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Question Number 66109 by Rio Michael last updated on 09/Aug/19

Commented by Rio Michael last updated on 09/Aug/19

$T h e d i a g r a m a b o v e s h o w s a u n i f o r m s e m i - c i r c u l a r l a m i n a o f r a d i u s 2 a$ $, c e n t e r O . T h e d i s t a n c e o f t h e c e n t r e o f m a s s f o r m P, v e r t i c a l l y a b o v e O i s$ $A \frac{6 a π - 8 a}{3 π}$ $B \frac{6 a π + 8 a}{3 π}$ $C \frac{8 a - 6 a π}{3 π}$ $D \frac{6 a π - 4 a}{3 π}$
	Answered by mr W last updated on 09/Aug/19

	$R = 2 a$ $\int y d A = 2 \int_{0}^{\frac{π}{2}} \frac{2}{3} R \sin θ \frac{R^{2} d θ}{2} = \frac{2 R^{3}}{3} \int_{0}^{\frac{π}{2}} \sin θ d θ = \frac{2 R^{3}}{3}$ $O C = y_{C} = \frac{\int y d A}{A} = \frac{2 R^{3}}{3 \times \frac{π R^{2}}{2}} = \frac{4 R}{3 π} = \frac{8 a}{3 π}$ $\Rightarrow O P = 2 a - O C = \frac{6 a π - 8 a}{3 π}$ $\Rightarrow a n s w e r A$
	Commented by mr W last updated on 10/Aug/19

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