6.6 Moments and Centers of Mass

Moments and Centers of Mass

Moments and centers of mass let you figure out where an object's mass is effectively concentrated. This matters for predicting how objects balance, rotate, and respond to forces. In Calculus I, you'll use integrals to find these values for continuous distributions along a line and across thin plates, and you'll learn shortcuts using symmetry and Pappus' theorem.

Moments and Centers of Mass

Center of mass for linear distributions

The center of mass is the single point where you could concentrate all of an object's mass and get the same balancing behavior. Think of it as the "balance point."

For a system of discrete particles (like beads on a wire), the center of mass is a weighted average of positions:

$\bar{x} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}$

where $m_i$ is the mass of the $i$-th particle and $x_i$ is its position. The numerator $\sum m_i x_i$ is called the first moment of the system about the origin.

For a continuous object like a thin rod with variable density, you replace the sum with an integral. If the rod lies along the $x$-axis from $x = a$ to $x = b$ and has linear mass density $\rho(x)$ (mass per unit length), the center of mass is:

$\bar{x} = \frac{\int_a^b x \, \rho(x) \, dx}{\int_a^b \rho(x) \, dx}$

The denominator gives the total mass $M$, and the numerator gives the first moment about the origin.

Example walkthrough: Suppose a rod extends from $x = 0$ to $x = 2$ with density $\rho(x) = 3x$ kg/m.

1. Find total mass: $M = \int_0^2 3x \, dx = \frac{3x^2}{2}\Big|_0^2 = 6$ kg
2. Find the first moment: $\int_0^2 x \cdot 3x \, dx = \int_0^2 3x^2 \, dx = x^3\Big|_0^2 = 8$
3. Compute center of mass: $\bar{x} = \frac{8}{6} = \frac{4}{3} \approx 1.33$ m

The center of mass is shifted toward the heavier end, which makes sense since the density increases with $x$.

Center of mass for thin plates

For a flat, two-dimensional region (a thin plate or lamina), you need two coordinates for the center of mass: $\bar{x}$ and $\bar{y}$.

If the plate has variable density $\rho(x, y)$, the formulas involve double integrals:

$\bar{x} = \frac{\iint x \, \rho(x, y) \, dA}{\iint \rho(x, y) \, dA} \qquad \bar{y} = \frac{\iint y \, \rho(x, y) \, dA}{\iint \rho(x, y) \, dA}$

In Calculus I, you'll most often work with plates of uniform (constant) density. When density is constant, it cancels from numerator and denominator, and the center of mass depends only on geometry. In this case it's called the centroid, and the formulas simplify.

For a region bounded by $y = f(x)$ (top) and $y = g(x)$ (bottom) from $x = a$ to $x = b$, with uniform density:

• Total area: $A = \int_a^b [f(x) - g(x)] \, dx$
• $\bar{x} = \frac{1}{A}\int_a^b x[f(x) - g(x)] \, dx$
• $\bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}[f(x)^2 - g(x)^2] \, dx$

The $\bar{y}$ formula comes from the fact that the average height of a vertical strip is the midpoint between $f(x)$ and $g(x)$.

To set these up:

1. Sketch the region and identify the bounding curves
2. Determine the limits of integration (where the curves intersect)
3. Compute the area $A$
4. Evaluate the integrals for $\bar{x}$ and $\bar{y}$

Symmetry in centroid calculations

The centroid is the geometric center of a shape. For uniform-density objects, it's the same as the center of mass. Symmetry can save you a lot of work.

The key principle: if a region is symmetric about a line, the centroid lies on that line. This means:

• If a plate is symmetric about the y-axis, then $\bar{x} = 0$
• If a plate is symmetric about the x-axis, then $\bar{y} = 0$
• If a plate is symmetric about both axes, the centroid is at $(0, 0)$

For example, a semicircular plate of radius $r$ sitting on top of the $x$-axis is symmetric about the $y$-axis, so you know $\bar{x} = 0$ immediately. You only need to compute $\bar{y}$, which cuts your work in half.

Always check for symmetry before setting up integrals. Even partial symmetry (about just one axis) eliminates one of the two coordinates.

Pappus' theorem for solids of revolution

Pappus' theorem connects centroids to volumes of revolution. It says: when a plane region is revolved around an external axis, the volume of the resulting solid equals the area of the region times the distance its centroid travels.

$V = 2\pi \bar{d} \cdot A$

where $A$ is the area of the region and $\bar{d}$ is the distance from the centroid to the axis of revolution. (If revolving around the $x$-axis, $\bar{d} = \bar{y}$; if around the $y$-axis, $\bar{d} = \bar{x}$.)

Classic example: A circle of radius $r$ with center at distance $R$ from an axis (where $R > r$) is revolved around that axis to form a torus (donut shape). The centroid of the circle is at its center, so $\bar{d} = R$. The area is $\pi r^2$. By Pappus' theorem:

$V = 2\pi R \cdot \pi r^2 = 2\pi^2 R r^2$

Try getting that result with the disk or shell method and you'll appreciate how much simpler Pappus' theorem can be.

To apply Pappus' theorem:

1. Find the area $A$ of the generating region
2. Find the centroid of the region ($\bar{x}$ and $\bar{y}$)
3. Determine which coordinate gives the distance from the centroid to the axis of revolution
4. Compute $V = 2\pi \bar{d} \cdot A$

This theorem also works in reverse: if you already know the volume (say, from the disk method), you can solve for the centroid.

Moments of Inertia

This topic goes well beyond Calculus I and is covered in depth in physics and engineering courses. For now, just be aware of the basic idea:

• The moment of inertia measures how much a body resists rotational acceleration, analogous to how mass resists linear acceleration
• It depends not just on total mass but on how that mass is distributed relative to the axis of rotation. Mass farther from the axis contributes more to the moment of inertia

If your Calculus I course covers this, the key formula for a continuous distribution is $I = \int r^2 \, dm$, where $r$ is the distance from each mass element to the axis of rotation. Your instructor will clarify how deep you need to go.