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Question Number 28171 by ajfour last updated on 21/Jan/18

Commented by ajfour last updated on 21/Jan/18

$Q . 28113 a n o t h e r s o l u t i o n :$
	Answered by ajfour last updated on 21/Jan/18

	$T R = (\frac{3 {M R}^{2}}{2}) (\frac{a_{2}}{R})$ $\Rightarrow \frac{T}{a_{2}} = \frac{3 M}{2} (m a s s i n e r t i a o f d i s c)$ $a_{c m} o f \frac{3 M}{2} a n d M = \frac{M g}{(5 M / 2)} = \frac{2 g}{5}$ $f o r c e o n M f r o m c e n t r e o f m a s s$ $f r a m e = M g - \frac{2 M g}{5} = \frac{3 M g}{5}$ $s p r i n g c o n s t a n t o f s e c t i o n o f$ $s p r i n g a t t a c h e d t o M i s$ $= k (\frac{5 M / 2}{3 M / 2}) = \frac{5 k}{3}$ $u n d e r m a x . s t r e t c h i n g o f t h i s$ $s e c t i o n o f s p r i n g :$ $\frac{1}{2} (\frac{5 k}{3}) e_{2}^{2} = (\frac{3 M g}{5}) e_{2}$ $\Rightarrow e_{2} = \frac{18 M g}{25 k}$ $m a x s t r e t c h i n g i n f u l l s p r i n g$ $= \frac{5}{3} \times \frac{18 M g}{25 k} = \frac{6 M g}{5 k} .$ $f o r m i n i m u m f r i c t i o n c o e f f i c i e n t$ $b e t w e e n d i s c a n d p l a t f o r m s o t h a t$ $d i s c d o e s n o t s l i p,$ $T_{m a x} - μ (M g) = M [\frac{T_{m a x} R}{(\frac{3 {M R}^{2}}{2})}] R$ $w i t h T_{m a x} = \frac{6 M g}{5}$ $μ M g = \frac{6 M g}{5} - M (\frac{2}{3 M} \times \frac{6 M g}{5})$ $\Rightarrow μ = \frac{6}{5} - \frac{4}{5} = \frac{2}{5} .$
	Commented by ajfour last updated on 21/Jan/18
	Thank you Sir.
	Commented by mrW2 last updated on 21/Jan/18

	$v e r y t r i c k y!$
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