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A heavy rope of mass m and length 2L is ...

A heavy rope of mass m and length `2L` is hanged on a smooth little peg with equal lengths on two sides of the peg. Right part of the rope is pulled a little longer and released. The rope begins to slide under the action of gravity. There is a smooth cover on the peg (so that the rope passes through the narrow channel formed between the peg and the cover) to prevent the rope from whiplashing.
(a) Calculate the speed of the rope as a function of its length (x) on the right side.
(b) Differentiate the expression obtained in (a) to find the acceleration of the rope as a function of x.
(c) Write the rate of change of momentum of the rope as a function of x. Take downward direction as positive
(d) Find the force applied by the rope on the peg as a function of x.
(e) For what value of x, the force found in (d) becomes zero? What will happen if there is no cover around the peg ?

Text Solution

Verified by Experts

The correct Answer is:
(a) `v=sqrt((g)/(L))(x-L)`
(b) `a=(g)/(L)(x-L)`
(c) `(dp)/(dt)=2mg((x-L)/(L))^(2)`
(d) `F=mg[1-2((x-L)/(L))^(2)]`
(e) `x=L+(L)/(sqrt(2))`
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