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I'm having a bit of trouble following the derivation for Newtons $2^{nd}$ law in a rotating frame, mainly just finding the second derivative of position in terms of the inertial fram. I am reading from Taylor's Classical Mechanics.

We start off with the relation between derivatives in a non-inertial & inertial reference frames. Let $S_0$ denote the non-inertial frame, and $S$ denote the inertial frame. $$\left(\frac{d\vec{r}}{dt}\right)_{S_0}=\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}$$ We can differentiate this again \begin{equation} \begin{split} \left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}&=\left(\frac{d}{dt}\right)_{S_0} \left[\left(\frac{d\vec{r}}{dt}\right)_{S_0}\right]\\ &=\left(\frac{d}{dt}_{S_0} \right)\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right] \end{split} \end{equation}

This next step is where I'm confused. The textbook states:

Applying (9.30) to the outside derivative on the right, we find $$\left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt} \right)_S \left[\left( \frac{d\vec{r}}{dt} \right)_S + \vec{\Omega} \times \vec{r}\right]+ \vec{\Omega} \times \left[ \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}\right]$$

Where (9.30) is the aforementioned relation in its general form: \begin{equation} \left(\frac{d\vec{Q}}{dt}\right)_{S_0}=\left(\frac{d\vec{Q}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{Q} \end{equation}

How do we get to the form shown in the book? I think the wording of the step confuses me.

Qmechanic
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brandon_ducks
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2 Answers2

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In the term $(1)\left(\frac{d}{dt}\right) _{S_0}\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right]$, treat Q in 9.30 as $\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right]$, and apply 9.30 to term (1) to express this term using $\left(\frac{d}{dt}\right) _{S}$.

John Darby
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The general rule is

$$ \left(\frac{d\vec{Q}}{dt}\right)_{S_0}=\left(\frac{d\vec{Q}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{Q} $$

where $\vec{Q}$ is any vector defined as riding on the rotating frame.

When you have $\vec{Q} = \vec{r}$ then you arrive at the first equation

$$ \left(\frac{d\vec{r}}{dt}\right)_{S_0}=\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r} $$

But use $\vec{Q} = \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}$ to arrive at the second equation.

$$\left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt} \right)_S \left[\left( \frac{d\vec{r}}{dt} \right)_S + \vec{\Omega} \times \vec{r}\right]+ \vec{\Omega} \times \left[ \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}\right]$$

John Alexiou
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