The part of chain that reaches the ground looses tension and stops accelerating. Suppose that at some instant of time the length of chain that has still not fallen is given by x as shown in the figure. Let the total mass of the chain be m. The mass of x length of the chain will be mx/l. The length of hanging part of the chain is always constant and equal to h. This hanging part of the chain is responsible to pulling the rest of chain downwards. The force of gravity acting on the chain at any instant is given by mgh/l. Let velocity of the chain at some instant be v. So we have,
Now v is also the rate of decrease of length of the remaining part of the chain and so,
Note that the upper limit of x was chosen as h since that will be the remaining part of the chain when its just slipping out of the tube.
1 comment:
sir
why force has been taken as
F=mx/l*dv
dt
while it should be
F=d(mx/l*v)
dt
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