velocity

12 Answers

33
Abhishek Priyam ·

Is √gR(2/Î +Î /4)

24
eureka123 ·

priyam,,,plz tell the method.....or give an outline of solution.....

1
skygirl ·

is it √2gR(1/√2 + Î /4) ??

1
skygirl ·

loss in pot energy of com = gain in kinetic energy of com

vel of com= vel of tip of the chian..

24
eureka123 ·

ans is √gr[(Î /2) +(4/Î )]

1
skygirl ·

!!!

33
Abhishek Priyam ·

kya???

mine just opposite but how can be this ....

surely answer is wrong in the book :(
:(
:(

someone else plz verify...

I am not posting my method just now as it will make many others to bring my (wrong) answer :(

plz others verify....

33
Abhishek Priyam ·

Any one trying.........

24
eureka123 ·

come on priyam ......post ur solution...........the book may be wrong also.......

33
Abhishek Priyam ·

Give me some time...

33
Abhishek Priyam ·

Initially COM of the chain(when θ=00) is at 2R/Πabove the base.

Finally (when θ=900) COM is ΠR/4 below base so 1/2mv2=mg(2R/Π+ΠR/4)..

and hence the answer :)

1
Diptosh ·

It is said that the chain starts at θ = 0

So at that position , chain's potential energy is ρgr^2
Final position , it is having a displacement of - πr/4
So final potential energy = [ -( π^2)(r^2)ρg ]/8
Hence change is potential energy = kinetic energy
So ,
r^2ρg(1+π ^2/8) = 1/2 . (π. r )/2 . ρ.v^2
Solving this we should get the required result ,

v = √rg(Ï€/2+4/Ï€)

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