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Conservation of angular and linear momentum

  1. Oct 25, 2012 #1
    a rigid massless rod of length L joins two particles each of mass M. The rod lies on
    a frictionless table, and is struck by a particle of mass M and velocity v0. After the collision, the projectile moves straight back. Find the
    angular velocity of the rod about its center of mass after the collision, assuming that
    mechanical energy is conserved.

    i tried to solve this question using the conservation of linear momentum first.
    using ((m-2m)/3m)v0 to find the speed of the mass M after collision.
    (2m/3m)v0, to find the speed of the rod after collision.
    translational speed of the rod = (2/3v0 - wr)
    angular speed of the rod = w

    hence i used the conservation of energy method to solve for w.
    however, my answer is incorrect. may i know which part, i am wrong
     
  2. jcsd
  3. Oct 25, 2012 #2

    ehild

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    You got the velocity of the particle after collision correctly: it is -Vo/3.
    The velocity of the rod is also correct, but it is the velocity of the CM : Vcm=2/3 Vo.
    I do not understand what you mean on "translational speed of the rod". You need to use conservation of angular momentum.

    ehild
     
  4. Oct 25, 2012 #3
    hi sir, if Vcm= 2/3Vo. then according to the conservation of mechanical energy, rotation of the rod could nt happen.
    the translation speed is Vcm.
     
  5. Oct 25, 2012 #4

    tiny-tim

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    isn't it the other way round? …

    the rod is heavier, so it'll move less :smile:
     
  6. Oct 25, 2012 #5

    ehild

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    Sorry, I misunderstood the problem. Your results are correct if the projectile hits the rod at its end, parallel to it, like in the first picture. Then the angular momentum really stays zero. Is not said anything where the incoming particle hits the rod and about the direction of Vo with respect to the rod? Was not it meant as in the second picture?

    ehild
     

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    Last edited: Oct 25, 2012
  7. Oct 25, 2012 #6
    sorry sir, i should have upload the diagram earlier. the projectile hit the rod at 45 degree
     

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  8. Oct 25, 2012 #7

    ehild

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    That is a difference! Set up a coordinate system, as in the picture for example. The linear momentum of he incoming particle has only x component before and after the collision, and it has some angular momentum with respect to the origin. The angular momentum is also conserved in the collision. Write up the equation for the angular momentum of the whole system in addition to the equations for linear momentum and KE.


    ehild
     

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  9. Oct 26, 2012 #8
    hi sir,
    i think i made a big conceptual mistake, thinking that the linear collision is elastic which is not, because some of the inital linear kinetic energy is converted into rotational energy, therefore the final linear speed of the projectile should not be calculated using elastic collision formula. i gt my answer now, thank a lot for the help.
     
  10. Oct 26, 2012 #9

    ehild

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    I am pleased that you solved the problem. Conservation of energy is valid, but the rotational energy has to be included.
    For the sake of other people reading this thread could you please show your solution ?

    ehild
     
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