Free Center of Mass Calculator

Last updated October 10, 2026 · Report an error

Thanks for rating!

The center of mass of a system of point masses is the mass-weighted average of their positions — the single balance point where, if you supported the whole system there, it would balance perfectly under gravity. Heavier masses pull the center of mass toward themselves, so it always sits closer to the more massive points and farther from the lighter ones, unlike a plain geometric average of positions, which treats every point equally regardless of weight. This concept is central to physics and engineering: it determines whether a vehicle, aircraft, or ship is stable (a center of mass too high or too far to one side risks tipping), it's the point used to simplify complex rigid-body motion into the motion of a single equivalent point in mechanics, and structural and mechanical engineers use it to balance rotating machinery, load cargo, and design stable furniture or equipment.

How it works

Enter the number of point masses (2 to 6), then each one's mass and (x, y) position. The calculator computes the center of mass coordinates as (Σmᵢxᵢ/Σmᵢ, Σmᵢyᵢ/Σmᵢ) — that is, it multiplies each point's x-coordinate by its own mass, adds all those products together, and divides by the total mass to get the x-coordinate of the center of mass; the y-coordinate is found the same way using the y-values. Because every term is weighted by its own mass, a very heavy point pulls the result strongly toward its own position, while a very light point barely moves the result at all.

Enter number of point masses.
Enter point 1 — mass.
Enter point 1 — x.
Enter point 1 — y.
Enter point 2 — mass.
Enter point 2 — x.
Enter point 2 — y.
Enter point 3 — mass.
Enter point 3 — x.
Enter point 3 — y.
Enter point 4 — mass.
Enter point 4 — x.
Enter point 4 — y.
Enter point 5 — mass.
Enter point 5 — x.
Enter point 5 — y.
Enter point 6 — mass.
Enter point 6 — x.
Enter point 6 — y.
Click Calculate to see your results.

Examples

Two 1kg masses at (0,0) and (10,0)

Equal masses give a center of mass exactly at the midpoint: ((1×0+1×10)/2, (1×0+1×0)/2) = (5, 0).

1kg at (0,0) and 3kg at (8,0)

x = (1×0+3×8)/(1+3) = 24/4 = 6. y = 0 throughout since both points sit on the x-axis. The center of mass at (6, 0) sits much closer to the heavier 3kg mass than to the lighter 1kg one.

Three masses: 2kg at (0,0), 3kg at (4,0), 5kg at (0,6)

Total mass = 2+3+5 = 10. x = (2×0+3×4+5×0)/10 = 12/10 = 1.2. y = (2×0+3×0+5×6)/10 = 30/10 = 3. The center of mass sits at (1.2, 3), pulled most strongly toward the 5kg mass.

Common mistakes to avoid

Entering a total mass of zero, which makes the formula undefined due to division by zero.
Entering a negative mass, which has no physical meaning for this calculation and distorts the weighted average unpredictably.
Confusing center of mass (weighted by mass) with a simple geometric centroid (an unweighted average of positions) — they only match when all masses are equal.
Mixing up x and y values between different points when entering several masses, which silently shifts the whole result.

Frequently asked questions

Center of mass is weighted by each point's mass, so heavier points pull it toward themselves. The Centroid Calculator instead finds a polygon's unweighted, area-based geometric center, treating every part of the shape's interior equally regardless of any "weight."

When every mass is identical, the mass terms cancel out of the formula entirely, and the center of mass reduces to the simple, unweighted average of the points' positions — equivalent to their geometric centroid.

No — the center of mass is always a weighted average of the given points' coordinates, so it must lie within the convex hull (the smallest enclosing shape) of those points; it can never fall strictly outside all of them.

The formula divides by the sum of all masses (Σmᵢ). If that sum is zero, the division is undefined — physically, a system with zero total mass has no meaningful "weighted average position" to compute.

Yes — combining masses at the same location is equivalent to adding their masses together as a single heavier point there, which shifts the weighted average further toward that shared position.

In a uniform gravitational field (which is an extremely good approximation for objects near Earth's surface), yes, they coincide exactly. They can differ only in unusual situations involving extremely large objects or non-uniform gravitational fields.