Consider a uniform chain of mass `m` and length `L`, the chain is released when hanging length is `L_0`. Find the speed with which the chain leaves the table.

Explanation

(a)
The part of the chain which is horizontal has zero potential energy as it is lying at the datum level. For the vertical part of chain, consider a small section of the chain at a distance x below the datum. The thickness of this section is dx and its mass is dm.
For this section, `dU=-(dm) g x`
Now, `dm=m/Ldx`
`therefore dU=-(mg)/(L)xdx.`
Integrating both sides, within proper limits,
`underset(0)overset(U)intdU=-(mg)/(L)underset(0)overset(I)intxdyimpliesU=-(mgl^(2))/(2L)`

As the end of chain just leaves the horizontal surface, chain becomes vertical. let its speed be v at this instant. So, its kinetic energy is `1/2mv^(2).` The potential energy can be calculated as above, by taking limits of integration from zero to L.
`U=-(mg)/(L)underset(0)overset(L)intxdx`
`impliesimpliesU=-(mgL)/(2)`
As the table is smooth, there is no friction. Also workdone by normal reaction between chain and horizontal surface is zero. So machanical energy of the chain is conserved. We hav e
`U_(i)+K_(i)=U_(f)+K_(f)`
`-(mgl^(2))/(2L)+0=-(mgL)/(2)+1/2mv^(2)`
`impliesv=sqrt(g(L-(l^(2))/(L)))`