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Linear and rotational motion

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data
    In this given pole, mass is "M" and length is "L" and it is on a frictionless surface as the picture describes. "G" is the center of gravity and "I" is the inertia.
    A ball with a mass of "m" comes as the picture and hits the pole with a velocity of "v" and turns the opposite side and leaves at the same velocity. Then the pole moves under both rotational and linear motions.


    2. Relevant equations
    (a)
    i. Write an equation for the momentum of the ball before the collision.
    ii. Considering only the linear motion of the pole, write an equation for the velocity of the pole "V".

    (b) Now consider the rotational motion.
    i. If the ball hits x distance from point "G", write an equation for the angular momentum of the ball around point "G".
    ii. Write an equation for the angular velocity of the pole around point "G"



    3. The attempt at a solution
    (a)
    i. P = mv
    ii. → mv + M0 = -mv + MV
    ∴ V = [itex]\frac{2mv}{M}[/itex]

    (b)
    i. L = mvx
    ii. → mvx + 0 = -mvx + Iω
    ∴ ω = [itex]\frac{2mvx}{I}[/itex]
     

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  3. Jun 22, 2012 #2

    tiny-tim

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    Hi Knightycloud! :smile:

    (btw, we'd say "opposite way" rather than "opposite side" :wink:)

    Yes, that's fine …

    what is worrying you about that?​
     
  4. Jun 22, 2012 #3
    Yeah!!! :D
    well did I get that b part correct???
     
  5. Jun 22, 2012 #4

    tiny-tim

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  6. Jun 22, 2012 #5
    Oh, So did i get that angular velocity correct???
    then the question asks at a certain value of x (lets say it's y) the point A becomes still. How to calculate that value Y???
     
  7. Jun 22, 2012 #6

    tiny-tim

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    let's see :rolleyes: … A is at distance -L/2 from the centre

    ok, from (a) and (b), what is the initial velocity of A? :smile:
     
  8. Jun 22, 2012 #7
    Velocity of point A is mvxL/I
    and thank you, I found the solution to the rest of the question :D I've been a bit lossing my self confidence!!! You know even if I had the answer, I believed it wrong!!!

    That Y distance is = 2I/ML by the way!!!
    Thanks Tim!!!
     
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