Launch with Chains

  • #1
I've read the FAQ and this is not homework, it is just a hard problem I can't solve.

A mass M attached to an end of a very long chain of mass per unit length [tex]\lambda[/tex]
, is thrown vertically up with velocity [tex]v_{0}[/tex].
Show that the maximum height that M can reach is:

[tex]h=\frac{M}{\lambda}\cdot \left [ \sqrt[3]{1+\frac{3\cdot \lambda\cdot v_{o}^{2}}{2\cdot M\cdot g}}-1 \right ][/tex]

and that the velocity of M when it returns to the ground is [tex]v=\sqrt{2\cdot g\cdot h}[/tex]
Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to v

I start by setting up that the total mass at a position y is:
[tex]M_{total}=M+\lambda\cdot y[/tex] and thus the momentum at any position is given by:

[tex]p=(M+\lambda\cdot y)\cdot v[/tex] but I can't figure out an expression for v and using

[tex]F=\frac{dp}{dt}[/tex] I get an differential equation I can't solve.

Any help would be appreciated.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi benf.stokes! :smile:
Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to v

No, you can use conservation of energy …

the question asks for the maximum height, which would be in the limiting case of no energy loss. :wink:

As for the question itself, I don't understand what position the mass and chain are in at the start. :confused:
 
  • #3
Hi tiny-tim. I think that conservation of energy can't be used is a part of the problem,my bad, and if you used conservation of energy wouldn't you get at most a square root and not a cubic one?
At the start both the mass and the chain are at rest on the floor. Thanks for the reply
 
  • #4
i already arrived at the solution. Thanks anyway tiny-tim :smile:
 

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