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Angular Momentum vs Linear Momentum

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data
    The following figure shows an overhead view of a thin rod of mass M=2.0 kg and length L = 2.0 m which can rotate horizontally about a vertical axis through the end A. A particle of mass m = 2.0 kg travelling horizontally with a velocity $$v_i=10 j \space m/s$$ strikes the rod (which was initially at rest) at point B. the particle rebounds with a velocity $$v_f=-6 j\space m/s$$. Find the angular speed of the rod just after the collision.

    2. Relevant equations

    $$(I\omega)_i=(i\omega)_f$$

    3. The attempt at a solution

    I have tried to solve this question using the previous equation , but I'm stuck with the momentum of the ball . Should it be linear or angular?

    I mean ;
    which of the following equation should i use :
    1)
    $$(mvl)_{ball}=(I\omega)_{rod} +mvl$$
    $$(2 \times 2 \times 10j)=(\frac{2 \times 2^2}{12}+2 \times 1^2)+2\times 2 \times -6j$$
    The moment of inertia of the rod is:
    $$\frac{ml^2}{12}+mh^2$$
    where h is the distance from the center of mass to the axis of rotation.

    2)
    $$(mv)_{ball}=(I\omega)_{rod}+mv$$
    $$ (2 \times 2)=(\frac{2 \times 2^2}{12}+2 \times 1^2)+(2 \times -6j)$$

    The equation 2 seems to lead to the correct answer , but , why i should take the linear momentum instead of angular momentum!!


    Help please!!!
     
  2. jcsd
  3. Dec 4, 2014 #2

    BvU

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    Which equation has the same dimension for all terms ?
     
  4. Dec 4, 2014 #3
    Yeah , ..
    How it comes that I didn't pay attention to this point ..

    Thanks ..
     
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