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Homework Help: Cube collision (with a pivot) and Angular momentum

  1. Jun 19, 2009 #1
    Hi everybody some help needed here!
    A cube, mass M move with v0 collide with small object "fixed" to the surface in Point P that make it pivot over its side

    So far I think I understand:
    During the impact there is no conservation of KE
    L Angular Momentum : there is an unknown force F (Impulse) on the pivot point
    p Linear Momentum : there is a Impulse = F*t in the pivot point (or I should have to consider the mass of the earth)
    Known data:
    mass M, velocity v0 or v, Side 2a (note that is 2a just for simplification), cm center of mass, rcm vector distance from P to cm, rp distance from P to P (only for completeness)
    Before (around Point P of pivot):
    [tex]\vec{L} = \vec{r_{cm}}\times M\vec{v_{0}}[/tex]
    in this case [tex]|r_{cm}||v_{0}|sin(\theta)=av_0[/tex]
    [tex] L_{i} = aMv_{0}[/tex]


    [tex]\vec{L_{f}} = \vec{r_{p}} \times \vec{F}*t + \vec{r_{cm}} \times M\vec{v_{cm}} +I\vec{\\w}[/tex]

    Where I is the Moment of Inertia of the cube and

    [tex]v=\\wr_{cm}[/tex] and [tex]r_{cm} = a\sqrt{2} [/tex]
    Note that [tex] r_{p} =0[/tex] so

    [tex]L_{f} = 0*F*t + a\sqrt{2}M\\wa\sqrt{2}+I\\w[/tex] So

    [tex]L_{i} = L_{f} [/tex]
    [tex] aMv_{0} = 2a^2M\\w +I\\w [/tex]

    \\w = \frac{aMv_{0}}{2a^2M+I}

    After the collision It is posible to use Conservation of energy:
    E = KE+ V = mgh and KE is 0 and the higher point (the min velocity).

    [tex] \frac{1}{2}Mv_{cm}+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)[/tex]

    It that ok?
    Last edited: Jun 19, 2009
  2. jcsd
  3. Jun 19, 2009 #2


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  4. Jun 20, 2009 #3
    I have doubts with KE after the collision.
    Is posible that in
    [tex] \frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)[/tex]
    The term
    [tex] \frac{1}{2}Mv_{cm}^2[/tex] is OK? or It's enough with [tex] \frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)[/tex]?

    I am not clear so some help would be great
  5. Jun 21, 2009 #4
    I think I have the answer.
    It depends on I (moment of inertia)
    If I is taken from the point of pivot P then I is including all the efect of the movement of the CM (center of mass). On the other hand, if I is refered to the CM then It is necessary to use :
    [tex] \frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\\w^2 = Mga(\sqrt{2}-1)[/tex]
    including the KE of the CM.

    same is true for the angular momentum L:
    [tex]\vec{L_{f}} = +I\vec{\\w}[/tex]
    If I is refered to P then [tex]\vec{r_{cm}} \times M\vec{v_{cm}}[/tex] is not necessary (It is included in I)
    Notice that [tex]v_{cm}=a\sqrt{2}\\w[/tex] and adding it in the [tex]\vec{L_f}[/tex] and in KE It should be, by Parallel axis theorem, equivalent. So
    [tex]I_p = I_{cm} + Md^2\ in\ this \ case \ I_p =I_{cm}+M2a^2}[/tex]
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