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A chain of mass m and length l is lumped over the hole in a horizontal plate and hangs over the when chain is at rest. if it is released from this position thrn find the velocity when last link of the chain leaves the hole ??

A chain of mass m and length l is lumped over the hole in a horizontal plate and hangs over the when chain is at rest. if it is released from this position thrn find the velocity when last link of the chain leaves the hole ??

Grade:12

1 Answers

Physics Expert
22 Points
2 years ago
Let at an instant of time the hanging part of the chain is of length x. If linear mass density of the chain is ‘λ’ then the momentum of the hanging part at this instant is  (λxv) of the  there are following two forces acting on the hanging part of the chain,
  1. Weight of the chain λxg (downward).
  2. An upward thrust force λv2. According to the law of action an reaction, this same force pulls the links of the chain on the table.
Now the net external force on the hanging part is equal to the rate of change of momentum of the hanging part. Thus the FBD of the hanging part of the chain says,
\lambda xg-\lambda v^{^{2}}=\frac{d(\lambda xv)}{dt}
\Rightarrow gx- v^{^{2}}=x\frac{dv}{dt}+v\frac{dx}{dt}
\Rightarrow gx- v^{^{2}}=x\frac{dv}{dt}+v^{2}
\Rightarrow gx- 2v^{^{2}}=x\frac{dv}{dx}\frac{dx}{dt}
\Rightarrow gx= 2v^{^{2}}+xv\frac{dv}{dx}
multiplying each term of the above equation by 2x3 we get,
2gx^{^{4}}= 2v^{^{2}}x^{^{3}}+2x^{4}v\frac{dv}{dx}
\Rightarrow 2gx^{^{4}}= \frac{d(v^{^{2}}x^{^{4}})}{dx}
Integrating both sides,
\int (2gx^{^{4}})dx= \int \frac{d(v^{^{2}}x^{^{4}})}{dx}dx+c
\Rightarrow \frac{2gx^{5}}{5}= (v^{2}x^{4})+c
as for x=0, v=0 so c=0
so we get, 
v=\sqrt{\frac{2gx}{5}}
when the last link of the chain leaves the table we have x=l, so the speed of the chain at this moment is,
v=\sqrt{\frac{2gl}{5}}

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