A chain AB of length l is located on a smooth horizontal table so that its fraction of length h hangs 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?

Explanation

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) and T=(l-x)ma...... (2)
Adding these equations, we get
x mg = [x m + (l -x) m ]a = l ma ( or) ` a=(x)/(l) g , " As a " (dv)/(dt) implies (dv)/(dt) = (x)/(l) g `
Multiplying both sides by `(dx)/(dt) , ` we get ` (dx)/(dt) (dv)/(dt) = (x)/(l) g(dx)/(dt) or vdv = (x)/(l) g 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)})`