Manipulation of Cartesian Tensors

1. Nov 6, 2006

Hoplite

I have a question relating to a particle rotating around a point with velocity $$u = \Omega \times r$$, where $$\Omega$$ is the angular velocity and r is the position relative to the pivot point.

I need to prove that the acceleration is given by,

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

I figured it should follow from the fact that,

$$a = \frac{du}{dt} = \frac{\partial u}{\partial t} + u \cdot \nabla u = u \cdot \nabla u$$

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. Nov 6, 2006

dextercioby

$$\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}\left(\vec{\Omega}\times \vec{r}\right) =\frac{d\vec{\Omega}}{dt}\times\vec{r} +\vec{\Omega}\times\frac{d\vec{r}}{dt}$$

Daniel.

3. Nov 6, 2006

dextercioby

$$\frac{d\vec{\Omega}}{dt}=\left(\vec{\Omega}\times \vec{r}\cdot\nabla\right)\vec{\Omega}$$

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

In the end i get the plus sign.

Daniel.

Last edited: Nov 6, 2006
4. Nov 6, 2006

Hoplite

Thanks, Daniel.

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

$$\vec u \cdot \nabla \vec u = \frac{1}{2} \nabla (\vec u \cdot \vec u) + (\nabla \times \vec u) \times \vec u$$