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Question Number 88623 by ajfour last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

$I f t h e s o l i d c y l i n d e r h a s t o r o l l$ $o n t o t h e i n c l i n e d p l a n e w i t h o u t$ $a j u m p, f i n d u_{m a x} .$
Commented by mr W last updated on 13/Apr/20

${i t}^{'} s o k i f i t j u s t h a p p e n d b y a c c i d e n t .$
Commented by Zainal Arifin last updated on 13/Apr/20

$I am sorry sir$
Commented by Zainal Arifin last updated on 14/Apr/20

$you are right sir . {it}^{'} s by accidence .$
	Answered by mr W last updated on 12/Apr/20

	Answered by ajfour last updated on 12/Apr/20

	Commented by ajfour last updated on 12/Apr/20
	$(1/2)((3/2)mR^2 )(u_(max) ^2 /R^2 )+mgR(1−cos α) = (1/2)((3/2)mR^2 )(v^2 /R^2 ) ((mv^2 )/R)=mgcos α−N_2 for no jump and u_(max) , N_2 =0 ⇒ (3/4)mu_(max) ^2 +mgR(1−cos α) = (3/4)mgRcos α ⇒ u_(max) ^2 =(4/3){(7/4)gRcos α−gR} u_(max) =(√(((4gR)/3)((7/4)cos α−1))) . (And it is right answer, Sir!)$
	$\frac{1}{2} (\frac{3}{2} {m R}^{2}) \frac{u_{m a x}^{2}}{R^{2}} + m g R (1 - \cos α)$ $= \frac{1}{2} (\frac{3}{2} {m R}^{2}) \frac{v^{2}}{R^{2}}$ $\frac{{m v}^{2}}{R} = m g \cos α - N_{2}$ $f o r n o j u m p a n d u_{m a x}, N_{2} = 0$ $\Rightarrow \frac{3}{4} {m u}_{m a x}^{2} + m g R (1 - \cos α)$ $= \frac{3}{4} m g R \cos α$ $\Rightarrow u_{m a x}^{2} = \frac{4}{3} {\frac{7}{4} g R \cos α - g R}$ $u_{m a x} = \sqrt{\frac{4 g R}{3} (\frac{7}{4} \cos α - 1)} .$ $(A n d i t i s r i g h t a n s w e r, S i r!)$
	Commented by mr W last updated on 12/Apr/20

	$t h i s i s a l s o w h a t i t h o u g h t : t h e$ $c y l i n d e r s h o u l d b e a b l e t o r o t a t e$ $a b o u t t h e e d g e w i t h o u t l o s i n g t h e$ $c o n t a c t f o r c e .$
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