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Homework Help: Rotating reference system - centrifugal force

  1. Jul 23, 2010 #1
    Hi!

    Can someone explain this expression for the centrifugal force?

    F = -mw x w x r, where the underline refers to vectors. (sorry for notation)

    w is the angular veolcity, m mass, a r is the radial vector.

    I cant really see where the expression comes from. Can you indentify and explain the expression thoroughly for me?

    Looking forward to answers! :)
     
  2. jcsd
  3. Jul 23, 2010 #2
    You'll probably be able to find derivations on the net. Here is one description (with no derivation):

    http://www.scar.utoronto.ca/~pat/fun/NEWT3D/PDF/CORIOLIS.PDF [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Jul 23, 2010 #3

    K^2

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    Do you understand why in a rotating reference frame the following holds for any vector quantity?

    [tex]\frac{d \vec{u}}{dt} = \vec{\omega }\times \vec{u} + \frac{\partial \vec{u}}{\partial t}[/tex]

    If so, the acceleration of a point fixed at position r is trivially

    [tex]\vec{a} = \frac{d^2 \vec{r}}{dt^2} = \vec{\omega} \times (\vec{\omega} \times \vec{r})[/tex]

    And that gives rise to a fictitious force you describe.

    If you don't understand where the first equation comes from, let us know, and somebody can explain it.
     
  5. Jul 24, 2010 #4
    Thank you for answering. Well, yeah, I sort of understand it. For the velocity ist like the velocity in the rotating reference frame + the velovity due to rotation.

    So for instance, if I sit in a merry-go-round, will the partial derivative then be 0?

    And no I cant really see how we get the acceleration that way. Could need some explanation on that. If we derivate the expression for the velocity, wont we get a lot of other terms as well?
     
  6. Jul 24, 2010 #5

    K^2

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    Sure, but all these other terms will contain velocity and acceleration of your point relative to the rotating frame or acceleration of the rotation. All of these terms are zero. The ωxωxr term is the only one that's non-zero. Go ahead and expand the equation for acceleration. It's a good exercise. You'll see not only centrifugal, but also the 2ωxv term come out, which is the Coriolis effect, and a couple of other terms you should be able to identify.
     
  7. Jul 24, 2010 #6
    Also, read Arthur C. Clarkes "Rendezvous with Rama". It contains an account of what happens when the explorers climb "down" a ladder going from the center to the outer shell inside a rotating cylinder (artificial gravity in a space ship). Also, it describes the weather system inside the rotating cylinder... Quite fun.

    I think it was that book, but might also be one of the follow-ups in the series.
     
  8. Jul 25, 2010 #7
    So I almost got this now. I have tried to derive the expression for the acceleration now, however, theres one term missing for me. I have w x (dr/dt)(rotating frame), all the other terms are OK. I should have a 2 in front of that so there's something missing. So im guessing this term might come from the differentiation of the velocity, (dr/dt)(rotating frame), but I cant see how? Please help me show where that 2 comes from.
     
  9. Jul 25, 2010 #8

    K^2

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    [tex]\frac{d^2}{dt^2}\vec{r} = \frac{d}{dt}\left( \vec{\omega} \times \vec{r} + \frac{\partial \vec{r}}{\partial t} \right)[/tex]

    [tex] = \left( \frac{d}{dt}\vec{\omega} \right) \times \vec{r} + \vec{\omega} \times \left( \frac{d}{dt}\vec{r} \right) + \frac{d}{dt} \frac{\partial \vec{r}}{\partial t}[/tex]

    [tex] = \left( \vec{\omega} \times \vec{\omega} + \frac{\partial \vec{\omega}}{\partial t} \right) \times \vec{r} + \vec{\omega} \times \left( \vec{\omega} \times \vec{r} + \frac{\partial \vec{r}}{\partial t} \right) + \vec{\omega} \times \frac{\partial \vec{r}}{\partial t} + \frac{\partial^2 \vec{r}}{\partial t^2}[/tex]

    [tex] = \frac{\partial \vec{\omega}}{\partial t} \times \vec{r} + \vec{\omega} \times \left( \vec{\omega} \times \vec{r} \right) + 2 \vec{\omega} \times \frac{\partial \vec{r}}{\partial t} + \frac{\partial^2 \vec{r}}{\partial t^2}[/tex]

    The contributions to Coriolis term come from two places, hence the factor of 2. Ans so you have terms from local acceleration, Centrifugal force, Coriolis force, and angular acceleration.
     
  10. Jul 25, 2010 #9
    Thanks for answer. I guess Im stupid but hope you still want to answer. I cant really see how you get all the terms. For example this part of the expression I dont understand.

    [tex] \left( \vec{\omega} \times \vec{\omega} + \frac{\partial \vec{\omega}}{\partial t} \right) \times \vec{r}[/tex]

    I mean w = w*z(hat) so I'm thinking the derivative would be zero. What technique do you use when taking the derivative of vectors? I mean you have for example [tex] \vec{\omega} \times \vec{\omega} [/tex] which is ofcourse zero, but how did you get that it should be there?

    Also, this expression I dont really understood how you got when you took derivative of [tex] \frac{d}{dt} \frac{\partial \vec{r}}{\partial t}[/tex] ? I mean what is the derivative of the unit vector r?



    [tex] \vec{\omega} \times \frac{\partial \vec{r}}{\partial t} + \frac{\partial^2 \vec{r}}{\partial t^2}[/tex]

    So what I probably need is some explanation on how to differentiate unit vectors I guess. Do you think about it logically only or do you have some method to ensure you get all terms right?
     
    Last edited: Jul 25, 2010
  11. Jul 25, 2010 #10

    K^2

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    Forget about breaking vectors into a scalar-unit product. You don't need it. You take a derivative of a vector exactly the same way you do of scalar quantity. If you have to actually work out components, you have to take it all apart, but in this derivation, just pretend they are scalars (except for cross products).

    Any time I have a d/dt of ANYTHING, I apply the rule I first stated in this thread. It becomes a cross product with omega plus a partial derivative. So let me take one particular step apart, and hopefully you'll see how the rest fall in place. I need to take d/dt of the r vector twice. So I start by taking it once and have the following expression.

    [tex]\frac{d}{dt}\left( \vec{\omega} \times \vec{r} + \frac{\partial \vec{r}}{\partial t}\right)[/tex]

    That's a derivative of the sum, which is a sum of derivatives.

    [tex]\left(\frac{d}{dt}\left(\vec{\omega} \times \vec{r}\right) \right) + \left(\frac{d}{dt}\frac{\partial \vec{r}}{\partial t}\right)[/tex]

    Now let me continue with just the first term in the sum above. Just like with ordinary products, d/dt(axb) = (da/dt)xb + ax(db/dt) for any vector quantities a and b. So here is what I get.

    [tex]\frac{d}{dt}\left(\vec{\omega} \times \vec{r}\right) = \left(\frac{d}{dt}\vec{\omega}\right) \times \vec{r} + \vec{\omega} \times \left(\frac{d}{dt}\vec{r}\right)[/tex]

    So now I have d/dt acting on vectors, which again I can decompose using the rule from the above. Again, let me just take the first term of this sum.

    [tex]\left(\frac{d}{dt}\vec{\omega}\right) \times \vec{r} = \left(\vec{\omega} \times \vec{\omega} + \frac{\partial \vec{\omega}}{\partial t}\right) \times \vec{r}[/tex]

    As you correctly pointed out, ωxω term is zero. So only one term from this makes it to the final result.

    [tex]\left( \frac{\partial \vec{\omega}}{\partial t}\right) \times \vec{r}[/tex]

    What does it mean? Let me denote partial derivatives with respect to time like this, ω'. For simplicity, lets assume ω' has the same direction as ω. That means that only the magnitude of ω changed, and not the direction. The direction of this term will be same as direction of ωxr, that is, perpendicular to the r and ω. Or in other words, along the rotation. So all that term tells you is that if you are standing inside a rotating space station, and it suddenly starts to rotate further, you'll get thrown against the rotation.

    This is the only term that depends on changes in ω. All others depend either on changes in r or just on location. Lets look at them.

    ω'xr - we just discussed.
    ωx(ωxr) - centrifugal term. That should be pretty clear.
    2ωxr' - Coriolis effect. It's easiest to understand if you simply think about r moving towards or away from the center. Then r' is in the same direction as r, and ωxr' is again aligned with direction of rotation. It just says that if you move towards center, you'll feel as if you are pulled along with rotation, and if you are moving away, you are pulled against the rotation. (Remember, fictitious force acting on body at r is -m(d²r/dt²))
    r'' - This is a term from the r actually accelerating relative to local frame. If you are on the rotating station, along with all other effects, if you simply decide to walk around, you will fee effects of your own acceleration. That's what this term tells you.

    I hope that clears up most of the problems. If anything is still not clear, let me know, and I'll try to get into more detail. If you have problems with full vs partial derivatives, we can talk about that too, because this is a good illustration.

    I do try to picture what the situation actually looks like, and what forces are going to be in place. Sometimes it helps to picture yourself in the situation, because your body is pretty good at predicting accelerations. But you cannot completely rely on this. Some of the effects are not obvious. For example, if you start running against rotation of a station, you get lighter. That's not immediately apparent, but it is accounted for in the Coriolis term. It does become clear in the extreme case, though. If you are moving against rotation at the same rate that the station rotates, you are no longer rotating with the station, and so no longer experience centrifugal effect, but the centrifugal term is still there. The Coriolis term simply cancels it.

    So you can use these sort of things to check your work, and look for suspicious terms, but don't over-rely on it, and make sure you derive everything step-by-step.
     
    Last edited: Jul 25, 2010
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