# Transforming to a Rotating Reference Frame

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1. Feb 28, 2017

### bananabandana

1. The problem statement, all variables and given/known data
Let $\mathbf{r}$ be the position of a point in a rigid body relative to some origin $O$. Let $\mathbf{R}$ be the position of the centre of mass from that origin. $\mathbf{r^{*}} = (\mathbf{r}-\mathbf{R})$. $d\boldsymbol{\phi}$ is the infitesimal vector directed along the axis of rotation, with magnitude of the rotation angle $d\phi$. Then, show:

$$\mathbf{v} = \mathbf{V}+\boldsymbol{\Omega} \times \mathbf{r^{*}}$$,

Where $$\mathbf{v} = \frac{d\mathbf{r}}{dt} \quad \mathbf{V}=\frac{d\mathbf{R}}{dt} \quad \boldsymbol{\Omega} = \frac{d\boldsymbol{\phi}}{dt}$$

Further, assume $\frac{d\boldsymbol{\Omega}}{dt} = \vec{0}$ and prove:

$$\frac{d\mathbf{v}}{dt} =\frac{d\mathbf{V}}{dt}+ 2 \boldsymbol{\Omega} \times \mathbf{V} + \boldsymbol{\Omega} \times \big( \boldsymbol{\Omega} \times \boldsymbol{r^{*}} \big)$$

2. Relevant equations

3. The attempt at a solution

I think it is possible I have got my origins muddled, here. The result for the velocity I can get to, after Landau [See Landau, Course of Theoretical Physics Vol 1, pg 96-97]

We realize that any displacement of the body (I think this is due to Euler's rotation theorem) can be thought of as a combination of a fixed axis rotation about the centre of mass, and a displacement of the centre of mass itself. So:

$$d\mathbf{r} =d \mathbf{R} + d\boldsymbol{\phi} \times \mathbf{r^{*}}$$
$$\mathbf{v} = \mathbf{V} + \boldsymbol{\Omega} \times \mathbf{r^{*}}$$

As required. However, I then run into trouble when I want to look at $d\mathbf{v}$. Can I write the following:

$$d\mathbf{v} = d\mathbf{V} + d\boldsymbol{\Omega} \times \mathbf{r^{*}} + \boldsymbol{\Omega} \times d(\mathbf{r}-\mathbf{R})$$

Then:
$$\frac{d\mathbf{v}}{dt} = \frac{d\mathbf{V}}{dt} + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{r^{*}} + \boldsymbol{\Omega} \times \bigg( \mathbf{V} + \boldsymbol{\Omega} \times \mathbf{r^{*}}\bigg) - \boldsymbol{\Omega} \times \frac{d\mathbf{R}}{dt}$$

Which using the assumption about $\boldsymbol{\dot{\Omega}}$ and the definition $\mathbf{V} = \frac{d\mathbf{R}}{dt}$ simplifies to:

$$\frac{d\mathbf{v}}{dt} = \frac{d\mathbf{V}}{dt} + \boldsymbol{\Omega}\times \big( \boldsymbol{\Omega} \times \mathbf{r^{*}} \big)$$

Which is completely missing the Coriolis term!! Where has it gone?

Thanks!

2. Feb 28, 2017

I'm not sure I agree with what this problem is attempting to prove. (I'd really need to look it over much more carefully, but a preliminary evaluation seems to suggest it may be incorrect.) The way I know it, any vector $\vec{A}$ in a rotating frame has $[\frac{d \vec{A}}{dt}]_o=\frac{d \vec{A}}{dt}+\Omega \times \vec{A}$.(The subscript zero implies the true rate of change in the vector A, relative to the stationary frame. The unscripted rate of change of the vector A is that measured in the rotating frame.) In this case, the velocity of the particle is part of a rigid body, so that the velocity in the rotating frame of the particle is equal to zero. The Coriolis term, as I know it, involves the velocity in the rotating frame, and the velocity in this term is not the velocity $V$ of the origin of the rotating frame relative to the origin of the fixed frame. I need to look this over further, but I'm not so sure the original problem is correctly stated. $\\$ Editing: The two dimensional case of a moving particle referenced to polar coordinates is worth considering. In this case the coordinate system rotates at $\omega=\dot{\theta}$ and the particle moves along the rotating frame axis x'. $\vec{v}=\dot{r} \hat{a}_r+r \dot{\theta} \hat{a}_{\theta}$. The acceleration is computed and is $\vec{a}=(\ddot{r} -r \dot{\theta}^2) \hat{a}_r+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{a}_{\theta}$. (Notice the $2 \dot{r} \dot{\theta}$ Coriolis term.) In the problem at hand, $\dot{r}=0$ because the particle is part of a rigid body. If $V$ is responsible for some $\Omega$, then you might get a Coriolis term out of it. If there was some $v$ in the rotating frame, e.g. a non-zero $\dot{r}$, then a Coriolis term would appear... $\\$ Additional editing: As a counterexample as to why the expression can not be correct as given (with a $2 \Omega \times V$ term), consider a finite and large $V$ and a large $\Omega$ with the particle of interest close to the origin of the rotating frame=(close to center of mass, etc.). It then has a small velocity in the rotating frame, but the computed Coriolis term in the formula as given in the OP formula will be huge.