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Lesson 1 of 9

System of Particles and Rotational Motion · Lesson 1 of 9

Rigid Bodies and Centre of Mass

“Describe rigid-body motion and locate the centre of mass of particles, continuous bodies and composite shapes.”

Learning Objectives

• Distinguish pure translation, fixed-axis rotation and combined motion. • Calculate mass-weighted centre-of-mass coordinates. • Use symmetry and integrals for continuous bodies. • Solve the PDF’s triangle, L-plate, molecule and cut-disc examples.

Rigid bodies and types of motion

Why size now matters

A point-particle model can describe a car travelling between towns, but it cannot describe how the wheels turn or why a door opens more easily when pushed at its edge. An extended body contains many particles at different positions. Their individual motions need not be identical. We therefore describe the motion of the body as a whole and the motion of its parts relative to one another.

An ideal rigid body maintains the distance between every pair of its particles. It can move and change orientation, but cannot bend, stretch or vibrate. Real bodies deform, so rigidity is an approximation. It works when those deformations have negligible influence on the question. A steel beam in a balance problem may be treated as rigid; its bending under a heavy load requires a more detailed model.

Translation preserves orientation

In pure translation, every point of the body has the same velocity at any given instant. A block sliding down an incline without turning is an example. The shared velocity may change with time, so translation need not mean uniform motion. A marked line on the body remains parallel to its original direction. The paths of different points are translated copies of one another; translation can also follow a curved route while orientation remains fixed.

Rotation gives circular paths about an axis

For rotation about a fixed axis, each point away from the axis moves in a circle. The circle lies in a plane perpendicular to the axis and its centre lies on the axis. The circle’s radius is the perpendicular distance to the axis. Points on the axis remain stationary. All particles share the same angular motion, although points farther from the axis travel faster along their larger circles. A ceiling fan and a potter’s wheel illustrate this model.

A spinning top can have a fixed contact point while its axis changes direction, sweeping a cone. This change in the axis direction is called precession. An oscillating table fan combines spinning blades with a side-to-side change of the blade axis. These show why fixing one point is different from fixing an entire axis. Most calculations in this chapter use the simpler fixed-axis model.

Translation and rotation can occur together

A rolling cylinder moves down an incline while its orientation changes. The centre translates, and other points rotate relative to it. Different points then have different ground velocities. For rolling without slipping on stationary ground, the instantaneous contact-point velocity is zero while the top point moves faster than the centre. That point is not permanently at rest: a different part of the rim becomes the contact point as rolling continues. Here rolling illustrates combined motion; detailed rolling dynamics is beyond the supplied chapter’s treatment.

A marked book

Problem
Slide a book across a table, then move it while turning it. What distinguishes the motions?

  1. 1.In the first motion, keep a marked edge parallel to itself. Every book point has the same instantaneous velocity: pure translation.
  2. 2.In the second, the marked edge changes direction. A point near an edge has an additional velocity relative to the centre, so translation and rotation coexist.
  3. 3.The centre can follow the same path in both demonstrations. The centre path alone does not reveal the complete motion.
TranslationTranslation + rotation
Translation preserves orientation; rotation changes it— The marked rectangular body keeps its orientation in translation and changes it in combined motion.

Centre of mass of particle 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.

Two particles on a lineLaTeX
Measure both signed coordinates from the same origin and express both masses in matching units.

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.

A system in spaceLaTeX
The sigma symbol means add over all particles. Compute each coordinate independently, using the same masses and total mass.

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.

An HCl molecule

Problem
The nuclei are 1.27 Å apart and chlorine has 35.5 times the hydrogen mass. Locate the CM.

  1. 1.Put hydrogen at x = 0 and chlorine at x = 1.27 Å. Neglect the electron contribution as in the molecular approximation.
  2. 2.X = 35.5 × 1.27 / 36.5 = 1.235 Å from hydrogen.
  3. 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.

Three unequal masses on an equilateral triangle

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. 1.The total mass is 450 g. Using grams consistently is sufficient because the mass units cancel.
  2. 2.X = [100(0) + 150(0.5) + 200(0.25)]/450 = 5/18 m ≈ 0.278 m.
  3. 3.Y = [200(√3/4)]/450 = √3/9 m ≈ 0.192 m.
  4. 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.
m3mCMJoining distance d; CM is 3d/4 from m
The centre of mass lies closer to the heavier particle— The two masses are idealised particles. The CM divides the joining segment inversely to their masses.

Continuous and composite bodies

From particles to a continuous mass distribution

A macroscopic rod contains too many atoms to sum individually. Divide it into small pieces with masses Δm and known approximate positions. As the pieces become smaller, the sum approaches an integral. The element dm means a small amount of mass; it is not automatically a length or an area. For a uniform thin rod dm = λ dx, where λ is mass per length. For a uniform lamina dm = σ dA, and for a volume distribution dm = ρ dV.

Continuous CMLaTeX
Integration adds the contributions of all small mass elements.

For a uniform rod from x = 0 to L, its constant linear density is M/L. Integrating x(M/L) dx from 0 to L and dividing by M gives X = L/2. This calculation shows explicitly how the integral reproduces the familiar midpoint. If density varies along the rod, it must stay inside the integral and the midpoint need not be the CM.

Symmetry removes unnecessary calculation

For a uniform rod centred at zero, each mass element at x has an equal partner at −x, so their weighted positions cancel. The same reflection argument places the CM at the geometric centre of a uniform sphere, cylinder, ring or cube. Symmetry of shape alone is insufficient if density is nonuniform. A ring’s CM lies at its empty centre: the CM need not lie in the material of the body.

For a uniform triangular lamina, divide the area into narrow strips parallel to one side. Each strip has its CM at its midpoint. These midpoints lie on the median from the opposite vertex. The whole lamina’s CM must therefore lie on that median. Repeating with the other two directions places it at the intersection of the medians, the centroid. This is an argument about a continuously filled triangle, distinct from three unequal vertex masses.

Centre of mass of a triangular lamina

Problem
Locate the centre of mass of a uniform triangular lamina, as in source Example 6.2.

  1. 1.Divide the lamina into narrow parallel strips. Reflection symmetry puts each strip’s CM at its midpoint.
  2. 2.The midpoints lie on the median from the opposite vertex, so the full lamina’s CM lies on that median.
  3. 3.Repeat with another side. The CM must lie at the intersection of the medians, the centroid. This reasoning assumes uniform areal density.

Replace whole components by mass points

A uniform L-shaped plate

Problem
An L consists of three unit squares with centres (0.5,0.5), (1.5,0.5) and (0.5,1.5) m. Each square has mass 1 kg.

  1. 1.Treat each square as its mass concentrated at its own CM for the purpose of the weighted sum.
  2. 2.X = (0.5 + 1.5 + 0.5)/3 = 5/6 m, and Y = 5/6 m.
  3. 3.Reflection in the line y = x interchanges the two arms, so X = Y is expected. If component masses differ, restore their separate weights instead of averaging the coordinates.
Three equal squaresHole shifts CM to the left
Composite plate and removed disc— The L-plate consists of unit squares centred at (0.5,0.5), (1.5,0.5) and (0.5,1.5), with the origin at the lower-left corner. Its marked CM is (5/6,5/6). The removed disc has radius R/2 with centre R/2 right of the original centre; the remaining CM is R/6 to the left.

Subtracting a hole

A missing componentLaTeX
Negative mass is bookkeeping for subtraction, not a claim that physical negative matter is present.
An off-centre circular hole

Problem
A uniform disc of radius R loses a circular piece of radius R/2 whose centre lies R/2 to the right of the original centre.

  1. 1.Let the original disc have mass M and its centre be x = 0. Equal thickness and density make the removed mass M/4, because area scales with radius squared.
  2. 2.The remaining mass is 3M/4. Its first moment is 0 − (M/4)(R/2).
  3. 3.X = [−MR/8]/(3M/4) = −R/6. Y = 0 by reflection symmetry. The CM shifts away from the missing mass.

Check your understanding

Quiz

Quick check

Which statement defines a rigid body?

Explanation

Rigidity constrains internal distances, not the body’s overall motion.

Quick check

Two equal masses lie at x = −2 m and x = 6 m. Where is their CM?

Explanation

Equal masses give the arithmetic average: (−2 + 6)/2 = 2 m.

Quick check

Can CM be defined in gravity-free space?

Explanation

It depends on mass distribution, not gravity.

Quick check

Where is the CM of a uniform ring?

Explanation

Symmetric contributions cancel at the centre even without material there.

Quick check

The CM of a uniform triangular lamina lies where?

Explanation

Strip midpoints place it on each median.

Key Takeaways

Key Takeaways

• Rigidity preserves all pairwise particle distances; translation and rotation describe different aspects of motion. • The centre of mass is mass weighted and need not lie in the material. • Use symmetry only when the mass distribution shares that symmetry. • Treat component masses at their own centres and subtract missing pieces.