Deriving acceleration in rotating reference frames

[Mr.Miyagi]

Hi, everyone:)

In my mechanics book a derivation is given for acceleration in rotating reference frames. However, there is one step I don't understand.

First of all, it is derived that v=v'+\omega\times r' or \left( \frac{dr}{dt}\right) _{fixed}=\left( \frac{dr'}{dt}\right) _{rot}+\omega\times r'
The unprimed vectors are relative to the fixed reference frame and the primed vectors to the rotating one. Then it is claimed that this should hold for any vector:
\left( \frac{dQ}{dt}\right) _{fixed}=\left( \frac{dQ}{dt}\right) _{rot}+\omega\times Q
In particular, it holds for the velocity vector:
\left( \frac{dv}{dt}\right) _{fixed}=\left( \frac{dv}{dt}\right) _{rot}+\omega\times v
In this equation v can be substituted as v=v'+\omega\times r' and then, after some algebraic manipulation, one can find the fictitious forces.

What i don't understand is why the primes are suddenly being dropped. Why are vectors relative to the rotating reference frame suddenly being replaced by ones relative to the fixed frame.

 

[D H]

First of all, it is derived that v=v'+\omega\times r' or \left( \frac{dr}{dt}\right) _{fixed}=\left( \frac{dr'}{dt}\right) _{rot}+\omega\times r'

The displacement vector is the same vector in the rotating and inertial frames. I'll qualify what I mean by the same vector. Suppose you have some vector. While this vector might have different coordinate representations in different reference frames, those are just different ways of representing what is conceptually one thing.

On the other hand, the time derivatives of this vector as observed in the rotating and inertial frames are different vectors. For example, the time derivative of the vector from the origin to a point fixed in the rotating frame has a zero time derivative in the rotating frame but a non-zero time derivative (\omega \times r) in the inertial frame. They are not the same vector.

Your text (Marion?) hand-waves this very important relation:

\left( \frac{dQ}{dt}\right) _{inertial}=\left( \frac{dQ}{dt}\right) _{rot}+\omega\times Q

This is only valid if Q represents one vector. There is no reason to denote Q as primed or unprimed because it must be the same vector on the left and right hand sides. You can derive the fictitious forces by replacing Q with either v or v':

\left( \frac{dv}{dt}\right) _{inertial}=\left( \frac{dv}{dt}\right) _{rot}+\omega\times v

or

\left( \frac{dv'}{dt}\right) _{inertial}=\left( \frac{dv'}{dt}\right) _{rot}+\omega\times v'

(Go ahead and try it. Either way works.) What you cannot do is mix and match v and v'. The vector has to be the same vector. v and v' are not the same vector.

 

[Mr.Miyagi]

I still feel I don't quite understand it.

The displacement vector is the same vector in the rotating and inertial frames. I'll qualify what I mean by the same vector. Suppose you have some vector. While this vector might have different coordinate representations in different reference frames, those are just different ways of representing what is conceptually one thing.

Agreed. A change of base doesn't change the vector itself.

On the other hand, the time derivatives of this vector as observed in the rotating and inertial frames are different vectors. For example, the time derivative of the vector from the origin to a point fixed in the rotating frame has a zero time derivative in the rotating frame but a non-zero time derivative (\omega \times r) in the inertial frame. They are not the same vector.

Alright.

Your text (Marion?) hand-waves this very important relation:

\left( \frac{dQ}{dt}\right) _{inertial}=\left( \frac{dQ}{dt}\right) _{rot}+\omega\times Q

This is only valid if Q represents one vector. There is no reason to denote Q as primed or unprimed because it must be the same vector on the left and right hand sides. You can derive the fictitious forces by replacing Q with either v or v':

\left( \frac{dv}{dt}\right) _{inertial}=\left( \frac{dv}{dt}\right) _{rot}+\omega\times v

or

\left( \frac{dv'}{dt}\right) _{inertial}=\left( \frac{dv'}{dt}\right) _{rot}+\omega\times v'

So this implies that
\left( \frac{dr}{dt}\right) _{inertial}=\left( \frac{dr'}{dt}\right) _{rot}+\omega\times r'=\left( \frac{dr}{dt}\right) _{rot}+\omega\times r
making the displacement vector a special case? I'll try to draw it out and convince myself.

The book is 'Analytical Mechanics' by Fowles and Halliday.

 

[D H]

I still feel I don't quite understand it.

Your lack of understanding may well result because the relation

\left( \frac{dQ}{dt}\right) _{inertial}=\left( \frac{dQ}{dt}\right) _{rot}+\omega\times Q

is just a hand-wave in your text. I gave a partial derivation in this thread, [thread=280326]Help understanding rotation relations in different frame[/thread] (see post #2). Take a look at that thread and see if it helps.

So this implies that
\left( \frac{dr}{dt}\right) _{inertial}=\left( \frac{dr'}{dt}\right) _{rot}+\omega\times r'=\left( \frac{dr}{dt}\right) _{rot}+\omega\times r
making the displacement vector a special case? I'll try to draw it out and convince myself.

Correct.

The book is 'Analytical Mechanics' by Fowles and Halliday.

You might want to look at some other texts that cover the same material. Two are Classical Dynamics of Particles and Systems by Thornton and Marion (or Marion and Thornton, or just Marion, depending on the edition) and Classical Mechanics by Goldstein. Marion at least does the same hand-wave as Fowles, but it does not distinguish r and r'.

 

[Mr.Miyagi]

Aha! It finally makes sense. I'll definitely check up on the texts you recommended.

Thank you for your help.

 

[D H]

You're welcome. Glad to be of assistance.