Acceleration in a rotating frame

[Nylex]

I'm getting confused by this. I have a handout from a lecture that has a derivation that ends with

"\vec{a} = \vec{a'} + 2\vec{\omega} \times \vec{v'} + \vec{\omega} \times (\vec{\omega} \times \vec{r})

Multiplying through by mass, m

m\vec{a} = \vec{F_{ext}} = m\vec{a'} + 2m\vec{\omega} \times \vec{v'} + m\vec{\omega} \times (\vec{\omega} \times \vec{r})

We preserve Newton II in rotating frame by writing \vec{F'_{net}} = m\vec{a'} where \vec{F'_{net}} is the net force measured by observer in rotating frame.

ie. \vec{F'_{net}} = \vec{F_{ext}} - 2m(\vec{\omega} \times \vec{v'}) - m[\vec{\omega} \times (\vec{\omega} \times \vec{r})]"

It's really the last line that's confusing me. The expressions for the Coriolis and centrifugal forces are

\vec{F_{Cor}} = -2m(\vec{\omega} \times \vec{v'}) and \vec{F_{cent}} = -m\vec{\omega} \times (\vec{\omega} \times \vec{r}), so why isn't it

\vec{F'_{net}} - \vec{F_{Cor}} - \vec{F_{cent}} = \vec{F_{ext}}?

 

[dextercioby]

Hold on a second,who's \vec{F}_{ext}...?

Daniel.

 

[Nylex]

"Sum of real forces (electrical, magnetic, gravitational, etc); only these forces are observed in stationary frame".

All I'm getting confused about is the signs of those forces.

 

[dextercioby]

Why shouldn't the NET force be the sum of all external forces...?Afer all,both Coriolis & centrifugal are inertial forces,they're not external forces.

Daniel.

 

[Nylex]

Grr, I know that, but:

m\vec{a} = \vec{F_{ext}} = m\vec{a'} + 2m\vec{\omega} \times \vec{v'} + m\vec{\omega} \times (\vec{\omega} \times \vec{r})

ie. \vec{F'_{net}} = \vec{F_{ext}} - 2m(\vec{\omega} \times \vec{v'}) - m[\vec{\omega} \times (\vec{\omega} \times \vec{r})]"

The first and second lines aren't the same. If \vec{F'_{net}} = m\vec{a'}, then the first line is m\vec{a} = \vec{F_{ext}} = \vec{F'_{net}} - \vec{F_{Cor}} - \vec{F_{cent}} .