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Rigid Body Rotation

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data

    This is not a HW problem, but something I am trying to prove/disprove for my own knowledge. I have a tube pinned at both ends and inclined in the x-y plane. The pin locations can both move freely in space but subject to the constraint that the tube will not stretch or contract.

    Assuming that the top of the tube displaces by [itex]\vec{u}^{top}[/itex] and the bottom displaces by [itex]\vec{u}^{bot}[/itex] we end up with a tube that has been rotated and translated in space.

    Let the [itex]\vec{R}[/itex] be the vector directed along the tube centerline (CL) from bottom-pin to top-pin and let [itex]\vec{R'}[/itex] be the new tube CL after translation & rotation.

    I am conjecturing that the direction of [itex]\vec{R'}[/itex] is that same the direction of the "relative displacement vector" shown below in (1):

    [itex]\vec{u}= (u^{top}_x - u^{bot}_x )_x + (u^{top}_y - u^{bot}_y )_y\qquad(1)[/itex]

    i.e., I am thinking that:

    direction([itex]\vec{R'}[/itex]) = direction([itex]\vec{u})\qquad(2)[/itex]

    2. Relevant equations

    Vector Rules/Identities

    3. The attempt at a solution
    I am having a little trouble starting this one off. That could very well be because it is not true, we'll have to find out.

    We can write (2) as:

    [itex]\tan^{-1}\frac{R'_y}{R'_x} = \tan^{-1}\frac{u_y}{u_x}[/itex]

    so that the ratios must be equal in order to be true:

    [itex]\frac{R'_y}{R'_x} =\frac{u_y}{u_x}\qquad(3)[/itex]

    If we let the bottom pin of R be at the origin, we can write:

    \vec{R'} = R'_x + R'_y

    \Rightarrow \vec{R'} = (R_x +u^{top}_x) - u^{bot}_x)_x + (R_y + u^{top}_y) - u^{bot}_y)_y


    \Rightarrow \vec{R'} = (R_x +(u^{top}_x - u^{bot}_x))_x + (R_y + (u^{top}_y - u^{bot}_y))_y

    (4) is as far as I seem to get. I am not sure if I made any errors or if there is a way to show that:

    \frac{(R_y + (u^{top}_y - u^{bot}_y))}{(R_x +(u^{top}_x - u^{bot}_x))} =\frac{R'_y}{R'_x}

    Any thoughts?
    Thank you.
  2. jcsd
  3. Apr 11, 2014 #2
    Well you can disprove your conjecture by contradiction immediately, assuming:

    \vec{u}^{top} = a\hat{x} = \vec{u}^{bot}

    \text{where } a\in\mathbb{R}

    This would give [itex]\vec{u}=0[/itex] and:


    Which means [itex]\vec{R}'[/itex] does not share the same direction as [itex]\vec{u}[/itex]. Your equation in (4) I believe is correct however.

    EDIT: I assumed "detached" vectors which are free to move through space.
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