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-circula

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)

\dots . (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 \boldsymbol u = \sqrt{\frac{(3 \sqrt{3} + 4) \boldsymbol 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

\boldsymbol 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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