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Question Number 109268 by mr W last updated on 22/Aug/20

Commented by mr W last updated on 22/Aug/20

$b l o c k m i s r e l e a s e d a t h e i g h t h .$ $f i n d t h e t i m e i t n e e d s t o r e a c h t h e$ $g r o u n d .$ $a l l m o t i o n i s f r i c t i o n l e s s .$
Commented by Dwaipayan Shikari last updated on 22/Aug/20

$m g - T = m a$ $2 T = (M + m) A (2 T = t o t a l t e n s i o n i n x d i r e c t i o n f o r b l o c k (M + m)$
Commented by mr W last updated on 22/Aug/20

$n o t c o r r e c t s i r!$ $y o u a s s u m e d t h a t m a n d M h a v e t h e$ $s a m e a c c e l e r a t i o n i n x d i r e c t i o n, b u t$ $t h i s i s o n l y t h e c a s e w h e n t h e s t r i n g$ $h o l d i n g m i s i n c l i n e d, b u t t h i s m e a n s$ $t h a t t h e a c c e l e r a t i o n o f m i n y$ $d i r e c t i o n i s n o t e q u a l t o a a s y o u$ $a s s u m e d .$ $b e s i d e s y o u w r o t e 2 T = (M + m) a,$ $h o w d i d y o u g e t 2 T ?$
Commented by mr W last updated on 22/Aug/20

$f o r b l o c k m t h e r e i s n o t t e n s i o n f o r c e$ $i n x d i r e c t i o n .$
Commented by Dwaipayan Shikari last updated on 22/Aug/20

$I h a v e a s s u m e d (M + m) a s s y s t e m . i f M g o e s i n x d i r e c t i o n t h e n$ $m a s s w i l l a l s o m o v e i n x d i r e c t i o n$
	Answered by mr W last updated on 22/Aug/20

	Commented by ajfour last updated on 22/Aug/20

	$T (1 - \sin θ) = M (\frac{d^{2} s}{{d t}^{2}})$ $m g \cos θ + m (\frac{d^{2} s}{{d t}^{2}}) \sin θ - T = m [\frac{d^{2} s}{{d t}^{2}} - R {(\frac{d θ}{d t})}^{2}]$ $m (\frac{d^{2} s}{{d t}^{2}}) \cos θ - m g \sin θ = m s (\frac{d^{2} θ}{{d t}^{2}})$ $- - - - - - - - - - - - - - - - - -$ $l e t \frac{M}{m} = k \Rightarrow$ $g \cos θ + (\frac{d^{2} s}{{d t}^{2}}) \sin θ - \frac{k}{(1 - \sin θ)} (\frac{d^{2} s}{{d t}^{2}})$ $= \frac{d^{2} s}{{d t}^{2}} - R {(\frac{d θ}{d t})}^{2} &$ $\frac{d^{2} s}{{d t}^{2}} \cos θ - g \sin θ = s (\frac{d^{2} θ}{{d t}^{2}})$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$
	Commented by mr W last updated on 22/Aug/20

	$t h a n k s s i r!$ $a s o l u t i o n e v e n f o r t h i s s i m p l e c a s e$ $s e e m s n o t t o b e e a s y .$
	Commented by ajfour last updated on 22/Aug/20
	$mg−Tcos θ=m((d^2 y/dt^2 )) T(1−sin θ)=M((d^2 s/dt^2 )) m((d^2 s/dt^2 ))−Tsin θ=m((d^2 X/dt^2 )) tan θ=(X/y) −−−−−−−−−−−−−−−− g−(T/m)((y/s))=(d^2 y/dt^2 ) .....(i) (T/m)(1−(X/s))=k((d^2 s/dt^2 )) ......(ii) (d^2 s/dt^2 )−(T/m)((X/s))=(d^2 X/dt^2 ) ......(iii) −−−−−−−−−−−−−−−− from (ii) & (iii) (d^2 s/dt^2 ){1−(k/(((s/X)−1)))}=(d^2 X/dt^2 ) let (s/X) = z ⇒ d^2 (Xz){1−(k/(z−1))}=d^2 X (1−(k/(z−1)))d(zdX+Xdz)=d(dX) .......$
	$m g - T \cos θ = m (\frac{d^{2} y}{{d t}^{2}})$ $T (1 - \sin θ) = M (\frac{d^{2} s}{{d t}^{2}})$ $m (\frac{d^{2} s}{{d t}^{2}}) - T \sin θ = m (\frac{d^{2} X}{{d t}^{2}})$ $\tan θ = \frac{X}{y}$ $- - - - - - - - - - - - - - - -$ $g - \frac{T}{m} (\frac{y}{s}) = \frac{d^{2} y}{{d t}^{2}} . . . . . (i)$ $\frac{T}{m} (1 - \frac{X}{s}) = k (\frac{d^{2} s}{{d t}^{2}}) . . . . . . (i i)$ $\frac{d^{2} s}{{d t}^{2}} - \frac{T}{m} (\frac{X}{s}) = \frac{d^{2} X}{{d t}^{2}} . . . . . . (i i i)$ $- - - - - - - - - - - - - - - -$ $f r o m (i i) & (i i i)$ $\frac{d^{2} s}{{d t}^{2}} {1 - \frac{k}{(\frac{s}{X} - 1)}} = \frac{d^{2} X}{{d t}^{2}}$ $l e t \frac{s}{X} = z$ $\Rightarrow d^{2} (X z) {1 - \frac{k}{z - 1}} = d^{2} X$ $(1 - \frac{k}{z - 1}) d (z d X + X d z) = d (d X)$ $. . . . . . .$
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