform chain of length L and mass M is hanging from a table of height L as shown below. If a student pulls the chain up onto the table, how much work does the student do in increasing the chain’s gravitational potential energy?
A.Work Method Using Calculus
BUsing Center of Mass to Simplify Calculations
EXAMPLE 2 : At t = 0 a ball of mass M is traveling at speed V in the +x direction inside a box of mass M and length 2L which is at rest as shown below. Graph as a function of time a) the position of the ball; b) the position of the box; and c) the position of the center of mass.
B.Energy for Systems of Particles
It is important in dealing with systems of particles to be careful and work from first principles and not just memorize. For instance, many students incorrectly assume that the kinetic energy of a system of particles is just the kinetic energy of the center of mass.
An object whose center-of-mass is at rest may still have particles who have speed, but whose net velocity is zero. For instance, consider the two 1 kg balls moving below:
The center-of-mass velocity is zero so the problem’s center-of-mass motion is equivalent to the following:
Thus, the kinetic energy of the center of mass is zero while the total kinetic energy of the system is 4 J!!
The actual kinetic energy of a system can be related to the kinetic energy of the center of mass by using our knowledge of reference frames. The velocity of the ith particle, , can be written in terms of the velocity of the center of mass, , and the velocity of the ith particle with respect to the center-of-mass reference frame, , as
We can then find the total kinetic energy as
Applying the dot product and remembering that the order of the vectors doesn’t matter for dot products, we have after some algebra
The last sum is a sum of linear momentum as measured in the center-of-mass reference frame and is always zero by the definition of the center-of-mass!!
Thus, the total energy is reduced to sum of two kinetic energy terms which can be interpreted as the kinetic energy of the center of mass and the kinetic energy with respect to the center of mass. This second term is usually called internal kinetic energy.
In our previous, example the system had 0 J of center-of-mass kinetic energy and 4 J of Internal Kinetic Energy.
A similar discussion can be performed for potential energy leading to
The center-of-mass potential energy is what we have been dealing with throughout this semester. We have neglected the energy which can be stored inside an object due to chemical, nuclear, and other interactions between the atoms and molecules which make up a real block, pulley, or other macroscopic object. The later chapters in this book cover a more complete discussion of energy leading to the fields of Statistical Mechanics and Thermal Physics (also called Thermodynamics). Due to time constraints and our belief in a less is more approach to physics education, we don’t cover 20+ chapters in PHYS1224 like some universities so. Thus, we will limit our discussion in this subject to the time available. You will be required to take advanced courses in these subjects if you are an engineer or physicists. The important thing to remember is that a strong understanding of classical mechanics is necessary for success in a wide range of related subjects like thermodynamics and not just for solving blocks moving on planes.
In thermodynamics, we usually separate quantities based upon macroscopic motion (center-of-mass) and microscopic motion (with respect to the center of mass). Here are several important terms of thermodynamics and their origins in mechanics:
1)Microscopic kinetic energy is called heat, Q.
2)Temperature, T, is a measure of the average kinetic energy of an atom or molecule. The correct temperature scale in physics problems is the Kelvin scale as you can’t have negative temperatures as they imply imaginary speeds!!
3)The total energy of a system can be written as
The change in the center-of-mass energy can be accounted by work done on the system by external forces, W. Thus, we can write the
energy equation as
This equation basically says that the total energy of a system can be changed by doing work on the system, adding heat energy or increasing the microscopic internal energy of the system.