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Question Number 40551 by ajfour last updated on 24/Jul/18

Commented by ajfour last updated on 24/Jul/18

$A c h a i n o f m a s s m, l e n g t h l$ $h a n g s t h e t a b l e e d g e o n l y a s m u c h$ $s o t h a t i t j u s t s t a r t s s l i p p i n g$ $d o w n . F r i c t i o n c o e f f i c i e n t$ $b e t w e e n c h a i n a n d t a b l e b e i n g μ .$ $F i n d s p e e d o f c h a i n a s e n d B o f$ $c h a i n r e a c h e s t h e t a b l e c o r n e r .$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18
$let when time=t y length of chain suspended l−y length on the tablle so wt of hanging chain is (m/l)yg↓ force on chain on table are 1)(m/l)(l−y)g↓ 2)friction μ((m/l))(l−y)g 3)forward force which cause the chain to move=(m/l)yg so acc=(((m/l)yg−μ((m/l))(l−y)g)/m)=v(dv/dy) (g/l)y−((μg)/l)(l−y)=v(dv/dy) ∫(g/l)ydy−∫μgdy+((μg)/l)∫ydy=∫vdv now initial suspended lenth of chain say y_0 so as per question (m/l)y_0 g=μ(m/l)(l−y_0 )g y_0 =μ(l−y_0 ) y_0 =((μl)/(1+μ)) wait i am intregating... (g/l)∫_((μl)/(1+μ)) ^l ydy −μg∫_((μl)/(1+μ)) ^l dy+((μg)/l)∫_((μl)/(1+μ)) ^l ydy=∫_0 ^v vdv (g/(2l))∣(y^2 )∣_((μl)/(1+μ)) ^l −μg(l−((μl)/(1+μ)))+((μg)/(2l)){(l^2 −(((μl)/(1+μ)))^2 }=(v^2 /2) (g/(2l)){l^2 −((μ^2 l^2 )/((1+μ)^2 ))}−μg((l/(1+μ)))+((μg)/(2l)){l^2 −((μ^2 l^2 )/((1+μ)^2 ))} (g/(2l))(1+μ)l^2 (((1+2μ+μ^2 −μ^2 )/((1+μ)^2 )))−μg((l/(1+μ)))=(v^2 /2) (g/(2l))l^2 (((1+2μ)/(1+μ)))−μg((l/(1+μ)))=(v^2 /2) ((gl(((1+2μ)/(1+μ)))−2μg((l/(1+μ)))=v^2 )/) ((gl+2μgl−2μgl)/(1+μ))=v^2 v=(√((gl)/(1+μ))) pls check...$
$l e t w h e n t i m e = t y l e n g t h o f c h a i n s u s p e n d e d$ $l - y l e n g t h o n t h e t a b l l e$ $s o w t o f h a n g i n g c h a i n i s \frac{m}{l} y g ↓$ $f o r c e o n c h a i n o n t a b l e a r e$ $1) \frac{m}{l} (l - y) g ↓$ $2) f r i c t i o n μ (\frac{m}{l}) (l - y) g$ $3) f o r w a r d f o r c e w h i c h c a u s e t h e c h a i n t o$ $m o v e = \frac{m}{l} y g$ $s o a c c = \frac{\frac{m}{l} y g - μ (\frac{m}{l}) (l - y) g}{m} = v \frac{d v}{d y}$ $\frac{g}{l} y - \frac{μ g}{l} (l - y) = v \frac{d v}{d y}$ $\int \frac{g}{l} y d y - \int μ g d y + \frac{μ g}{l} \int y d y = \int v d v$ $n o w i n i t i a l s u s p e n d e d l e n t h o f c h a i n s a y y_{0}$ $s o a s p e r q u e s t i o n$ $\frac{m}{l} y_{0} g = μ \frac{m}{l} (l - y_{0}) g$ $y_{0} = μ (l - y_{0})$ $y_{0} = \frac{μ l}{1 + μ}$ $w a i t i a m i n t r e g a t i n g . . .$ $\frac{g}{l} \int_{\frac{μ l}{1 + μ}}^{l} y d y - μ g \int_{\frac{μ l}{1 + μ}}^{l} d y + \frac{μ g}{l} \int_{\frac{μ l}{1 + μ}}^{l} y d y = \int_{0}^{v} v d v$ $\frac{g}{2 l} ∣ (y^{2}) ∣_{\frac{μ l}{1 + μ}}^{l} - μ g (l - \frac{μ l}{1 + μ}) + \frac{μ g}{2 l} {(l^{2} - {(\frac{μ l}{1 + μ})}^{2}} = \frac{v^{2}}{2}$ $\frac{g}{2 l} {l^{2} - \frac{μ^{2} l^{2}}{{(1 + μ)}^{2}}} - μ g (\frac{l}{1 + μ}) + \frac{μ g}{2 l} {l^{2} - \frac{μ^{2} l^{2}}{{(1 + μ)}^{2}}}$ $\frac{g}{2 l} (1 + μ) l^{2} (\frac{1 + 2 μ + μ^{2} - μ^{2}}{{(1 + μ)}^{2}}) - μ g (\frac{l}{1 + μ}) = \frac{v^{2}}{2}$ $\frac{g}{2 l} l^{2} (\frac{1 + 2 μ}{1 + μ}) - μ g (\frac{l}{1 + μ}) = \frac{v^{2}}{2}$ $\frac{g l (\frac{1 + 2 μ}{1 + μ}) - 2 μ g (\frac{l}{1 + μ}) = v^{2}}{}$ $\frac{g l + 2 μ g l - 2 μ g l}{1 + μ} = v^{2}$ $v = \sqrt{\frac{g l}{1 + μ}} p l s c h e c k . . .$
Commented by ajfour last updated on 24/Jul/18

$T h a n k y o u T a n m a y S i r; b u t m r W$ $s i r h a s s o l v e d i t b e a u t i f u l l y .$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18

$y e s y o u a r e r i g h t . . . i h a v e r e a c h e d d e s t i n a t i o n$ $b u t a l o n g t h e l o n g p a t h . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jul/18

$i h a v e p u t t h e v a l u e o f y_{0} e a r l y . . . {M W r}_{3} p u t t h e$ $v a l u e i n t h e f i n a l s t a g e a n d i h a v e c a l c u l a t e d$ $m a s s o f c h a i n b y \frac{m}{l} f a c t o r s o i n c a l c u l a t i o n$ $i t t a k e s t i m e . . . b u t h e p u t s i m p l y ρ = \frac{m}{l}$ $a g a i n p u t t h e v a l u e o f ρ i n f i n a l s t a g e . . .$
	Answered by MrW3 last updated on 24/Jul/18

	$l e n g t h o f h a n g i n g p a r t = y$ $ρ = \frac{m}{l}$ $y_{0} ρ g = μ (l - y_{0}) ρ g$ $\Rightarrow y_{0} = \frac{μ}{1 + μ} l$ $y ρ g - μ (l - y) ρ g = m a = l ρ v \frac{d v}{d y}$ $y - μ (l - y) = \frac{l}{g} v \frac{d v}{d y}$ $\int_{y_{0}}^{l} [(1 + μ) y - μ l] d y = \frac{l}{g} \int_{0}^{v} v d v$ ${[(1 + μ) \frac{y^{2}}{2} - μ l y]}_{y_{0}}^{l} = \frac{l}{g} \frac{v^{2}}{2}$ $\frac{1}{2} (1 + μ) (l^{2} - y_{0}^{2}) - μ l (l - y_{0}) = \frac{l}{g} \frac{v^{2}}{2}$ $[(1 + μ) (l + y_{0}) - 2 μ l] (l - y_{0}) = \frac{l}{g} v^{2}$ $[(1 + μ) (1 + \frac{μ}{1 + μ}) - 2 μ] (1 - \frac{μ}{1 + μ}) l g = v^{2}$ $\frac{l g}{1 + μ} = v^{2}$ $\Rightarrow v = \sqrt{\frac{l g}{1 + μ}}$
	Commented by MrW3 last updated on 24/Jul/18

	$T h a n k s f o r c h e c k i n g!$
	Commented by ajfour last updated on 24/Jul/18

	$I a p o l o g i s e S i r; i t s i n d e e d c o r r e c t .$ $T h a n k s!$
	Commented by MrW3 last updated on 24/Jul/18

	$i f μ = 0 w e g e t v = \sqrt{l g}, a n d t h i s i s r i g h t,$ $s i n c e v = \sqrt{2 g h} = \sqrt{2 g \times \frac{l}{2}} = \sqrt{g l}$
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