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Question Number 146026 by ajfour last updated on 10/Jul/21

Commented by ajfour last updated on 10/Jul/21

$l e t s a s s u m e n o f r i c t i o n . . s o$ $s l i d e s .$
Commented by ajfour last updated on 10/Jul/21

$A p a r a b o l i c w e d g e r e s t s o n a$ $f r i c t i o n l e s s g r o u n d . N o w a$ $c y l i n d e r o f a n e q u a l m a s s i s$ $r e l e a s e d o n t o t h e w e d g e - t o p .$ $F i n d m a x i m u m c o m p r e s s i o n$ $o f t h e s p r i n g (b y t h e t i m e t h e$ $c y l i n d e r l o s e s c o n t a c t w i t h$ $t h e c u r v e d s u r f a c e t h e f i r s t$ $t i m e) .$
Commented by mr W last updated on 10/Jul/21

$c y l i n d e r r o l l s o r s l i d e s a l o n g t h e$ $w e d g e ?$
	Answered by mr W last updated on 10/Jul/21

	Commented by mr W last updated on 11/Jul/21

	$a t t i m e t$ $d i s p l a c e m e n t o f w e d g e i s s$ $P (p, {k p}^{2}) w i t h k = 1$ $\tan θ = 2 k p$ $U = \frac{d s}{d t}$ $u = \frac{d p}{d t}$ $x_{C} = s + p - r \sin θ = s + p - \frac{2 r k p}{\sqrt{1 + {(2 k p)}^{2}}}$ $y_{C} = {k p}^{2} + r \cos θ = {k p}^{2} + \frac{r}{\sqrt{1 + {(2 k p)}^{2}}}$ $v_{x} = U + [1 - \frac{2 r k}{\sqrt{1 + {(2 k p)}^{2}}} + \frac{8 {r k}^{3} p^{2}}{{(1 + {(2 k p)}^{2})}^{\frac{3}{2}}}] u$ $v_{y} = 2 k p [1 - \frac{2 r k}{{(1 + {(2 k p)}^{2})}^{\frac{3}{2}}}] u$ $. . . . .$
	Answered by ajfour last updated on 10/Jul/21

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