Point Particle Systems

This page describes point particle systems and how they can be used to model certain aspects of a system's motion.

The Main Idea

A point particle system is a physical system, usually composed of multiple parts, modeled as though it were a single particle at its center of mass.

When work is done on a system, the energy imparted on it may take on multiple forms. These include Translational, Rotational and Vibrational Energy, Potential Energy, and Thermal Energy. Translational kinetic energy is kinetic energy due to the movement of the system's center of mass. All other types of energy the system has are considered "internal" types of energy because they do not affect the system's motion through its environment, but rather indicate its local properties. Sometimes, when work is done on a system, it imparts both translational kinetic energy and internal types of energy. For example, consider a ball rolling down a ramp. Gravity does work on the ball and increases its energy. It gains some translational kinetic energy (its center of mass gains a velocity down the ramp) and some rotational kinetic energy (its rolling motion causes it to rotate about its center of mass).

The purpose of modeling a system as a point particle system is to isolate the work done on it that raises its translational kinetic energy, which may only be a fraction of the total work done on the system. This is because point particles cannot have any internal types of energy. Modeling a system as a point particle system can make it much easier to calculate its motion through its environment, but offers no insights about the internal state of the system.

Point particle systems are in contrast to Real Systems (also known as extended systems), which analyze each part of a system individually instead of reducing it to a single point. Real systems can model a system's internal behavior in addition to its motion through its environment, although they can be very complicated and difficult to analyze quantitatively. Often, to obtain a complete model of a system's behavior, both point-particle and extended models are used.

A Mathematical Model

For a point particle system, the work-energy theorem is still true (see Work/Energy).

The work done on a point particle system is defined as

[math]\displaystyle{ W = \int \vec{F_{net}} \cdot d \vec{r} }[/math].

Let us assume that the net force acting on the particle is constant, so that we can get rid of the integral, and that the net force acts in the direction of the particle's motion, so that we can replace the dot product with regular multiplication. This will be the case for most point particle system problems. With these assumptions, the work done on a point particle is given by

[math]\displaystyle{ W = F_{net} d }[/math]

where [math]\displaystyle{ F_{net} }[/math] is the magnitude of the net force acting on the particle and [math]\displaystyle{ d }[/math] is the distance traveled by the particle.

The work-energy theorem states that work done on a system increases that system's energy. A point particle cannot experience an increase in internal energy. For example, the moment of inertia of a point particle about its center of mass is 0, so it can have no rotational kinetic energy. This means that all of the work done on a point particle system becomes translational kinetic energy.

In other words, for a point particle system,

[math]\displaystyle{ \Delta KE = F_{net} d }[/math]

and

[math]\displaystyle{ KE = \frac{1}{2} m v^2 }[/math].

The two equations above are the basis for answering questions using point particle systems.

Examples

1. (Simple)

A 60kg person jumps straight up in the air from a crouching position. From the time the person begins to push off of the ground to the time their feet leave the ground, their center of mass moves up 2m, and the normal force between the ground and the person's feet has a constant magnitude of twice the person's weight. Find the velocity of the jumper at the moment their feet leave the ground. Use 10m/s² for g.

Solution:

Let us model the person as a point particle system. A force diagram for the person would look like this:

The net force acting on the person is the vector sum of the upward normal force and the downward gravitational force, which is a 600N force upwards. This force is constant and in the same direction as the person's motion, so multiplying it by the 2m displacement of the center of mass yields the work done on the person by their muscles:

[math]\displaystyle{ W = 600 * 2 = 1200 }[/math]Nm.

The person begins at rest, so their initial kinetic energy is 0J and their final energy is 1200J.

[math]\displaystyle{ \frac{1}{2} m v^2 = KE }[/math]

[math]\displaystyle{ \frac{1}{2} * 60 * v^2 = 1200 }[/math]

[math]\displaystyle{ v = \sqrt{40} }[/math]m/s.

In this problem, the system that was reduced to a single particle consisted of only one body that has no changes in internal energy, so treating the system like a point particle is not particularly innovative. The middling and difficult problems use point particle systems in a more useful way that might be less familiar to you.

2. (Middling)

A giant 20kg yo-yo floats at rest in space. Its string (whose mass is negligible compared to the mass of the yoyo) is pulled with a constant force of 8N. What is the speed of the yoyo when it has travelled 5m?

(picture)

Solution:

Solving this problem using energy while treating the system as a real system would be difficult; it would require knowledge of rotational physics and would require many steps. Let us therefore analyze it as a point particle system. This allows us to find only the work done on the yo-yo that is converted to translational kinetic energy.

Treating the yo-you as a point particle, the work done on it is given by the magnitude of the force times the particle's displacement:

[math]\displaystyle{ W = 8 * 5 = 40 }[/math]Nm

The point particle's kinetic energy was originally 0, so it is now 40J.

[math]\displaystyle{ KE = \frac{1}{2} m v^2 }[/math]

[math]\displaystyle{ 40 = \frac{1}{2} 20 v^2 }[/math]

[math]\displaystyle{ v = 2 }[/math]m/s.

The point particle represents the center of mass of the yo-yo, so it must be moving at 2m/s.

It may not be immediately obvious, but if we modeled the yo-yo as a real system, we would have found a greater value for work, because instead of multiplying the magnitude of the force by the yo-yo's displacement, we would have had to find the distance over which the force was exerted, taking into account that as the yo-yo turns, it dispenses more string. The distance over which the force was exerted would have been the displacement of the yo-yo plus the length of string that unraveled because that is the distance the end of the string would have traveled under the influence of the force. This higher value for work makes sense, because when the yo-yo is modeled as a real system, it has a final rotational kinetic energy in addition to its final translational kinetic energy. This problem is an example of how the point particle model can simplify problems if we are only interested in its translational behavior.

3. (Middling)

A 50kg mass is suspended in the inside of a large cubic box of negligible mass by six rubber bands- one attaching the mass to each face of the box. The box and the mass are at rest. A 200N rightward force is applied to the side of the box, causing it to accelerate. At time [math]\displaystyle{ t_1 }[/math], the box has been displaced by 10m to the right. At this time, the mass is no longer exactly in the center of the box; it is 1m to the left of the center of the box due to its inertia. What is the speed of the mass at time [math]\displaystyle{ t_1 }[/math]?

(picture)

Solution:

4. (Difficult)

A pair of 10kg masses are connected by a spring of unknown spring constant. They begin at rest at the spring's equilibrium length. A 30N rightward force is applied to the mass on the right for 4s. At the end of the 4 second period, the force ceases to act and the mass on the right has traveled 20m. The system continues to travel to the right, and the distance between the two masses oscillates sinusoidally. How much oscillatory kinetic energy does the system have? (Oscillatory kinetic energy is the energy of the mass' oscillatory movement, as opposed to their translational kinetic energy, which is the energy of the rightward movement of their center of mass.)

(picture)

Solution:

Let us first analyze this system as a real system. The work done by the rightward force on the real system is given by its magnitude times the distance over which it was exerted:

[math]\displaystyle{ W = 30 * 20 = 600 }[/math]Nm.

Now, let us treat the system as a point particle system to find the particle's kinetic energy (the real system's translational kinetic energy). We are now modeling the system of two masses as a single point at its center of mass. That is, we are now treating the system as a single mass of 20kg. A 30N force is applied to the particle for 4 seconds, so its acceleration is given by

[math]\displaystyle{ \vec{a} = \frac{\vec{f}}{m} = \frac{3}{2} }[/math]m/s.

and its displacement at the end of the 4 second period is given by

[math]\displaystyle{ d = \frac{1}{2} a t^2 = \frac{1}{2} * \frac{3/2} * 4^2 = 12 }[/math]m.

The work done by the rightward force on the point particle system is given by its magnitude times the displacement of that particle:

[math]\displaystyle{ W = 30 * 12 = 360 }[/math]Nm.

Recall that the work done on the point particle system represents the increase in translational kinetic energy of the real system, so the system now has 360J of translational kinetic energy.

It was found earlier that the force imparted a total of 600J of energy to the real system, so [math]\displaystyle{ 600-360=240 }[/math] of the system's joules are oscillatory energy.

Connectedness

1. How is this topic connected to something that you are interested in?

I find the entire realm of physics fascinating, and I find it interesting how physicists are constantly coming up with new ways to solve problems and use formulas. The point particle system is a perfect example of that. It can turn a complicated force problem into something easy to approve. I also find it interesting how you can find other forms of energy, such as chemical energy, by using point particle and real systems. Something so small and seemingly unattainable can be found using this method.

1. How is it connected to your major?

I am majoring in Materials Science and Engineering, and there are many ways all different types of physics can be used in MSE. The engineering of materials, specifically being able to calculate the amount and types of energy (and thus the cost) to produce something is absolutely crucial. Point particle systems can make this easier to do, while also adding precision to the calculations.

1. An interesting industrial application

As seen in the above examples, there are many real life applications to point particle systems, such as the energy in a person falling, or in a yoyo. This system can also be applied to industry and manufacturing, with the use of various machines that may require gears, levers, or other objects that rotate. Using point particle and real systems, you can calculate the amount of internal energy happening in a moving machine, and therefore how much energy is lost.

See also

Further reading

For more help, a helpful page is: http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:pp_vs_real

A helpful video lecture: https://www.youtube.com/watch?v=T780lL5FlLg&index=41&list=PL9HgJKLOnKxedh-yIp7FDzUTwZeTeoR-Y

External links

See also Real Systems for further information on using Point Particle Systems to solve for the Real Systems.

References

Chabay, Ruth W., and Bruce A. Sherwood. "9." Matter & Interactions. N.p.: n.p., n.d. N. pag. Print.

Purdue Physics. https://www.physics.purdue.edu/webapps/index.php/course_document/index/phys172/1160/42/5399.

Yo-yo Clipart: https://www.clipartbest.com