Study Guide: Moments and Centers of Mass — Calculus II: Integral Calculus - Integration, Series, and Parametric Equations Study Notes | BrainyBee

Learning Objectives

Calculate the moment and center of mass for a system of discrete point masses on a 1D line.
Understand how to calculate moments with respect to the $x$ - and $y$ -axes for point masses in a 2D plane.
Set up and evaluate integrals to find the mass, moments, and centroid of a continuous planar lamina bounded by a single function and the $x$ -axis.
Compute the coordinates of the center of mass for a planar region bounded by two continuous functions.

Key Terms & Glossary

Point Mass: An idealized mathematical object that has a specified mass but no physical dimensions or volume.
Moment: A quantitative measure of the tendency of a mass to cause rotation about a specific axis or fulcrum.
Center of Mass: The unique point in a system where the weighted relative position of the distributed mass sums to zero; the perfect "balancing point."
Lamina: A very thin, flat continuous plate of material of uniform thickness.
Centroid: The geometric center of a 2D lamina, which aligns perfectly with the center of mass when the lamina's density is strictly uniform.
Density ( $\\rho$ ): The mass per unit area of a 2D lamina, assumed constant when purely finding a geometric centroid.

The "Big Idea"

The foundational concept of a center of mass starts with the physical intuition of balancing a seesaw. If you place different weights (point masses) at different distances along a rigid board, the board will only balance at a specific fulcrum point where the rotational tendencies (the moments) perfectly cancel each other out.

Calculus scales this idea from a discrete collection of weights to a continuous object. By slicing a 2D shape (a lamina) into an infinite number of infinitesimally thin vertical rectangles, we can treat each rectangle as a tiny point mass. Integrating the moments of all these "point masses" allows us to find the exact balancing coordinates $(\\bar{x}, \\bar{y})$ of any continuous shape, no matter how complex the bounding curves are.

Formula / Concept Box

Concept	Mass Equation ($m)	Moment Equations (M_x $,$ M_y $,$ M_0$)	Center of Mass Coordinate(s)
1D Point Masses	$m = \\sum_{i=1}^n m_i$	$M_0 = \\sum_{i=1}^n m_i x_i$	$\\bar{x} = \\frac{M_0}{m}$
2D Point Masses	$m = \\sum_{i=1}^n m_i$	$M_x = \\sum_{i=1}^n m_i y_i$ , $M_y = \\sum_{i=1}^n m_i x_i$	$\\bar{x} = \\frac{M_y}{m}, \\bar{y} = \\frac{M_x}{m}$
Lamina (1 Curve)	$m = \\rho \\int_a^b f(x) \\, dx$	$M_x = \\rho \\int_a^b \\frac{1}{2}[f(x)]^2 \\, dx$ , $M_y = \\rho \\int_a^b x f(x) \\, dx$	$\\bar{x} = \\frac{M_y}{m}, \\bar{y} = \\frac{M_x}{m}$
Lamina (2 Curves)	$m = \\rho \\int_a^b [f(x) - g(x)] \\, dx$	$M_x = \\rho \\int_a^b \\frac{1}{2}([f(x)]^2 - [g(x)]^2) \\, dx$ $M_y = \\rho \\int_a^b x[f(x) - g(x)] \\, dx$	$\\bar{x} = \\frac{M_y}{m}, \\bar{y} = \\frac{M_x}{m}$

[!IMPORTANT] When the question strictly asks for the centroid of a geometric region rather than the center of mass of a physical plate, the constant uniform density $\\rho will eventually cancel out in the division \\frac{M}{m}$ .

Hierarchical Outline

1. Systems of Discrete Point Masses
- 1.1. On a 1D Number Line
  - Sum of individual masses yields total mass ( $m$ ).
  - Sum of (mass $\\times coordinate) yields Moment w.r.t the origin (M_0$ ).
- 1.2. In a 2D Coordinate Plane
  - Uses $x-coordinates to find distance from y$ -axis ( $M_y$ ).
  - Uses $y-coordinates to find distance from x$ -axis ( $M_x$ ).
2. Center of Mass of a Continuous Lamina
- 2.1. Region Bounded by One Curve ( $f(x)$ )
  - Uses representative vertical rectangles.
  - Height is $f(x), centroid of the rectangle is at (x, \\frac{f(x)}{2})$ .
- 2.2. Region Bounded by Two Curves ( $f(x)$ and $g(x)$ )
  - Region is between a top curve $f(x) and a bottom curve g(x)$ .
  - The rectangle height becomes $f(x) - g(x)$ .
  - The rectangle's $y-centroid shifts to the average height: \\frac{f(x) + g(x)}{2}$ .

Visual Anchors

1. The Geometry of a 2D Lamina Centroid

Here is a geometric visualization of a lamina bounded by two functions, $f(x)$ on top and $g(x)$ on the bottom.

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2. Workflow for Calculating a Centroid

The mental algorithm for calculating the center of mass using integration.

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Definition-Example Pairs

Moment with respect to the $y$ -axis ( $M_y$ )
- Definition: A measurement of how heavily a shape or mass distribution pulls horizontally, computed by multiplying the mass by its distance from the vertical axis ( $x$ -coordinate).
- Real-World Example: When opening a heavy door, pushing closer to the handle (a large $x-distance from the hinges on the y-axis) creates a larger moment M_y$ , making it easier to rotate.
Centroid
- Definition: The arithmetic mean position of all the points in a 2D shape.
- Real-World Example: If you cut an arbitrary shape out of a piece of stiff cardboard (a uniform lamina), the centroid is the exact spot where you could balance the cardboard horizontally on the tip of a pencil.

Worked Examples

▶Example 1: Finding the Center of Mass along a 1D Line

Problem: Suppose four point masses are placed on a number line. Let the total mass of the system be $m = 12 kg, and the total moment of the system with respect to the origin be M_0 = -6$ kg $\cdot$ m. Find the center of mass.

Step-by-Step Breakdown:

Identify Given Values:
- Total mass, $m = 12$
- Total moment, $M_0 = -6$
Apply the Center of Mass Formula:
- The center of mass $\\bar{x}$ is the moment divided by the total mass.
- $\\bar{x} = \\frac{M_0}{m}$
Calculate:
- $\\bar{x} = \\frac{-6}{12} = -\\frac{1}{2}$

Conclusion: The center of mass is located $1/2$ m to the left of the origin.

▶Example 2: Centroid of a Region Bounded by Two Functions

Problem: Find the centroid of the region bounded above by $f(x) = x$ and below by $g(x) = x^2$ over the interval $[0, 1]$ . Assume density $\$ \rho = 1$$ for simplicity.

Step-by-Step Breakdown:

Find the Mass ( $m$ ):
- $m = \\int_0^1 (x - x^2) \\, dx$
- $m = \\left[ \\frac{x^2}{2} - \\frac{x^3}{3} \\right]_0^1 = \\frac{1}{2} - \\frac{1}{3} = \\frac{1}{6}$
Find the Moment w.r.t the $y$ -axis ( $M_y$ ):
- $M_y = \\int_0^1 x(x - x^2) \\, dx = \\int_0^1 (x^2 - x^3) \\, dx$
- $M_y = \\left[ \\frac{x^3}{3} - \\frac{x^4}{4} \\right]_0^1 = \\frac{1}{3} - \\frac{1}{4} = \\frac{1}{12}$
Find the Moment w.r.t the $x$ -axis ( $M_x$ ):
- $M_x = \\frac{1}{2} \\int_0^1 ([x]^2 - [x^2]^2) \\, dx = \\frac{1}{2} \\int_0^1 (x^2 - x^4) \\, dx$
- $M_x = \\frac{1}{2} \\left[ \\frac{x^3}{3} - \\frac{x^5}{5} \\right]_0^1 = \\frac{1}{2} \\left( \\frac{1}{3} - \\frac{1}{5} \\right) = \\frac{1}{2} \\left( \\frac{2}{15} \\right) = \\frac{1}{15}$
Calculate Coordinates:
- $\\bar{x} = \\frac{M_y}{m} = \\frac{1/12}{1/6} = \\frac{1}{2}$
- $\\bar{y} = \\frac{M_x}{m} = \\frac{1/15}{1/6} = \\frac{6}{15} = \\frac{2}{5}$

Conclusion: The centroid of the region is $(\$ \frac{1}{2} $, \$ \frac{2}{5} $)$ .

Checkpoint Questions

Why do we use the $x-coordinate to calculate the moment with respect to the y$ -axis ( $M_y$ )?

Self-Check Hint: Remember the definition of a moment. A moment is mass times the perpendicular distance to the axis of rotation. The distance from any point $(x,y)$ to the $y-axis is horizontally measured by its x$ -coordinate.
If you are tasked with finding the geometric centroid of a lamina, what happens to the density variable $\$ \rho$$ in your calculations?

Self-Check Hint: Since a centroid assumes uniform density, $\$ \rho$$ is a constant. It appears in the numerator (moments) and the denominator (mass), ultimately canceling out completely.
When setting up the integral for $M_x for a region bounded by two curves, why is the integrand \$ \frac{1}{2} $( [f(x)]^2 - [g(x)]^2 )$ instead of $\$ \frac{1}{2} $[f(x) - g(x)]^2$ ?

Self-Check Hint: We subtract the moment of the empty space below the region from the moment of the region extending all the way down to the $x$ -axis. $(a^2 - b^2)$ is the difference of the squares, not the square of the difference.