Talk:PlanetPhysics/Centre of Mass of Half Disc

Original TeX Content from PlanetPhysics Archive
%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: center of mass of half-disc %%% Primary Category Code: 45.40.-f %%% Filename: CentreOfMassOfHalfDisc.tex %%% Version: 3 %%% Owner: pahio %%% Author(s): bci1, pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Let $E$ be the upper half-disc of the disc\, $x^2+y^2 \leqq R$\, in $\mathbb{R}^2$ with a constant surface-density 1. By the symmetry, its \htmladdnormallink{centre of mass}{http://planetphysics.us/encyclopedia/CenterOfGravity.html} locates on its medium radius, and therefore we only have to calculate the ordinate $Y$ of the centre of mass. For doing that, one can use in this \htmladdnormallink{two-dimensional}{http://planetphysics.us/encyclopedia/CoriolisEffect.html} case instead a triple integral the double integral $$Y = \frac{1}{\nu(E)}\int\!\!\int_E y\,dx\,dy,$$ where\, $\nu(E) = \frac{\pi R^2}{2}$\, is the area (and the \htmladdnormallink{mass}{http://planetphysics.us/encyclopedia/Mass.html}) of the half-disc. The region of integration is defined by $$E = \{(x,\,y)\in\mathbb{R}^2\,\vdots\;\; -R\leqq x \leqq R,\; 0 \leqq y \leqq \sqrt{R^2-x^2}\}.$$ Accordingly, we may write $$Y = \frac{2}{\pi R^2}\!\int_{-R}^R\!dx\int_0^{\sqrt{R^2-x^2}}\!y\,dy = \frac{2}{\pi R^2}\!\int_{-R}^R\frac{R^2\!-\!x^2}{2}\,dx = \frac{2}{\pi R^2}\sijoitus{-R}{\quad R}\left(\frac{R^2x}{2}-\frac{x^3}{6}\right) = \frac{4R}{3\pi}.$$ Thus the centre of mass is the point\, $(0,\,\frac{4R}{3\pi})$.

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