A chain AB of length l is located on a smooth horizontal table so that its fraction of length h hange freely with end B on the table . At a certain moment , the end A of the chain is set free . With what velocity with this end of the chain slip off the table ?

A chain AB of length l is located on a smooth horizontal table so that

1.	A chain AB of length l is located on a smooth horizontal table so that its fraction of length h hange freely with end B on the table . At a certain moment , the end A of the chain is set free . With what velocity with this end of the chain slip off the table ?
Answer» Solution :Let m be the mass per unit length of the chain . Let at any instant X be the length of hanging part of chain and T the tension in chain . Then `x mg -T=x ma`….(1)andT=(l-x)ma…..(2) ADDING these equations , we get `x mg=[XM+(l-x)m]a=l ma` (or) `a=x/lg , As a=(dv)/(dt) rArr (dv)/(dt) =x/lg` Multiplying both sides by `(dx)/(dt) , ` we get `(dx)/(dt) (dv)/(dt) =x/l g (dx)/(dt)"or"vdv= x/lg dx` Integrating both sides, we get `int_(0)^(v) v dv =g/l int_(H)^(l) x dx` `[v^(2)/2]_(0)^(v)= g/l[x^(2)/2]_(h)^(l) =g/l[(l^(2)-h^(2))/2]:. v=sqrt{(g(l^(2)-h^(2)))/l}.