Asosiy tarkibga o'tish

8.3-bo'lim : Massa markazi

1. Birinchi kvadrantda joylashgan \(y = 4 - {x^2}\) bilan chegaralangan mintaqaning massa markazini toping.

Yechim

Keling, nuqta bilan ko'rsatilgan massa markazi bilan mintaqaning qisqacha chizmasidan boshlaylik (bu nuqtaning koordinatalari, albatta, oxirgi bosqichda aniqlanishi kerak ....).

A sketch of the region bounded by \(y = 4 - {x^2}\) in the first quadrant.  The parabola falls from \(\left( {0,4} \right)\) to \(\left( {2,0} \right)\) and the region under it is shaded in.  A dot inside the region marks the center of mass, whose coordinates are still to be determined.

Bizga ushbu hududning maydoni ham kerak bo'ladi, shuning uchun avval uni topamiz.

\[A = \int_{0}^{2}{{4 - {x^2}\,dx}} = \left. {\left( {4x - \frac{1}{3}{x^3}} \right)} \right|_0^2 = \frac{{16}}{3}\]
2-qadam

Keyinchalik, ikkita momentni hisoblashimiz kerak. Quyidagi hisob-kitoblarga zichlikni kiritmadik, chunki u faqat oxirgi bosqichda bekor qilinadi.

\[\begin{align*}{M_x} & = \int_{0}^{2}{{\frac{1}{2}{{\left( {4 - {x^2}} \right)}^2}\,dx}} = \int_{0}^{2}{{\frac{1}{2}\left( {16 - 8{x^2} + {x^4}} \right)\,dx}} = \left. {\frac{1}{2}\left( {16x - \frac{8}{3}{x^3} + \frac{1}{5}{x^5}} \right)} \right|_0^2 = \frac{{128}}{{15}}\\ {M_y} & = \int_{0}^{2}{{x\left( {4 - {x^2}} \right)\,dx}} = \int_{0}^{2}{{4x - {x^3}\,dx}} = \left. {\left( {2{x^2} - \frac{1}{4}{x^4}} \right)} \right|_0^2 = 4\end{align*}\]
3-qadam

Nihoyat, massa markazining koordinatalari,

\[\overline{x} = \frac{{{M_y}}}{M} = \frac{{\rho \left( 4 \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{3}{4}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\overline{y} = \frac{{{M_x}}}{M} = \frac{{\rho \left( {\frac{{128}}{{15}}} \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{8}{5}\]

U holda massa markazi: \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\frac{3}{4},\frac{8}{5}} \right)}}\).