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A chain AB of length l is located in a smooth horizontal tube so that its fraction h hangs freely and touches the surface of the table with its end B. At a certain moment, the end a of the chain is set free. With what velocity will the end A of chain slip out of the tube? Can anyone please provide a detailed solution, because I cannot understand the irodov solution?

A chain AB of length l is located in a smooth horizontal tube so that its fraction h hangs freely and touches the surface of the table with its end B. At a certain moment, the end a of the chain is set free. With what velocity will the end A of chain slip out of the tube? Can anyone please provide a detailed solution, because I cannot understand the irodov solution?

Grade:Select Grade

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
7 years ago
Dear student,

We know that the forces acting on the hanging part of the chain are the weight of the hanging part, the tension (T) due to the part of chain in the tube and the normal reaction from ground(N). Let the length of chain remaining in tube be x and mass per unit of chain be λ.
Therefore when we apply Newton’s second law on the hanging part of the chain, we get
λhg−T−N=λha
Newton's second law for part in the tube,
T=λxa
Now if you substitute N = 0, you would be able to get the value of acceleration which would further give the velocity with which the end A will slip out.
Regards

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