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[parent] center of mass of half-disc (Example)

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 centre of mass 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 two-dimensional case instead a triple integral the double integral

$\displaystyle Y = \frac{1}{\nu(E)}\int\!\!\int_E y\,dx\,dy,$
where  $\nu(E) = \frac{\pi R^2}{2}$  is the area (and the mass) of the half-disc. The region of integration is defined by

$\displaystyle E = \{(x,\,y)\in\mathbb{R}^2\,\vdots\;\; -R\leqq x \leqq R,\; 0 \leqq y \leqq \sqrt{R^2-x^2}\}.$
Accordingly, we may write

$\displaystyle Y = \frac{2}{\pi R^2}\!\int_{-R}^R\!dx\int_0^{\sqrt{R^2-x^2}}\!y\... ...\!\!-R}^{\,\quad R}\left(\frac{R^2x}{2}-\frac{x^3}{6}\right) = \frac{4R}{3\pi}.$
Thus the centre of mass is the point  $(0,\,\frac{4R}{3\pi})$.



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See Also: centre of mass of polygon


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Cross-references: mass, two-dimensional, centre of mass

This is version 3 of center of mass of half-disc, born on 2007-07-02, modified 2009-04-18.
Object id is 253, canonical name is CentreOfMassOfHalfDisc.
Accessed 1020 times total.

Classification:
Physics Classification: 45.40.-f (Dynamics and kinematics of rigid bodies)
 45.50.Dd (General motion)
 02.40.Yy (Geometric mechanics )
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