Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 350: 52

Answer

$\rm (20\;cm,0\;cm)$

Work Step by Step

It is obvious that the $y_{cm}$ of our steel plate is zero due to the symmetry since it has the shape of the isosceles triangle. $$y_{cm}=\color{red}{\bf 0} \;\rm cm \tag 1$$ Now we need to find its $x_{cm}$. We know that $$x_{cm}=\int \dfrac{xdm}{M}$$ where $M$ is the whole mass of the plate. Thus, $$x_{cm}=\dfrac{1}{M}\int xdm\tag 2$$ As we see from the figure below, the dark-yellow rectangle has a mass of $dm$ where $dx$ is its width while $y$ is its height. Thus, its area is given by $$dA=y dx\tag 3$$ This area is from $[A=(y)(0)=0]$ to $[A=(20)(30)=600$ cm$^2]$. Solving for $dx$; And since the mass is uniformly distributed, $\sigma=\dfrac{dm}{dA}=\dfrac{M}{A}$. Solving for $dm$; $$dm=\dfrac{MdA}{A}$$ Plugging from (3); $$dm=\dfrac{M ydx}{A}$$ Plugging into (2); $$x_{cm}=\dfrac{1}{M}\int x dm = \dfrac{1}{ \color{red}{\bf\not}M}\int x\dfrac{ \color{red}{\bf\not}M ydx}{A}$$ $$x_{cm}= \dfrac{1}{A}\int x ydx \tag 4$$ Now we need to find $y$; from the geometry of the figure below (the similarity between the big triangle and the small trinagle inside it), we can see that $$\dfrac{y}{0.2}=\dfrac{x}{0.3}$$ Thus, $$y=\dfrac{2x}{3}$$ Plugging into (4); $$x_{cm}= \dfrac{1}{A}\int x \dfrac{2x}{3}dx = \dfrac{2}{3A}\int^{0.3}_{0} x^2dx $$ $$x_{cm}= \dfrac{2}{3A} \dfrac{x^3}{3}\bigg|_0^{0.3} $$ $$x_{cm}= \dfrac{2}{3\times \frac{1}{2}(0.3)(0.2)} \dfrac{0.3^3}{3}=\color{black}{\bf 0.2} \;\rm m $$ $$x_{cm}=\color{red}{\bf 20} \;\rm cm $$
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