2.6 Moments and Centers of Mass

Moments and Centers of Mass

The center of mass is the single point where you could balance an object perfectly. Calculating it requires integrating mass distributions across a shape, and it connects integration techniques you've already learned to real physical problems like balance, rotation, and structural design.

In Calc II, you'll mostly work with one-dimensional distributions (rods, wires) and two-dimensional regions (thin plates). Symmetry often cuts the work in half, and Pappus's theorem gives you a clever shortcut for volumes of revolution.

Center of Mass for Linear Distributions

The center of mass is the point where an object's total mass can be thought of as concentrated. For a system of discrete particles along a line, you find it with a weighted average:

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

Here $m_i$ is the mass of particle $i$ and $x_i$ is its position. The numerator is called the first moment of the system. The denominator is the total mass.

For continuous objects like a rod or wire, mass isn't concentrated at points. Instead, you describe it with a linear density function $\lambda(x)$, which gives mass per unit length at position $x$. A rod with varying thickness, for instance, has higher $\lambda(x)$ where it's thicker.

The continuous version of the center of mass formula is:

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

• The denominator $\int_a^b \lambda(x) \, dx$ gives the total mass $M$.
• The numerator $\int_a^b x \, \lambda(x) \, dx$ gives the moment about the origin $M_0$, which measures how the mass is distributed relative to $x = 0$.

Example: Suppose a rod extends from $x = 0$ to $x = 4$ with linear density $\lambda(x) = 3x$. The total mass is $\int_0^4 3x \, dx = 24$. The moment is $\int_0^4 x \cdot 3x \, dx = \int_0^4 3x^2 \, dx = 64$. So $\bar{x} = \frac{64}{24} = \frac{8}{3} \approx 2.67$. 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

A thin plate (or lamina) is a flat, two-dimensional object with negligible thickness. Its mass distribution is described by a surface density (or area density) $\sigma(x, y)$, which gives mass per unit area.

The center of mass $(\bar{x}, \bar{y})$ for a plate occupying region $R$ is:

$\bar{x} = \frac{\iint_R x \, \sigma(x, y) \, dA}{\iint_R \sigma(x, y) \, dA}, \qquad \bar{y} = \frac{\iint_R y \, \sigma(x, y) \, dA}{\iint_R \sigma(x, y) \, dA}$

When the plate has constant (uniform) density, $\sigma$ cancels out of the fraction, and the formulas simplify to:

$\bar{x} = \frac{\iint_R x \, dA}{\iint_R dA}, \qquad \bar{y} = \frac{\iint_R y \, dA}{\iint_R dA}$

The denominator is just the area of $R$. In this uniform case, the center of mass is called the centroid since it depends only on geometry, not mass.

In many Calc II problems, you won't set up full double integrals. Instead, for a region between curves $f(x)$ and $g(x)$ (where $f(x) \geq g(x)$) from $x = a$ to $x = b$ with uniform density, you can use:

• Total area: $A = \int_a^b [f(x) - g(x)] \, dx$
• Moment about the y-axis: $M_y = \int_a^b x[f(x) - g(x)] \, dx$
• Moment about the x-axis: $M_x = \int_a^b \frac{1}{2}[f(x)^2 - g(x)^2] \, dx$

Then $\bar{x} = M_y / A$ and $\bar{y} = M_x / A$. The $M_x$ formula comes from the fact that the average height of a thin vertical strip between the two curves is $\frac{1}{2}[f(x) + g(x)]$.

Symmetry in Centroid Location

Symmetry is your best friend for these problems because it can eliminate half the computation.

• If a region is symmetric about the y-axis, then $\bar{x} = 0$.
• If a region is symmetric about the x-axis, then $\bar{y} = 0$.
• If a region is symmetric about both axes (or the origin), both coordinates are 0.

How to use symmetry in practice:

1. Sketch the region and look for lines of symmetry.
2. For each axis of symmetry, set the corresponding centroid coordinate to the value on that axis. (If symmetric about $x = 0$, then $\bar{x} = 0$.)
3. Only compute the integral for the remaining coordinate(s).

For example, the region between $y = x^2$ and $y = 1$ is symmetric about the y-axis, so $\bar{x} = 0$ immediately. You only need to calculate $\bar{y}$.

Pappus's Theorem for Revolution Solids

Pappus's theorem provides an elegant connection between centroids and volumes of revolution: the volume of a solid formed by rotating a plane region about an external axis equals the area of the region times the distance its centroid travels.

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

Here $\bar{d}$ is the distance from the centroid to the axis of rotation, and $A$ is the area of the region. The quantity $2\pi \bar{d}$ is the circumference of the circle traced by the centroid.

Steps to apply Pappus's theorem:

1. Identify the plane region and the axis of rotation. (The axis must not pass through the region.)
2. Find the area $A$ of the region using geometry or integration.
3. Find the centroid of the region, then determine $\bar{d}$, the perpendicular distance from the centroid to the axis of rotation.
4. Compute $V = 2\pi \bar{d} \cdot A$.

Example: A semicircular region of radius $r$ has area $A = \frac{1}{2}\pi r^2$ and its centroid sits at distance $\bar{d} = \frac{4r}{3\pi}$ from the diameter. Rotating about the diameter gives:

$V = 2\pi \cdot \frac{4r}{3\pi} \cdot \frac{1}{2}\pi r^2 = \frac{4}{3}\pi r^3$

That's the volume of a sphere, which confirms the theorem works.

Pappus's theorem also works in reverse: if you know the volume of a solid of revolution and the area of the generating region, you can solve for the centroid location.

Moments of Inertia (Second Moments)

While the center of mass uses first moments ($\int x \, dm$), moments of inertia use second moments ($\int x^2 \, dm$). The moment of inertia measures how mass is distributed relative to an axis and determines an object's resistance to rotational acceleration.

• Moment of inertia about the y-axis: $I_y = \int_a^b x^2 \, \lambda(x) \, dx$ (for a linear distribution)
• Moment of inertia about the x-axis: $I_x = \iint_R y^2 \, \sigma(x,y) \, dA$ (for a thin plate)

Moment of inertia is to rotation what mass is to linear motion. A larger moment of inertia means the object is harder to spin. This concept connects the integration techniques from this unit to rotational dynamics in physics, where torque, angular momentum, and rotational equilibrium all depend on how mass is distributed relative to an axis.