IIT JEE , AIEEE Forum - Physics , Mechanics

Greatdreams

The position of instantaneous axis of rotation should be (l/2 sin

, l/2 cos

)

l/2 = x ( see in figure)

Now applying conservation of mechanical energy ,

Decrease in potential energy of the rod = increase in rotational energy about instantaneous axis of rotation

m g l/2 ( 1 - sin

) = 1/2 I

²

Now I = m l²/ 12 + m x ² = ml ² / 12 + ml ² /4 = ml ² /3

So now putting the value of I ,
we have

mg l/2 ( 1 - sin

) = 1/2 ( m l ²/ 3 )

²

So

² = 3g ( 1 - sin

) / l

So

=

3g / l ( 1 - sin

)

I think answer is as given by Ramyani didi

.I hope thats a nishpap solution.

Sorry for daring to touch mechanics section even after guys like Anchitsaini being around.

D.C.Pandey seems to be a good book.But better books are available.

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