IIT JEE , AIEEE Forum - Physics , Mechanics

ramyani

Q. A uniform bar of length l stands vertically touching a wall OA, when slightly displaced, its lower end begins to slide along the floor. obtain an expression for the angular velocity

of the bar as a function of

. neglect friction everywhere.

[ D C PANDEY page 27, introductory exercise 9.5, problem 2 ]

rates assured!!

amanks

i think you are sick with ICR

any way here is the solution

from the concept of ICR

draw normals to Vx Vy

the poc of normals is ICR

Y/X=tan@

diffn. wrt t

Vy =tan@ Vx

wx=Vy

wy=Vx

squaring both the eqns

wl=(Vx)sec@

w=(Vx/l)sec@

ramyani

the answer is not correct. the correct answer is

[ 3g/l * (1 -- sin

) ]

rated 4 ur efforts

amanks

let me try

the IAR of the system lies at L/2 from the mp of hypotenuse.

moment of inertia about IAR=ML²/3

conserving energy

change in p.e.=rotational energy

(Mgl/2)(1- sin@)=1/2 I_A^_W2

^{W=SQ. ROOT 3g/l (1-sin@)}

^cheers

^{attached file for clear concept}

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.

ramyani

please explain how u found out the co ordinates of IC.

ramyani

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

, l/2 cos

)

plz explain

the co-ordinate of c.o.m is ( l/2 cos theta, l/2 sin theta )

i don't bother if the question is silly .

karthik2007

@Ramyani - ICR is found by taking any two points, and by drawing perpendiculars to their velocity vectors. The point where the perpendiculars meet gives the ICR.

ramyani

ooof ! I understand now !

thanks.

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