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Collision and rotation

  1. Sep 4, 2014 #1
    Hi there, right now I am making my first steps towards physics, and I would appreciate your help to solve some of my problems. As you surely have already noticed, I'm not english and I guarantee I'll make thousands of errors. Should I make any, please, try to always advise me, and if it's not too annoying, a correction would be neat.

    1. The problem statement, all variables and given/known data
    Two identical poles of length L and mass M are secured at one extremity they ave in common, so that they move and rotate integrally and they make an angle of [itex]\frac{\pi}{2}[/itex].
    At the beginning, they are located so that the common vertex is at the origin, the poles are oriented along the positive x and y axis and they may only rotate over the origin (around the Z axis).
    A particle of mass m and initial velocity [itex]\vec{v_i} = 2.3\ \hat{x}\ \frac{m}{s}[/itex]
    is directed toward the free extremity of the pole along the y axis. After the collision, its velocity is [itex]\vec{v_f} = 0.7\ \hat{x}\ \frac{m}{s}[/itex].
    Friction does not exist (or at least not in this problem ;) )
    Calculate the angular velocity [itex]\omega[/itex] of the poles after the collision and the mechanic energy dissipated during the collision.

    [itex]M = 0.45 kg[/itex]
    [itex]L=0.30 m[/itex]
    [itex]m=0.12 kg[/itex]

    2. Relevant equations
    Energy conservation..

    3. The attempt at a solution
    From kinetic energy conservation I think this should be enough to resolve:
    [itex]\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2+\frac{1}{2}I_{O}\omega^2[/itex]
    with no energy dissipated during the collision.
    Of course I'm wrong and there IS energy dispersion, though I have no idea how to account of it..
     
  2. jcsd
  3. Sep 5, 2014 #2

    haruspex

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    Work is not conserved, but some other physical properties are. What conservation laws can you apply here?
     
  4. Sep 5, 2014 #3
    Well, maybe I can apply laws of conservation of momentum.
    Considering the nature of the system (poles+particle), we can't state that linear momentum is conserved, since the poles are not free to translate, they can only rotate. I know I should consider angular momentum, but first please help me clear this dilemma that is bugging me.
    The system has zero external force, meaning the derivative of the linear momentum with respect to time is always zero. This means that linear momentum is actuallly a constant, and therefore conserved. Yet the poles can only rotate, so linear momentum must not be conserved..
    What is this supposed to mean? How does the formula [itex]\displaystyle \frac{d\vec{p}}{dt}(t) = \sum \vec{F}_{ext}[/itex] account of the change in linear momentum? Maybe some external forces appear only during collision?
     
  5. Sep 5, 2014 #4
    Ok I think I have a clear idea now. Please confirm my reasoning.
    The bind with the "floor" is not part of my system and therefore it produces an external force during the collision, binding the poles to the origin.
    If the collision lasts [itex]\Delta t[/itex], then, during the collision:
    [itex]\displaystyle \frac{d\vec{p}}{dt}(t)_{avg} = \frac{\vec{p}_f-\vec{p}_i}{\Delta t}[/itex]
    [itex]\displaystyle \sum\vec{F}_{ext,avg} = \frac{\vec{p}_f-\vec{p}_i}{\Delta t}[/itex].
    Now, [itex]\vec{p}_i = m\vec{v}_i[/itex] and [itex]\vec{p}_f = m\vec{v}_i+2M\vec{v}_t[/itex], where [itex]\vec{v}_t[/itex] is the linear velocity of the poles after the collision, which I guess can be calculated by detecting the distance [itex]r[/itex] of the mass center of the two poles from the origin:
    [itex]v_t = r\omega[/itex], and of course the direction is tangent to the circumference of the motion of the mass center.
    If I find out [itex]\omega[/itex], I have the average force the bind produces over the poles, and this force is what produces a work:
    [itex]\frac{1}{2}mv_i^2-(\frac{1}{2}mv_f^2+\frac{1}{2}I_O\omega^2)[/itex], which is the energy dissipated (the answer to the second question).

    [itex]\omega[/itex] can be easily found applying the law of conservation of angular momentum, which holds since the external force is applied on the origin and thus makes no twisting moment.
     
  6. Sep 5, 2014 #5

    BvU

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    Actually, ##\vec{p}_f = m\vec{v}_i+2M\vec{v}_t## is not correct. The poles move in opposite directions, so their linear momenta cancel! And for the particle you want to use ##\vec{v}_f##. Looks as if momentum conservation is violated, which isn't happening, however. As you note, the rotation axle (or whatever) exercises force too. Phew....

    And now for the elegant, smart approach you propose: Take the z axis as axis of rotation and there is no external torque from this axle! Angular momentum conservation helps you find ##\omega##. The bulk of the work is to get ##I_0## expressed in L and M..

    Just to be clear: after the collision the axle also has to exercise a force to keep the -- now unbalanced -- contraption rotating around the fixed axis ! (Your sentence is correct, but might induce others to think that there only is a force there during the inelastic collision, so I thought I should mention this.)​
    The "energy lost" expression looks OK to me.
     
  7. Sep 5, 2014 #6

    haruspex

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    No, they move orthogonally, so the linear momenta do not cancel. But as noted, conservation of linear momentum is unhelpful here because there's an unknown impulse from the axle.
     
  8. Sep 5, 2014 #7
    Yeah, they move orthogonally, consider the two poles as a unique extended body.
    And yeah, I intended [itex]\vec{p}_f = m\vec{v}_f + 2M\vec{v}_t[/itex].
    [itex]I_O[/itex] is pretty simple to calculate:
    [itex]\displaystyle I_O = 2\int_{0}^{L}l^2\frac{M}{L}dl = 2\frac{M}{L}\frac{L^3}{3} = \frac{2}{3}ML^2[/itex].
    Thanks everybody!
     
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