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Question Number 131729 by ajfour last updated on 07/Feb/21

Commented by ajfour last updated on 07/Feb/21

$F i n d r a d i u s o f s p h e r e o f d e n s i t y$ $ρ t h a t r e c e i v e s m a x i m u m$ $n o r m a l r e a c t i o n f r o m w a l l .$
	Answered by mr W last updated on 08/Feb/21

	$l e t λ = \frac{r}{R}$ $m = \frac{4 π r^{3} ρ}{3} = \frac{4 π R^{3} ρ λ^{3}}{3}$ $\cos θ = \frac{R - r}{R + r} = \frac{1 - λ}{1 + λ}$ $N = \frac{m g}{\tan θ} = \frac{4 π ρ {g R}^{3} λ^{3}}{3 \sqrt{{(\frac{1 + λ}{1 - λ})}^{2} - 1}} = \frac{4 π ρ {g R}^{3}}{3} f (λ)$ $f (λ) = \frac{λ^{3}}{\sqrt{{(\frac{1 + λ}{1 - λ})}^{2} - 1}}$ $f_{m a x} \approx 0.0616 a t λ \approx 0.7143$ $N_{m a x} \approx 0.0616 M g$ $M = m a s s e o f s p h e r e w i t h r a d i u s R$ $a n d s a m e d e n s i t y ρ$
	Commented by mr W last updated on 08/Feb/21

	$g r e a t! y o u g o t i t e x a c t l y!$
	Commented by ajfour last updated on 08/Feb/21

	Commented by ajfour last updated on 08/Feb/21
	$cos θ=((1−λ)/(1+λ)) λ=((1−cos θ)/(1+cos θ))=tan^2 (θ/2) N=((4/3)πρR^3 )g{((tan^6 (θ/2)(1−tan^2 (θ/2)))/(2tan (θ/2)))} say tan (θ/2)=x ⇒ N=((2/3)πρR^3 g)(x^5 −x^7 ) (dN/(d(tan (θ/2))))=0 ⇒ 5x^4 =7x^6 x^2 =(5/7)=λ=(r/R)≈0.7143 N_(max) =((Mg)/2)(((25)/(49))−((125)/(343)))(√(5/7)) N_(max) =((25(√5))/(343(√7)))Mg =((25(√5))/(343(√7)))((1/λ^3 ))mg N_(max) =((mg)/( (√(35)))) N_(max) ≈0.0616Mg$
	$\cos θ = \frac{1 - λ}{1 + λ}$ $λ = \frac{1 - \cos θ}{1 + \cos θ} = \tan^{2} \frac{θ}{2}$ $N = (\frac{4}{3} π ρ R^{3}) g {\frac{\tan^{6} \frac{θ}{2} (1 - \tan^{2} \frac{θ}{2})}{2 \tan \frac{θ}{2}}}$ $s a y \tan \frac{θ}{2} = x$ $\Rightarrow N = (\frac{2}{3} π ρ R^{3} g) (x^{5} - x^{7})$ $\frac{d N}{d (\tan \frac{θ}{2})} = 0 \Rightarrow 5 x^{4} = 7 x^{6}$ $x^{2} = \frac{5}{7} = λ = \frac{r}{R} \approx 0.7143$ $N_{m a x} = \frac{M g}{2} (\frac{25}{49} - \frac{125}{343}) \sqrt{\frac{5}{7}}$ $N_{m a x} = \frac{25 \sqrt{5}}{343 \sqrt{7}} M g$ $= \frac{25 \sqrt{5}}{343 \sqrt{7}} (\frac{1}{λ^{3}}) m g$ $N_{m a x} = \frac{m g}{\sqrt{35}}$ $N_{m a x} \approx 0.0616 M g$
	Commented by ajfour last updated on 08/Feb/21

	$T h a n k y o u S i r, e x a c t a n s w e r$ ${i s n}^{'} t c o m p l i c a t e d, S i r!$
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