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Question Number 52825 by ajfour last updated on 13/Jan/19

Commented by ajfour last updated on 13/Jan/19

$F i n d θ . A s s u m e n o f r i c t i o n a n d$ $a s t a t i c s i t u a t i o n .$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19

Commented by ajfour last updated on 14/Jan/19

Commented by mr W last updated on 14/Jan/19

$p e r f e c t l y i l l u s t r a t e d! t h a n k s a l o t s i r!$
	Answered by mr W last updated on 14/Jan/19
	$let η=(m/M), λ=(L/h) α=angle between string and vertical a=distance from ring to lower end of rod N=contact force between ring and rod ((MgL cos θ)/2)=Na ((Mg)/(sin (π−θ)))=((mg)/(sin (θ−α)))=(N/(sin α)) sin (θ−α)=(m/M) sin θ=η sin θ ⇒θ−α=sin^(−1) (η sin θ) ⇒α=θ−sin^(−1) (η sin θ) N=((Mg sin α)/(sin θ)) a=((h sin α)/(sin ((π/2)+θ−α)))=((h sin α)/(cos (θ−α)))=((h sin α)/(√(1−η^2 sin^2 θ))) ⇒((MgL cos θ)/2)=((Mg sin α)/(sin θ))×((h sin α)/(√(1−η^2 sin^2 θ))) ⇒((cos θ)/2)×(L/h)=((sin^2 α)/(sin θ (√(1−η^2 sin^2 θ)))) ⇒((λ sin θ cos θ (√(1−η^2 sin^2 θ)))/2)=sin^2 α ⇒λ sin 2θ (√(1−η^2 sin^2 θ))=2(1−cos 2α) ⇒λ sin 2θ (√(1−η^2 sin^2 θ))=2[1−cos 2{θ−sin^(−1) (η sin θ)}] examples: λ=(L/h)=2, η=(m/M)=0.2⇒θ=52.7639° λ=(L/h)=2, η=(m/M)=0⇒θ=45°$
	$l e t η = \frac{m}{M}, λ = \frac{L}{h}$ $α = a n g l e b e t w e e n s t r i n g a n d v e r t i c a l$ $a = d i s t a n c e f r o m r i n g t o l o w e r e n d o f r o d$ $N = c o n t a c t f o r c e b e t w e e n r i n g a n d r o d$ $\frac{M g L \cos θ}{2} = N a$ $\frac{M g}{\sin (π - θ)} = \frac{m g}{\sin (θ - α)} = \frac{N}{\sin α}$ $\sin (θ - α) = \frac{m}{M} \sin θ = η \sin θ$ $\Rightarrow θ - α = \sin^{- 1} (η \sin θ)$ $\Rightarrow α = θ - \sin^{- 1} (η \sin θ)$ $N = \frac{M g \sin α}{\sin θ}$ $a = \frac{h \sin α}{\sin (\frac{π}{2} + θ - α)} = \frac{h \sin α}{\cos (θ - α)} = \frac{h \sin α}{\sqrt{1 - η^{2} \sin^{2} θ}}$ $\Rightarrow \frac{M g L \cos θ}{2} = \frac{M g \sin α}{\sin θ} \times \frac{h \sin α}{\sqrt{1 - η^{2} \sin^{2} θ}}$ $\Rightarrow \frac{\cos θ}{2} \times \frac{L}{h} = \frac{\sin^{2} α}{\sin θ \sqrt{1 - η^{2} \sin^{2} θ}}$ $\Rightarrow \frac{λ \sin θ \cos θ \sqrt{1 - η^{2} \sin^{2} θ}}{2} = \sin^{2} α$ $\Rightarrow λ \sin 2 θ \sqrt{1 - η^{2} \sin^{2} θ} = 2 (1 - \cos 2 α)$ $\Rightarrow λ \sin 2 θ \sqrt{1 - η^{2} \sin^{2} θ} = 2 [1 - \cos 2 {θ - \sin^{- 1} (η \sin θ)}]$ $e x a m p l e s :$ $λ = \frac{L}{h} = 2, η = \frac{m}{M} = 0.2 \Rightarrow θ = 52.7639 °$ $λ = \frac{L}{h} = 2, η = \frac{m}{M} = 0 \Rightarrow θ = 45 °$
	Commented by mr W last updated on 13/Jan/19

	Commented by ajfour last updated on 14/Jan/19

	$A l r i g h t a l l - a l o n g S i r . N o t s o e a s y .$ $G o d b l e s s y o u!$
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