6.6 Moments and Centers of Mass

Big idea

A flat plate balances on a single point. Which point? Answering that needs a way of measuring not just how much mass there is but where it sits, and the measurement is called a moment. A moment is mass times a signed distance from a reference line, so it grows when mass moves away from the line and changes sign when mass crosses to the other side. The balance point is where the moments cancel.

For finitely many point masses on a line the answer is arithmetic: add up mass times position, divide by total mass, and you have a weighted average of the positions. A continuous plate has no sum to write, so the same three-step move as the rest of this chapter takes over. Slice the plate into thin vertical strips. One strip is nearly a rectangle, whose mass is density times area and whose mass behaves, by symmetry, as if concentrated at its midpoint. Add the strips’ moments and let them thin out, and the sums become integrals.

The step that surprises people is the vertical moment. A strip at position $x$ running from $y = g(x)$ up to $y = f(x)$ has its centre at height $\tfrac{1}{2}[f(x) + g(x)]$, so its moment about the horizontal axis is mass $\rho[f(x) - g(x)]\,dx$ times that height, which multiplies out to $\tfrac{\rho}{2}\big[f(x)^2 - g(x)^2\big]dx$. That is why a difference of squares appears in one coordinate and a plain difference in the other; it is not a separate rule.

When the density is constant it cancels out of every ratio and the balance point depends only on the shape. That geometric point is the centroid. Two things make centroid problems short: symmetry, which locates one coordinate for free, and the fact that all of this is built from the same area integrals used for the area between two curves.

Decoder

The centre of mass is the point at which the whole mass could be concentrated without changing any moment, and each of its coordinates is the corresponding moment divided by the total mass.

Read the definition backwards and it becomes a recipe. Compute the total mass, compute the moment about each axis, divide. The moment about the $y$-axis gives the horizontal coordinate $\bar{x}$, and the moment about the $x$-axis gives $\bar{y}$. That crossing of names is a persistent source of confusion: distance from the $y$-axis is measured in the $x$ direction.

The words doing the work are “without changing any moment.” That requirement is what makes the centre of mass a physical point rather than an arbitrary average. It is also why moments are the primary object: moments add when you combine two objects, so a composite shape is handled by adding moments and masses separately, then dividing once at the end.

The standard mistake is averaging the wrong thing, computing $\bar{y}$ as the average value of $f(x)$ or reading it off a sketch. The correct $\bar{y}$ weights each height by how much mass sits at it, which is what the difference of squares encodes.

Definitions and results

Moment of a point mass. A mass $m$ at position $x$ on a line has moment $mx$ about the origin. Positions on opposite sides contribute opposite signs, so moments cancel.

Centre of mass of a system on a line. For masses $m_i$ at positions $x_i$, with total moment $M = \sum m_i x_i$ and total mass $m = \sum m_i$,

$$ \bar{x} = \frac{M}{m} = \frac{\sum m_i x_i}{\sum m_i} $$

It always lies between the smallest and largest position, and it moves toward the heavier masses.

Moments in the plane. For masses $m_i$ at points $(x_i, y_i)$, the moment about the $y$-axis is $M_y = \sum m_i x_i$ and the moment about the $x$-axis is $M_x = \sum m_i y_i$. The centre of mass is $\left( M_y / m,\; M_x / m \right)$.

A thin plate of constant density. Let a plate occupy the region between $y = f(x)$ and $y = g(x)$ on $[a, b]$, with $f \ge g$ and constant density $\rho$. Its mass is $\rho A$, where $A = \int_a^b [f(x) - g(x)]\,dx$, and

$$ M_y = \rho \int_a^b x \big[ f(x) - g(x) \big]\, dx, \qquad M_x = \frac{\rho}{2} \int_a^b \Big( \big[f(x)\big]^2 - \big[g(x)\big]^2 \Big) dx $$

The centroid. Dividing the moments by the mass cancels $\rho$ entirely:

$$ \bar{x} = \frac{1}{A} \int_a^b x \big[ f(x) - g(x) \big] dx, \qquad \bar{y} = \frac{1}{2A} \int_a^b \Big( \big[f(x)\big]^2 - \big[g(x)\big]^2 \Big) dx $$

These depend only on the shape, so the point is called the centroid rather than the centre of mass.

The symmetry principle. If a region is symmetric about a line, its centroid lies on that line. A region symmetric about the $y$-axis has $\bar{x} = 0$; one symmetric about $y = c$ has $\bar{y} = c$. Checking for symmetry first often removes half the work, and a region with two lines of symmetry has its centroid at their intersection with no integration at all.

Additivity. Moments and masses of disjoint pieces add, so a shape built from two rectangles is handled by computing each piece’s mass and moment, adding, and dividing once.

The theorem of Pappus for volume. If a plane region of area $A$ is rotated about a line in its plane that does not cut through it, the solid has volume $V = 2\pi d A$, where $d$ is the distance from the axis to the centroid. It gives a fast independent check on a volume computed by shells or washers.

Worked examples

Balancing three masses on a line

Masses of $2$, $3$ and $5$ kilograms sit at positions $-1$, $2$ and $4$ on a line. Find the centre of mass.

The total mass is $2 + 3 + 5 = 10$ kilograms and the total moment about the origin is $2(-1) + 3(2) + 5(4) = 24$, so $\bar{x} = \tfrac{24}{10} = 2.4$.

The answer should sit toward the heavy end, and it does: half the mass is at $x = 4$, and the balance point lands between $2$ and $4$ rather than at the plain average of the positions, which is $\tfrac{5}{3}$.

Centroid of a region under a parabola

Find the centroid of the region bounded by $y = x^2$, $y = 0$ and $x = 2$.

The area is $A = \int_0^2 x^2\,dx = \tfrac{8}{3}$. The upper curve is $f(x) = x^2$ and the lower is $g(x) = 0$, so

$$ \bar{x} = \frac{1}{A}\int_0^2 x \cdot x^2\, dx = \frac{3}{8} \left[ \frac{x^4}{4} \right]_0^2 = \frac{3}{8} \cdot 4 = \frac{3}{2} $$

$$ \bar{y} = \frac{1}{2A} \int_0^2 (x^2)^2\, dx = \frac{3}{16} \left[ \frac{x^5}{5} \right]_0^2 = \frac{3}{16} \cdot \frac{32}{5} = \frac{6}{5} $$

The centroid is $\left( \tfrac{3}{2}, \tfrac{6}{5} \right)$. Both coordinates pass an eyeball test: the region’s area is concentrated to the right, so $\bar{x}$ should exceed the midpoint $1$, and $\bar{y}$ should sit well below half the maximum height $4$.

Check with the theorem of Pappus. Rotating this region about the $y$-axis gives, by shells, $V = 2\pi\int_0^2 x \cdot x^2 dx = 8\pi$, and Pappus predicts $V = 2\pi \bar{x} A = 2\pi \cdot \tfrac{3}{2} \cdot \tfrac{8}{3} = 8\pi$.

Using symmetry

Find the centroid of the region between $y = 4 - x^2$ and the horizontal axis.

The curve meets the axis at $x = \pm 2$, and the region is unchanged by replacing $x$ with $-x$, so the symmetry principle gives $\bar{x} = 0$ with no integration.

For the area, $A = \int_{-2}^{2}(4 - x^2)dx = 2\left[4x - \tfrac{x^3}{3}\right]_0^2 = 2\left(8 - \tfrac{8}{3}\right) = \tfrac{32}{3}$. For the vertical coordinate,

$$ \bar{y} = \frac{1}{2A}\int_{-2}^{2} (4 - x^2)^2\, dx = \frac{3}{64} \cdot 2\int_0^2 (16 - 8x^2 + x^4)\, dx $$

using evenness. The inner integral is $\left[16x - \tfrac{8x^3}{3} + \tfrac{x^5}{5}\right]_0^2 = \tfrac{480 - 320 + 96}{15} = \tfrac{256}{15}$, so

$$ \bar{y} = \frac{3}{64} \cdot \frac{512}{15} = \frac{8}{5} $$

The centroid is $\left(0, \tfrac{8}{5}\right)$, which is below the halfway height $2$, as it must be for a region that is widest at the bottom.

A region between two curves

Find the centroid of the region enclosed by $y = x$ and $y = x^2$.

The curves meet where $x = x^2$, at $x = 0$ and $x = 1$, and the line is above the parabola in between. The area is

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

$$ \bar{x} = 6\int_0^1 x(x - x^2)\, dx = 6\left( \frac{1}{3} - \frac{1}{4} \right) = 6 \cdot \frac{1}{12} = \frac{1}{2} $$

$$ \bar{y} = \frac{6}{2}\int_0^1 \big( x^2 - x^4 \big)\, dx = 3\left( \frac{1}{3} - \frac{1}{5} \right) = 3 \cdot \frac{2}{15} = \frac{2}{5} $$

The centroid is $\left( \tfrac{1}{2}, \tfrac{2}{5} \right)$, and it lies inside the region: at $x = \tfrac{1}{2}$ the region runs from $y = \tfrac{1}{4}$ to $y = \tfrac{1}{2}$, and $\tfrac{2}{5}$ falls between those. Containment is not automatic for every shape, but for a sliver like this one it is a genuine test of the arithmetic.

Practice

Begin with the integrals themselves. Every moment computation ends in a definite integral of a polynomial or simple algebraic expression, and the difference of squares makes those integrands slightly heavier than usual.

Then the regions. Locating a centroid starts with exactly the setup used for the area between two curves: crossings, which curve is on top, and the limits. Get that right and the moment integrals write themselves.

Quiz

Five items on moments of point masses, centroids of plane regions, and the symmetry shortcut.