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Question Number 145193 by mr W last updated on 03/Jul/21

Commented by mr W last updated on 03/Jul/21

$a n u n i f o r m r o d i s r e l e a s e d h o r i z o n t a l l y$ $a t h e i g h t h o v e r t h e t o p o f a t a l l w a l l .$ $f i n d t h e p o s i t i o n w h e r e t h e r o d$ $s t r i k e s t h e w a l l f o r t h e s e c o n d t i m e$ $i f t h i s h a p p e n s .$
	Answered by mr W last updated on 04/Jul/21

	Commented by mr W last updated on 04/Jul/21

	$u = v e l o c i t y j u s t b e f o r e t h e f i r s t h i t$ $v = v e l o c i t y j u s t a f t e r t h e f i r s t h i t$ $ω = a n g u l a r v e l o c i t y j u s t a f t e r t h e f i r s t h i t$ $P = i m p u l s e g i v e n b y t h e f i r s t h i t$ $u = \sqrt{2 g h}$ $\frac{m}{2} (u^{2} - v^{2}) = \frac{1}{2} \times \frac{{m L}^{2}}{12} \times ω^{2}$ $\Rightarrow 12 (u^{2} - v^{2}) = L^{2} ω^{2} . . . (i)$ $m u - P = m v$ $P (\frac{L}{2} - a) = \frac{{m L}^{2}}{12} ω$ $m (u - v) (\frac{L}{2} - a) = \frac{{m L}^{2}}{12} ω$ $\Rightarrow 6 (u - v) (L - 2 a) = L^{2} ω . . . (i i)$ $(i) / (i i) :$ $ω L = \frac{2 (u + v)}{1 - \frac{2 a}{L}}$ $p u t i n t o (i i) :$ $6 (u - v) (L - 2 a) = L \frac{2 (u + v)}{1 - \frac{2 a}{L}}$ $3 {(1 - \frac{2 a}{L})}^{2} (u - v) = u + v$ $\Rightarrow v = \frac{3 {(1 - \frac{2 a}{L})}^{2} - 1}{3 {(1 - \frac{2 a}{L})}^{2} + 1} u = β u$ $v + u = \frac{6 {(1 - \frac{2 a}{L})}^{2}}{3 {(1 - \frac{2 a}{L})}^{2} + 1} u$ $\Rightarrow ω L = \frac{12 (1 - \frac{2 a}{L})}{3 {(1 - \frac{2 a}{L})}^{2} + 1} u = α u$ $θ = ω t$ $y = v t + \frac{1}{2} {g t}^{2}$ $y = β u \frac{θ}{ω} + \frac{g θ^{2}}{2 ω^{2}}$ $y = \frac{β θ L}{α} + \frac{g θ^{2} L^{2}}{2 α^{2} u^{2}}$ $\Rightarrow y = \frac{β θ L}{α} + \frac{θ^{2} L^{2}}{4 α^{2} h}$ $w h e n t h e r o d s t r i k e s t h e w a l l f o r t h e$ $s e c o n d t i m e,$ $θ = \frac{π}{2} + \sin^{- 1} (1 - \frac{2 a}{L})$ $H = y + \sqrt{{(\frac{L}{2})}^{2} - {(\frac{L}{2} - a)}^{2}} = y + \sqrt{(L - a) a}$ $w i t h$ $α = \frac{12 (1 - \frac{2 a}{L})}{3 {(1 - \frac{2 a}{L})}^{2} + 1}$ $β = \frac{3 {(1 - \frac{2 a}{L})}^{2} - 1}{3 {(1 - \frac{2 a}{L})}^{2} + 1}$
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