7.1 Locate the centre of mass of a particle system | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Systems of Particles

A mass-weighted position in one or two dimensions

The centre of mass is a single position that summarizes how mass is distributed among the particles in a system. It is found by averaging particle positions, but weighting each position according to its particle’s mass. A heavier particle therefore pulls the centre of mass closer to its own position. In this lesson, the system is the specified collection of particles, and an observer uses one fixed reference frame to describe their positions. We focus on locating the centre of mass, not predicting motion.

What you will learn

Define a particle system and choose a reference frame, origin, and positive coordinate directions.
Calculate the centre of mass from particle masses and positions.
Use signed coordinates correctly in one- and two-dimensional systems.
Check a result using units, particle locations, and limiting cases.

1. Define the system and position coordinates

A particle is an idealized body whose size and shape are ignored for the calculation; its mass is treated as concentrated at a point. A particle system is the specified collection of particles being considered. The observer is the person or measuring setup describing their positions. The reference frame is the coordinate system used by that observer.

Choose an origin and positive directions before recording positions. For a planar system, one common choice is positive

x

to the right and positive

y

upward. A coordinate is signed: a particle left of the origin has a negative

x

coordinate, and a particle below it has a negative

y

coordinate. Use the same frame and origin for every particle.

A position vector describes a location relative to the origin; it is not the distance travelled along a path. The centre of mass is generally not the geometric midpoint of the particles. It is a mass-weighted average, so particles of unequal mass do not contribute equally.

\mathbf r_G=\frac{\sum_i m_i\mathbf r_i}{\sum_i m_i}

Include only the particles specified as the system.
Use one consistent origin and set of axes.
Record coordinates with their signs and positions in metres.

2. Calculate each coordinate component

In the vector relation,

m_i

is the mass of particle

i

,

\mathbf r_i

is its position vector, and

\mathbf r_G

is the centre-of-mass position vector. The denominator is the total mass of the system. Since masses are positive, each particle’s position is weighted by a positive share of that total.

For a line of particles, calculate one coordinate. For a planar system, calculate the horizontal and vertical coordinates separately. In either case, multiply each particle’s coordinate by its mass, add those products, and divide by the total mass. This component calculation is just the vector equation written in coordinates.

Signed coordinates matter. A particle on the negative side of an axis contributes a negative mass-coordinate product for that component. Replacing a signed coordinate with a positive distance can put the answer on the wrong side of the origin. Do not divide by the number of particles unless all masses are equal and you have shown that this is equivalent to dividing by total mass.

x_G=\frac{\sum_i m_i x_i}{\sum_i m_i},\qquad y_G=\frac{\sum_i m_i y_i}{\sum_i m_i}

Use total mass, not particle count, as the denominator.
Calculate each coordinate component from its own weighted sum.
Retain negative signs for positions in negative coordinate directions.

3. Interpret and verify the result

Before calculating, list each particle’s mass and coordinates, then state which centre-of-mass coordinate or coordinates are unknown. Write the weighted-average relation before substituting values. This makes it easier to check that every particle is included and that all components use the same total mass.

Check dimensions: each product of mass and position has units of kilogram-metres; dividing by kilograms leaves metres. Check the sign and location against the particle layout. For positive masses, a coordinate of the centre of mass must lie between the smallest and largest particle coordinates for that axis.

Two limiting cases are useful. For equal masses, the centre-of-mass coordinates are the ordinary averages of the coordinates. If one particle’s mass is much larger than all the others, the centre of mass should be close to that particle’s position. A change of origin changes the numerical coordinates, but not the physical point being described.

[m_i x_i]=\mathrm{kg\,m},\qquad [x_G]=\mathrm{m}

Check units, signs, and whether the result is plausible from the layout.
For positive masses, each component lies between the extreme particle coordinates.
Use equal-mass and dominant-mass cases as reasonableness checks.

Worked example

Two particles on a line

A

2.0\,\mathrm{kg}

particle is at

x=0.50\,\mathrm{m}

, and a

6.0\,\mathrm{kg}

particle is at

x=2.50\,\mathrm{m}

. Find the centre-of-mass position. The system consists of these two particles. The observer uses a fixed frame with origin at

x=0

and positive

x

to the right.

Particle positions on the x-axis

Position vectors are measured from the stated origin.

Identify the knowns and unknown
The system contains two particles. Their masses and signed $x$ coordinates are known; the unknown is $x_G$ . Both coordinates are positive because both particles are to the right of the origin.
Apply the weighted-position relation
Multiply each coordinate by its particle’s mass, add the products, and divide by the total mass. The heavier particle has the greater influence on the result.
$x_G=\frac{(2.0\,\mathrm{kg})(0.50\,\mathrm{m})+(6.0\,\mathrm{kg})(2.50\,\mathrm{m})}{2.0\,\mathrm{kg}+6.0\,\mathrm{kg}}$
Calculate and check
The numerator is $16.0\,\mathrm{kg\,m}$ and the total mass is $8.0\,\mathrm{kg}$ . The answer has units of metres and should lie closer to the heavier particle.
$x_G=2.00\,\mathrm{m}$

Answer: The centre of mass is at

x_G=2.00\,\mathrm{m}

, measured from the origin in the positive

x

direction.

Check: The coordinate lies between

0.50\,\mathrm{m}

and

2.50\,\mathrm{m}

and is closer to the

6.0\,\mathrm{kg}

particle.

Worked example

Particles on opposite sides of the origin

A

3.0\,\mathrm{kg}

particle is at

x=-1.0\,\mathrm{m}

, and a

1.0\,\mathrm{kg}

particle is at

x=3.0\,\mathrm{m}

. Find the centre-of-mass coordinate. The system contains only these particles; the observer uses a fixed origin between them and positive

x

to the right.

Positions on opposite sides

The position vectors point away from the origin in their respective directions.

Keep the coordinate signs
The particle to the left of the origin has a negative coordinate; the particle to the right has a positive coordinate. Preserve these signs in the weighted sum.
Substitute the signed positions
The two mass-coordinate products are equal in magnitude and opposite in sign. Using positive distances for both would fail to represent their positions in this frame.
$x_G=\frac{(3.0\,\mathrm{kg})(-1.0\,\mathrm{m})+(1.0\,\mathrm{kg})(3.0\,\mathrm{m})}{3.0\,\mathrm{kg}+1.0\,\mathrm{kg}}$
Interpret the result
The weighted products cancel, so the centre of mass is at the origin. Its coordinate lies between the two particle coordinates, as required for positive masses.
$x_G=0\,\mathrm{m}$

Answer: The centre of mass is at the origin:

x_G=0\,\mathrm{m}

.

Check: The weighted positions sum to zero, confirming the result and its direction.

Worked example

A planar three-particle system

Three particles are located at

(0,0)\,\mathrm{m}

with mass

2.0\,\mathrm{kg}

, at

(4.0,0)\,\mathrm{m}

with mass

1.0\,\mathrm{kg}

, and at

(0,2.0)\,\mathrm{m}

with mass

3.0\,\mathrm{kg}

. Find the centre-of-mass coordinates. The observer uses a fixed frame with positive

x

right and positive

y

upward, and the system consists of all three particles.

Three particles in a plane

The 2.0 kg particle is at the origin; the other two have the shown coordinate directions.

Find the total mass
All three particles belong to the system. Add their masses before calculating either coordinate so both components use the same denominator.
$m_{\mathrm{tot}}=2.0\,\mathrm{kg}+1.0\,\mathrm{kg}+3.0\,\mathrm{kg}=6.0\,\mathrm{kg}$
Calculate the horizontal coordinate
Only the particle at $(4.0,0)\,\mathrm{m}$ has a nonzero $x$ coordinate. Its mass-coordinate product is divided by the total mass.
$x_G=\frac{(2.0)(0)+(1.0)(4.0)+(3.0)(0)}{6.0}=0.667\,\mathrm{m}$
Calculate the vertical coordinate
Only the particle at $(0,2.0)\,\mathrm{m}$ has a nonzero $y$ coordinate. The result is positive because that particle is above the origin.
$y_G=\frac{(2.0)(0)+(1.0)(0)+(3.0)(2.0)}{6.0}=1.00\,\mathrm{m}$

Answer: The centre of mass is at

(x_G,y_G)=(0.667,1.00)\,\mathrm{m}

.

Check: Both coordinates lie between the corresponding minimum and maximum particle coordinates. The vertical coordinate is weighted strongly by the

3.0\,\mathrm{kg}

particle above the origin.

Common mistakes and how to avoid them

Taking an ordinary average when particle masses differ.

Correction: Weight each coordinate by its particle’s mass and divide by total mass.

Replacing a negative coordinate with a positive distance.

Correction: Use signed coordinates measured from the stated origin.

Dividing by the number of particles.

Correction: Divide by the sum of the particle masses.

Calculating only one coordinate for a planar system.

Correction: Calculate every component needed to specify the centre-of-mass position.

Lesson summary

Specify the system, observer’s reference frame, origin, and positive directions.
For each coordinate, divide the sum of mass-times-coordinate products by total mass.
Preserve signed coordinates and report the result in metres.
Check units, coordinate bounds, and whether the result shifts toward larger masses.

Check your understanding

Question 1

A

1.0\,\mathrm{kg}

particle is at

x=0\,\mathrm{m}

and a

3.0\,\mathrm{kg}

particle is at

x=4.0\,\mathrm{m}

. Where is the centre of mass?

$x_G=1.0\,\mathrm{m}$
$x_G=3.0\,\mathrm{m}$
$x_G=2.0\,\mathrm{m}$
$x_G=4.0\,\mathrm{m}$

Show answer and explanation

x_G=3.0\,\mathrm{m}

The weighted coordinate is

[(1.0)(0)+(3.0)(4.0)]/(1.0+3.0)=3.0\,\mathrm{m}

. It is closer to the heavier particle.

Question 2

Two equal-mass particles are at

x=-2.0\,\mathrm{m}

and

x=2.0\,\mathrm{m}

. What is their centre-of-mass coordinate?

$x_G=-2.0\,\mathrm{m}$
$x_G=2.0\,\mathrm{m}$
$x_G=0\,\mathrm{m}$
$x_G=4.0\,\mathrm{m}$

Show answer and explanation

x_G=0\,\mathrm{m}

Equal masses at opposite coordinates have equal and opposite mass-coordinate products, so the weighted sum is zero.

Key terms

Particle system: A specified collection of particles, each represented by its mass and position.
Reference frame: The observer’s coordinate system used to describe positions.
Centre of mass: The mass-weighted average position of the particles in a system.
Position vector: A vector from the chosen origin to a particle or to the centre of mass.

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7.1 · Locate the centre of mass of a particle system

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

A mass-weighted position in one or two dimensions

What you will learn

1. Define the system and position coordinates

2. Calculate each coordinate component

3. Interpret and verify the result

Two particles on a line

Particles on opposite sides of the origin

A planar three-particle system

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review