# Help with derivation - involves cross & dot products

1. Jun 27, 2007

### lets_resonate

Hello,

I'm reading a classical mechanics book. In it, they show a derivation of the centrifugal force equation:

$$\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)$$

I understood the derivation up to that point. However, they have a couple additional steps after that whereby they derive another equation based on this one. It is valid if the axis of rotation is chosen to lie along the z-axis:

$$\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho$$

Here, $\vec \rho$ is the "cylindrical-radius vector to the particle from the z-axis".

The derivation only shows a couple steps:

$$\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)$$
$$\vec{ F_{\textrm{cf}}} = -m \left[ \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \omega^{2} \right]$$
$$\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \left( x \mathbf{\hat{x}} + y \mathbf{\hat{y}} \right)$$
$$\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho$$

Attempt at a solution

If the first two steps in the derivation are valid, it would imply this:

$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \| \vec \omega \|^{2}$$

So I decided to work with the left side of this equation and hope that it would yield the right side. I used this definition of the cross product:

$$\vec{ a } \times \vec{ b } = \left< a_y b_z - a_z b_y, a_z b_x - a_x b_z, a_x b_y - a_y b_x \right>$$

Then I wrote out $\vec{\omega}$ in component form. Since it is chosen to lie along the z-axis, its x and y components are zero:

$$\vec{ \omega } = \left< 0, 0, \| \vec{ \omega } \| \right>$$

Then I wrote out $\vec r$. I simply named its components x, y, and z:

$$\vec{ r } = \left< x, y, z \right>$$

Then $\vec{ \omega } \times \vec{ r }$ is:

$$\vec{ \omega } \times \vec{ r } = \left< -\| \vec{ \omega } \| y, \| \vec{ \omega } \| x, 0 \right>$$

And $\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)$ is:

$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \left< -\| \vec{ \omega } \|^2 x, -\| \vec{ \omega } \|^2 y, 0 \right>$$

$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = - \vec{ r } \| \vec{ \omega } \|^2$$

However, this is only the second term of the equation above (the first one under my attempt). The first term is missing. What did I do wrong?

Last edited: Jun 27, 2007
2. Jun 27, 2007

### Dr Transport

How about $$\vec{a} \times ( \vec{b} \times \vec{c}) = \vec{b}(\vec{a} * \vec{c}) - \vec{c}(\vec{a} * \vec{b})$$.

Last edited: Jun 27, 2007
3. Jun 27, 2007

### nrqed

Yes, this is a vector identity (it's easy to prove using the Levi-Civita symbol if you have seen that)
It's your very last step that you did incorrectly. The line before is NOT equal to $- \vec{ r } \| \vec{ \omega } \|^2$
Do you see why?

4. Jun 27, 2007

### lets_resonate

Gah! I got it! But I might have one more small question:

$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \left< -\| \vec{ \omega } \|^2 x, -\| \vec{ \omega } \|^2 y, 0 \right>$$
$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = -\| \vec{ \omega } \|^2 \left< x, y, 0 \right>$$
$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = -\| \vec{ \omega } \|^2 \left( \vec{r} - z \right)$$
$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \| \vec{ \omega } \|^2 z - \vec{r} \| \vec{ \omega } \|^2$$
$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \| \vec{ \omega } \| \left( \| \vec{ \omega } \| z \right) - \vec{r} \| \vec{ \omega } \|^2$$
$$\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \| \vec{ \omega } \| \left( \vec \omega \cdot \vec r \right) - \vec{r} \| \vec{ \omega } \|^2$$

In the very last line: I have the magnitude of $\vec \omega$ outside the parantheses in the first term, but in the book they have the actual vector. Did I miss a small detail somewhere that led to this mistake?

====

And also, a bigger question. This classical mechanics book that I'm reading -- it seems to move along at a pretty vigorous pace. In many cases, there is very little explanation that accompanies the derivations.

Now, I'm only a pithy little high school student. I took first year physics and calculus. What other prerequisites are there for a typical classical mechanics course? Where did you guys learn your "Levi-Civita symbols" (whatever those are)? Or am I actually ready for this with my background?

5. Jun 27, 2007

### nrqed

It's not quite right to write this. I know what you mean, but you are not subtracting z (which is a scalar) but you are subtracting the *vector* $z \vec{k}$

What you must then use is that z by itself is

$$z = \frac{1}{\|\vec{\omega} \|} \vec{\omega} \cdot \vec{r}$$

and the unit vector $\vec{k}$ can be written as $\frac{\vec{\omega}}{\| \vec{\omega} \|}$

Then you will get the desired result.

But notice that this is not a very satisfying approach since a lots of steps are involved and it's for a very special case (omega along z). It's much much better to do it for the general case with the general identity. But I understand the need to do a specific case the long way.

What book are you using? Is it for self-study or are you taking a class?

The key point is that in order to do more advanced mechancis, you need to build some background in maths, mostly in vector analysis and vector calculus. So At this point it might be good to pick up a math book (oriented toward physicists or engineers. I would suggest Boas or Arfken) and to just focus on these topics before going back to Mechanics.

Last edited: Jun 27, 2007