System of Particles and Rotational Motion · Lesson 2 of 15
Centre of Mass of Particle Systems
“Locate a system’s mass-weighted representative point in one, two and three dimensions.”
• Calculate a centre of mass using consistent coordinates. • Explain why it lies closer to the heavier mass. • Use symmetry and equal-mass averages. • Apply the method to molecular and triangular systems.
A mass-weighted location
The centre of mass, or CM, is the position obtained by weighting each particle’s location by its mass. Heavy particles influence it more strongly. For two equal masses it lies midway between them. For unequal masses it lies nearer the heavier mass. It is a geometric property of the mass distribution and can be defined even where there is no gravity.
To see the division of the joining segment, put the first particle at zero and the second at distance d. Then X = m₂d/(m₁ + m₂). The distances from the CM satisfy m₁X = m₂(d − X). Thus the heavier mass has the shorter lever arm. This also shows that with positive masses the CM lies between the two particle positions.
Changing the origin changes every coordinate and the CM coordinate consistently; it does not move the physical CM relative to the particles. If the origin is placed at the CM, the mass-weighted vector sum Σmᵢrᵢ is zero. This does not mean the particles are all at zero. Their positive and negative weighted coordinates cancel.
Problem
The nuclei are 1.27 Å apart and chlorine has 35.5 times the hydrogen mass. Locate the CM.
- 1.Put hydrogen at x = 0 and chlorine at x = 1.27 Å. Neglect the electron contribution as in the molecular approximation.
- 2.X = 35.5 × 1.27 / 36.5 = 1.235 Å from hydrogen.
- 3.It is only 1.27 − 1.235 ≈ 0.0348 Å from chlorine. The location is close to the much heavier nucleus. One angstrom is 10⁻¹⁰ m.
Masses at the corners of a triangle
For three equal particles at triangle vertices, the CM coordinates are the arithmetic averages of the vertex coordinates. That point is the triangle’s centroid. Unequal vertex masses generally shift the CM away from the geometric centroid. Do not replace the mass-weighted formula by a simple average unless the masses really are equal.
Problem
Place 100 g at (0,0), 150 g at (0.5,0), and 200 g at (0.25, √3/4) metres. Find the CM.
- 1.The total mass is 450 g. Using grams consistently is sufficient because the mass units cancel.
- 2.X = [100(0) + 150(0.5) + 200(0.25)]/450 = 5/18 m ≈ 0.278 m.
- 3.Y = [200(√3/4)]/450 = √3/9 m ≈ 0.192 m.
- 4.The triangle’s geometric centroid is (0.25, √3/12) m. The heavier upper mass shifts the CM upward, and the unequal lower masses also shift it right.
Quiz
Two equal masses lie at x = −2 m and x = 6 m. Where is their CM?
Equal masses give the arithmetic average: (−2 + 6)/2 = 2 m.
If the right-hand mass increases while positions stay fixed, the CM moves where?
The larger weight in the coordinate average pulls the CM toward that position.
Which expression vanishes when the origin is the CM?
The mass-weighted position sum is zero about the CM.
For three equal vertex masses, the CM is which point?
Coordinate averages locate the centroid of a triangle.
Can CM be defined in gravity-free space?
It depends on mass distribution, not gravity.
Key Takeaways
• CM is a weighted average of positions. • Use one coordinate system and matching mass units. • Equal masses permit arithmetic averaging. • The CM origin makes Σmᵢrᵢ vanish.