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Hello,

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

[tex]\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/tex]

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:

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho[/tex]

Here, [itex]\vec \rho[/itex] is the "cylindrical-radius vector to the particle from the z-axis".

The derivation only shows a couple steps:

[tex]\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/tex]

[tex]\vec{ F_{\textrm{cf}}} = -m \left[ \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \omega^{2} \right][/tex]

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \left( x \mathbf{\hat{x}} + y \mathbf{\hat{y}} \right)[/tex]

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho[/tex]

I'm confused about how they went from the first step to the second step. Can anyone please help?

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

[tex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \| \vec \omega \|^{2}[/tex]

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:

[tex]\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>[/tex]

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

[tex]\vec{ \omega } = \left< 0, 0, \| \vec{ \omega } \| \right>[/tex]

Then I wrote out [itex]\vec r[/itex]. I simply named its components x, y, and z:

[tex]\vec{ r } = \left< x, y, z \right>[/tex]

Then [itex]\vec{ \omega } \times \vec{ r }[/itex] is:

[tex]\vec{ \omega } \times \vec{ r } = \left< -\| \vec{ \omega } \| y, \| \vec{ \omega } \| x, 0 \right>[/tex]

And [itex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/itex] is:

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

[tex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = - \vec{ r } \| \vec{ \omega } \|^2[/tex]

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?

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

[tex]\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/tex]

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:

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho[/tex]

Here, [itex]\vec \rho[/itex] is the "cylindrical-radius vector to the particle from the z-axis".

The derivation only shows a couple steps:

[tex]\vec{ F_{\textrm{cf}}} = -m \vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/tex]

[tex]\vec{ F_{\textrm{cf}}} = -m \left[ \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \omega^{2} \right][/tex]

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \left( x \mathbf{\hat{x}} + y \mathbf{\hat{y}} \right)[/tex]

[tex]\vec{ F_{\textrm{cf}}} = m \| \vec{ \omega } \| ^ 2 \vec \rho[/tex]

I'm confused about how they went from the first step to the second step. Can anyone please help?

**Attempt at a solution**If the first two steps in the derivation are valid, it would imply this:

[tex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = \vec{ \omega } \left( \vec{ \omega } \cdot \vec{ r } \right) - \vec{r} \| \vec \omega \|^{2}[/tex]

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:

[tex]\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>[/tex]

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

[tex]\vec{ \omega } = \left< 0, 0, \| \vec{ \omega } \| \right>[/tex]

Then I wrote out [itex]\vec r[/itex]. I simply named its components x, y, and z:

[tex]\vec{ r } = \left< x, y, z \right>[/tex]

Then [itex]\vec{ \omega } \times \vec{ r }[/itex] is:

[tex]\vec{ \omega } \times \vec{ r } = \left< -\| \vec{ \omega } \| y, \| \vec{ \omega } \| x, 0 \right>[/tex]

And [itex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right)[/itex] is:

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

[tex]\vec{ \omega } \times \left( \vec{ \omega } \times \vec{ r } \right) = - \vec{ r } \| \vec{ \omega } \|^2[/tex]

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?

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