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Question Number 31894 by Tinkutara last updated on 16/Mar/18

Commented by Tinkutara last updated on 16/Mar/18
Can you explain with diagrams please? I can't understand in which direction forces are balanced.
Commented by mrW2 last updated on 16/Mar/18

$m {(v \cos θ)}^{2} / r = m g$ $\Rightarrow θ = \cos^{- 1} \sqrt{\frac{g r}{v^{2}}}$
Commented by mrW2 last updated on 16/Mar/18

Commented by mrW2 last updated on 16/Mar/18

$s i n c e t h e s u r f a c e i s s m o o t h, t h e$ $c y l i n d e r c a n o n l y a p p l y a n o r m a l f o r c e$ $t o w a r d t h e c e n t e r l i n e o f c y l i n d e r .$
Commented by Tinkutara last updated on 16/Mar/18

$B u t v \cos θ i s n o t a l o n g t h e d i r e c t i o n$ $o f r a d i u s j o i n i n g t h e p a r t i c l e t h e n$ $w h y \frac{m {(v \cos θ)}^{2}}{r} ?$
Commented by mrW2 last updated on 16/Mar/18

$v \cos θ i s t h e v e l o c i t y o f p a r t i c l e i n$ $h o r i z o n t a l d i r e c t i o n . i n h o r i z o n t a l$ $d i r e c t i o n t h e p a r t i c l e m a k e s a$ $c i r c u l a r m o t i o n w i t h r a d i u s r . t h e$ $c o r r e s p o n d i n g n o r m a l f o r c e o n$ $c y l i n d e r w a l l i s \frac{m {(v \cos θ)}^{2}}{r} w h i c h$ $e q u a l s m g a t t h e m o m e n t .$
Commented by Tinkutara last updated on 16/Mar/18
But why the particle makes a circular turn in the direction of horizontal component of velocity?
Commented by mrW2 last updated on 16/Mar/18

$i t i s s a i d t h a t t h e p a r t i c l e m o v e s a l o n g$ $t h e i n n e r w a l l o f c y l i n d e r . t h i s m o t i o n$ $h a s t w o c o m p o n e n t s :$ $c i r c u l a r m o t i o n w i t h v e l o c i t y v \cos θ a n d$ $m o t i o n p a r a l l e l l y t o t h e a x i s o f$ $c y l i n d e r w i t h v e l o c i t y v \sin θ$
Commented by Tinkutara last updated on 17/Mar/18
Thank you very much Sir! I got the answer. ��
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