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Question Number 20764 by Tinkutara last updated on 02/Sep/17

A s m a l l b e a d i s s l i p p e d o n a h o r i z o n t a l

r o d o f l e n g t h l . T h e r o d s t a r t s m o v i n g

w i t h a h o r i z o n t a l a c c e l e r a t i o n a i n a

d i r e c t i o n m a k i n g a n a n g l e α w i t h t h e

l e n g t h o f t h e r o d . A s s u m i n g t h a t

i n i t i a l l y t h e b e a d i s i n t h e m i d d l e o f

t h e r o d, f i n d t h e t i m e e l a p s e d b e f o r e

t h e b e a d l e a v e s t h e r o d . C o e f f i c i e n t o f

f r i c t i o n b e t w e e n t h e b e a d a n d t h e r o d

i s μ . (N e g l e c t g r a v i t y) .

Answered by ajfour last updated on 02/Sep/17

Commented by ajfour last updated on 02/Sep/17

N = m a \sin α

l e t a_{r} b e r e l a t i v e a c c e l e r a t i o n o f

b e a d a l o n g r o d .

m a \cos α - μ N = {m a}_{r}

\Rightarrow {m a}_{r} = m a \cos α - μ (m a \sin α)

a_{r} = a (\cos α - μ \sin α)

\frac{l}{2} = \frac{1}{2} a_{r} t^{2}

\Rightarrow t = \sqrt{\frac{l}{a_{r}}} = \sqrt{\frac{l}{a (\cos α - μ \sin α)}} .

Commented by Tinkutara last updated on 02/Sep/17

Why the masses of rod and bead same ?

Commented by ajfour last updated on 02/Sep/17

i h a v e n o t a s s u m e d a n y s u c h t h i n g . .

Commented by Tinkutara last updated on 02/Sep/17

In 4^{th} line, you have assumed {m a}_{r} and

m a \cos α, is m the mass of the bead and

rod, both ?

Commented by ajfour last updated on 02/Sep/17

r o d i s g i v e n t o b e a c c e l e r a t i n g,

i h a v e m a r k e d f o r c e s o n b e a d .

m a i s t h e p s e u d o f o r c e o n b e a d

f r o m f r a m e o f r o d . m i s t h e

m a s s o f b e a d o n l y .

Commented by Tinkutara last updated on 02/Sep/17

Thank you very much Sir!