Applications to Physics and Engineering

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 6.4 – Work
• Section 7.7 – Approximate Integration
• Section 8.3 – Applications to Physics and Engineering

Expected Educational Results

• Objective 14–1: Given the velocity of an object along a straight line, I can set up the integral to compute the displacement and total distance traveled of the object.
• Objective 14–2: Given the acceleration of an object along a straight line, I can set up the integral to compute the velocity, speed, displacement, and total distance traveled of the object.
• Objective 14–3: I can estimate a definite integral using a Riemann Sum.
• Objective 14–4: I can use the shape of a function to determine if a Riemann Sum is an over- or under-estimate of a definite integral.
• Objective 14–5: I can set up the integral to compute the moments and center of mass of a lamina.
• Objective 14–6: I can set up and evaluate the integral to compute the work on an object.
• Objective 14–7: I can set up and evaluate the integral using Hooke's Law to compute the work on a spring.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Center of Mass

Lamina

Definition: Lamina

A lamina is a flat plate with uniform density, $\rho$$\rho$.

Mass of a Lamina

Let ${\displaystyle A={\int }_a^b[f(x)-g(x)]dx}$$\displaystyle A=\int_a^b{\left[f(x)-g(x)\right]}\,dx$ be the area of some given lamina between $f(x)$$f(x)$ and $g(x)$$g(x)$.

Then the mass of the lamina is ${\displaystyle m=\rho {\int }_a^b[f(x)-g(x)]dx}$$\displaystyle m=\rho\int_a^b{\left[f(x)-g(x)\right]}\,dx$.

Moments of a Lamina

Definition: Moment

The moment of the lamina about the $y$$y$-axis is ${\displaystyle M_y=\rho {\int }_a^{b}x[f(x)-g(x)]dx}$$\displaystyle M_y=\rho\int_a^b{x\left[f(x)-g(x)\right]}\,dx$. The moment of the lamina about the $x$$x$-axis is ${\displaystyle M_x=\rho {\int }_a^b\frac{1}{2}[f(x{)}^2-g(x{)}^2]dx}$$\displaystyle M_x=\rho\int_a^b{\frac{1}{2}\left[f(x)^2-g(x)^2\right]}\,dx$.

Center of Mass

Definition: Center of Mass

The center of mass, a.k.a., centroid, of the lamina bounded by $f(x)$$f(x)$ and $g(x)$$g(x)$ is ${\displaystyle (\overline{x},\overline{y})}$$\displaystyle (\bar{x},\bar{y})$, where $f(x)\ge g(x)$$f(x)\geq g(x)$ is

$\overline{x}=\frac{M_y}{m}=\frac{1}{A}{\int }_a^{b}x[f(x)-g(x)]dx$$\bar{x}=\frac{M_y}{m}=\frac{1}{A}\int_a^b{x\left[f(x)-g(x)\right]}\,dx$

and

$\overline{y}=\frac{M_x}{m}=\frac{1}{A}{\int }_a^b\frac{1}{2}[f(x{)}^2-g(x{)}^2]dx$$\bar{y}=\frac{M_x}{m}=\frac{1}{A}\int_a^b{\frac{1}{2}\left[f(x)^2-g(x)^2\right]}\,dx$

Example 02: Find the center of mass of the lamina bounded by $y=3x+5$$y=3x+5$ and $y=x^2-5$$y=x^2-5$.

Solution 02:

The embedded DESMOS graph above shows the region between the two curves. Note that the red curve ($y=3x+5$$y=\color{red}3x+5$) is “above” the blue curve ($y=x^2-5$$y=\color{blue}x^2-5$).

The intersections of the two curves if found by

$3x+5\stackrel{set}{=}{x}^2-5\Rightarrow x^2-3x-10=0\Rightarrow x=-2,5$${\color{red}3x+5}\stackrel{set}{=}{\color{blue}x^2-5}\Rightarrow x^2-3x-10=0\Rightarrow x=-2, 5$

[Alternatively, to find the intersections of the two functions, you can click on the intersections in the DESMOS graph.]

The area of the region bounded by the curves is

$A={\int }_{-2}^5[3x+5-(x^2-5)]dx={(-{\displaystyle \frac{x^3}{3}}+{\displaystyle \frac{3x^2}{2}}+10x)|}_{-2}^5={\displaystyle \frac{343}{6}}$$A=\int_{-2}^5{\left[{\color{red}3x+5}-({\color{blue}x^2-5})\right]\,dx}=\left.\left(-\dfrac{x^3}{3} + \dfrac{3 x^2}{2} + 10 x\right)\right|_{-2}^5={\color{\purple}\dfrac{343}{6}}$

The $x$$x$-coordinate of the center of mass is (need to distribute the two factors in the integrand)

$\overline{x}=\frac{1}{343/6}{\int }_{-2}^{5}x[3x+5-(x^2-5)]dx=\frac{1}{343/6}{\int }_{-2}^5[-x^3+3x^2+10x]dx$$\bar{x}=\frac{1}{{\color{\purple}343/6}}\int_{-2}^5{{\color{brown}x}\left[{\color{red}3x+5}-({\color{blue}x^2-5})\right]\,dx}=\frac{1}{{\color{\purple}343/6}}\int_{-2}^5{\left[-x^3+3x^2+10x\right]\,dx}$

$=\frac{1}{343/6}{(-{\displaystyle \frac{x^4}{4}}+x^3+5x^2)|}_{-2}^5={\displaystyle \frac{6}{343}}\left({\displaystyle \frac{343}{4}}\right)={\displaystyle \frac{3}{2}}$$=\frac{1}{\color{purple}343/6}\left.\left(-\dfrac{x^4}{4} + x^3 + 5 x^2\right)\right|_{-2}^5={\color{purple}\dfrac{6}{343}}\left(\dfrac{343}{4}\right)=\dfrac{3}{2}$

and

$\overline{y}=\frac{1}{2}\left(\frac{1}{343/6}\right){\int }_{-2}^5[(3x+5{)}^2-(x^2-5{)}^2]dx=\frac{1}{2}\left(\frac{1}{343/6}\right){\int }_{-2}^5[-x^4+19x^2+30x]dx$$\bar{y}=\frac{1}{2}\left(\frac{1}{\color{purple}343/6}\right)\int_{-2}^{5}{\left[({\color{red}3x+5})^2-({\color{blue}x^2-5})^2\right]\,dx}=\frac{1}{2}\left(\frac{1}{\color{purple}343/6}\right)\int_{-2}^{5}{\left[-x^4+19 x^2+30 x\right]\,dx}$

$=\frac{1}{2}\left(\frac{1}{343/6}\right){(-{\displaystyle \frac{x^5}{5}}+{\displaystyle \frac{19x^3}{3}}+15x^2)|}_{-2}^5={\displaystyle \frac{1}{2}}{\displaystyle \frac{6}{343}}\left({\displaystyle \frac{7889}{15}}\right)={\displaystyle \frac{23}{5}}$$=\frac{1}{2}\left(\frac{1}{\color{purple}343/6}\right)\left.\left(-\dfrac{x^5}{5} + \dfrac{19 x^3}{3} + 15 x^2\right)\right|_{-2}^5=\dfrac{1}{2}{\color{purple}\dfrac{6}{343}} \left(\dfrac{7889}{15}\right)=\dfrac{23}{5}$

Thus, the center of mass of the lamina is ${\displaystyle (\overline{x},\overline{y})=(\frac{3}{2},\frac{23}{5})}$$\displaystyle\left(\bar{x}, \bar{y}\right)=\left(\frac{3}{2}, \frac{23}{5}\right)$

Investigation 02

1. Find the center of mass of the lamina bounded by $y=2+x$$y=2+x$, $y=0$$y=0$, $x=0$$x=0$, and the $x=2$$x=2$.
2. Find the center of mass of the lamina bounded by $y=3-x$$y=3-x$, $y=x+1$$y=x+1$, and the $y$$y$-axis.
3. Find the center of mass of the lamina bounded by $y=4-(x-2{)}^2$$y=4-(x-2)^2$ and the $x$$x$-axis.
4. Find the center of mass of the lamina bounded by $y=\sin (2x)$$y=\sin{(2x)}$ and the $x$$x$-axis from $x=0$$x=0$ to $x=\frac{\pi }{2}$$x=\frac{\pi}{2}$.
5. Estimate (to two (2) decimal places) the location of the center of mass of the lamina bounded by $y={\sec }^2(x)$$y=\sec^2{(x)}$ and $y=x+2$$y=x+2$ from $x=-0.5$$x=-0.5$ to $x=0.5$$x=0.5$.

CC BY-NC-SA 4.0

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Created: Monday, 19 October 2020 8:23 EDT Last Modified: Friday, 15 July 2022 - 10:30 (EDT)