Conservation of angular and linear momentum

  • Thread starter Nikitin
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  • #1
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Homework Statement



http://home.phys.ntnu.no/brukdef/undervisning/tfy4145/ovinger/Ov10.pdf
Look at the picture in "oppgave 1".

Suppose you have an incoming mass which hits the very thin rod straight on in a completely inelastic collision. the incoming mass is ##m##, the rod has a mass of ##M##, and the little mass hits the rod at a length ##l## from the top.

According to the text, the linear momentum right before and right after are NOT preserved, while the angular momentum is.


The Attempt at a Solution



I calculated the total linear momentum before and after, and indeed I got:

[tex] p_0 = m v[/tex]
and
[tex] p_1 = \frac{m v + M L/2l}{MvL^2 /3ml^2 +1}[/tex]

so the two momentums are seemingly unpreserved. Why is this so? I realize it's a translational motion going over to a rotational one, but what does this have to do with linear momentum not being preserved??
 
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Answers and Replies

  • #2
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You forgot to mention that the rod is hinged at its top. What is the implication of that?
 
  • #3
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That the motion after the collision is rotational?
 
  • #4
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That too. But on a more fundamental level, conservation of momentum works only when no external forces act on the system. Is that the case with the hinged rod?
 
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  • #5
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ahh, of course. How stupid of me. thanks for the help :)
 

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