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Coriolis & Centrifugal forces

by PhMichael
Tags: centrifugal, coriolis, forces
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PhMichael
#1
Aug12-10, 01:31 AM
P: 125


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1. The problem statement, all variables and given/known data
The picture speaks for itself: one man "A" is standing of a rotating disk whose angular velocity is [tex] \omega [/tex] CCW, and another one "B" is standing still on earth. I'm riquired to find the coriolis and centrifugal forces that act on "B" as seen from "A".


3. The attempt at a solution

Coriolis:

[tex] \vec{F}_{cor} = -2m \vec{\omega} \times \vec{v}_{rel} [/tex]

Centrifugal:

[tex] \vec{F}_{cen} = -m \vec{\omega} \times (\vec{\omega} \times \vec{r}) [/tex]

where,

the angular velocity should be taken in the CW sense, since "A" sees "B" doing circles in that direction, so:

[tex] \vec{\omega}=-\omega \hat{z} [/tex]

about the relative velocity of "B" with respect to "A" i'm not quite sure because the coriolis force appears only when there's movement in a rotating frame, now "A" clearly doesn't move, but the question is: can it see "B" moving in the following velocity:

[tex] \vec{v}_{rel} = \omega r \hat{\theta} [/tex]

?!?!

and the radius vector is:

[tex] \vec{r} = R \hat{r} [/tex]

any clarification would be appreciated =)
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hikaru1221
#2
Aug12-10, 02:03 AM
P: 799
Earlier you stated that "A sees B doing circles", so there is relative motion, correct? Then if there is relative motion, then why isn't there relative velocity?
By the way, if you arrive at the correct centrifugal force and Coriolis force, adding them together, you will see something very intuitive.
PhMichael
#3
Aug12-10, 02:12 AM
P: 125
wow, thanks alot ... in two words you've just cleared some stuff that i've been breaking my head over for the last couple of hours :D

and yeah, adding them both would yield the centripetal force that is applied on bodies in circular motions, isn't it?

hikaru1221
#4
Aug12-10, 02:27 AM
P: 799
Coriolis & Centrifugal forces

Yep
PhMichael
#5
Aug12-10, 02:36 AM
P: 125
one more thing that i want to verify about the relative velocity calculation ... All I did to obtain this is to differentiate the position vector with respect to time so that:

[tex] \frac{d \vec{r}}{dt}=R \frac{d \hat{r}}{dt} = R \omega \hat{\theta} [/tex]

First of all, is true to get it that way (in all circumstances) ?

Is there another "physical", rather that mathematical, way to obtain this result? because frankly I can't imagine the picture that would lead me to this consequence of the realtive velocity.
hikaru1221
#6
Aug12-10, 03:39 AM
P: 799
When you differentiate the position vector or any vector w.r.t. time, you must state which reference frame you are considering.
_ In the reference frame of the ground or B, [tex]\vec{r}_B=const[/tex]. It doesn't change at all. So differentiate it w.r.t. time, you get zero, which is intuitive. But you will see that in this reference frame, A is moving, so [tex]d\vec{r}_A/dt\neq 0[/tex].
_ In the reference frame of the disk or A, you will A stays at rest, while B is moving in the opposite direction of the circular motion of A. So in this reference frame, [tex]\vec{r}_A=const[/tex] and [tex]d\vec{r}_B/dt\neq 0[/tex].
The motions of A and B can be described via the two coordinate systems sticking to A and B. For A, it's the red one, and for B, it's the blue one. The blue one sees the red one rotating (which corresponds to B's viewpoint), and reversely, the red one sees the blue one rotating in the opposite direction (A's viewpoint).
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PhMichael
#7
Aug12-10, 12:29 PM
P: 125
But then a problem of choosing the right sign for the angular velocity is raised...

because:

[tex] \frac{d \hat{r}}{dt} = \vec{\omega} \times \hat{r} [/tex]

so according to A, the angular velocity would be [tex]\vec{\omega}=-\omega \hat{z} [/tex] (the one which we substitute in the relative velocity equation), which has to be wrong because it won't give the final correct answer of the velocity ... In other words, which [tex]\vec{\omega}[/tex] need I choose and why?
hikaru1221
#8
Aug12-10, 12:53 PM
P: 799
Did you notice that I put a minus sign before [tex]\omega[/tex] in the picture? The rotations in the 2 frames obviously are in opposite directions.

EDIT: I didn't see your correction when posting. Your problem is that you use the notation [tex]\vec{\omega}[/tex] interchangeably between the 2 frames. If [tex]\vec{\omega}[/tex] is the angular velocity of A in the ground's frame, then the angular velocity of B in A's frame fortunately is [tex]\vec{\omega}_B=-\vec{\omega}[/tex]. Therefore, in A's frame:
[tex]\vec{v}_{rel}=d\vec{r}/dt = \vec{\omega}_B\times \vec{r}=-\vec{\omega}\times\vec{r}[/tex]
PhMichael
#9
Aug12-10, 01:00 PM
P: 125
nono, I understand your picture ... what I mean is that the angular velocity that I used in the calculation of the relative velocity is [tex] \vec{\omega}=\omega \hat{z} [/tex] , however, for the expressions of the coriolis and centrifugal forces I used the opposite one, [tex] \vec{\omega} = -\omega \hat{z} [/tex] and it seemed correct because it produced the correct answer ... but, now, i'm not quite sure why different sign omegas are used, as one equation yields the other ... can u please clear for me this difference?
hikaru1221
#10
Aug12-10, 01:19 PM
P: 799
Okay. Let's make it clear:
_ The angular velocity of A in the ground's frame: [tex]\vec{\omega}_A=\vec{\omega}[/tex]
_ The angular velocity of B in A's frame: [tex]\vec{\omega}_B=-\vec{\omega}[/tex]
_ Relative velocity of B in A's frame: [tex]\vec{v}_{rel}=\vec{\omega}_B\times\vec{r}=-\vec{\omega}\times\vec{r}[/tex]
We have:
[tex]\vec{F}_{coriolis}=-2m\vec{\omega}_A\times\vec{v}_{rel}=2m\vec{\omega}\times(\vec{\omega}\t imes\vec{r})[/tex]
[tex]\vec{F}_{centrifugal}=-m\vec{\omega}_A\times(\vec{\omega}_A\times\vec{r})=-m\vec{\omega}\times(\vec{\omega}\times\vec{r})[/tex]
The total force on B in A's frame:
[tex]\vec{F}=\vec{F}_{coriolis}+\vec{F}_{centrifugal}=m\vec{\omega}\times(\v ec{\omega}\times\vec{r})=-m\omega^2\vec{r}[/tex] which corresponds to the centripetal force.
PhMichael
#11
Aug12-10, 03:40 PM
P: 125
Thanks pal!


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