# Dynamics of a particle system: examples, exercises

The **dynamics of a particle system** consists of the application of Newton’s laws of motion to a set of particles, which can be discrete (the particles can be counted) or form part of an extended object, in this case the system it is continuous.

To explain the motion of a system of particles, it is inconvenient to analyze each one separately and see what forces act on it.Â Instead, a representative point of the set is defined, called theÂ *center of massÂ ***.**

Describing the motion of the center of mass offers a very accurate overview of the overall motion of the whole, it also allows applying Newton’s laws in a way analogous to when the object is considered as a dimensionless particle.

This last model, calledÂ *the particle model*Â , is good for describing translations and also when it is not necessary to consider the dimensions of the object.Â But ordinary objects have size and if they also have rotational motion, it is necessary to take into account the points on which the forces are applied.

**Examples**

**The Earth and the Moon**

A set of discrete particles mÂ _{1}Â , mÂ _{2}Â , mÂ _{3}Â … that eventually moves with respect to the origin of a coordinate system due to some resultant force acting on them is a good example of a particle system.

The Earth can be considered as one particle and the Moon another, so both constitute a system of 2 particles under the action of theÂ Sun’s force of gravity.

**Extended objects**

A person, an animal or any object in the environment can also be considered as a system of particles, only that these are so small that they cannot be counted one by one.Â This is a continuous system, but taking into account certain considerations, its treatment is the same as for a discrete system.

**The center of mass of a system of particles**

To begin the study of a particle system, we must find the center of mass (CM), which is the point where all the mass of the system is concentrated.

For the discrete system of Figure 1, withÂ *n*Â particles, each has a position vector directed from the origin O of the coordinate system to the point P (x, y, z) where the particle is.Â These vectors are denoted asÂ **rÂ **_{1}Â ,Â **rÂ **_{2}Â ,Â **rÂ **_{3}Â â€¦Â **rÂ **_{n}.

The coordinates of the CM are calculated using the following equations:

Where each of the masses of the set is represented as mÂ _{1}Â , mÂ _{2}Â , mÂ _{3}Â … mÂ _{n}Â .Â Note that the summation âˆ‘ mÂ _{i is}Â equal to the total mass M of the set.Â If the system is continuous, the summations are replaced with integrals.

Each of the perpendicular directions is represented by the unit vectorsÂ **i**Â ,Â **j**Â andÂ **k**Â , therefore, the position vector of the CM, denotedÂ **rÂ **_{CM}Â , can be expressed by:

**rÂ **_{CM}Â = xÂ _{CMÂ }**i**Â + yÂ _{CMÂ }**j**Â + zÂ _{CMÂ }**k**

**CM movement**

Once the location of the center of mass is known, the known equations of motion apply.Â The velocity of the CM is the first derivative of the position with respect to time:

In this case, the system has a total momentumÂ **P**Â which is calculated as the product of the total mass of the system and the velocity of the center of mass:

**P**Â = M âˆ™Â **vÂ **_{CM}

Alternatively, the total momentum of the system can be calculated directly:

**P**Â = mÂ _{1Â }**vÂ **_{1}Â + mÂ _{2Â }**vÂ **_{2}Â + mÂ _{3Â }**vÂ **_{3}Â +â€¦.Â = âˆ‘ mÂ _{iÂ }**vÂ **_{i}

While the acceleration of the CM is the derivative of the velocity:

**Force on CM**

The forces acting on a system of particles can be:

- Internal forces, due to interactions between the same particles.
- External forces, caused by agents external to the system.

As the internal forces are presented in pairs, of equal magnitude and direction, but opposite directions, according to Newton’s third law, it is true that:

âˆ‘Â **FÂ **_{int}Â =Â **0**

Therefore, the internal forces do not alter the movement of the whole, but they are very important to determine the internal energy.

If the system is isolated and there are no external forces, according to Newton’s first law the center of mass is at rest or moves with uniform rectilinear motion.Â Otherwise, the center of mass experiences an acceleration given by:

âˆ‘Â **FÂ **_{ext}Â = M âˆ™Â **aÂ **_{CM}

Where M is the total mass of the system.Â The above equation can be written like this:

And it means that the external force is equivalent to the temporal variation in the momentum, another way of expressing Newton’s second law and the same one that the famous English physicist used in his bookÂ *Principia*Â .

**Exercise resolved**

The center of mass of a 2-particle system is on the x axis at a certain moment, at the position x = 2.0 m and moving with velocity 5.0 m / s in the same direction and in a positive direction.Â If one of the particles is at the origin and the other, of mass 0.1 kg, is at rest at x = 8.0 m, calculate:

a) The mass of the particle that is at the origin.

b) Amount of movement of the system

c) What is the speed of the particle at the origin?

**Solution to**

From the equation for the position of the center of mass:

**rÂ **_{CM}Â = xÂ _{CMÂ }**i**Â + yÂ _{CMÂ }**j**Â + zÂ _{CMÂ }**k =**Â 2.0 mÂ **i**

As the CM has only x-coordinate, only the first equation of the previously given trio is used:

Now the coordinates are substituted, if the particle at the origin is denoted as number 1 and the other as number 2, the numerical data are:

xÂ _{1}Â = 0 m, xÂ _{2}Â = 8.0 m, mÂ _{2}Â = 0.1 kg, xÂ _{CM}Â = 2.0 m

Remaining:

**Solution b**

The amount of movement of the system is calculated by:

**P**Â = M âˆ™Â **vÂ **_{CM}

The total mass M is equal to:

M = 0.3 kg + 0.1 kg = 0.4 kg

Therefore:

**P**Â = 0.4 kg âˆ™ 5.0 m / sÂ **i**Â = 2 kg.m / sÂ **i**

**Solution c**

From the equation forÂ **P**Â of a system of two particles,Â **vÂ **_{1}Â is clearedÂ , since the other data are known, since theÂ statementÂ says that the particle 2 is at rest, therefore:

**vÂ **_{2}Â =Â **0Â **

AndÂ **P**Â is simply as:

**P**Â = mÂ _{1Â }**vÂ **_{1}

**vÂ **_{1}Â =Â **P**Â / mÂ _{1}Â = 2 kg.m / sÂ **i**Â / 0.3kgÂ **=**Â 6.67 m / sÂ **i**

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