A particle will leave a vertical circle of radius r, when its velocity at the lowest point of the circle (v_L ) is (a) (√(2gr)) (b) (√(5gr)) (c) (√(3gr)) (d) (√(6gr))


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Question Number 21604 by Tinkutara last updated on 29/Sep/17

$A particle will leave a vertical circle of$ $radius r, when its velocity at the lowest$ $point of the circle (v_{L}) is$ $(a) \sqrt{2 g r}$ $(b) \sqrt{5 g r}$ $(c) \sqrt{3 g r}$ $(d) \sqrt{6 g r}$
	Answered by ajfour last updated on 29/Sep/17

	$a t t h e t o p i f i t l e a v e s t h e$ $c i r c l e, t h e n$ $m g = m \frac{v_{H}^{2}}{r} o r {m v}_{H}^{2} = m g r . . . (i)$ $f r o m c o n s e r v a t i o n o f e n e r g y :$ $2 m g r = \frac{1}{2} m (v_{L}^{2} - v_{H}^{2})$ $\Rightarrow 4 g r = {m v}_{L}^{2} - m g r [s e e (i)]$ $v_{L} = \sqrt{5 g r} .$ $w h e b e v e r v_{L} < \sqrt{5 g r} p a r t i c l e$ $l e a v e s t h e v e r t i c a l c i r c l e .$ $S o (a) a n d (c) .$
	Commented by Tinkutara last updated on 29/Sep/17

	$But answer is given only (c) not (a) .$
	Commented by ajfour last updated on 29/Sep/17

	$b u t w i t h v_{L} = \sqrt{2 g r} i t s w i n g s o n l y$ $i n t h e l o w e r h a l f v e r t i x a l c i r c l e;$ $n o c h a n c e o f l e a v i n g t h e c i r c l e .$ $s o j u s t (c) a n d n o t (a) . s o r r y$ $f o r t h e m i s t a k e .$
	Commented by Tinkutara last updated on 29/Sep/17

	$Thank you very much Sir!$
	Commented by Tinkutara last updated on 29/Sep/17

	$Can you please guide Q . 21307 ?$
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