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Is this angular momentum question correct?

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A ball travelling in a straight line, colides with the end of a pole on a centre pivot, ie, the pole initially has inertia given by equation (ML^2)/12.After the colision, the ball sticks to the pole and the two rotate together. What is needed to be found is the angular speed

    use the variables m for mass of ball, d for length of pole, omega symbol for angular speed, v for speed of ball before colision.
    2. Relevant equations
    cross product of vectors for ball initially are sued to generate the balls angular momentum.
    so, angluar momentum=(displacement vector) *(momentum vector) = L = r*p
    i take this as L=-d/2 *m*v
    this is also angular momentum intital, as the conservation of momentum is used to calulate the resulting angular speed. the final angular momentum = inertia *omega
    final inertia is equal to the sum of the two moment of inertias about the axis. this is
    mr^2 + (ML^2)/12

    3. The attempt at a solution

    using
    -d/2 *m*v=(omega)(mr^2 + (ML^2)/12)
    rearange to isolate omega results in
    omega= [-d/2 *m*v]/[(mr^2) + (ML^2)/12]

    is this right though? im not sure ive done the cross product correctly.
     
  2. jcsd
  3. Aug 5, 2010 #2

    Filip Larsen

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    Gold Member

    Assuming that the velocity of the ball is perpendicular to the pole when it hits, it looks ok to me.

    Though, you probably want to express r and L (in ML^2) in terms of d to avoid confusing yourself later (especially since you also used L to mean angular momentum).
     
  4. Aug 5, 2010 #3
    yeah, ok. thanks for your help
     
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