Centre of Mass & Centre-of-Mass Motion

For particles of mass mᵢ at positions rᵢ, the centre of mass is r_cm = Σ mᵢrᵢ / M with M = Σ mᵢ, worked one component at a time. Put 1.0 kg at the origin, 2.0 kg at (3.0 m, 0) and 3.0 kg at (0, 2.0 m). Then x_cm = (2.0 × 3.0) ÷ 6.0 = 1.0 m and y_cm = (3.0 × 2.0) ÷ 6.0 = 1.0 m; every mass sitting on an axis drops out of one sum, which is the reason to choose the origin with care. Because every weight is positive, the centre of mass lies inside the outline joining the outermost masses and is pulled towards the heavier ones, although it need not sit on any material. Moving the origin changes every xᵢ by the same amount and x_cm by that same amount, so the centre of mass is a physical point, not a property of the coordinates you happened to choose.

A rod or a plate has no list of particles, so cut it into pieces of mass dm and let the sum become an integral: x_cm = (1/M) ∫ x dm with M = ∫ dm. Along a rod, dm = λ(x) dx, where λ is the linear density in kg m⁻¹, and the limits are the rod's two ends. A uniform rod returns L/2, as symmetry promises. Now let the density grow from zero at one end, λ = bx. Then M = ∫₀ᴸ bx dx = bL²/2 and ∫₀ᴸ x(bx) dx = bL³/3, so x_cm = (bL³/3) ÷ (bL²/2) = 2L/3. The constant b cancels, as any overall density scale must, and the answer lies past the midpoint towards the dense end. A flat plate needs no double integral at this level: slice it into strips of width dx and height h(x), so dm = σh(x) dx with σ in kg m⁻², and the same single integral does the work.

A composite body never needs a new integral. Replace each part by its own mass placed at its own centre of mass, then combine the parts as particles. A hole is simply a part with negative mass. Cut a 0.10 m square hole, centred 0.08 m right of centre, from a uniform 0.30 m square plate whose full mass is M₀. By area the missing piece is 1/9 of M₀, so the plate is a whole plate at x = 0 plus a mass −M₀/9 at x = 0.08 m: x_cm = (0 − (M₀/9)(0.08 m)) ÷ (M₀ − M₀/9) = −0.010 m. The centre of mass moves 10 mm away from the hole, which is the sanity check to run every time: removing material pushes the balance point away from where it was taken. Symmetry settles the other coordinate before any arithmetic. The hole sits on the horizontal centre line, so y_cm = 0.

Differentiate M r_cm = Σ mᵢrᵢ with every mass fixed and M v_cm = Σ mᵢvᵢ = P. However the parts move, the total momentum is that of one particle of mass M riding at the centre of mass. A 0.40 kg glider at +1.5 m s⁻¹ and a 0.60 kg glider at −0.50 m s⁻¹ have P = 0.60 − 0.30 = 0.30 kg m s⁻¹, so v_cm = 0.30 m s⁻¹, and on a level frictionless track it stays 0.30 m s⁻¹ through any collision between them. Start a system from rest with no horizontal external force and v_cm stays zero, so the centre of mass cannot move. A 55 kg student walking 4.0 m forward along a 165 kg floating pontoon drives it backwards (water drag neglected). With ground displacements, Σ mᵢΔxᵢ = 0 reads 55(Δxₚ + 4.0 m) + 165 Δxₚ = 0, so Δxₚ = −1.0 m and the student moves 3.0 m relative to the water, not 4.0 m.

A second derivative gives M a_cm = Σ mᵢaᵢ = Σ Fᵢ, the sum of every force on every particle. Split each into external forces and the internal forces exerted by the other particles of the system. Internal forces come in third-law pairs, Fᵢⱼ = −Fⱼᵢ, so they cancel pairwise and leave Σ Fₑₓₜ = M a_cm = dP/dt. Nothing about the internal forces survives: not their size, not their timing, not whether they come from springs, muscles or explosives. A diver whose centre of mass leaves the board at 3.2 m s⁻¹ upward peaks 3.2² ÷ (2 × 9.81) = 0.52 m above take-off whether she tucks, pikes or stays straight, because gravity is the only external force in flight. By the same argument, a shell that bursts in mid-air sends the centre of mass of its fragments along the original parabola, for as long as every fragment is still in the air.

The theorem is useful because it ignores internal detail, and that is also its limit. Kinetic energy splits as K = ½ M v_cm² + Kᵣₑₗ, where Kᵣₑₗ is measured in the centre-of-mass frame. The gliders above carry K = ½(0.40)(1.5)² + ½(0.60)(0.50)² = 0.525 J, but ½(1.00)(0.30)² = 0.045 J belongs to the centre-of-mass motion, so no collision between them can dissipate more than 0.48 J. The centre of mass need not be on the body: a ring's sits at its empty centre, and a high jumper's can pass beneath the bar. It matches the centre of gravity only where g is uniform. And the derivation held every mass fixed, so a system that gains or sheds mass, such as a rocket tracked without its exhaust, needs the variable-mass treatment of the next topic.