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Question Number 50101 by ajfour last updated on 13/Dec/18

Commented by ajfour last updated on 14/Dec/18

$I f s y s t e m o f r i n g a n d b l o c k b e$ $r e l e a s e d a s s h o w n, a n d l a t e r t h e$ $r i n g h i t s t h e b l o c k a s s h o w n i n$ $d a s h e d l i n e s, d e t e r m i n e a .$ $F i n d a l s o t h e v e l o c i t y o f r i n g t h e n .$ $A s s u m e f r i c t i o n s u f f i c i e n t s u c h$ $t h a t r i n g r o l l s p u r e l y a s i t m o v e s .$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$p l s m e n t i o n t h e s o u r c e o f q u e s t i o n a n d a l s o$ $a t t a c h t h e f i n a l a n s w e r . . . s o t h a t w e c a n$ $f o l l o w t h e r i g h t p a t h t o g e t t h e a n s w e r . . .$ $b e c a u s e q u e s t i o n i t s e l f i s n o t c l e a r . . .$ $1) i f t h e t h r e a d g e t s u n w i n d e d . . . t h e r i n g h a s$ $a t e n d e n c y t o r o t a t e a n t i c l o c k w i s e . . .$ $f u r t h e r w e i g h t o f b l o c k M g c a u s e t h e r i n g$ $t o g e t u n w i n d e d . . .$
Commented by ajfour last updated on 14/Dec/18

$w h i c h w a y d o y o u t h i n k n e t t o r q u e$ $o n r i n g s h a l l b e, T a n m a y S i r;$ $i s e l d o m p o s t q u e s t i o n s f r o m$ $s o u r c e s .$
Commented by mr W last updated on 14/Dec/18

$t o m e t h e q u e s t i o n i s c l e a r a n d u n i q u e . t h e$ $t h r e a d k e e p s i n t e n s i o n .$
	Answered by mr W last updated on 14/Dec/18
	$let θ=angle between string and ground length of string at t=0: a+(R/(tan (θ_0 /2)))=(h/(tan θ_0 ))=((h(1−tan^2 (θ_0 /2)))/(2tan (θ_0 /2))) δ_0 =tan (θ_0 /2) a+(R/δ_0 )=((h(1−δ_0 ^2 ))/(2δ_0 )) a=((h(1−δ_0 ^2 )−2R)/(2δ_0 )) δ_0 ^2 +2((a/h))δ_0 −(1−((2R)/h))=0 with α=(a/h) and γ=(R/h) δ_0 ^2 +2αδ_0 −(1−2γ)=0 ⇒δ_0 =(√(1+α^2 −2γ))−α=tan (θ_0 /2) l_0 =a+(h/(sin θ_0 ))−(R/(tan (θ_0 /2))) l_0 =a+((h−2R+hδ_0 ^2 )/(2δ_0 )) length of string at t=t_1 : R+(R/(tan (θ_1 /2)))=(h/(tan θ_1 ))=((h(1−tan^2 (θ_1 /2)))/(2tan (θ_1 /2))) R+(R/δ_1 )=((h(1−δ_1 ^2 ))/(2δ_1 )) R=((h(1−δ_1 ^2 )−2R)/(2δ_1 )) 2Rδ_1 =h−2R−hδ_1 ^2 δ_1 ^2 +2γδ_1 −(1−2γ)=0 ⇒δ_1 =(√(γ^2 +1−2γ))−γ=tan (θ_1 /2) l_1 =h−R+(h/(sin θ_1 ))−(R/(tan (θ_1 /2))) l_1 =h−R+((h−2R+hδ_1 ^2 )/(2δ_1 )) l_1 =l_0 +a−R h−R+((h−2R+hδ_1 ^2 )/(2δ_1 ))=a+((h−2R+hδ_0 ^2 )/(2δ_0 ))+a−R h+((h−2R+hδ_1 ^2 )/(2δ_1 ))=2a+((h−2R+hδ_0 ^2 )/(2δ_0 )) 1+((1−2γ+δ_1 ^2 )/(2δ_1 ))=2α+((1−2γ+δ_0 ^2 )/(2δ_0 )) ((1−2γ)/δ_1 )+δ_1 −((1−2γ)/δ_0 )−δ_0 =2(2α−1) ((1−2γ)/((√(γ^2 +1−2γ))−γ))+(√(γ^2 +1−2γ))−γ−((1−2γ)/((√(1+α^2 −2γ))−α))−(√(1+α^2 −2γ))+α=4α−2 ⇒3α+((1−2γ)/((√(1+α^2 −2γ))−α))+(√(1+α^2 −2γ))=2−γ+((1−2γ)/((√(γ^2 +1−2γ))−γ))+(√(γ^2 +1−2γ)) γ is constant, α is unknown ⇒solve for α=.... ⇒a=αh=..... at time t: position of block is y (above ground) position of ring is x x+(R/(tan (θ/2)))=(h/(tan θ))=((h(1−tan^2 (θ/2)))/(2tan (θ/2))) x+(R/δ)=((h(1−δ^2 ))/(2δ)) x=((h−2R−hδ^2 )/(2δ)) λ=(x/h)=((1−2γ−δ^2 )/(2δ)) δ^2 +2λδ−(1−2γ)=0 ⇒δ=(√(1+λ^2 −2γ))−λ (dδ/dt)=((λ/(√(1+λ^2 −2γ)))−1)(1/h)×(dx/dt) l=h−y+(h/(sin θ))−(R/(tan (θ/2))) l=h−y+((h−2R+hδ^2 )/(2δ))=l_0 +a−x 1−(y/h)+((1−2γ+δ^2 )/(2δ))=2α+((1−2γ+δ_0 ^2 )/(2δ_0 ))−λ v_y =−(dy/dt)=velocity of block v=−(dx/dt)=velocity of ring (v_y /h)+[1−((1−2γ+δ^2 )/(2δ^2 ))](dδ/dt)=(v/h) (v_y /h)−(((δ^2 −1+2γ)/(2δ^2 )))((λ/(√(1+λ^2 −2γ)))−1)(v/h)=(v/h) v_y ={1+(((δ^2 −1+2γ)/(2δ^2 )))((λ/(√(1+λ^2 −2γ)))−1)}v at t=t_1 : δ=δ_1 , x=R, λ=(R/h)=γ v_y ={1+((((√(γ^2 +1−2γ))−γ)^2 −1+2γ)/(2((√(γ^2 +1−2γ))−γ)^2 ))((γ/(√(1+γ^2 −2γ)))−1)}v ⇒v_y ={1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}v (1/2)Mv^2 +(1/2)(MR^2 )ω^2 +(1/2)Mv_y ^2 =Mg(h−a−R) 2v^2 +v_y ^2 =2gh(1−α−γ) [2+{1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}^2 ]v^2 =2gh(1−α−γ) ⇒v=(√((2gh(1−α−γ))/(2+{1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}^2 )))$
	$l e t θ = a n g l e b e t w e e n s t r i n g a n d g r o u n d$ $l e n g t h o f s t r i n g a t t = 0 :$ $a + \frac{R}{\tan \frac{θ_{0}}{2}} = \frac{h}{\tan θ_{0}} = \frac{h (1 - \tan^{2} \frac{θ_{0}}{2})}{2 \tan \frac{θ_{0}}{2}}$ $δ_{0} = \tan \frac{θ_{0}}{2}$ $a + \frac{R}{δ_{0}} = \frac{h (1 - δ_{0}^{2})}{2 δ_{0}}$ $a = \frac{h (1 - δ_{0}^{2}) - 2 R}{2 δ_{0}}$ $δ_{0}^{2} + 2 (\frac{a}{h}) δ_{0} - (1 - \frac{2 R}{h}) = 0$ $w i t h α = \frac{a}{h} a n d γ = \frac{R}{h}$ $δ_{0}^{2} + 2 α δ_{0} - (1 - 2 γ) = 0$ $\Rightarrow δ_{0} = \sqrt{1 + α^{2} - 2 γ} - α = \tan \frac{θ_{0}}{2}$ $l_{0} = a + \frac{h}{\sin θ_{0}} - \frac{R}{\tan \frac{θ_{0}}{2}}$ $l_{0} = a + \frac{h - 2 R + h δ_{0}^{2}}{2 δ_{0}}$ $l e n g t h o f s t r i n g a t t = t_{1} :$ $R + \frac{R}{\tan \frac{θ_{1}}{2}} = \frac{h}{\tan θ_{1}} = \frac{h (1 - \tan^{2} \frac{θ_{1}}{2})}{2 \tan \frac{θ_{1}}{2}}$ $R + \frac{R}{δ_{1}} = \frac{h (1 - δ_{1}^{2})}{2 δ_{1}}$ $R = \frac{h (1 - δ_{1}^{2}) - 2 R}{2 δ_{1}}$ $2 R δ_{1} = h - 2 R - h δ_{1}^{2}$ $δ_{1}^{2} + 2 γ δ_{1} - (1 - 2 γ) = 0$ $\Rightarrow δ_{1} = \sqrt{γ^{2} + 1 - 2 γ} - γ = \tan \frac{θ_{1}}{2}$ $l_{1} = h - R + \frac{h}{\sin θ_{1}} - \frac{R}{\tan \frac{θ_{1}}{2}}$ $l_{1} = h - R + \frac{h - 2 R + h δ_{1}^{2}}{2 δ_{1}}$ $l_{1} = l_{0} + a - R$ $h - R + \frac{h - 2 R + h δ_{1}^{2}}{2 δ_{1}} = a + \frac{h - 2 R + h δ_{0}^{2}}{2 δ_{0}} + a - R$ $h + \frac{h - 2 R + h δ_{1}^{2}}{2 δ_{1}} = 2 a + \frac{h - 2 R + h δ_{0}^{2}}{2 δ_{0}}$ $1 + \frac{1 - 2 γ + δ_{1}^{2}}{2 δ_{1}} = 2 α + \frac{1 - 2 γ + δ_{0}^{2}}{2 δ_{0}}$ $\frac{1 - 2 γ}{δ_{1}} + δ_{1} - \frac{1 - 2 γ}{δ_{0}} - δ_{0} = 2 (2 α - 1)$ $\frac{1 - 2 γ}{\sqrt{γ^{2} + 1 - 2 γ} - γ} + \sqrt{γ^{2} + 1 - 2 γ} - γ - \frac{1 - 2 γ}{\sqrt{1 + α^{2} - 2 γ} - α} - \sqrt{1 + α^{2} - 2 γ} + α = 4 α - 2$ $\Rightarrow 3 α + \frac{1 - 2 γ}{\sqrt{1 + α^{2} - 2 γ} - α} + \sqrt{1 + α^{2} - 2 γ} = 2 - γ + \frac{1 - 2 γ}{\sqrt{γ^{2} + 1 - 2 γ} - γ} + \sqrt{γ^{2} + 1 - 2 γ}$ $γ i s c o n s t a n t, α i s u n k n o w n$ $\Rightarrow s o l v e f o r α = . . . .$ $\Rightarrow a = α h = . . . . .$ $a t t i m e t :$ $p o s i t i o n o f b l o c k i s y (a b o v e g r o u n d)$ $p o s i t i o n o f r i n g i s x$ $x + \frac{R}{\tan \frac{θ}{2}} = \frac{h}{\tan θ} = \frac{h (1 - \tan^{2} \frac{θ}{2})}{2 \tan \frac{θ}{2}}$ $x + \frac{R}{δ} = \frac{h (1 - δ^{2})}{2 δ}$ $x = \frac{h - 2 R - h δ^{2}}{2 δ}$ $λ = \frac{x}{h} = \frac{1 - 2 γ - δ^{2}}{2 δ}$ $δ^{2} + 2 λ δ - (1 - 2 γ) = 0$ $\Rightarrow δ = \sqrt{1 + λ^{2} - 2 γ} - λ$ $\frac{d δ}{d t} = (\frac{λ}{\sqrt{1 + λ^{2} - 2 γ}} - 1) \frac{1}{h} \times \frac{d x}{d t}$ $l = h - y + \frac{h}{\sin θ} - \frac{R}{\tan \frac{θ}{2}}$ $l = h - y + \frac{h - 2 R + h δ^{2}}{2 δ} = l_{0} + a - x$ $1 - \frac{y}{h} + \frac{1 - 2 γ + δ^{2}}{2 δ} = 2 α + \frac{1 - 2 γ + δ_{0}^{2}}{2 δ_{0}} - λ$ $v_{y} = - \frac{d y}{d t} = v e l o c i t y o f b l o c k$ $v = - \frac{d x}{d t} = v e l o c i t y o f r i n g$ $\frac{v_{y}}{h} + [1 - \frac{1 - 2 γ + δ^{2}}{2 δ^{2}}] \frac{d δ}{d t} = \frac{v}{h}$ $\frac{v_{y}}{h} - (\frac{δ^{2} - 1 + 2 γ}{2 δ^{2}}) (\frac{λ}{\sqrt{1 + λ^{2} - 2 γ}} - 1) \frac{v}{h} = \frac{v}{h}$ $v_{y} = {1 + (\frac{δ^{2} - 1 + 2 γ}{2 δ^{2}}) (\frac{λ}{\sqrt{1 + λ^{2} - 2 γ}} - 1)} v$ $a t t = t_{1} : δ = δ_{1}, x = R, λ = \frac{R}{h} = γ$ $v_{y} = {1 + \frac{{(\sqrt{γ^{2} + 1 - 2 γ} - γ)}^{2} - 1 + 2 γ}{2 {(\sqrt{γ^{2} + 1 - 2 γ} - γ)}^{2}} (\frac{γ}{\sqrt{1 + γ^{2} - 2 γ}} - 1)} v$ $\Rightarrow v_{y} = {1 + \frac{γ (1 - 2 γ)}{(1 - γ) (\sqrt{γ^{2} + 1 - 2 γ} - γ)}} v$ $\frac{1}{2} {M v}^{2} + \frac{1}{2} ({M R}^{2}) ω^{2} + \frac{1}{2} {M v}_{y}^{2} = M g (h - a - R)$ $2 v^{2} + v_{y}^{2} = 2 g h (1 - α - γ)$ $[2 + {1 + \frac{γ (1 - 2 γ)}{(1 - γ) (\sqrt{γ^{2} + 1 - 2 γ} - γ)}}^{2}] v^{2} = 2 g h (1 - α - γ)$ $\Rightarrow v = \sqrt{\frac{2 g h (1 - α - γ)}{2 + {1 + \frac{γ (1 - 2 γ)}{(1 - γ) (\sqrt{γ^{2} + 1 - 2 γ} - γ)}}^{2}}}$
	Commented by ajfour last updated on 14/Dec/18

	$P l e n t i f u l o f m i n d a n d e f f o r t S i r;$ $t h a n k s, l e t m e t i m e t o f o l l o w i t$ $e n t i r e l y . .$
	Commented by mr W last updated on 14/Dec/18

	Commented by mr W last updated on 14/Dec/18

	$a t t = 0 : x = a, θ = θ_{0}$ $a t t = t_{1} : x = R, θ = θ_{1}$
	Commented by ajfour last updated on 16/Dec/18

	$V e r y b r i l l i a n t S i r, AWESOME!$
	Answered by ajfour last updated on 14/Dec/18

	Commented by ajfour last updated on 14/Dec/18

	$l e t i n i t i a l l y θ = α, x = a, y = a, z = l$ $f i n a l l y θ = β, x = R, y = h - R, z = s$ $x + R \cos θ = z \sin θ$ $R + R \sin θ = h - z \cos θ$ $\Rightarrow R + R \sin θ = h - z (\frac{z \sin θ - x}{R})$ $\Rightarrow \sin θ = \frac{h - R + \frac{x z}{R}}{R + \frac{z^{2}}{R}}$ $________________________$ $\Rightarrow \sin α = \frac{h - R + \frac{a l}{R}}{R + \frac{l^{2}}{R}} . . . (i)$ $& \sin β = \frac{h - R + s}{R + \frac{s^{2}}{R}} . . . (i i)$ $l^{2} = {(h - R - R \sin α)}^{2} + a^{2} . . . (i i i)$ $s^{2} = {(h - R - R \sin β)}^{2} + R^{2} . . . . (i v)$ $f r o m l e n g t h o f s t r i n g,$ $(α R + l + a) - (β R + s + R) = a - R$ $. . . (v)$ $u n k n o w n s a r e : α, β, l, s, a$
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