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Question Number 178673 by Tawa11 last updated on 20/Oct/22

	Answered by mr W last updated on 20/Oct/22

	$I = \frac{{m r}^{2}}{2} + {m r}^{2} = \frac{3 {m r}^{2}}{2}$ $\frac{I ω^{2}}{2} = m g L \sin θ$ $ω = \frac{1}{r} \sqrt{\frac{2 g L \sin θ}{\frac{3}{2}}} = \frac{2}{r} \sqrt{\frac{g L \sin θ}{3}} = \frac{2}{0.1} \sqrt{\frac{9.81 \times 6 \times \sin 30 °}{3}} = 62.64 r a d / s$ $v = ω r = 6.264 m / s$ $t = \frac{d}{v \cos θ}$ $H = v \sin θ t + \frac{{g t}^{2}}{2}$ $H = v \sin θ \times \frac{d}{v \cos θ} + \frac{g}{2} \times \frac{d^{2}}{v^{2} \cos^{2} θ}$ $\Rightarrow \frac{g (1 + \tan^{2} θ)}{2 v^{2}} d^{2} + \tan θ d - H = 0$ $d = \frac{v^{2}}{g (1 + \tan^{2} θ)} [- \tan θ + \sqrt{\tan^{2} θ + \frac{2 g H (1 + \tan^{2} θ)}{v^{2}}}]$ $d = \frac{6 . 264^{2}}{9.81 (1 + \frac{1}{3})} [- \frac{1}{\sqrt{3}} + \sqrt{\frac{1}{3} + \frac{2 \times 9.81 \times 5 (1 + \frac{1}{3})}{6 . 264^{2}}}] \approx 4.01 m$
	Commented by Tawa11 last updated on 20/Oct/22

	$God bless you sir . I appreciate .$
	Answered by Acem last updated on 20/Oct/22

	$H e r e μ = 0 a n d w i t h o u t f r i c t i o n t h e c y l i n d e r {w o n}^{'} t$ $r o l l b u t w i l l s l i p p s . . . a n y w a y i t s e e m s t h e d i r e c t o r$ $w a n t s t o r o l l i t b y f o r c e!$ $a ∙ w = \frac{\sqrt{\frac{2 g L \sin θ}{1 + c}}}{R} = 62.64 r a d . s^{- 1}$ $; c = \frac{1}{2}, v = 6.264 m . s^{- 1}$ $b ∙ v_{0} = 6.264 m . s^{- 1} ∡ 30^{°}$ $H = v_{0} . \sin θ . t + 0.5 g . t^{2}$ $4.905 t^{2} + 3.132 t - 5 = 0 \Rightarrow t = 0.739 s$ $H o r i z o n t a l d i s t a n c e = v_{0} \cos θ . t = 4.008 m$
	Commented by Tawa11 last updated on 20/Oct/22

	$God bless you sir . I appreciate .$
	Commented by Acem last updated on 20/Oct/22

	$Y e s i w a s w r o n g, w e m u s t s a y t h e a c c e l e r a t i o n$ $h a s c h a n g e d f r o m g . s i n θ t o g ‘ ‘ n o t s u m . . . {e v e r}^{″}$
	Commented by Acem last updated on 20/Oct/22

	${Y o u}^{'} r e v e r y m u c h w e l c o m e . . . l e t m e k n o w h o w$ $m a n y c y c l e s t h e c y l i n d e r d i d w h e n r e a c h e d t h e e d g e$ $p l u s t h e a n g l e i f a n y .$ $(: t h i n k i t b e f o r e s t a r t i n g t h e s o l v e =)$
	Commented by Tawa11 last updated on 20/Oct/22

	$sir mrW distance is correct sir .$
	Commented by Tawa11 last updated on 20/Oct/22

	$Thanks for your time sir .$
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