Center of Mass
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
The center of mass is the mass-weighted average position: xcm = Σmixi/M, or ∫x dm/M for a continuous object. The whole system’s motion follows its center of mass: Fnet,ext = Macm.
1. Finding it
2. How it moves
vcm = Σmivi/M = Ptotal/M. Internal forces can’t change it; only external forces can.
A firework that explodes in midair: the pieces fly apart, but their center of mass keeps following the original parabola.
If there’s no net external force and everything starts at rest, the center of mass stays put. A 60 kg person walking 3.0 m along a 120 kg boat moves the boat 1.0 m the other way: 60(2.0) = 120(1.0).
3. Example, then try it
Masses of 2.0 kg, 3.0 kg and 5.0 kg sit at x = 0, 2.0 m and 4.0 m. Find the center of mass.
xcm = [(2.0)(0) + (3.0)(2.0) + (5.0)(4.0)] ÷ 10 = 26 ÷ 10 = 2.6 m.
A rod 2.0 m long has linear density λ = 3x kg/m (x in meters from one end). Find its mass and its center of mass.
Mass: M = ∫02 3x dx = [1.5x²] = 6.0 kg.
Moment: ∫x dm = ∫02 3x² dx = [x³] = 8.0 kg·m.
Center of mass: xcm = 8.0 ÷ 6.0 ≈ 1.33 m, which is 2L/3, past the midpoint.
Answer: M = 6.0 kg; xcm ≈ 1.33 m from the light end.
4. Common mistakes
A firework shell explodes into pieces at the top of its flight. Ignoring air resistance, what path does the center of mass of the pieces follow?
The explosion forces are internal. Gravity is the only external force, so the center of mass keeps following the original projectile path.
Ready to practice?
5 questions on Center of Mass, each with a hint and an explanation for every wrong answer.