IIT JEE , AIEEE Forum - Physics , Mechanics - Rotation

neeraj_agarwal_1990

The rod AB is hinged at end A . If it is released from the horizontal position, then find the loss in gravitational potential energy at the instant when the acceleration of the end B has the magnitude 'g' .The mass of the rod is 'M' and the length is 'l'.

(a) Mgl/2

(b) 3Mgl/4

(c) 2Mgl/3

(d) Mgl/3

neelesh.sahay

take any position of rod at an angle

.

for calculating

(Mg sin

)L/2=I

WHERE WE TAKE I ABOUT HINGE THUS I = ML^2/3

THUS

= (3gsin

)/2L

AND APPLYING ENERGY CONSERVATION WE GET

(MgL cos

)/2 = 1/2 I

^2

THUS

^2= (3g cos

)/L

NOW NET ACCELERATION OF POINT B WILL BE⁼

[(a_t)^2 +( a_r)^2]

a_t =

L = [(3g sin

)/2L]L = (3 g sin

)/2

a_r = (

^2)L = 3 g cos

thus( A_B) ^ 2= 9g^2 cos ^ 2

+ (9g ^2 sin ^ 2

) /4...............1

g^2= the above equation

but i m not getting the answer can u point the mistake?

avery absurd answer is coming none from the options given

neeraj_agarwal_1990

I also tried it this way...But I'm getting sin(theta)>1...

neelesh.sahay

SOLVE THE LAST EQN WHICH I WROTE U WILL COS

WHICH IS COMING VERY ABSURD

PLZ JUST TRY IT OUT..

neeraj_agarwal_1990

Well I think the acc. will be 'g' when the rod reaches the other end...

Since v=0 ,only a(tangential) = g is present

neeraj_agarwal_1990

Well I think the acc. will be 'g' when the rod reaches the other end...

Since v=0 ,only a(tangential) = g is present
But then loss in GPE is zero

neelesh.sahay

NO I DONT THINK SO BECAUSE THEN

= 3gL/2 THUS TANGENTIAL ACCN

= 3g/2 of point B..

amaron

let angle made by rod with horizontal be x
(Mg sin x)L/2=I

I is taken about hinge, I = ML² /3

THUS

= (3gsin x )/2L
but a=r

and a should be equal to g,implies

=g/r
equating,
sinx =2/3
implies loss in PE is mg(l/2)sinx= 1/3mgl
therefore D)
I think I am right.
That solves the problem!

<the missing pictures represent alpha(angular acceleration)>

neelesh.sahay

well amaron even i m getting cos^2

negative which is not possible...

so i think the situation will not come...when a = g

neeraj_agarwal_1990

Well..I do not know enough abt differentiation coz I'm inXI th..

But are you really below Xth...?

what I really mean is that I don't get it

amaron

hey,look at my edited post!
<I had made am mistake in my earlier post at a very early stage >

neelesh.sahay

well neeraj did u get my method?

neeraj_agarwal_1990

Hey...

In your Ist solution...mistake
You have taken total acc the vector sum of a(tan) and a(rad) but the total acceleration is given by A= L(alpha)

I'm getting sin(theta)= 2/3
Thus cos(theta) = root(5)/3

But again none of the options match

amaron

hey neeraj,I dont see the problem! if you have got sin(theta)=2/3
you have solved the problem as loss in PE is given by mglsin(theta)
therefore ans can easily be got <I think its mgl/3 >

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