- #1
lets_resonate
- 15
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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?
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?
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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