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.

varun.tinkle

RAMAYANI I EVEN HV A BETTER WAY TO SOLVETHE SUM
THE LOSS IN THE ENERGY OF THE ENERG OF THE SYSTEM IS EQUAL TO THE GAIN IN ENERGY ABT THE CENTRE OF MASSS THIS CONCEPT IS VERY CLEARLY EXPLAINED BY HCV
SO
MGL(1-SIN@)=1/2MV^2+I/2IW^2
SINCE THE ROD IS ROTATING V=RW
MGL(1-SIN@)=1/2MW^2R^2+1/2IW^2
SOLVING IT WE GET THE ANSWER
PLZ RATE ME IF U FIND ME USEFUL
!!!!!!CHEERS!!!!!!!!!!

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