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tSolution We recall a general vector time-derivative identity relating derivatives measured in an inertial frame (subscript $I$) and a rotating frame (subscript $R$) with constant rotation vector $\boldsymbol{\Omega}$. For any vector $\mathbf{A}$ (expressed in the rotating-frame basis),

$$\left(\frac{d\mathbf{A}}{dt}\right)_I = \left(\frac{d\mathbf{A}}{dt}\right)_R + \boldsymbol{\Omega}\times\mathbf{A}.$$ (3)

This identity follows because the rotating-frame basis vectors themselves change with time: $d\hat{e}_i/dt = \boldsymbol{\Omega}\times\hat{e}_i$ (for constant $\Omega$).

Apply () to the position vector $\mathbf{r}$. The inertial velocity is

$$\mathbf{V}_I = \left(\frac{d\mathbf{r}}{dt}\right)_I = \left(\fra... ...l{\Omega}\times\mathbf{r} = \mathbf{V}_R + \boldsymbol{\Omega}\times\mathbf{r}.$$ (4)

Differentiate again using () (applied to $\mathbf{V}_R$ and remembering $\boldsymbol{\Omega}$ is constant):

$$\left(\frac{d\mathbf{V}_I}{dt}\right)_I = \left(\frac{d}{dt}\right)_I(\mathbf{V}_R + \boldsymbol{\Omega}\times\mathbf{r})$$ (5)

$$= \left(\frac{d\mathbf{V}_R}{dt}\right)_I + \left(\frac{d}{dt}\right)_I(\boldsymbol{\Omega}\times\mathbf{r}).$$ (6)

Using () on the first term and on the vector $\boldsymbol{\Omega}\times\mathbf{r}$ yields

$$\left(\frac{d\mathbf{V}_I}{dt}\right)_I = \left[\left(\frac{d\mathbf{V}_R}{dt}\right)_R + \boldsymbol{\Om... ...bf{r}) + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})\right].$$ (7)

But $\boldsymbol{\Omega}$ is constant, so

$$\left(\frac{d}{dt}\right)_R(\boldsymbol{\Omega}\times\mathbf{r}) ... ...\left(\frac{d\mathbf{r}}{dt}\right)_R = \boldsymbol{\Omega}\times\mathbf{V}_R.$$

Therefore,

$$\left(\frac{d\mathbf{V}_I}{dt}\right)_I = \left(\frac{d\mathbf{V}_R}{dt}\right)_R + \boldsymbol{\Omega}\t... ...es\mathbf{V}_R + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})$$ (8)

$$= \left(\frac{d\mathbf{V}_R}{dt}\right)_R + 2\boldsymbol{\Omega}\... ...s\mathbf{V}_R + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r}).$$ (9)

Rearranging gives the desired relation (with $\mathbf{V}_g$ denoting the velocity measured in the rotating frame, which here we have called $\mathbf{V}_R$):

$$\boxed{\;\left(\frac{d\mathbf{V}_R}{dt}\right)_R = \left(\frac{d\... ...thbf{V}_g - \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})\; }.$$ (10)

This completes the derivation.