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Manipulation of Cartesian Tensors

  1. Nov 6, 2006 #1
    I have a question relating to a particle rotating around a point with velocity [tex]u = \Omega \times r[/tex], where [tex]\Omega[/tex] is the angular velocity and r is the position relative to the pivot point.

    I need to prove that the acceleration is given by,

    [tex]a = -\frac{1}{2} \nabla [(\Omega \times r)^2] [/tex]

    I figured it should follow from the fact that,

    [tex]a = \frac{du}{dt} = \frac{\partial u}{\partial t} + u \cdot \nabla u = u \cdot \nabla u [/tex]

    But I can't work out where to go from there. We are supposed to use Cartesian tensor methods to work it out.

    Could anyone help me out?
     
  2. jcsd
  3. Nov 6, 2006 #2

    dextercioby

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    [tex] \vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}\left(\vec{\Omega}\times \vec{r}\right) =\frac{d\vec{\Omega}}{dt}\times\vec{r} +\vec{\Omega}\times\frac{d\vec{r}}{dt}[/tex]

    Daniel.
     
  4. Nov 6, 2006 #3

    dextercioby

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    [tex] \frac{d\vec{\Omega}}{dt}=\left(\vec{\Omega}\times \vec{r}\cdot\nabla\right)\vec{\Omega} [/tex]

    [tex] \frac{d\vec{\Omega}}{dt}\times \vec{r}=\left[\left(\vec{\Omega}\times \vec{r}\cdot\nabla\right)\vec{\Omega}\right]\times \vec{r}= \frac{1}{2}\nabla\left[\left(\vec{\Omega}\times\vec{r}\right)\cdot\left(\vec{\Omega}\times\vec{r}\right)\right] -\vec{\Omega}\times\left(\vec{\Omega}\times\vec{r}\cdot\nabla\right)\vec{r}[/tex]

    In the end i get the plus sign.

    Daniel.
     
    Last edited: Nov 6, 2006
  5. Nov 6, 2006 #4
    Thanks, Daniel.

    Yeah, I suspect the negative sign was a typo on the question sheet, especially considering the identity,

    [tex]\vec u \cdot \nabla \vec u = \frac{1}{2} \nabla (\vec u \cdot \vec u) + (\nabla \times \vec u) \times \vec u [/tex]
     
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