Center of Mass

The center of mass is the balance point of an object. For a flat region of uniform density between y = f(x) on top and y = g(x) below, it is the centroid (x̄, ȳ): \bar{x} = \dfrac{1}{A}\int_a^b x\big(f(x) - g(x)\big)\,dx and \bar{y} = \dfrac{1}{A}\int_a^b \dfrac{1}{2}\big(f(x)^2 - g(x)^2\big)\,dx, where A is the region’s area.

Before this lesson: Area Between Curves and Definite Integrals

Balancing a flat plate

Two kids on a seesaw balance when each one’s weight times distance from the pivot is the same. That product, mass times distance, is called a moment. The center of mass is the point where the moments on every side cancel.

For a flat plate, cut it into thin vertical strips. Each strip’s mass sits at its middle. Adding the strips’ moments with integrals, then dividing by the total area, gives the balance point. When density is the same everywhere, the center of mass is called the centroid, and it depends only on the shape.

Center of mass formulas

For the region between y = f(x) (top) and y = g(x) (bottom) on [a, b]:

A = \int_a^b \big(f - g\big)\,dx, \qquad M_y = \int_a^b x\big(f - g\big)\,dx, \qquad M_x = \int_a^b \dfrac{1}{2}\big(f^2 - g^2\big)\,dx

M_y is the moment about the y-axis, so it gives x̄. M_x is the moment about the x-axis, so it gives ȳ. The 1/2 comes from each strip’s midpoint height, (f + g)/2, times its height f − g.

Center of mass examples

Example 1: Masses on a line

Masses of 3 kg, 5 kg and 2 kg sit at x = 1, x = 4 and x = −2. Find the center of mass.

\sum m_i = 3 + 5 + 2 = 10

\sum m_i x_i = 3(1) + 5(4) + 2(-2) = 3 + 20 - 4 = 19

\bar{x} = \dfrac{19}{10} = 1.9

Answer: x̄ = 1.9, closer to the heavy 5 kg mass.

Example 2: Masses in a plane

Masses of 2, 3 and 5 kg sit at (1, 1), (4, −1) and (−2, 3). Find the center of mass.

\sum m_i = 10

M_y = 2(1) + 3(4) + 5(-2) = 4

M_x = 2(1) + 3(-1) + 5(3) = 14

\bar{x} = \dfrac{4}{10} = 0.4, \; \bar{y} = \dfrac{14}{10} = 1.4

Answer: (0.4, 1.4).

Example 3: The region under a parabola

Find the centroid of the region under y = x² from x = 0 to x = 2.

A = \int_0^2 x^2\,dx = \dfrac{8}{3}

M_y = \int_0^2 x\cdot x^2\,dx = \Big[\dfrac{x^4}{4}\Big]_0^2 = 4

\bar{x} = \dfrac{4}{8/3} = \dfrac{3}{2}

M_x = \int_0^2 \dfrac{1}{2}(x^2)^2\,dx = \dfrac{1}{2}\cdot\dfrac{32}{5} = \dfrac{16}{5}

\bar{y} = \dfrac{16/5}{8/3} = \dfrac{16}{5}\cdot\dfrac{3}{8} = \dfrac{6}{5}

Answer: the centroid is (1.5, 1.2).

Example 4: Between two curves

Find the centroid of the region between y = x (top) and y = x² (bottom).

x = x^2 \Rightarrow x = 0, 1

A = \int_0^1 (x - x^2)\,dx = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}

M_y = \int_0^1 x(x - x^2)\,dx = \dfrac{1}{3} - \dfrac{1}{4} = \dfrac{1}{12}

\bar{x} = \dfrac{1/12}{1/6} = \dfrac{1}{2}

M_x = \int_0^1 \dfrac{1}{2}(x^2 - x^4)\,dx = \dfrac{1}{2}\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{1}{15}

\bar{y} = \dfrac{1/15}{1/6} = \dfrac{6}{15} = \dfrac{2}{5}

Answer: the centroid is (1/2, 2/5) = (0.5, 0.4).

Example 5: Balancing a half-disk sign

A metal sign is a half-disk of radius 3 ft, with its straight edge along the bottom. Where is its balance point?

\bar{x} = 0

A = \dfrac{1}{2}\pi(3)^2 = \dfrac{9\pi}{2}

M_x = \int_{-3}^{3} \dfrac{1}{2}\left(\sqrt{9 - x^2}\right)^2 dx = \dfrac{1}{2}\int_{-3}^{3} (9 - x^2)\,dx

= \dfrac{1}{2}\Big[9x - \dfrac{x^3}{3}\Big]_{-3}^{3} = \dfrac{1}{2}(18 - (-18)) = 18

\bar{y} = \dfrac{18}{9\pi/2} = \dfrac{36}{9\pi} = \dfrac{4}{\pi} \approx 1.27

Answer: the balance point is on the center line, about 1.27 ft above the straight edge. In general, a half-disk’s centroid is 4r/(3π) above its diameter.

Common mistakes

Related methods

Pappus’s theorem

Spinning a region around an axis makes a solid whose volume is the area times the distance the centroid travels: V = A · 2πd. For Example 4 around the y-axis, V = (1/6) · 2π · (1/2) = π/6, the same answer as the shell method gives.

Centroids and fluid force

The force of water on an upright plate equals ρ times the centroid’s depth times the area. See hydrostatic force.

Rods with changing density

If a rod’s density is ρ(x), its mass is \int_a^b \rho(x)\,dx and its moment about 0 is \int_a^b x\rho(x)\,dx. Divide to get x̄. Practice problem 8 is an example.

Practice problems

1. Masses of 4 kg and 6 kg sit at x = 0 and x = 5. Find the center of mass.
2. Masses of 1, 2 and 3 kg sit at (0, 0), (3, 0) and (0, 4). Find the center of mass.
3. Use the formulas to find the centroid of the rectangle 0 ≤ x ≤ 4, 0 ≤ y ≤ 2.
4. Find the centroid of the triangle under y = x from x = 0 to x = 3.
5. Find the centroid of the region under y = √x from x = 0 to x = 4.
6. Find the centroid of the region under y = 4 − x² and above the x-axis.
7. Find the centroid of the region between y = 2x (top) and y = x² (bottom).
8. A rod from x = 0 to x = 4 m has density ρ(x) = 1 + x kg/m. Find its center of mass.
9. Find the centroid of the quarter-disk x² + y² ≤ 4 in the first quadrant.
10. Use Pappus’s theorem and Example 3 to find the volume when the region under y = x² from 0 to 2 spins around the x-axis.

\bar{x} = \dfrac{4(0) + 6(5)}{4 + 6} = \dfrac{30}{10}

= 3

\sum m_i = 6, \; M_y = 1(0) + 2(3) + 3(0) = 6, \; M_x = 1(0) + 2(0) + 3(4) = 12

\bar{x} = \dfrac{6}{6} = 1, \; \bar{y} = \dfrac{12}{6} = 2

A = 8, \; M_y = \int_0^4 2x\,dx = 16, \; M_x = \int_0^4 \dfrac{1}{2}(2)^2\,dx = 8

\bar{x} = \dfrac{16}{8} = 2, \; \bar{y} = \dfrac{8}{8} = 1

A = \dfrac{9}{2}, \; M_y = \int_0^3 x^2\,dx = 9, \; M_x = \int_0^3 \dfrac{x^2}{2}\,dx = \dfrac{9}{2}

\bar{x} = \dfrac{9}{9/2} = 2, \; \bar{y} = \dfrac{9/2}{9/2} = 1

A = \int_0^4 x^{1/2}\,dx = \dfrac{2}{3}(8) = \dfrac{16}{3}

M_y = \int_0^4 x^{3/2}\,dx = \dfrac{2}{5}(32) = \dfrac{64}{5}, \; \bar{x} = \dfrac{64}{5}\cdot\dfrac{3}{16} = \dfrac{12}{5}

M_x = \int_0^4 \dfrac{x}{2}\,dx = 4, \; \bar{y} = 4\cdot\dfrac{3}{16} = \dfrac{3}{4}

\bar{x} = 0, \; A = \int_{-2}^{2} (4 - x^2)\,dx = \dfrac{32}{3}

M_x = \dfrac{1}{2}\int_{-2}^{2} (16 - 8x^2 + x^4)\,dx = \dfrac{1}{2}\left(64 - \dfrac{128}{3} + \dfrac{64}{5}\right) = \dfrac{256}{15}

\bar{y} = \dfrac{256}{15}\cdot\dfrac{3}{32} = \dfrac{8}{5}

A = \int_0^2 (2x - x^2)\,dx = 4 - \dfrac{8}{3} = \dfrac{4}{3}

M_y = \int_0^2 (2x^2 - x^3)\,dx = \dfrac{16}{3} - 4 = \dfrac{4}{3}, \; \bar{x} = 1

M_x = \dfrac{1}{2}\int_0^2 (4x^2 - x^4)\,dx = \dfrac{1}{2}\left(\dfrac{32}{3} - \dfrac{32}{5}\right) = \dfrac{32}{15}, \; \bar{y} = \dfrac{32}{15}\cdot\dfrac{3}{4} = \dfrac{8}{5}

m = \int_0^4 (1 + x)\,dx = 4 + 8 = 12

\int_0^4 x(1 + x)\,dx = 8 + \dfrac{64}{3} = \dfrac{88}{3}

\bar{x} = \dfrac{88/3}{12} = \dfrac{22}{9} \approx 2.44

\bar{x} = \bar{y}, \; A = \dfrac{1}{4}\pi(2)^2 = \pi

M_x = \int_0^2 \dfrac{1}{2}(4 - x^2)\,dx = \dfrac{1}{2}\left(8 - \dfrac{8}{3}\right) = \dfrac{8}{3}

\bar{y} = \dfrac{8/3}{\pi} = \dfrac{8}{3\pi} \approx 0.849

V = A\cdot 2\pi\bar{y} = \dfrac{8}{3}\cdot 2\pi\cdot\dfrac{6}{5}

= \dfrac{32\pi}{5}

Center of mass FAQ

What is the center of mass?

It is the point where an object balances. For point masses, it is the weighted average of their positions, with each position weighted by its mass.

What is the difference between center of mass and centroid?

The centroid is the geometric center of a shape. The center of mass takes density into account. When the density is the same everywhere, they are the same point.

How do you find the centroid of a region between two curves?

Find the area A, then M_y = ∫ x(f − g) dx and M_x = ∫ ½(f² − g²) dx. The centroid is (M_y/A, M_x/A).

Can the center of mass be outside the object?

Yes. A ring or a boomerang balances at a point that is not on the object itself.

Summary

• Center of mass = total moment ÷ total mass.
• Point masses: x̄ = Σmᵢxᵢ / Σmᵢ, and the same for ȳ.
• Region between f (top) and g (bottom): x̄ = (1/A)∫ x(f − g) dx and ȳ = (1/A)∫ ½(f² − g²) dx.
• Use symmetry to skip integrals.
• Pappus: volume of revolution = area × 2π × (distance from the centroid to the axis).

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