A uniform rope of mass m per unit length, hangs vertically from a support so that the lower end touches - Physics - Systems of Particles and Rotational Motion

Question

A uniform rope of mass m per unit length, hangs vertically from a
support so that the lower end touches table top If it is released,
then at that time a length y of the rope has fallen, the force on
the table is equivalent to the weight of the length Ky of the rope
Find the value of K [3]

Sanjay Singh · Accepted Answer

Dear student
Force exerted by length of wire y that has fallen on the table=mass of length y ×g×mForce exerted by length of wire that is falling into the table=rate of change of momentum that is change of momentum in one second
Each segment of wire falls with acceleration g so velocity of each part of string after a length  y has fallen can be calculated asv2-u2=2gyv2=2gy  
1Length of string falling per unit time is equal to velocity =vmass of string falling per unit time is equal=(length falling per second)×(mass per unit length)=v×mForce exerted by length of wire that is falling into the table=Momentum falling on table per second=(mass falling per second)×(velocity)=(v×m)×v=m(2gy)  
(putting value of v 2 from eqn 1)Total force on table=Force exerted by length of wire y that has fallen on the table+Force exerted by length of wire that is falling into the tableF=m×y×g+m(2gy)=3mgyWeight of length  Ky=KymgAccording to question3mgy=KymgK=3Regards

A uniform rope of mass m per unit length, hangs vertically from a support so that the lower end touches table top. If it is released, then at that time a length y of the rope has fallen, the force on the table is equivalent to the weight of the length Ky of the rope. Find the value of K. [3]