Centre of Mass
On this page
Direct answer
The centre of mass is the mass-weighted mean position of a system: x_cm = (m1x1 + m2x2 + ...)/(m1 + m2 + ...), with matching expressions for y and z. It behaves like a single particle carrying the entire mass — M a_cm = F_ext — so internal forces cannot move it. For two particles it divides the joining line in the inverse ratio of the masses, and for a uniform rod, ring or disc it sits at the geometric centre. This one idea answers the NEET staples: a man walking on a boat, an exploding projectile whose fragments' COM still follows the parabola, and the shift of COM when part of a body is removed.
What you must remember
- Definition: R_cm = Σm_i r_i / M; the point moves as if all mass were concentrated there and all external forces acted on it.
- Two-body split: the COM lies between the masses, dividing the separation in inverse ratio — closer to the heavier mass; if a 1 kg and a 3 kg body are 4 m apart, the COM is 1 m from the 3 kg body.
- Newton's law for systems: M a_cm = F_ext; internal forces cancel in pairs by Newton's third law, so they shift parts but never the COM of an isolated system.
- Standard locations: centre for a uniform rod, ring, disc, sphere and cube; for a semicircular wire the COM is 2R/π and for a semicircular disc 4R/3π from the centre along the axis of symmetry.
- Explosion in flight: a shell bursting at the top of its trajectory leaves fragments scattered, but their COM continues along the original parabola as if nothing had happened.
- Shift on removing mass: when a hole is cut, treat it as negative mass at the hole's COM and take the weighted difference — the classic disc-with-a-hole construction.
The boat-man question, worked honestly
A man of mass 50 kg stands at one end of a 200 kg boat of length 4 m floating on still water; he walks to the other end. The water is assumed frictionless, so no horizontal external force acts and the system's COM cannot move. As he walks 4 m relative to the boat rightward, the boat must drift leftward. The man's displacement relative to ground is (4 - x) if the boat shifts x: writing fixed COM positions, 200(x) = 50(4 - x), giving 250x = 200, so x = 0.8 m. The man's ground displacement is 3.2 m, not the 4 m he feels — the floor itself retreated. Every such problem is the same one-line equation: Σm Δx = 0, because internal walking forces cannot shift the COM. The identical logic solves a girl walking on a trolley and a monkey climbing a rope hung over a frictionless pulley with a counterweight.
How the examiner exploits the concept
NEET rarely asks for the COM coordinate directly; it tests the consequence — that the COM of an isolated system is immovable by internal action. The distractor answers in the boat question are 4 m (forgetting the boat moves) and 0.8 m for the man's displacement (handing the boat's answer to the man). A second framing is the exploding projectile: fragments land at different points, and the question asks where their COM lands — on the original parabola, since gravity acts externally throughout and the explosion forces are internal. A third is the negative-mass trick: a circular disc of radius R with a hole of radius R/2 touching the edge has its COM shifted toward the intact side by (R/6) from the centre, computed by weighting the removed disc as negative — a construction worth practising once because it recurs with squares and cubes.
Frequently asked questions
Can internal forces accelerate the centre of mass of a system?
No — they occur in equal and opposite pairs, so M a_cm = F_ext requires a net external force; only external forces move the COM.
Where does the centre of mass of a two-particle system lie?
On the line joining them, dividing it in the inverse ratio of the masses, so it always sits nearer the heavier particle.
How far does a boat move when a person walks its length?
The boat recoils by m d/(M + m) on frictionless water — the displacement that keeps the system's COM fixed while the person covers d relative to the boat.
Why does the centre of mass of an exploded shell follow the original trajectory?
Explosion forces are internal; gravity remains the only external force, so the fragments' COM traces the unbroken parabola of the intact shell.
How is a hole in a disc handled in COM problems?
Treat the hole as negative mass placed at its own centre of mass and compute the weighted average of the full disc plus the negative hole.