Derivation of acceleration in rotating coordinates

1. Dec 5, 2015

I was just trying to write out the derivation for an object's trajectory from an inertial coordinate system if the object is rotating in another coordinate system (e.g. finding Coriolis, centrifugal acceleration). I seem to have gotten something close to what I was looking for, but after checking the formula, it shows a coefficient of 2 on the Coriolis term online, yet that is not what I derived. I have let the Euler acceleration = 0 by letting $\frac {d\vec{\omega}}{dt} = 0$, but by doing this am I losing a term? Have I done anything wrong with my derivation in general?

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2. Dec 5, 2015

Mercy

In the last line where you cancel out $$\frac {d}{dt} (\omega x r)$$

Use the product rule on this term to obtain $$(\dot{\omega} x r) + (\omega x \dot{r})$$

the former term is the one that you throw away and the latter, combined with the last term in your second to last line, form $$2\omega x v_r$$

hope that makes sense :)

Last edited: Dec 5, 2015
3. Dec 5, 2015

Thank you for the response.

The only thing is, that term is strictly $\frac {d(\omega)}{dt} \times r$ since it was derived from using the product rule earlier. In case I'm missing something, it is only the time derivative of $\omega$ in the last line, and not both cross product of both terms (i.e. not the time derivative of $\omega \times r$).

4. Dec 5, 2015

Mercy

Then did you forget to include the last term in the second to last line? It looks like it disappeared

5. Dec 5, 2015

Sorry for the confusion since I rearranged some terms from the second last to last line. Essentially all I did was state that $\frac{d\vec{r}}{dt}_r = \vec{v}_r$ and so the $\omega \times \frac{d\vec{r}}{dt}_r$ term just appears as $\omega \times \vec{v}_r$. Sorry for the slight sloppiness in that line, but I believe the math works out fine.

6. Dec 5, 2015

Mercy

$$v_{fix} = \frac{dr}{dt}_{rot} + (\omega x r)$$

$$\big( \frac{dv}{dt} \big)_{fix} = \big( \frac{dv_{rot}}{dt} \big)_{fix} + (\dot{\omega} x r) + \omega x \big( \frac{dr}{dt} \big)_{fix}$$

$$a_f = a_r + (\omega x v_r) + (\dot{\omega} x r) + (\omega x (\omega x r) ) + (\omega x v_r)$$

$$a_f = a_r + (\dot{\omega} x r) + (\omega x (\omega x r) ) + 2(\omega x v_r)$$
subscripts: f = fixed frame, r = rotating frame

Here's a quick derivation, you just have to expand the two "fix" terms in the second line properly and then everything should work out. I'd explain a bit more but I need to go catch the bus home :) Hope you can compare to yours and figure out where the mistake is. Cheers

7. Dec 5, 2015

Thank you for the derivation. I think the one part I am not quite catching is $\big( \frac{dv_{rot}}{dt} \big)_{fix}$.

How is that expression equal to $a_r + (\dot{\omega} x r)$? Isn't it simply $a_r$?

8. Dec 6, 2015

DrGreg

By the way, the way to do vector products in LATEX is to use "\times" instead of "x":$$v_{fix} = \frac{dr}{dt}_{rot} + \omega \times r$$

9. Dec 6, 2015

Mercy

haha thank you it was bugging me that I couldn't figure it out :)

10. Dec 6, 2015

Also, just to clarify, did you mean to state: $\big( \frac{dv_{rot}}{dt} \big)_{fix} = a_r + (\dot{\omega} \times r)$ or did I misinterpret your post?

11. Dec 6, 2015

Mercy

No, what I wrote is what is in my textbook (Thorton/Marion pg 389 - 392).

Fixed frame quantities expand this way:

$$\big( \frac{dr}{dt} \big)_{fixed} = \big( \frac{dr}{dt} \big)_{rot}+ (\omega \times r)$$

So from the second line:

$$\big( \frac{dv_{rot}}{dt} \big)_{fixed} =a_r+ (\omega \times v_r)$$

$$\omega \times \big( \frac{dr}{dt} \big)_{fixed} = (\omega \times v_r) + \omega \times (\omega \times r))$$

So from expanding the second line in my earlier post, you can see that two $$(\omega \times v_r)$$ terms come out.

Last edited: Dec 6, 2015
12. Dec 6, 2015

Mercy

The $$(\dot{\omega} \times r)$$ term comes from the product rule from the time derivative of the last term in the first line.

13. Dec 6, 2015

Okay, that seems to be where my confusion arose. But is that true for higher order terms (e.g. acceleration, jerk, etc.)?

14. Dec 6, 2015

Mercy

I would assume so, you would just be taking another set of time derivatives, although I'm not 100% sure on that.

15. Dec 6, 2015