What is the centre of mass?
A system of particles means a group of small bits of matter that we study together. A rigid body is a body whose shape never changes: the distance between any two of its bits stays fixed.
The centre of mass (CM) is the special point of a system that moves as if all the mass were there and all outside forces acted there. It is a mass-weighted average position.
Why do we need it? A spinning bat thrown in the air moves in a messy way. But its centre of mass follows a neat parabola, just like a small ball. So the CM lets us treat a big body like one point.
Centre of mass of two particles
Put mass m₁ at x₁ and mass m₂ at x₂ on a line. Then
x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
- If m₁ = m₂, x_cm = (x₁ + x₂)/2, the exact middle.
- The CM always lies on the line joining the two masses, between them.
- It is closer to the heavier mass. Distances from the CM are in the inverse ratio of masses: r₁/r₂ = m₂/m₁.
You can put the origin anywhere. The CM point in space stays the same; only its number changes.
Centre of mass of many particles and of a rigid body
For n particles, do the weighted average in each direction:
x_cm = Σmᵢxᵢ / M, y_cm = Σmᵢyᵢ / M, z_cm = Σmᵢzᵢ / M, where M = Σmᵢ.
In vector form, R_cm = Σmᵢrᵢ / M.
A rigid body has a huge number of tiny bits, so the sum becomes an integral: x_cm = (1/M)∫x dm. For bodies that are uniform and symmetric (sphere, ring, disc, cube, rod) the CM is at the geometric centre. The CM may even lie outside the material, like the centre of a ring or a bangle.
Centre of mass of a uniform rod
Take a thin uniform rod of mass M and length L along the x-axis from x = 0 to x = L. Uniform means every centimetre has the same mass. Mass per unit length λ = M/L.
Step 1: take a tiny slice of length dx at position x. Its mass dm = λ dx.
Step 2: x_cm = (1/M)∫₀ᴸ x λ dx = (λ/M)·(L²/2).
Step 3: put λ = M/L. x_cm = (M/L)(L²/2)/M = L/2.
So the CM of a uniform rod is its midpoint. The symmetry reason: every slice at distance d left of centre has an equal slice at distance d right of centre, and they cancel.
Motion of the centre of mass
Newton's second law for a whole system is M a_cm = F_ext. Internal forces come in equal and opposite pairs (third law), so they cancel. That means:
- If F_ext = 0, v_cm stays constant. A CM at rest stays at rest.
- Total momentum P = M v_cm.
This is why a man walking on a boat floating in still water pushes the boat backward: the CM of man + boat does not move.
Try it at home
Balance a 30 cm scale on one finger. It balances at 15 cm. Now tape two coins at the 25 cm mark and find the new balance point. It moves towards the coins. Use the formula with the scale's own mass to check your answer.
Key formulas and definitions
- x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
- R_cm = Σmᵢrᵢ / Σmᵢ
- x_cm = (1/M)∫x dm (continuous body)
- Uniform rod of length L: x_cm = L/2
- r₁/r₂ = m₂/m₁ (two particles, distances from CM)
- M v_cm = Σmᵢvᵢ = total momentum P
- M a_cm = F_ext
Worked examples
1. Two balls of 2 kg each are placed at x = 0 and x = 6 m. Find x_cm.
Equal masses, so the CM is in the middle. x_cm = (2×0 + 2×6)/(2 + 2) = 12/4 = 3 m.
2. Masses 1 kg at x = 0 and 3 kg at x = 4 m. Find the centre of mass.
x_cm = (1×0 + 3×4)/(1 + 3) = 12/4 = 3 m. It is 3 m from the lighter mass and 1 m from the heavier one. Check: r₁/r₂ = 3/1 = m₂/m₁. ✔
3. Earth (6 × 10²⁴ kg) and Moon (7.4 × 10²² kg) are 3.84 × 10⁸ m apart. How far is their CM from Earth's centre?
Take Earth at x = 0. x_cm = m_moon·d/(m_earth + m_moon) = (7.4 × 10²² × 3.84 × 10⁸)/(6.074 × 10²⁴) ≈ 4.68 × 10⁶ m ≈ 4680 km. This is inside the Earth (radius 6400 km).
4. Three particles 1 kg at (0, 0), 2 kg at (4, 0) and 1 kg at (0, 4) (metres). Find the CM.
M = 4 kg. x_cm = (1×0 + 2×4 + 1×0)/4 = 2 m. y_cm = (1×0 + 2×0 + 1×4)/4 = 1 m. CM = (2 m, 1 m).
5. Three equal masses m sit at the corners of an equilateral triangle of side 2 m, with corners at (0,0), (2,0) and (1, √3). Find the CM.
x_cm = (0 + 2 + 1)/3 = 1 m. y_cm = (0 + 0 + √3)/3 = 0.577 m. It is the centroid of the triangle, as symmetry says.
6. A uniform rod 1 m long has mass 2 kg. A 1 kg ball is fixed at its right end. Where is the CM of rod + ball from the left end?
Replace the rod by its whole mass at its midpoint (0.5 m). x_cm = (2×0.5 + 1×1)/(2 + 1) = 2/3 ≈ 0.67 m from the left end.
7. A 60 kg man stands at one end of a 40 kg boat 5 m long, at rest on still water. He walks to the other end. How far does the boat move? (Ignore water drag.)
No outside horizontal force, so the CM does not move. Man moves 5 m relative to the boat. Let the boat move d backward. Man's shift on the ground = 5 − d. 60(5 − d) = 40d → 300 = 100d → d = 3 m. The boat moves 3 m backward; the man moves 2 m forward.
8. A shell moving horizontally at 20 m/s bursts into two equal pieces. One piece falls straight down from rest (speed 0 just after burst). Find the other piece's speed.
Momentum is conserved (the burst is internal). M×20 = (M/2)×0 + (M/2)×v → v = 40 m/s in the same direction. The CM keeps moving at 20 m/s.
Common mistakes
- Thinking the centre of mass must be inside the body. For a ring or a bangle it is at the empty centre.
- Using the geometric middle when masses are unequal. The CM is closer to the heavier mass.
- Forgetting the rod's own mass in rod + ball problems. Treat the rod as its mass placed at L/2.
- Thinking internal forces (a bomb burst, walking on a boat) can move the CM. Only an external force changes v_cm.