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A 3-D collision

  1. Nov 29, 2004 #1

    arildno

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    Consider a cube of uniform density, mass M, sidelengths 2a resting on a frictionless plane.
    Origin is placed in the cube's center.

    A rod of length L, attached to the ceiling z=a+L, mass m, hits with its tip the corner (-a,a,a) on the side x=-a with velocity [tex]V_{0}\vec{i}[/tex].

    Determine the state of motion after elastic collision(s).

    To give a hint:
    Neither angular nor linear momenta are conserved..
     
  2. jcsd
  3. Nov 29, 2004 #2
    trick question! the rod is moving too slowly and is too light to effect the cube by any meaningful amount!

    just kidding. but the cube does spin a bit, right?
     
    Last edited: Nov 29, 2004
  4. Nov 29, 2004 #3

    NoTime

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    Collision?
    It seems just as fair to say the rod is sitting on the cube as it does to say it was attached to the ceiling.
    V=0
    So conservation is not an issue.
     
  5. Nov 29, 2004 #4

    arildno

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    The rod may rotate about its pivot point.
    And the cube will start sliding, plus rotating about a couple of axes
    (the [tex]\vec{k},\vec{j}[/tex] axes, actually).
     
  6. Nov 29, 2004 #5

    NoTime

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    The rod might be in a highly unstable position, but barring any external forces why should it topple rather than just stand on the corner of the cube :confused:
     
  7. Nov 30, 2004 #6

    Gokul43201

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    The recoil of the rod is not restricted to [itex]\hat{i} [/itex], since it's hitting a corner ?

    This is a toughie...
     
  8. Nov 30, 2004 #7

    arildno

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    Not THAT evil, it undergoes only uniaxial rotation..
    Consider the impulse between the rod&cube to be in the [tex]\vec{i}[/tex] direction.
     
  9. Nov 30, 2004 #8

    arildno

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    Consider it to hit the cube at the side x=-a at a point only slightly (insignificantly) displaced from the corner.
     
  10. Nov 30, 2004 #9

    NoTime

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    Because it wasn't stated in the original problem :biggrin:
    Also a corner would be at sqrt(a^2 + a^2).
    Somewhat more than a.
     
  11. Nov 30, 2004 #10

    arildno

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    Nope. the distance from the center to the corner is [tex]\sqrt{3}a[/tex] its (vector) position is (-a,a,a), measured from the origin (i.e, the cube's center).
     
  12. Nov 30, 2004 #11

    NoTime

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    Good point.
    Still don't see why it should fall over. :smile:
     
  13. Nov 30, 2004 #12

    Gokul43201

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    There's no reason to assume it won't. After all the impulse does provide a torge about [itex]\hat {j} [/itex].
     
  14. Nov 30, 2004 #13

    NoTime

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    Well, z=a+L and was not given as z=a+L+d.
    So being attached or not to the ceiling doesn't seem like a particularly useful bit of information. Although there are certainly considerations that would make it otherwise, none of these seem to be specified. I see no reason to assume them.
    My main question from the start is -> What impulse?
    Or why isn't it just as correct to say the rod is sitting on the cube?
    What reason is there to believe that this situation should change?
    Perhaps I am just being overly picky, but it is a brain teaser :biggrin:

    Suppose the rod was made of lead with a diameter of 5a with the cube being soft butter. Rotation or sliding seem unlikely in this case. :smile:
     
  15. Dec 1, 2004 #14

    Gokul43201

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    The rod is not sitting on the cube. I think you've misunderstood the problem.

    There's this cubical box on the floor. On the ceiling, exactly above one of the corners of the box is a rod hanging from a pivot, such that the bottom end of the rod just touches a top corner of the box. The rod is pulled back along the XZ plane and released. The tip of the rod strikes the YZ face of the box very near this corner, at a velocity v_0.
     
  16. Dec 1, 2004 #15

    arildno

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    A very good objection!
    The cube and the rod are to be modelled as ideal rigid bodies; the rod has only length (which was the only dimension I gave), and undergoes constrained motion about its pivot point in the ceiling.
    That is, at all times, it is to be considered as a physical pendulum, with an associated angular velocity [tex]\omega\vec{j}[/tex]
    hence, initially, [tex]\omega_{0}=-\frac{V_{0}}{L}[/tex]
    since we then have:
    [tex]\omega_{0}\vec{j}\times{(-L\vec{k})}=V_{0}\vec{i}[/tex]
     
  17. Dec 1, 2004 #16

    NoTime

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    Ok, the picture is now radically different.
    No mention of pivots, rigid or perfectly elastic in the original statement.

    Is there a diagram for this setup?

    Are there other forces acting on the cube, like gravity?
    Or restating, is there a constraint that one face of the cube will remain parallel to the plane?
     
  18. Dec 1, 2004 #17

    Gokul43201

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    Yes there's gravity. No, there's no other constraint.
     
  19. Dec 1, 2004 #18

    arildno

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    The "constraints" are those which gives us a reasonable model for analyzing a "real-life" collision.
    First, most collisons we see between practically rigid bodies takes place over an extremely short period of time.
    How can we maximally utilize the insight of an extremely short collision period to develop a good, simple model of collisions?
    Secondly, it is physically unreasonable that the cube is going to enter the table/plane and get stuck deep within it.
    How must we take account of this, does it for example mean that we necessarily have to assume that the table/plane imparts an impulse on the cube (and if so, what can we say of that impulse on basis of the information that the plane is frictionless)?

    My basic reason for posting this problem, is that I would think someone might be interested in how to extend our ideas/problem-solving skills beyond the treatment of simplistic 2-D collisions ordinarily taught (in which, for the most part, we make do with conservation laws).
     
    Last edited: Dec 1, 2004
  20. Dec 1, 2004 #19

    NoTime

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    Interesting ideas.
    It seems to me that if the cube was free floating in a 0g environment and struck on the corner of the x face that it would rotate around the z and y axis with some drift.

    Just a thought.
    The surface might elastically deform enough to form a step that would latch the edge.
     
  21. Dec 2, 2004 #20

    arildno

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    Let's see how the idea of extremely short duration of collision may be used to our advantage concerning external forces of the gravity and air-resistance kind:
    Clearly, the magnitude of these forces will be roughly proportional to the duration interval, i.e, we may reasonably expect them to be negligible in comparison to the huge collision forces associated with the actual deformations.

    Hence, to obtain a reasonably accurate&simple model of collisions, we should be justified in neglecting the impulses from these forces.
    Agreed?
     
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