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 easier to calculate its translational motion through its environment. When the work done on a system is calculated while modeling it as a point particle, only the forces that affect the system's translational kinetic energy can be found. Newton's Second Law: the Momentum Principle can also be used on a point particle to calculate its translational acceleration. Furthermore, the work done by any forces acting on a point particle system can strictly only affect the system's translational kinetic energy. Often, this work done is only a fraction of the work done on the real system, which can affect multiple types of energy at once. Modeling a system as a point particle offers no insights about the internal state of the system because only forces and work affecting its translational motion can be found.

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. It is important to remember that when modeling a system as a point particle system, the calculating the work done on the system as described above does not find all of the work done on the system, but rather just the work that becomes translational kinetic energy.

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.

Note: 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?

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.

Note: 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 metal sphere is suspended in the inside of a large cubic box of negligible mass by six rubber bands- one attaching the sphere to each face of the box. The box and the sphere are initially 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 sphere 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 sphere at time [math]\displaystyle{ t_1 }[/math]?

Solution:

Let us model the sphere-box system as a point particle system. The center of mass of the system is simply the center of the sphere because the box has negligible mass. The work done on the particle is given by the force acting on the particle times its displacement:

[math]\displaystyle{ W = 200 * 9 = 1800 }[/math]Nm. The displacement of the particle is only 9m because although the box was displaced 10m, the sphere was only displaced 9m due to its inertia and the non-rigidity of its suspension within the box.

The particle initially had no energy, so its energy at time [math]\displaystyle{ t_1 }[/math] is 1800J.

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

[math]\displaystyle{ 1800 = \frac{1}{2} * 50 * v^2 }[/math]

[math]\displaystyle{ v = \sqrt{72} }[/math]m/s.

Note: If this system were modeled as a real system, we would have found that the force exerted 2000Nm of work on the sphere-box system. This is because when we model it as a real system, we have to consider that the force was acting on the box, not the sphere, and was therefore exerted over all 10m. When using the point particle system, we find only the work that gets turned into translational kinetic energy; the remaining 200Nm of work got turned into potential energy. More specifically, it became the potential energy of the stretched rubber bands since the sphere is no longer in the center of the box, which is its lowest-energy position.

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.)

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

Scenario: Acrobatics

Consider an acrobat who has jumped from a trampoline and is now on a trajectory through the air. The acrobat is performing a flip and is therefore turning, and they can tuck in or extend their arms in order to affect their moment of inertia and thereby their rotational rate and rotational kinetic energy. Modeling the trajectory of the acrobat may seem complicated, but it can be easily found by modeling the acrobat as a point particle system. When the acrobat is reduced to a point particle, the only force acting on them (ignoring air resistance) is gravity. This means gravity is the only force that can affect the acrobat's translational motion and therefore trajectory, so the acrobat obeys regular Projectile Motion. Any forces that are present in the real system but not in the point particle system, such as the acrobat's muscles, can only affect the internal energies of the acrobat system, so any movement the acrobat does in the air does not affect their trajectory. This is something acrobats know well; to land in a certain place, they must correctly push off when jumping, and then they are free to do whatever tricks they want in the air.

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