Center of Mass
- Locate the center of mass of a system of discrete particles
- Compute the center of mass of a continuous body using x_cm = (1/M)∫x dm
- Relate the motion of the center of mass to the net external force
The mass-weighted average position
The center of mass is the single point that moves as if all the mass were concentrated there. For discrete particles it is the mass-weighted average of their positions: each position counts in proportion to its mass, so the center of mass always sits closer to the heavier objects.
The center of mass obeys Newton's second law
Internal forces never move the center of mass — only external forces do: ΣF_ext = M a_cm. That is why a wrench tossed spinning across a table has a center of mass that travels in a smooth parabola even as the wrench tumbles. For a continuous body with nonuniform density, the sum becomes the integral x_cm = (1/M)∫x dm, with dm written in terms of the local density.
A thin rod of length L lies along the x-axis from 0 to L with a linear mass density that increases as λ(x) = cx. Find the center of mass.
- 1.Write the mass element: dm = λ dx = cx dx.
- 2.Total mass: M = ∫₀ᴸ cx dx = c[x²/2]₀ᴸ = cL²/2.
- 3.First moment: ∫₀ᴸ x dm = ∫₀ᴸ x(cx) dx = c[x³/3]₀ᴸ = cL³/3.
- 4.Divide: x_cm = (cL³/3)/(cL²/2) = (2/3)L. The rod is denser toward the far end, so the center of mass lies past the midpoint.
Two particles lie on the x-axis: 1 kg at x = 0 and 3 kg at x = 8 m. Where is the center of mass?
A rod lies along the x-axis from x = 0 to L with linear density λ(x) = cx. Its center of mass is located at:
For a continuous body, the recipe is always the same: write dm using the density, integrate x dm for the numerator and dm for the total mass M, then divide. Never average the endpoints — that only works for a uniform object.
Answer the 2 checkpoints as you read.
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