A particle is suspended vertically from point O by ideal string of length L. It is given horizontal velocity ′v′. There is vertical line AB at a distance (L/8) from P. At some point, it leaves circular motion and follows projectile motion. At the instant it crosses AB, its velocity is horizontal. Find u


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Question Number 24739 by Tinkutara last updated on 25/Nov/17

$A p a r t i c l e i s s u s p e n d e d v e r t i c a l l y$ $f r o m p o i n t O b y i d e a l s t r i n g o f l e n g t h$ $L . I t i s g i v e n h o r i z o n t a l v e l o c i t y^{'} v^{'} .$ $T h e r e i s v e r t i c a l l i n e A B a t a d i s t a n c e$ $\frac{L}{8} f r o m P . A t s o m e p o i n t, i t l e a v e s$ $c i r c u l a r m o t i o n a n d f o l l o w s p r o j e c t i l e$ $m o t i o n . A t t h e i n s t a n t i t c r o s s e s A B,$ $i t s v e l o c i t y i s h o r i z o n t a l . F i n d u$
Commented by Tinkutara last updated on 25/Nov/17

Commented by ajfour last updated on 25/Nov/17

$u = \sqrt{\frac{(3 \sqrt{3} + 4) g L}{2}} .$
Commented by Tinkutara last updated on 25/Nov/17

$H o w ?$
Commented by ajfour last updated on 25/Nov/17

Commented by ajfour last updated on 25/Nov/17

$\frac{1}{2} m (u^{2} - v^{2}) = m g y = m g L (1 + \sin θ)$ $. . . . (i)$ $\frac{{m v}^{2}}{L} = m g \sin θ \Rightarrow v^{2} = g L \sin θ . . (i i)$ $x = \frac{R}{2} = \frac{v^{2} \sin [2 (90 - θ)]}{2 g} = L \cos θ - \frac{L}{8}$ $\Rightarrow g L \sin θ \times \frac{2 \sin θ \cos θ}{2 g} = L (\cos θ - \frac{1}{8})$ $\Rightarrow \cos θ (1 - \cos^{2} θ) = \cos θ - \frac{1}{8}$ $o r \cos θ = \frac{1}{2}$ $N o w f r o m (i) : v^{2} = \frac{\sqrt{3} g L}{2}$ $u s i n g t h i s i n (i) :$ $u^{2} = v^{2} + 2 g L (1 + \sin θ)$ $= \frac{\sqrt{3} g L}{2} + 2 g L (1 + \frac{\sqrt{3}}{2})$ $o r u = \sqrt{\frac{(3 \sqrt{3} + 4) g L}{2}} .$
Commented by Tinkutara last updated on 25/Nov/17

$x = \frac{R}{2} = \frac{v^{2} \sin [2 (90 - θ)]}{2 g} = L \cos θ - \frac{L}{8}$ $H o w t h i s L (\cos θ - \frac{1}{8}) ?$
Commented by ajfour last updated on 25/Nov/17

$s e e d i a g r a m, x = L \cos θ - \frac{L}{8}$ $s i n c e v e l o c i t y i s h o r i z o n t a l$ $w h e n a s a p r o j e c t i l e m a s s^{'} m^{'}$ $c r o s s e s A B s o f r o m t h e p o i n t$ $w h e r e c i r c u l a r m o t i o n e n d s$ $a n d t i l l l i n e A B i s r e a c h e d t h e$ $h o r i z o n t a l d i s t a n c e$ $x = v_{x} △ t = (v \sin θ) (\frac{v \cos θ}{g}) .$
Commented by Tinkutara last updated on 26/Nov/17

$Thank you Sir!$
Commented by ajfour last updated on 27/Nov/17

$θ, f o r a r e a s o n i s a c u t e a n g l e d . .$
Commented by ajfour last updated on 28/Nov/17

$m g L (1 - \cos ϕ) = \frac{1}{2} m (u^{2} - v_{1}^{2})$ $F o r T e n s i o n T t o v a n i s h,$ $m g \cos ϕ = - \frac{{m v}^{2}}{L}$ $\Rightarrow ϕ > 90 ° .$
Commented by mrW1 last updated on 28/Nov/17

$T h a n k y o u s i r!$ $N o w I u n d e r s t a n d t h e q u e s t i o n t o t a l l y .$ $C a n y o u f i n d t h e p o i n t w h e r e t h e$ $p a r t i c l e c o m e s b a c k ?$
Commented by mrW1 last updated on 28/Nov/17

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