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Inelastic collision between ball and rotating system.

  1. Mar 9, 2014 #1
    dAt7puO.png

    1. The problem statement, all variables and given/known data

    A homogenous, thin stick with the mass m and the length l is hanging from the ceiling and can rotate freely around the point P.

    A small ball with the mass of M is attached to the end of the stick.

    The small ball on the end of the stick gets hit by another small ball with the mass of M in a completely inelastic collision. Before the two balls hit, the ball not touching the stick is moving with the speed of u.

    What is the sum of the angular momentum before and after the collision?

    2. Relevant equations

    The moment of inertia for the stick itself is I = 1/3*m*l2.

    I figured the total moment of inertia for the system after the balls collide would be (1/3*m+2M)*l2.

    3. The attempt at a solution

    My initial attempt was to simply say that "Conservation of angular momentum must mean, that since the ball initially has the angular moment L = l*M*u around the point P. Therefore the resulting angular momentum must be equal that."

    I however realized that I didn't understand what I was answering.

    From this: https://www.physicsforums.com/showthread.php?t=473302
    I found that the linear AND angular momentum must be conserved, but since the ball moves in exactly the same direction as it's angular momentum and the resulting system not moving in the direction of u, what happened to the linear momentum?

    Of course, I would've happily solved it myself and found my own solution to these questions, but I am at a loss.
     
  2. jcsd
  3. Mar 9, 2014 #2

    SammyS

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    Is that the exact wording of what you are being asked ?

    .
     
  4. Mar 9, 2014 #3
    It's in another language originally, but I assume this is what it would translate to.

    Does it sound odd?
     
  5. Mar 9, 2014 #4

    SammyS

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    I would expect something a bit more like:
    What is the sum of the angular momenta before the collision and the sum after the collision?​
     
  6. Mar 9, 2014 #5
    That works too, I think. I'm not so sure what the difference between the two questions is, but I can try to make the question again:

    "Specify the collected angular momentum for the system consisting of the two balls and the stick with regards to the point P, both before and after the collision, expressed by l, m, M, u and v."

    But my main problem with this question was that it was "too easy" which is usually a good sign that I did something wrong. Is the angular momentum not conserved?
     
  7. Mar 9, 2014 #6

    SammyS

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    Yes angular momentum is conserved.

    The angular momentum prior to the collision is quite easy to arrive at.

    Expressing the overall angular momentum after the collision, in terms of the final speed of the balls is not quite as easy to get.
     
  8. Mar 9, 2014 #7
    Can you guide me in a direction as to how this is solved then? Are there any identities that I'm not taking into account when I simply equate the angular momentum from both before and after?
     
  9. Mar 9, 2014 #8

    SammyS

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    What is the overall problem to be solved here?
     
  10. Mar 9, 2014 #9
    You have me absolutely stunned here with that question, since I believe that I already made this clear. Apparently I don't even know what I'm asking.

    I'll try to explain what I've done until now.

    The angular momentum before the collision, with regards to the point P, must be L = l*M*u, since M is the mass of the moving ball and u is a movement vector perpendicular to the arm on where it will collide, l.

    When the two balls collide on the end of the stick, the moment of inertia for the stick stick-system changes. Because of this, the way we calculate the angular momentum changes, but since it is conserved, it is still equal to the previous L.

    There's supposed to be three parts in solving this task and I'm sitting with one that isn't even expressed in all the variables.

    I suppose the first thing to ask is:

    What is the sum of the angular momentum before the collision?

    What is the sum of the angular moment after the collision?
     
  11. Mar 9, 2014 #10

    SammyS

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    If you were to be asked for the velocity, v, of the two balls immediately after the collision (in terms of m, M, u, and ), then that would give us a clue.

    ... or you might be asked for the angular velocity of the stick right after the collision. ...


    etc.
     
  12. Mar 9, 2014 #11
    The question after this sounds: "Determine the velocity, v, of the two balls immediately after collision, expressed by m, M, l, and u."

    I didn't add this question, because the two are defined within their own "questions", but I'm sorry for not adding it.

    To make sure that I deliver all the information from the questions, I will add the two questions following.

    "For the system consisting of the two balls and the stick, write down the collected impuls p_before, immediately before the collision, and the collected impulse immediately after, p_after."

    "What is largest? The impulse before the collision or after the collision? What should m be for there to be impulse conservation?"
     
  13. Mar 9, 2014 #12

    SammyS

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    WOW !
    I got that almost word for word!

    I'm not quite sure what to make of those last two questions.
    They would make sense if we interpret "impulse" as being linear momentum. In fact I'm pretty sure, that must be what is meant.​
     
  14. Mar 9, 2014 #13
    I was much surprised at how close you were too.

    I think you might be right about the two last questions, but I've not looked on them, considering I can barely manage to answer the first one. Does this new information provide any insight in what I'm doing wrong?
     
  15. Mar 9, 2014 #14

    SammyS

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    This is one approach.

    Then relate the angular velocity, ω, of this system to the velocity, v, of the two balls.

    ... or ...

    The angular momentum after the collision is Iω + 2Mv . Would still have to relate v and ω .
     
  16. Mar 9, 2014 #15
    Are you trying to solve for the speed here?

    In case you are, I think you might also have taken a few steps I don't understand.

    I understand there is a connection between angular velocity and velocity/radius, but there are still some elements missing.

    I would also like to point out that I still don't understand the first question.

    (I can make a list of the questions in order if you like, but the order is in the way I presented them.)
     
  17. Mar 9, 2014 #16

    SammyS

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    Yes, I was talking about solving for v .


    Is angular momentum (about point, P) conserved during the collision ?
     
  18. Mar 9, 2014 #17
    I would very much assume it is, since nothing else has been said.

    I don't know if I made it clear that it is a completely inellastic collision and the stick the ball is attached to has a mass, if this has anything to do with it.

    With regards to finding the velocity, if L_before = l*M*u and L_after = I*omega, do I have to do any conversion work on the units in order to equate them? Can I simply say L_before = L_after?

    Considering that the balls are at the end of the stick, the angular velocity is the same for all distances, I'm assuming that omega = v/l.

    If I can equate the two, and this relation between angular velocity and velocity at a point is also true, I would assume that I could find the velocity, v, by solving for v in the equation:

    l*M*u = I*v/l.

    Also, I'm sorry for the not-so-viewable formatting of the math. I normally use maple.
     
  19. Mar 9, 2014 #18

    SammyS

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    Mainly, LBefore = LAfter

    Yes, ω = v/ .

    Use either I = (1/3*m+2M)*2 and LAfter = Iω2

    or use

    I = (1/3)*m*2 and LAfter = 2Mv + Iω2 .
     
  20. Mar 9, 2014 #19
    So would you say that my way of solving the two questions is correct? I'm sorry to look for such a binary answer, but I can really not tell if I'm understanding this right.

    How would you have solved the first question, if that was the only one presented to you?
     
  21. Mar 9, 2014 #20

    SammyS

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    The answer to the first question was as simple as you first suspected.
     
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