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\title[]{MA140-Engineering Calculus}
\author[ Lecture 31 ]{\large{ Lecture 31} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\textcolor{red}{Centers of Mass:}}\\
{\textcolor{red}{Masses Along a Line:}}\\
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec7001.pdf}
		\end{center}
\end{frame}
\begin{frame}
Each mass $m_{k}$ exerts a downward force $m_{k}g$.\\
Each of these forces has a tendency to turn the axis about the origin This turning effect is called a
torque.\\
The torque of each mass is: $$m_{k}gx_{k}$$
$$System \quad torque=m_{1}gx_{1}+m_{2}gx_{2}+m_{3}gx_{3}$$
$$=g\cdot (m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3})$$
The number $m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}$ is called the moment of the system about the
origin.
 		
		
\end{frame}
\begin{frame}
We usually want to know where to place the fulcrum to make the system balance, that
is, at what point $\bar{x}$ to place it to make the torques add to zero.
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec7002.pdf}
		\end{center}
		
\end{frame}
\begin{frame}
The torque of each mass about the fulcrum in this special location is: $$(x_{k}-\bar{x})m_{k}g$$
So the torque of the new system is $$\sum_{k=1}^{3}(x_{k}-\bar{x})m_{k}g$$
Then: $$\sum_{k=1}^{3}(x_{k}-\bar{x})m_{k}g=0 \Rightarrow g\sum_{k=1}^{3}(x_{k}-\bar{x})m_{k}=0$$

$$\Rightarrow \sum_{k=1}^{3}(m_{k}x_{k}-m_{k}\bar{x})=0 \Rightarrow \sum_{k=1}^{3} m_{k}x_{k}-\sum_{k=1}^{3} m_{k}\bar{x}=0$$

$$\Rightarrow \sum_{k=1}^{3} m_{k}x_{k}=\bar{x}\sum_{k=1}^{3} m_{k}\Rightarrow  \bar{x}=\frac{\sum_{k=1}^{3} m_{k}x_{k}}{\sum_{k=1}^{3} m_{k}}$$

\end{frame}
\begin{frame}
This last equation tells us to find $\bar{x}$ by dividing the system's moment about the origin by
the system's total mass:
$$\Rightarrow  \bar{x}=\frac{\sum_{k=1}^{3} m_{k}x_{k}}{\sum_{k=1}^{3} m_{k}}=\frac{system\enspace moment \enspace about \enspace origin }{system \enspace mass} $$
\end{frame}
\begin{frame}

{\textcolor{red}{Wires and Thin strips:}}\\
\begin{center}
			\includegraphics[scale=0.35]{Figures/lec7003.pdf}
		\end{center}
\end{frame}
\begin{frame}
$$\bar{x}\cong \frac{system \enspace moment}{system \enspace mass}$$
The system mass is $$\sum_{k=1}^{n} \Delta m_{k}$$
the moment of each piece $x_{k}\Delta m_{k}$, so $$ System \enspace moment \cong \sum_{k=1}^{n} x_{k}\Delta m_{k}$$
if the density of the strip at $x_{k}$ is $\delta(x_{k})$, expressed in terms of mass per unit
length and if $\delta$ is continuous, then $\Delta m_{k}$ is approximately equal to $\delta(x_{k}) \Delta x_{k}$:
$$\Delta m_{k} \cong \delta(x_{k}) \Delta x$$  
\end{frame}
\begin{frame}
Therefore:
$$\bar{x}\cong \frac{system \enspace moment}{system \enspace mass}\cong \frac{\sum_{k=1}^{n} x_{k} \Delta m_{k}}{\sum_{k=1}^{n} \Delta m_{k}}\cong \frac{\sum_{k=1}^{n} x_{k} \delta(x_{k}) \Delta x}{\sum_{k=1}^{n} \delta(x_{k})  \Delta x}$$
The sums is a Riemann sum so when $\Delta x\to 0$ or $n\to \infty$, then $$\bar{x}=\frac{\int_{a}^{b} x\delta(x)dx}{\int_{a}^{b} \delta(x)dx}$$
\end{frame}
\begin{frame}
\begin{example}
Show that the center of mass of a straight, thin strip or rod of constant density lies halfway
between its two ends.
\end{example}
\begin{center}
			\includegraphics[scale=0.35]{Figures/lec7004.pdf}
		\end{center}
\end{frame}
\begin{frame}
We model the strip as a portion of the x-axis from $x=a$ to $x=b$.
We know that $$\bar{x}=\frac{\int_{a}^{b} x\delta(x)dx}{\int_{a}^{b} \delta(x)dx}$$
the density is constant so $\delta(x)=\delta$. The numerator is: $$\int_{a}^{b} x\delta dx=\delta \int_{a}^{b} xdx=\delta \frac{1}{2}x^{2} \Big ]_{a}^{b}=\frac{\delta}{2} (b^{2}-a^{2})$$
The denominator is: $$\int_{a}^{b} \delta dx=\delta \int_{a}^{b} dx=\delta x \Big ]_{a}^{b}=\delta (b-a)$$
So: $$\bar{x}=\frac{\int_{a}^{b} x\delta(x)dx}{\int_{a}^{b} \delta(x)dx}=\frac{a+b}{2}$$
\end{frame}
\begin{frame}
When the density is constant, we call the center of mass the centroid of the object. To find the centroid we set $\delta=1$

\end{frame}

\end{document}