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Question Number 28113 by ajfour last updated on 20/Jan/18

Commented by ajfour last updated on 20/Jan/18

$I f s y s t e m o f d i s c, s p r i n g, a n d$ $b l o c k o f m a s s m, b e r e l e a s e d f r o m$ $r e s t w i t h s p r i n g i n n a t u r a l l e n g t h,$ $(f r i c t i o n b e i n g s u f f i c e n t t o$ $p r e v e n t s l i p p i n g) f i n d m a x i m u m$ $d e f o r m a t i o n i n s p r i n g .$ $A l s o f i n d t h e m i n i m u m c o e f f i c i e n t$ $o f f r i c t i o n r e q u i r e d t o p r e v e n t$ $s l i p p i n g o f d i s c o n t h e f i x e d$ $p l a t f o r m .$
	Answered by mrW2 last updated on 21/Jan/18

	$x_{1} = d i s p l a c e m e n t o f b l o c k$ $a_{1} = a c c e l e r a t i o n o f b l o c k = \frac{d^{2} x_{1}}{{d t}^{2}}$ $x_{2} = d i s p l a c e m e n t o f d i s c$ $a_{2} = a c c e l e r a t i o n o f d i s c = \frac{d^{2} x_{2}}{{d t}^{2}}$ $T = f o r c e i n s p r i n g$ $e = d e f o r m a t i o n i n s p r i n g = x_{1} - x_{2}$ $S p r i n g :$ $T = k e = k (x_{1} - x_{2})$ $\Rightarrow \frac{d^{2} T}{{d t}^{2}} = k (a_{1} - a_{2})$ $B l o c k :$ $m g - T = {m a}_{1}$ $\Rightarrow a_{1} = g - \frac{T}{m}$ $D i s c :$ $T R = I_{2} α_{2} = (\frac{{m R}^{2}}{2} + {m R}^{2}) \times \frac{a_{2}}{R} = \frac{3 m R}{2} \times a_{2}$ $\Rightarrow a_{2} = \frac{2 T}{3 m}$ $\Rightarrow \frac{d^{2} T}{{d t}^{2}} = k (a_{1} - a_{2}) = k (g - \frac{T}{m} - \frac{2 T}{3 m}) = \frac{k}{m} (m g - \frac{5 T}{3})$ $\Rightarrow \frac{d^{2} T}{{d t}^{2}} = \frac{k}{m} (m g - \frac{5 T}{3})$ $l e t S = m g - \frac{5 T}{3}$ $\frac{d^{2} S}{{d t}^{2}} = - \frac{5}{3} \times \frac{d^{2} T}{{d t}^{2}}$ $\Rightarrow - \frac{3}{5} \frac{d^{2} S}{{d t}^{2}} = \frac{k}{m} S$ $\Rightarrow \frac{d^{2} S}{{d t}^{2}} + \frac{5 k}{3 m} S = 0$ $\Rightarrow S = c_{1} \sin ω t + c_{2} \cos ω t$ $w i t h ω = \sqrt{\frac{5 k}{3 m}}$ $\Rightarrow T = \frac{3}{5} (m g - S) = \frac{3}{5} (m g - c_{1} \sin ω t - c_{2} \cos ω t)$ $\Rightarrow \frac{d T}{d t} = \frac{3}{5} (- c_{1} \cos ω t + c_{2} \sin ω t) ω$ $a t t = 0 : T = 0 a n d \frac{d T}{d t} = k (v_{1} - v_{2}) = 0$ $m g - c_{2} = 0 \Rightarrow c_{2} = m g$ $\Rightarrow c_{1} = 0$ $\Rightarrow T = \frac{3 m g}{5} (1 - \cos ω t)$ $\Rightarrow T_{m a x} = \frac{6 m g}{5} a t ω t = (2 n - 1) π o r t = (2 n - 1) π \sqrt{\frac{3 m}{5 k}}$ $\Rightarrow e_{m a x} = \frac{6 m g}{5 k}$ $T - f = {m a}_{2}$ $f = T - {m a}_{2} = T - \frac{2 T}{3} = \frac{T}{3} ⩽ μ m g$ $\Rightarrow μ ⩾ \frac{T_{m a x}}{3 m g} = \frac{6 m g}{3 \times 5 m g} = \frac{2}{5} = 0.4$ $a_{1} = g - \frac{3 g}{5} (1 - \cos ω t) = \frac{g}{5} (2 + 3 \cos ω t)$ $v_{1} = \frac{g}{5} (2 t + \frac{3}{ω} \sin ω t)$ $\Rightarrow x_{1} = \frac{g}{5} [t^{2} + \frac{3}{ω^{2}} (1 - \cos ω t)]$ $a_{2} = \frac{2 g}{5} (1 - \cos ω t)$ $\Rightarrow v_{2} = \frac{2 g}{5} (t - \frac{1}{ω} \sin ω t)$ $\Rightarrow x_{2} = \frac{g}{5} [t^{2} - \frac{2}{ω^{2}} (1 - \cos ω t)]$ $w e s e e t h e s y s t e m i s i n v i b r a t i o n . t h e$ $f o r c e i n s p r i n g v a r i e s b e t w e e n 0 a n d$ $\frac{6 m g}{5}, a b o u t t h e m e a n v a l u e \frac{3 m g}{5} .$ $a f t e r t h e v i b r a t i o n i s v a n i s h e d, t h e$ $f o r c e i n s p r i n g r e m a i n s c o n s t a n t,$ $t h e s p r i n g a c t s a s n o r m a l s t r i n g,$ $b l o c k a n d d i s c h a v e t h e s a m e$ $a c c e l e r a t i o n :$ $a_{1} = a_{2}$ $g - \frac{T}{m} = \frac{2 T}{3 m}$ $\Rightarrow T = \frac{3 m g}{5} = m e a n v a l u e o f v i b r a t i o n$ $\Rightarrow e = \frac{3 m g}{5}$
	Commented by ajfour last updated on 21/Jan/18

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