Physics I: Mechanics › Momentum › full formula sheet

xcm = (m₁x₁ + m₂x₂ + …) / M

Say it: “the center of mass is the mass-weighted average position — the balance point of the system”

Center of mass

The balance point of any system — where it would balance on a pin, and the point that moves as if all the mass lived there.

mi are the masses, xi their positions on your coordinate axis, M = Σmi the total mass. xcm is a position — same units as x.

Where it comes from

Balance a baseball bat on your finger: there's exactly one point where it sits level. A dumbbell balances at its middle; a sledgehammer near its heavy head. That point — the center of mass — is where the mass “averages out.”

Before reading on: a 2 kg mass at x = 0 and a 6 kg mass at x = 4 m. Is the balance point at x = 2 (the midpoint), closer to the 6 kg mass, or closer to the 2 kg mass?

Closer to the 6 kg mass: x = 3 m. The heavy mass pulls the average toward itself — three times the mass, three times the pull. A plain midpoint would ignore the masses entirely:

xcm = (m₁x₁ + m₂x₂ + …) / Meach mass votes with its weight — the average position of all the massSay it: “the center of mass is the mass-weighted average position”

Derivation

Demand that the system balance at xcm — torques about that point must cancel:

balance
=
Σ mig(xi − xcm) = 0
Step 1 — zero torque. Each mass pulls down with mig at lever arm (xi − xcm). Balance means the torques cancel.
=
Σ mixi − xcmΣmi = 0
Step 2 — expand. g cancels (balance doesn't depend on gravity's strength — it works on the Moon too).
xcm
=
(Σ mixi) / M
Step 3 — solve. The mass-weighted average. For continuous bodies the sum becomes an integral: (1/M)∫x dm. ∎

How to use it

The procedure, every time:

  1. Set a coordinate axis. x = 0 wherever convenient — an end, a pivot, one of the masses.
  2. List (mi, xi). Every mass with its signed position. Masses left of origin get negative x.
  3. Compute Σmixi. Each mass times its position, added.
  4. Divide by M = Σmi. The total mass — forgetting it is the classic error.
  5. Sanity-check. xcm lies between the extreme masses, pulled toward the heavy ones. Outside the range = arithmetic error.
Common mistake: averaging the positions without the masses (the plain midpoint). The masses are the weights — “weighted average” means the mi multiply.

Worked examples

Four problems, easiest first. The heavy mass always pulls the average.

Example 1 — two masses: 2 kg at x = 0, 6 kg at x = 4 m

  1. Numerator. 2×0 + 6×4 = 24.
  2. Denominator. M = 8.
  3. Divide. xcm = 24/8 = 3 m — three-quarters of the way to the heavy mass.
Common mistake: xcm = 2 m (the midpoint). The 6 kg mass outvotes the 2 kg mass 3-to-1 — the balance point sits 3 m from the light end, 1 m from the heavy end.
Your turn — 3 kg at x = 0, 9 kg at x = 6 m. xcm = ?

Answer: 4.5 m. (0 + 54)/12 = 4.5 m.

Example 2 — three masses: 1 kg at 0, 2 kg at 3 m, 3 kg at 6 m

  1. Numerator. 1×0 + 2×3 + 3×6 = 0 + 6 + 18 = 24.
  2. Denominator. M = 6.
  3. Divide. xcm = 24/6 = 4 m.
Common mistake: dividing by 3 (the count of masses) instead of 6 (the total mass). M is kilograms, not “how many objects.”
Your turn — 2 kg at 0, 3 kg at 4 m, 5 kg at 10 m. xcm = ?

Answer: 6.2 m. (0 + 12 + 50)/10 = 6.2 m.

Example 3 — seesaw: 30 kg kid at x = −2 m; where does a 20 kg kid sit to balance?

  1. Balance condition. xcm = 0 (pivot at origin).
  2. Solve. (30×(−2) + 20×x)/50 = 0 → −60 + 20x = 0 → x = +3 m.
  3. Check. Lighter kid sits farther out — torque balance: 30×2 = 20×3. ✓
Common mistake: putting the lighter kid closer (“lighter needs less room”). Torque is force × distance — less weight needs MORE distance to balance.
Your turn — 40 kg at x = −3 m; where does a 30 kg kid sit?

Answer: +4 m. −120 + 30x = 0 → x = 4 m.

Example 4 — lopsided rod: 1 kg at x = 0, 4 kg at x = 1 m

  1. Numerator. 1×0 + 4×1 = 4.
  2. Denominator. M = 5.
  3. Divide. xcm = 4/5 = 0.8 m — very near the heavy end.
Before moving on: if the 4 kg mass slid to x = 2 m, would xcm double too? Predict the direction of the change (not the number).
Common mistake: xcm = 0.5 m (“middle of the rod”). The rod's geometry middle means nothing — the mass distribution decides.
Your turn — 2 kg at x = 0, 3 kg at x = 2 m. xcm = ?

Answer: 1.2 m. 6/5 = 1.2 m.

Memorization tips

  • Say it aloud: “the center of mass is the mass-weighted average position.”
  • Heavies pull: xcm leans toward the big masses. If your answer leans the wrong way, recheck.
  • Divide by MASS: M = Σmi in kilograms — never by the count of objects.
  • Range check: xcm always falls between the extreme positions. Outside = error.
  • Origin is free: put x = 0 wherever the arithmetic is easiest — the physics doesn't care.
  • Balance = zero torque: the derivation in one line. If you blank on the formula, re-derive from torques.

Final challenge

Five mixed questions — weighting, origins, and balance. Score 5/5 and center of mass is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

Is the center of mass always inside the object?

No! A donut's center of mass is in its hole; a boomerang's is outside the wood. The CM is a property of the mass distribution, not a material point.

What's the difference between center of mass and center of gravity?

In uniform gravity they're the same point. They differ only when g varies across the object (skyscrapers, satellites) — then gravity's average point shifts slightly from the mass average.

Why does the center of mass matter for motion?

A system's CM moves as if all the mass were there with all external forces applied (Fnet = Macm). Internal forces can't move it — that's why it's the key to CM velocity and momentum.

How do you find the CM of a continuous object?

Replace the sum with an integral: xcm = (1/M)∫x dm. Symmetry often does the work for free — a uniform rod's CM is its midpoint by symmetry alone.

Can the center of mass move without any external force?

No — that's the deep result. Internal forces (muscles, engines, explosions) cancel pairwise, so acm = 0 without external F. You can't lift yourself by your bootstraps, literally.

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