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Centre of mass questions come in two kinds. The first asks you to find it: for two particles, a rod with changing density, or a plate with a hole cut out. The second hides it inside a motion problem: a person walking on a boat, a shell bursting in mid-air. Students often solve the second kind the long way because they do not spot that the centre of mass makes it simple. This post covers both kinds with worked examples.
Where is centre of mass in the syllabus?
The JEE (Main) 2026 syllabus lists "centre of mass of a two-particle system, centre of mass of a rigid body" in its rotational motion unit, and the JEE (Advanced) 2026 brochure lists "systems of particles; centre of mass and its motion". The CBSE 2026-27 physics curriculum includes momentum conservation and centre of mass motion, and the centre of mass of a uniform rod. Check the current syllabus for your exam.
How do you find the centre of mass?
For particles, it is a mass-weighted average of positions: x_cm = (m₁x₁ + m₂x₂ + ...) / (m₁ + m₂ + ...).
Example 1. A 2 kg mass sits at x = 0 and a 3 kg mass at x = 5 m. x_cm = (2 × 0 + 3 × 5) / 5 = 3 m. The centre of mass is closer to the heavier mass, as it should be.
For a uniform body, NCERT's chapter on Systems of Particles and Rotational Motion uses symmetry: the centres of mass of homogeneous rings, discs, spheres and rods lie at their geometric centres. Notice that for a ring, that point has no mass at all. The centre of mass does not have to lie inside the body.
Example 2: a rod with changing density. A rod of length L has mass per unit length λ = kx, rising from zero at one end. Then x_cm = ∫x λ dx / ∫λ dx = (kL³/3) / (kL²/2) = 2L/3, measured from the light end. The answer leans towards the heavy end, which is a quick check.
Example 3: a disc with a hole. From a uniform disc of radius R, a circular hole of radius R/2 is cut, touching the edge, so the hole's centre is R/2 from the disc's centre. Mass goes as area, so if the full disc has mass M, the removed piece has mass M/4. Treat the hole as negative mass:
x_cm = [M × 0 − (M/4)(R/2)] / (M − M/4) = (−MR/8) / (3M/4) = −R/6
The centre of mass shifts R/6 away from the hole.
Why does the centre of mass simplify motion?
NCERT shows that the centre of mass of a system of particles moves as if all the mass were concentrated there and all the external forces were applied at that point. To find its motion, you need no knowledge of the internal forces. That is the key that turns hard problems into one-liners.
Example 4: walking on a boat. A 60 kg person walks 4 m along a 140 kg boat floating on still water. Ignore water resistance. How far does the boat move?
No horizontal external force acts, and everything starts at rest, so the centre of mass stays where it is. Let the boat move a distance b (positive in the direction the person walks). The person moves b + 4 relative to the water. Then 60(b + 4) + 140b = 0, so 200b = −240 and b = −1.2 m. The boat moves 1.2 m backwards and the person moves 2.8 m forwards relative to the water. Check: 60 × 2.8 = 168 and 140 × 1.2 = 168, so the mass-weighted shifts cancel.
Example 5: a shell explodes. A shell is fired at 20 m/s at 45° (take g = 10 m/s²), so its range would be v² sin 2θ / g = 400/10 = 40 m. At the top of its path it bursts into two equal pieces. One piece falls straight down from rest. Where does the other land?
The explosion forces are internal, so the centre of mass keeps following the original parabola. NCERT makes the same point about a projectile that explodes in mid-air. Right after the burst, both pieces have zero vertical velocity (the first piece has none, and total vertical momentum was zero at the top), so both land at the same time. The first lands at 20 m. The centre of mass lands at 40 m, so the second piece lands at 60 m, since (20 + 60) / 2 = 40.
Which mistakes should you avoid?
Assuming the centre of mass always stays still. It stays still only if the net external force is zero and it started at rest. Otherwise it moves with constant velocity, or accelerates under the external force.
Using it after a piece hits the ground. Once one fragment lands, the ground pushes on the system, so the parabola rule stops working. That is why the pieces in Example 5 landing together matters.
Measuring positions from different origins. Fix one origin and one positive direction for every mass.
Using area instead of mass for non-uniform plates. "Mass goes as area" holds only for uniform thickness and density.
The centre of mass also matters for collisions: in its frame, total momentum is zero. Our guide to collisions and the coefficient of restitution uses this. If you can follow the equations but cannot picture why the boat slides back, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: Five "find it" problems: two particles, three particles in a plane, a rod with variable density, and two shapes with holes.
Day 2: Four "it stays put" problems: boats, a person on a plank on ice, a gun and its recoil.
Day 3: Three explosion problems, each stating why the centre of mass rule applies or stops applying.
Day 4: A timed mixed set. For each, write one line first: "External force on the system: ...". For more on drawing the system before the algebra, see our post on drawing the diagram first.
This week, redo one boat or explosion problem you solved with forces, using only the centre of mass, and compare how many lines it takes.
