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Question Number 91167 by mr W last updated on 28/Apr/20

Commented by mr W last updated on 28/Apr/20

$a l s o$
Commented by mr W last updated on 28/Apr/20

$a l o n g r o d w i t h m a s s m a n d l e n g t h l$ $r e s t s o n a c y l i n d e r w i t h m a s s M a n d$ $r a d i u s R . a s m a l l i m p u l s i s g i v e n$ $a n d t h e r o d b e g i n s t o f a l l . f i n d t h e$ $t i m e i t n e e d s t i l l o n e e n d h i t s t h e$ $g r o u n d . f r i c t i o n b e t w e e n c y l i n d e r$ $a n d r o d a n d g r o u n d i s e n o u g h s u c h$ $t h a t n o s l i p p i n g o c c u r s .$
Commented by jagoll last updated on 28/Apr/20

$p h y s i c s s i r ?$
Commented by ajfour last updated on 29/Apr/20

$v e r y d i f f i c u l t o n e s i r, p o i n t o f$ $c o n t a c t o f c y l i n d e r a n d p l a n k (r o d)$ $s h a l l g o o n c h a n g i n g t o o, m a n y$ $i n t e r d e p e n d e n t t h i n g s . .!$
	Answered by mr W last updated on 28/Apr/20

	Commented by mr W last updated on 29/Apr/20

	$r o t a t i o n o f c y l i n d e r φ$ $r o t a t i o n o f r o d θ$ $O C = R φ$ $P S = R (φ - θ)$ $r o d h i t s t h e g r o u n d i f$ $[\frac{l}{2} + R (φ - θ)] \tan \frac{θ}{2} = R$ $x_{S} = R φ + R \sin θ + R (φ - θ) \cos θ$ $y_{S} = R \cos θ - R (φ - θ) \sin θ$ $v_{S x} = R [\frac{d φ}{d t} + \cos θ \frac{d θ}{d t} + (\frac{d φ}{d t} - \frac{d θ}{d t}) \cos θ - (φ - θ) \sin θ \frac{d θ}{d t}]$ $v_{S x} = R [\frac{d φ}{d t} (1 + \cos θ) - (φ - θ) \sin θ \frac{d θ}{d t}]$
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