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S4.2 Systems of particles

[A] What is a system of particles?

A system of particles is any set of particles whose properties we wish to consider collectively.

For example:

We can think of a rigid object as being made up of many particles, with fixed relative positions.
Graphic - No title - system1

Or alternatively,

a collection of gas molecules contained in a box is a system of particles
Graphic - No title - system2
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Note

Note the (sometimes confusing) nomenclature : the word ‘particles’ is often used generically for a collection of atoms, ions, molecules, electrons etc.
 

[B] Centre of Mass

Key Point 4.1

The Centre of Mass of a system of particles with positions $\vec{r}_i$ is defined as

\begin{eqnarray*} %%%Was makeeqnarraystar; label was com \vec{r}_{cm}&=&\frac{1}{M}\left[ m_1 \vec{r}_1 + m_2 \vec{r}_2 + m_3 \vec{r}_3 + ...\,\,m_n \vec{r}_n \right]\\ &=&\frac{1}{M} {\sum_i m_i \vec{r}_i}\\ \end{eqnarray*}

where the total mass of the system is

\[ M = \sum_i m_i \]

We can easily check some special cases:

You can think of the centre of mass as being a (mass) weighted average of the particles positions.

Graphic - No title - comdiag

Commentary

Commentary

Rigid bodies such as a juggling club consist of so many particles, we can best treat them as a continuous distribution of matter. The particles then become differential mass elements dm and the sums become integrals. However in this course you will not be expected to perform such integrals.
 

Worked example

Worked example

The distance between the centers of the hydrogen atom (mass 1.67×10-27kg) and the chlorine atom (mass 2.84×10-26kg) in a molecule of hydrogen chloride is 1.27×10-10m. Locate the center of mass (CoM) of the molecule.

Solution:

Choose the origin to coincide with the chlorine atom and let the x-axis lie along the interatomic vector to the hydrogen atom. Then

\begin{eqnarray*} %%%Was makeeqnarraystar; label was hcl x_{cm}&=&{\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{ \frac{1}{m_H+m_{Cl}}(m_{Cl} \times 0 + m_H\times 1.27 \times 10^{-10})} }\\ &=&7.43\times 10^{-12}m\;\; (3sf)\\ \end{eqnarray*}

\[ y_{cm}={\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{m_H+m_{Cl}}(m_{Cl} \times 0 + m_H\times 0)=0}} \]

Thus the CoM lies on the vector between the two atoms, a distance 7.43×10-12m from the chlorine atom.

 

Commentary

Commentary

Hints for problem solving:

Many objects have a point, line or plane of symmetry. The CoM of the objects then lies on that plane. For example:

  • The CoM of a tennis ball lies at its center,
  • The CoM of a cone lies on the axis of the cone,
  • The CoM of a doughnut lies at its center.

Sometimes we can find the CoM of an object having a complicated shape by decomposing it into a small number of simpler components of symmetrical shape. We can replace each component by a point mass located at its CoM and then find the CoM of the whole object by dealing only with these point masses.

For very irregular shaped objects, the CoM might be difficult to determine analytically. However one can still find it experimentally by taking advantage of the intuitive notion that an object suspended from a point will orient itself so that its center of mass lies vertically below that point. Doing this for two or more points will uniquely locate the CoM.

 

Demo

Demo

Demonstration : locating the center of mass of a (very) irregularly shaped object.
 

Worked example

Worked example

Three thin rods, each of length L are arranged in an inverted “U” as shown below. The two rods on the arms of the U each have mass M. The third rod has mass 3M. Where is the CoM of the system?

 

Graphic - No title - u

Solution: The object comprises 3 rods of length L. By symmetry, the CoM of each rod is at L/2 from one end. We can therefore imagine replacing each rod by a point mass located at its CoM The problem of finding the CoM of the whole system then reduces to finding the CoM of three point particles.

Introducing an x-y coordinate system as shown we have that the CoM is at

\[ x_{cm}={\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{5M}(M\times 0 + 3M\times L/2 + M\times L)= L/2}} \]

\[ y_{cm}={\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{1}{5M}(M\times L/2 + 3M\times L + M\times L/2)= 4L/5}} \]

 
IHanging in the air
IThe Physics of High Jumping
MThe Centre of Mass of n Points
TCentre of Mass of the Earth, Sun and Mars

Learning Resources

Textbook: HRW Chapter 9.1, 9.2. Ignore the material on calculating the centre of mass for continuous objects by integration. This is a straightforward, but tedious, extension of our definition from which you learn little apart from integration.
Course Questions:
Self-Test Questions: