Applications of Integration: Center of Mass, Mean Value, and Root Mean Square

Center of Mass

Center of Mass for Discrete Masses

• Consider a mechanical system with $n$ discrete masses located at positions along an axis.
• Index each mass with $i = 1$ to $n$.
• The $i^{th}$ body has mass $m<em>i$ and is located at position $x</em>i$ with respect to the origin.
• The location of the center of mass $\bar{x}$ is calculated as:
  $\bar{x} = \frac{\sum<em>{i=1}^{n} x</em>i m<em>i}{\sum</em>{i=1}^{n} m_i}$

Example: Discrete Masses

• Four masses are located at positions -2, -1, 1, and 2.
• The masses are 2, 1, 2, and 3, respectively.
• Calculate the center of mass:
  $\bar{x} = \frac{\sum<em>{i=1}^{4} x</em>i m<em>i}{\sum</em>{i=1}^{4} m<em>i}$$\bar{x} = \frac{x</em>1m<em>1 + x</em>2m<em>2 + x</em>3m<em>3 + x</em>4m<em>4}{m</em>1 + m<em>2 + m</em>3 + m_4}$
  $\bar{x} = \frac{(-2)(2) + (-1)(1) + (1)(2) + (2)(3)}{2 + 1 + 2 + 3}$
  $\bar{x} = \frac{-4 - 1 + 2 + 6}{8}$
  $\bar{x} = \frac{3}{8}$

Center of Mass for Continuous Systems

• For a one-dimensional object with linear density $\rho(x)$, where $\rho(x)$ gives the density at each point $x$ from the origin.
• Assume the density function is defined from $x = a$ to $x = b$.
• Divide the interval $[a, b]$ into $n$ equal subintervals of length $\Delta x = \frac{b - a}{n}$.
• Approximate the mass in each subinterval as $\rho(x_i) \Delta x$.
• Approximate the center of mass as:
  $\bar{x} \approx \frac{\sum<em>{i=1}^{n} x</em>i \rho(x<em>i) \Delta x}{\sum</em>{i=1}^{n} \rho(x_i) \Delta x}$
• The exact center of mass is found by taking the limit as $n \to \infty$:
  $\bar{x} = \frac{\int<em>{a}^{b} x \rho(x) dx}{\int</em>{a}^{b} \rho(x) dx}$

Example: Continuous System

• Find the center of mass of an object with linear density $\rho(x) = ax^2$, where a > 0 and $0 \le x \le L$.
• Apply the formula:
  $\bar{x} = \frac{\int<em>{0}^{L} x(ax^2) dx}{\int</em>{0}^{L} ax^2 dx}$
  $\bar{x} = \frac{a \int<em>{0}^{L} x^3 dx}{a \int</em>{0}^{L} x^2 dx}$
  $\bar{x} = \frac{\int<em>{0}^{L} x^3 dx}{\int</em>{0}^{L} x^2 dx}$
  $\bar{x} = \frac{\frac{x^4}{4} \Big|<em>0^L}{\frac{x^3}{3} \Big|</em>0^L}$
  $\bar{x} = \frac{\frac{L^4}{4}}{\frac{L^3}{3}}$
  $\bar{x} = \frac{3}{4}L$

Mean Value of a Function

• The mean value of a function $f(x)$ over an interval $[a, b]$ is needed to find a varying quantity.
• Approximate the mean value by subdividing the interval into $n$ equal parts and averaging the function values at each point $x<em>i$: $\frac{1}{n} \sum</em>{i=1}^{n} f(x_i)$
• Since $\Delta x = \frac{b - a}{n}$, we have $n = \frac{b - a}{\Delta x}$.
• The mean value can be approximated as:
  $\frac{1}{b-a} \sum<em>{i=1}^{n} f(x</em>i) \Delta x$
• The exact mean value $\bar{f}$ is:
  $\bar{f} = \lim<em>{n \to \infty} \frac{1}{b-a} \sum</em>{i=1}^{n} f(x<em>i) \Delta x = \frac{1}{b-a} \int</em>{a}^{b} f(x) dx$
• Geometrically, the area under the curve of $f(x)$ between $a$ and $b$ is equal to the area of a rectangle with height $\bar{f}$ and width $(b - a)$.

Example 1: Mean Value

• Find the mean value of $f(x) = x^2$ over the interval $[0, 1]$:
  $\bar{f} = \frac{1}{1 - 0} \int<em>{0}^{1} x^2 dx$$\bar{f} = \int</em>{0}^{1} x^2 dx$
  $\bar{f} = \frac{x^3}{3} \Big|_0^1$
  $\bar{f} = \frac{1}{3}$

Example 2: Mean Value

• Find the mean value of $f(x) = \sin(\pi x)$ over the interval $[-1, 1]$:
• The function is odd-symmetric, meaning $f(-x) = -f(x)$.
• The area under the curve cancels out, so the mean value should be zero.
  $\bar{f} = \frac{1}{1 - (-1)} \int<em>{-1}^{1} \sin(\pi x) dx$$\bar{f} = \frac{1}{2} \int</em>{-1}^{1} \sin(\pi x) dx$
  $\bar{f} = \frac{1}{2} \left[ -\frac{\cos(\pi x)}{\pi} \right]_{-1}^{1}$
  $\bar{f} = \frac{1}{2\pi} [-\cos(\pi) + \cos(-\pi)]$
  $\bar{f} = \frac{1}{2\pi} [-(-1) + (-1)] = 0$

Root Mean Square Value

• The mean value of a function is sometimes useless (e.g., when the function is anti-symmetric over the interval of integration).

• The root mean square (RMS) value is introduced for a function $f(x)$ defined on an interval $[a, b]$ to obtain more information when the average is uninformative.

• The RMS value is defined as the square root of the mean value of the square of the function:
  $RMS = \sqrt{\text{Mean Value of } f(x)^2}$

  $RMS = \sqrt{\frac{1}{b-a} \int_{a}^{b} [f(x)]^2 dx}$

Example: Root Mean Square Value

• Calculate the RMS value for the function $f(x) = 2\cos(x)$ over the interval $[0, 2\pi]$:

  $RMS = \sqrt{\frac{1}{2\pi - 0} \int<em>{0}^{2\pi} (2\cos(x))^2 dx}$$RMS = \sqrt{\frac{1}{2\pi} \int</em>{0}^{2\pi} 4\cos^2(x) dx}$
  $RMS = \sqrt{\frac{4}{2\pi} \int_{0}^{2\pi} \cos^2(x) dx}$

• Use the power reduction formula: $\cos^2(x) = \frac{1 + \cos(2x)}{2}$
  $RMS = \sqrt{\frac{4}{2\pi} \int<em>{0}^{2\pi} \frac{1 + \cos(2x)}{2} dx}$$RMS = \sqrt{\frac{4}{4\pi} \int</em>{0}^{2\pi} (1 + \cos(2x)) dx}$
  $RMS = \sqrt{\frac{1}{\pi} \left[ x + \frac{\sin(2x)}{2} \right]_{0}^{2\pi}}$
  $RMS = \sqrt{\frac{1}{\pi} (2\pi + 0 - 0 - 0)}$
  $RMS = \sqrt{\frac{2\pi}{\pi}}$
  $RMS = \sqrt{2}$