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Question Number 214656 by mr W last updated on 15/Dec/24

Commented by mr W last updated on 17/Dec/24

$s a m e c o n d i t i o n s a s i n Q 214449$ $f i n d t h e f i n a l s p e e d o f t h e l o w e r$ $c y l i n d e r .$
	Answered by mr W last updated on 15/Dec/24

	Commented by mr W last updated on 17/Dec/24
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	Commented by mr W last updated on 16/Dec/24

	$(X, Y) = c e n t e r o f l o w e r c y l i n d e r$ $(x, y) = c e n t e r o f u p p e r c y l i n d e r$ $Y = r_{2}$ $x = X + (r_{1} + r_{2}) \sin θ$ $y = r_{2} + (r_{1} + r_{2}) \cos θ$ $l e t ω = \frac{d θ}{d t}$ $V_{x} = - \frac{d X}{d t} = U (\leftarrow)$ $V_{y} = 0 (↓)$ $v_{x} = \frac{d x}{d t} = - U + (r_{1} + r_{2}) ω \cos θ (\to)$ $v_{y} = - \frac{d y}{d t} = (r_{1} + r_{2}) ω \sin θ (↓)$ $b o t h c y l i n d e r s o b t a i n n o r o t a t i o n .$ $l e t k = \frac{m_{2}}{m_{1}} + 1 > 1$ $m_{2} U = m_{1} v_{x}$ $\frac{m_{2} U}{m_{1}} = v_{x} = - U + (r_{1} + r_{2}) ω \cos θ$ $\Rightarrow U = \frac{(r_{1} + r_{2}) ω \cos θ}{k}$ $l o s s i n P . E . = g a i n i n K . E .$ $m_{1} g (r_{1} + r_{2}) (1 - \cos θ) = \frac{m_{2} U^{2}}{2} + \frac{m_{1} [- 2 U (r_{1} + r_{2}) ω \cos θ + U^{2} + {(r_{1} + r_{2})}^{2} ω^{2}]}{2}$ $2 g (r_{1} + r_{2}) (1 - \cos θ) = {k U}^{2} - 2 U (r_{1} + r_{2}) ω \cos θ + {(r_{1} + r_{2})}^{2} ω^{2}$ $2 g (1 - \cos θ) = (1 - \frac{\cos^{2} θ}{k}) (r_{1} + r_{2}) ω^{2}$ $\Rightarrow ω^{2} = \frac{2 k g (1 - \cos θ)}{(r_{1} + r_{2}) (k - \cos^{2} θ)}$ $\Rightarrow ω \frac{d ω}{d θ} = \frac{g k (k - 2 \cos θ + \cos^{2} θ) \sin θ}{(r_{1} + r_{2}) {(k - \cos^{2} θ)}^{2}}$ $A_{x} = \frac{{d V}_{x}}{d t} = ω \frac{d U}{d θ} = \frac{(r_{1} + r_{2}) ω}{k} (- ω \sin θ + \cos θ \frac{d ω}{d θ})$ $A_{x} = g \sin θ [\frac{2 k \cos θ - 2 k - 2 \cos^{3} θ + 2 \cos^{2} θ + k \cos θ - 2 \cos^{2} θ + \cos^{3} θ}{{(k - \cos^{2} θ)}^{2}}]$ $A_{x} = g \sin θ [\frac{- \cos^{3} θ + 3 k \cos θ - 2 k}{{(k - \cos^{2} θ)}^{2}}]$ $N \sin θ = m_{2} A_{x}$ $\Rightarrow \frac{N}{m_{2} g} = \frac{- \cos^{3} θ + 3 k \cos θ - 2 k}{{(k - \cos^{2} θ)}^{2}}$ $N = 0 :$ $\cos^{3} θ - 3 k \cos θ + 2 k = 0$ $\Rightarrow \cos θ = 2 \sqrt{k} \sin (\frac{1}{3} \sin^{- 1} \frac{1}{\sqrt{k}})$ $a g a i n w e s e e t h a t θ i s i n d e p e n d e n t$ $f r o m t h e r a d i i o f t h e c y l i n d e r s, b u t$ $o n l y f r o m t h e r a t i o o f t h e i r m a s s e s .$ $U = \frac{(r_{1} + r_{2}) ω \cos θ}{k}$ $\Rightarrow U = \cos θ \sqrt{\frac{2 (1 - \cos θ)}{k (k - \cos^{2} θ)}} \sqrt{g (r_{1} + r_{2})}$
	Commented by mr W last updated on 16/Dec/24

	Commented by mr W last updated on 16/Dec/24

	$t h i s p i c t u r e s h o w s w h a t m o t i o n s$ $t h e c y l i n d e r s m a k e .$ $\underset{―}{f r o m t = 0 t o t_{1} :}$ $t h e c y l i n d e r s h a v e c o n t a c t t o e a c h$ $o t h e r . t h e u p p e r c y l i n d e r p u s h e s$ $t h e l o w e r o n e t o w a r d s l e f t a n d$ $m a k e s t h i s t o m o v e f a s t e r a n d f a s t e r .$ $N > 0$ $\underset{―}{a t t = t_{1} :}$ $t h e c o n t a c t b e t w e e n t h e c y l i n d e r s$ $b r e a k s .$ $N = 0 .$ $\underset{―}{a f t e r t = t_{1} :}$ $t h e c y l i n d e r s m o v e i n d e p e n d e n t l y$ $f r o m e a c h o t h e r . t h e l o w e r o n e$ $m o v e s w i t h c o n t a n t s p e e e d t o w a r d s$ $l e f t . t h e u p p e r o n e m o v e s t o w a r d s$ $r i g h t a n d f a l l s o n t o t h e g r o u n d a n d$ $r e b o u n d s f r o m t h e g r o u n d a g a i n$ $a n d r e p e a t s t h i s o n a n d o n . . . . .$
	Commented by ajfour last updated on 16/Dec/24

	$L o o k s m a r v e l l o u s! w h a t a f i n e$ $t r e a t m e n t, i s h a l l l o o k i n t o t h e$ $d e t a i l s . g r e a t w o r k s i r .$
	Commented by mr W last updated on 16/Dec/24

	$i u s e d m y s t a n d a r d m e t h o d . i h o p e$ $y o u a l s o g e t t h i s r e s u l t u s i n g y o u r$ $m e t h o d .$
	Commented by ajfour last updated on 16/Dec/24
	https://youtu.be/ANHpcwzc_BM?si=ociiHdkc2qWXLw_5
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