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Question Number 44454 by ajfour last updated on 29/Sep/18

Commented by ajfour last updated on 29/Sep/18

$A s p h e r e i n p u r e r o l l i n g m o t i o n$ $o n a r o u g h g r o u n d, e n c o u n t e r s a s t a i r,$ $c o l l i d e s c o m p l e t e l y i n e l a s t i c a l l y$ $w i t h i t s c o r n e r . J u s t a f t e r i t$ $c o m e s a t o p t h e s t a i r, f i n d i t s$ $v e l o c i t y i n t e r m s o f u, h, a n d R .$
Commented by ajfour last updated on 30/Sep/18

Commented by ajfour last updated on 30/Sep/18

$I n i t i a l a n g u l a r m o m e n t u m$ $o f s p h e r e a b o u t c o r n e r P :$ $L_{0} = (\frac{2}{5} {M R}^{2}) \frac{u}{R} + M u (R - h)$ $j u s t a f t e r c o l l i s i o n$ $L_{1} = \frac{7}{5} {M R}^{2} (\frac{v_{1}}{R})$ $a b o u t P w e c a n c o n s e r v e a m g u l a r$ $m o m e n t u m, s o L_{1} = L_{0}$ $\Rightarrow \frac{7 v_{1} R}{5} = \frac{2 u R}{5} + u (R - h) . . . (i)$ $v_{1} = \frac{2 u}{7} + \frac{5 u}{7} - \frac{5}{7} (\frac{h}{R}) u$ $\Rightarrow v_{1} = (1 - \frac{5 h}{7 R}) u = λ u$ $N o w a s i t r i s e s a b o v e,$ $\frac{1}{2} (\frac{7}{5} {M R}^{2}) {(\frac{v_{1}}{R})}^{2} = M g h$ $+ \frac{1}{2} (\frac{7}{5} {M R}^{2}) {(\frac{v_{2}}{R})}^{2}$ $u s i n g (i) :$ $\frac{1}{2} (\frac{7}{5}) λ^{2} u^{2} = g h + \frac{1}{2} (\frac{7}{5}) v_{2}^{2}$ $\Rightarrow v_{2}^{2} = λ^{2} u^{2} - \frac{10}{7} g h$ $v_{2}^{2} = {(1 - \frac{5 h}{7 R})}^{2} u^{2} - \frac{10}{7} g h$ $\Rightarrow v_{2} = \sqrt{{(1 - \frac{5 h}{7 R})}^{2} u^{2} - \frac{10}{7} g h} .$
Commented by MrW3 last updated on 30/Sep/18

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