Manipulation of Cartesian Tensors

  • Thread starter Hoplite
  • Start date
  • #1
51
0
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?
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,002
552
[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.
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,002
552
[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:
  • #4
51
0
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]
 

Related Threads on Manipulation of Cartesian Tensors

  • Last Post
Replies
8
Views
2K
Replies
6
Views
4K
Replies
1
Views
7K
  • Last Post
Replies
5
Views
746
  • Last Post
Replies
1
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
995
Top