From a circular disc of radius R a circular hole of radius R/2 is cut out.The centre of the circular hole is at R/2 from the centre of the original disc.Locate the centre of gravity of the resulting flat body. please help me as fast as possible. Thanks!


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Question Number 31476 by NECx last updated on 09/Mar/18

$F r o m a c i r c u l a r d i s c o f r a d i u s R$ $a c i r c u l a r h o l e o f r a d i u s R / 2 i s$ $c u t o u t . T h e c e n t r e o f t h e c i r c u l a r$ $h o l e i s a t R / 2 f r o m t h e c e n t r e o f$ $t h e o r i g i n a l d i s c . L o c a t e t h e c e n t r e$ $o f g r a v i t y o f t h e r e s u l t i n g f l a t$ $b o d y .$ $p l e a s e h e l p m e a s f a s t a s p o s s i b l e .$ $T h a n k s!$
	Answered by mrW2 last updated on 09/Mar/18

	$\frac{R}{2} \times \frac{π R^{2}}{4} = e \times (π R^{2} - \frac{π R^{2}}{4})$ $e = \frac{R}{6}$
	Commented by NECx last updated on 09/Mar/18

	$p l e a s e c a n y o u e x p l a i n$
	Commented by mrW2 last updated on 09/Mar/18

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