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Moment of intertia of a chain

  1. Aug 29, 2012 #1
    Hi

    1. The problem statement, all variables and given/known data

    A chain rotates fast. Observation: the chain gets into a horizontal position. Why?



    2. Relevant equations

    [tex]L=I \omega[/tex] [tex] E= \frac 1 2 I \omega²[/tex] [tex] E=\frac 1 2 \frac {L²} I[/tex]



    3. The attempt at a solution

    Well, I have two equations for the energy. I know that I have to use the second one, because when I do the experiment I see that the chain "uses" its highest moment of intertia. But I can't I use the first equation for the energy which implies that the chain has to use the smallest moment of inertia to have the minimum energy?

    Where is my mistake?

    Thanks for help.
     
  2. jcsd
  3. Aug 29, 2012 #2

    berkeman

    User Avatar

    Staff: Mentor

    Is the problem asking about swinging a chain around in a circular motion overhead? If so, it never makes it to horizontal, right? It can get close if swung very fast, but it can never be horizontal if there is a vertical gravitational field.

    Can you just approach this problem using a free body diagram (FBD)? To model the chain (as opposed to a weight on a string), you would need to do a distributed FBD with something like 10 weights evenly distributed along the chain length, joined by short strings. With that kind of FBD, you can show how the "chain" tends to become more horizontal as it is swung faster and faster in a circle...
     
  4. Aug 30, 2012 #3
    I think I found my mistake. Is it true that the angular velocity is the lowest when the chain rotates around its biggest moment of inertia? If so, I get it ^^
     
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