velocity

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

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....

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

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

Give me some time...

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

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

and hence the answer :)

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€)