PHYS2250 Chap.6 Systems of Particles and Centre of Mass
Systems of Particles and Centre of Mass
Compress internal complexity without losing mass distribution
The centre of mass is a mass-weighted position whose translation responds only to external force. Discrete sums become integrals for continuous bodies, with the differential element chosen to match geometry. Symmetry helps only when both shape and mass distribution share it.
Internal forces may change configuration and rotation without altering centre-of-mass motion.
Reasoning route and exam method
For discrete particles the centre of mass is the mass-weighted average position, rCM = Σmᵢrᵢ/Σmᵢ. For a continuous body, the sum becomes an integral over dm; this relationship belongs in the diagnostic checklist for the chapter.
Symmetry can locate a centre of mass without integration, but only when both the geometry and the mass distribution possess that symmetry; this is a useful checkpoint before any numerical substitution. Once located, the centre of mass moves as though the total external force acted on the total mass there; the same result should survive the relevant limiting-case test.
Internal forces can change the configuration, rotation and internal energy while leaving centre-of-mass motion governed solely by external force; that distinction determines which quantities may be combined.
Choose a differential element that matches the symmetry: a strip for a lamina described between curves, a ring for an axisymmetric disk, or a shell for a spherical body; this observation fixes the physical interpretation of the algebra.
Express both dm and the coordinate of the element in one integration variable; the statement supplies a direct check on the final result. The denominator is the total mass and supplies a useful check; this condition should remain visible beside the governing equation.
A result outside the permitted coordinate range is impossible; a result on the wrong side of a symmetry axis exposes a sign or limit error.
A 2.0 kg particle is at x = 0 and a 5.0 kg particle is at x = 3.5 m. Then xCM = [2(0) + 5(3.5)]/(2 + 5) = 2.50 m. Relative to that point the lighter particle has moment 2(2.50) = 5.0 kg m and the heavier has 5(1.00) = 5.0 kg m in the opposite direction.
The equal first moments verify the answer. Averaging the coordinates to obtain 1.75 m would ignore mass weighting.
Answer. It continues with the velocity it had immediately before the explosion because the explosion forces are internal.
WEIGHTED POSITION Continuous body. Write xCM = ∫x dm / ∫dm.
If density is uniform it may cancel, but state the assumption before cancelling it.
The centre of mass lies closer to the heavier orange body because position is weighted by mass. It need not coincide with either particle or even lie inside the material of a curved object.
The centre of mass simplifies translation; it does not erase rotation or the distribution of mass.
What this chapter covers
- 01
weighted position
- 02
continuous distributions
- 03
symmetry arguments
- 04
centre-of-mass motion
- 05
internal and external forces
Worked example · free
Locate a two-particle balance point
- 2Write the weighted-position expression.
- 2Substitute the masses and coordinates.
- 1Check first moments about the result.
Key terms
- Centre Of Mass
- The mass-weighted mean position of a system.
- First Moment
- Mass multiplied by signed distance from a reference point.
- Mass Element
- A differential portion of a continuous body chosen to match its density and geometrical symmetry.
Systems of Particles and Centre of Mass FAQ
Can the centre of mass lie outside a body?
Yes. A ring has its centre of mass at an empty geometric centre, and curved bodies may place it outside their material while retaining symmetric mass moments about the relevant axes.
When does symmetry locate it without integration?
Only when geometry and density share the symmetry. A symmetric outline with nonuniform density does not earn the shortcut because equal geometric pieces may carry unequal mass and unequal first moments.
Do internal forces move the centre of mass?
They cannot change total centre-of-mass motion, though they can alter relative positions, rotation and internal energy while external force remains the translational control for the complete chosen system.
Exam move
Practise selecting strips, rings and shells as differential elements. Verify every result lies within the coordinate range allowed by the distribution or is justified by symmetry. Begin each continuous problem by drawing a representative element and writing dm in the integration variable before setting limits. Compute the total mass from the same element and use it as the denominator rather than importing a separate formula.
For composite bodies, use signed masses only for removed regions and check whether the final point lies on the expected side of the symmetry axis. Follow one explosion or internal-motion example to see that fragments rearrange while the centre of mass continues according to external force. Conclude by balancing first moments about the calculated point, which catches incorrect limits and missing density factors quickly.
For one nonuniform example, keep density symbolic until numerator and denominator are formed, preventing an unjustified cancellation that uniform-density cases can make habitual.
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