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Calculating angular velocity of a ratatable bar in inelastic collision

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    1 m long rotatable bar with a mass of 2 kg is held on vertical stick that goes through its center of gravity. A bullet with mass 0.2 kg is shot with a speed 1 m/s horizontally at the edge of the bar. After the collision the bullet lodges into the bar. Calculate the angular velocity of the bar after the collision, if the bar was at rest at the beginning?

    2. Relevant equations

    Conservation of the angular moment

    3. The attempt at a solution

    mRv= (MR + mR + I) ω

    ω= mRv / (MR² + mR² + MR²/12)
    ω= mv / (MR + mR + MR/12)
    ω= 0.17 m/s

    Are my calculations correct?
     
  2. jcsd
  3. Nov 27, 2009 #2

    kuruman

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    Why are there three terms in the denominator? You are calculating angular momentum about the pivot which is also the CM of the rod. In the denominator you need only two terms, the angular momentum of the rod about the pivot (or its CM) and the angular momentum of the bullet about the pivot. Is "R" the length of the rod or is it the distance of the bullet from the pivot?
     
  4. Nov 27, 2009 #3
    mRv= (mR + I) ω

    ω= mRv / (mR² + MR²/12)
    ω= 0.19 m/s

    "R" is half the length of the rod, in other words the distance of the bullet from the pivot.

    Are now my calculations correct?
     
  5. Nov 27, 2009 #4

    kuruman

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    The moment of inertia of a rod of mass M and total length L about its mid point is

    [tex]I=\frac{1}{12}ML^2[/tex]

    what should the formula look like if you used R instead of L?
     
  6. Nov 28, 2009 #5
    I think it should be like this ω= mRv / (mR² + M(2R)²/12)?

    And the answer is now:

    ω= 12 m/s
     
  7. Nov 28, 2009 #6

    kuruman

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    I didn't put in the numbers, but the algebraic expression you have is correct.
     
  8. Nov 28, 2009 #7
    Thank you very much!
     
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