System of Particles and Rotational Motion · Lesson 3 of 15
Centre of Mass of Continuous and Composite Bodies
“Use symmetry, small mass elements and signed mass bookkeeping to locate centres of extended bodies.”
• Relate particle sums to integrals. • Locate CM by reflection symmetry. • Combine component bodies by their own centres. • Use subtraction for a missing piece.
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.
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.
Replace whole components by mass points
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.Treat each square as its mass concentrated at its own CM for the purpose of the weighted sum.
- 2.X = (0.5 + 1.5 + 0.5)/3 = 5/6 m, and Y = 5/6 m.
- 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.
Subtracting a 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.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.The remaining mass is 3M/4. Its first moment is 0 − (M/4)(R/2).
- 3.X = [−MR/8]/(3M/4) = −R/6. Y = 0 by reflection symmetry. The CM shifts away from the missing mass.
Quiz
For a uniform rod, dm equals what?
Linear density converts a small length into a mass element.
Where is the CM of a uniform ring?
Symmetric contributions cancel at the centre even without material there.
A hole on the right shifts CM which way?
Removing right-hand mass leaves relatively more mass on the left.
A plate has symmetric shape but nonuniform density. Is its CM necessarily geometric centre?
Mass symmetry, rather than shape alone, is required.
The CM of a uniform triangular lamina lies where?
Strip midpoints place it on each median.
Key Takeaways
• An integral is the continuous counterpart of a particle sum. • Use mass symmetry to locate CM efficiently. • Composite bodies can be replaced by component masses at their CMs. • Subtract removed material’s mass and first moment consistently.