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I Proof derivative of a vector following precession motion

  1. Apr 10, 2016 #1
    I do not get some points of this proof about the time derivative of a unit vector $\hat{u}$ (costant magnitude) which is following a precession motion. The picture is the following.


    rrrrrrr.png
    I want to prove that $$\frac{d\hat{u}}{dt}=\vec{\Omega}\wedge \hat{u}.$$

    I'm ok with almost all the proof except the following points.

    Consider the red vector [itex]d\hat{u}[/itex].
    Since [itex]\hat{u}[/itex] has constant magnitude,[itex]d\hat{u}[/itex] must be orthogonal to [itex]\hat{u}[/itex].That's ok.

    But why must [itex]d\hat{u}[/itex] be orthogonal to [itex]\vec{\Omega}[/itex] too (i.e. be tangential to a circle orthogonal to [itex]\vec{\Omega})[/itex]?

    Secondly how can I geometrically prove that the angle [itex]d\theta[/itex] (the one in purple) that separates [itex]\hat{u}(t)[/itex] and [itex]\hat{u}(t+dt)[/itex] is the same that separates the projection of these two vectors on the circle orthogonal to [itex]\vec{\Omega}[/itex] (the projections are [itex]\hat{u}(t) Sin(\alpha) [/itex]and[itex]\hat{u}(t+dt) Sin(\alpha)[/itex])?
     
  2. jcsd
  3. Apr 10, 2016 #2

    andrewkirk

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    Define ##\vec v## as the vector from the origin (the lowest point drawn on the diagram) to the centre of the drawn circle, and ##\vec w(t),\vec w(t+dt)## as the two vectors you have drawn from the centre of the circle to its circumference. Then we have ##\vec u(t)=\vec v+\vec w(t)##, and ##\vec u(t+dt)=\vec v+\vec w(t+dt)##.

    A bit of geometry shows that the lengths of ##\vec u(t)## and ##\vec u(t+dt)## are the same. Let that length be ##l##. Then ##d\hat u=\hat u(t+dt)-\hat u(t)=\tfrac{1}{l}(\vec u(t+dt)-\vec u)=\tfrac{1}{l}(\vec w(t+dt)-\vec w)##. Both ##\vec w(t+dt)## and ##\vec w## are perpendicular to ##\vec\Omega##, so their difference must be too, by the linearity of the dot product.
     
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