IIT JEE , AIEEE Forum - Physics , Mechanics

sachin_gupta1991

Q. A chain is hanging on a table such that its part below the table hanging vertically is L/n and on the table horizontally, it is (L-L/n), where L is length of chain. Let mass of chain=M. Find velocity of chain when last link of chain leaves the table.

sachin_gupta1991

Doesn't anybody have a solution to this question?

nishant_88

Let us consider the table as the reference point of zero potential energy such that the part of the chain on the table has no potential energy.

For potential energy of the hanging part,
Cnsider an elementary part of length dl at a distance l from the lower tip of the hanging part
The PE of the element is
dU=gl dm=gl M/Ldl
or U=Mg/L integration(ldl)from L/n to 0
this gives U=-MgL/2n^2

When the last link leaves, The potential energy can be obtained in a similar fashion as above by integrating from L to 0.
This gives U'=-MgL/2

so Change in PE=MgL/2(1-1/n^2)

1/2Mv^2=MgL/2(1-1/2n^2)
v=root[gL(1-1/2n^2)]

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