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8.3-bo'lim: Massa markazi

Ushbu bo'limda biz topamiz massa markazi or markaziy bir xil zichlikdagi \(\rho \) yupqa plastinka. Massa markazi yoki mintaqaning markaziy qismi, agar bu nuqtadan to'xtatilgan bo'lsa, mintaqa gorizontal ravishda mukammal darajada muvozanatlanadigan nuqtadir.

Shunday qilib, plastinka \(\left[ {a,b} \right]\) oralig'ida \(f\left( x \right)\) va \(g\left( x \right)\) ikkita egri chiziq bilan chegaralangan mintaqa deb faraz qilaylik. Shunday qilib, biz quyida mintaqaning massa markazini topmoqchimiz.

This is the graph of two unknown functions on the domain a<x<b that are mostly in the 1st quadrant. The graph of f(x) is always over the graph of g(x) and the graph of g(x) dips briefly into the 4th quadrant just to make the point that the quadrant we are in does not matter. The area between the two functions has been shaded in.

Bizga birinchi navbatda bu plastinkaning massasi kerak bo'ladi. Massa,

\[\begin{align*}M & = \rho \left( {{\mbox{Area of plate}}} \right)\\ & = \rho \int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}\end{align*}\]

Keyinchalik, bizga kerak bo'ladi daqiqalar mintaqaning. \({M_x}\) va \({M_y}\) bilan belgilangan ikkita moment mavjud. Momentlar mintaqaning mos ravishda \(x\) va \(y\) o'qi atrofida aylanish tendentsiyasini o'lchaydi. Lahzalar tomonidan berilgan,

Momentlar tenglamalari

\[\begin{align*}{M_x} & = \rho \int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)\,dx}}\\ {M_y} & = \rho \int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \right) - g\left( x \right)} \right)\,dx}}\end{align*}\]

Massalar markazining koordinatalari \(\left( {\overline{x},\overline{y}} \right)\), u holda:

Massa koordinatalari markazi

\[\begin{align*}\overline{x} & = \frac{{{M_y}}}{M} = \frac{{\int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \right) - g\left( x \right)} \right)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}}} = \frac{1}{A}\int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \right) - g\left( x \right)} \right)\,dx}}\\ \overline{y} & = \frac{{{M_x}}}{M} = \frac{{\int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}}} = \frac{1}{A}\int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)\,dx}}\end{align*}\]

qayerda,

\[A = \int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}\]

E'tibor bering, plastinkaning zichligi, \(\rho \) bekor qilinadi va shuning uchun haqiqatan ham kerak emas.

Keling, bir nechta misollar bilan ishlaylik.

1-misol \(\left[ {0,\displaystyle \frac{\pi }{2}} \right]\) oralig'ida \(y = 2\sin \left( {2x} \right)\), \(y = 0\) bilan chegaralangan mintaqa uchun massa markazini aniqlang.
Yechim

Bu erda massa markazi nuqta bilan belgilangan mintaqaning eskizi.

This is the graph of $y=2\sin \left( 2x \right)$ on the domain 0<x<$\frac{\pi}{2}$ and the area between the curve and the x-axis is shaded. There is also a dot at the center of mass whose coordinates will be determined below.

Keling, avval hududning maydonini olaylik.

\[\begin{align*}A & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2\sin \left( {2x} \right)\,dx}}\\ & = \left. { - \cos \left( {2x} \right)} \right|_0^{\frac{\pi }{2}}\\ & = 2\end{align*}\]

Endi, lahzalar (zichliksiz, chunki u shunchaki tushib ketadi),

\[\begin{array}{*{20}{c}}\begin{aligned}{M_x} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2{{\sin }^2}\left( {2x} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{1 - \cos \left( {4x} \right)\,dx}}\\ & = \left. {\left( {x - \frac{1}{4}\sin \left( {4x} \right)} \right)} \right|_0^{\frac{\pi }{2}}\\ & = \frac{\pi }{2}\end{aligned}& \hspace{0.5in} &\begin{aligned}{M_y} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2x\sin \left( {2x} \right)\,dx}}\hspace{0.25in}{\mbox{integrating by parts}}...\\ & = - \left. {x\cos \left( {2x} \right)} \right|_0^{\frac{\pi }{2}} + \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\cos \left( {2x} \right)\,dx}}\\ & = - \left. {x\cos \left( {2x} \right)} \right|_0^{\frac{\pi }{2}} + \left. {\frac{1}{2}\sin \left( {2x} \right)} \right|_0^{\frac{\pi }{2}}\\ & = \frac{\pi }{2}\end{aligned}\end{array}\]

U holda massa markazining koordinatalari,

\[\begin{align*}\overline{x} & = \frac{{{}^{\pi }/{}_{2}}}{2} = \frac{\pi }{4}\\ \overline{y} & = \frac{{{}^{\pi }/{}_{2}}}{2} = \frac{\pi }{4}\end{align*}\]

Yana shuni yodda tutingki, biz zichlikni kiritmadik, chunki u bekor qilinadi.

Shunday qilib, ushbu mintaqa uchun massa markazi \(\left( {\frac{\pi }{4},\frac{\pi }{4}} \right)\).

2-misol \(y = {x^3}\) va \(y = \sqrt x \) bilan chegaralangan mintaqa uchun massa markazini aniqlang.
Yechim

Ikki egri chiziq \(x = 0\) va \(x = 1\) nuqtalarida kesishadi va bu erda massa markazi quti bilan belgilangan mintaqaning eskizi.

This is the graph of $y={{x}^{3}}$ and $y=\sqrt{x}$ on the domain 0<x<1. The graph of $y=\sqrt{x}$ is always larger than the graph of $y={{x}^{3}}$ and the area between the curves is shaded. There is also a dot at the center of mass whose coordinates will be determined below.

Biz birinchi navbatda mintaqaning maydonini olamiz.

\[\begin{align*}A & = \int_{{\,0}}^{{\,1}}{{\sqrt x - {x^3}\,dx}}\\ & = \left. {\left( {\frac{2}{3}{x^{\frac{3}{2}}} - \frac{1}{4}{x^4}} \right)} \right|_0^1\\ & = \frac{5}{{12}}\end{align*}\]

Endi lahzalar, yana zichliksiz

\[\begin{array}{*{20}{c}}\begin{aligned}{M_x} & = \int_{{\,0}}^{{\,1}}{{\frac{1}{2}\left( {x - {x^6}} \right)\,dx}}\\ & = \left. {\frac{1}{2}\left( {\frac{1}{2}{x^2} - \frac{1}{7}{x^7}} \right)} \right|_0^1\\ & = \frac{5}{{28}} \\ & \end{aligned}& \hspace{0.5in} &\begin{aligned}{M_y} & = \int_{{\,0}}^{{\,1}}{{x\left( {\sqrt x - {x^3}} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{{x^{\frac{3}{2}}} - {x^4}\,dx}}\\ & = \left. {\left( {\frac{2}{5}{x^{\frac{5}{2}}} - \frac{1}{5}{x^5}} \right)} \right|_0^1\\ & = \frac{1}{5}\end{aligned}\end{array}\]

U holda massa markazining koordinatalari,

\[\begin{align*}\overline{x} & = \frac{{{1}/{5}\;}}{{{5}/{{12}}\;}} = \frac{{12}}{{25}}\\ \overline{y} & = \frac{{{5}/{{28}}\;}}{{{5}/{{12}}\;}} = \frac{3}{7}\end{align*}\]

Massalar markazining koordinatalari \(\left( {\frac{{12}}{{25}},\frac{3}{7}} \right)\).