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I have a question that result equation is how it has derived.

When we have a following equation,

$$\frac{\mathrm dr}{\mathrm dt}=\frac{\delta r}{\delta t}+\omega \times r.$$ $\omega$ is angular velocity, $r$ is position vector.

It expresses that how movement of vector $r$ which belong to B coordinate is looks to view of A coordinate.

But, If do derivative one time more,

$$\frac{\mathrm d}{\mathrm dt}\frac{\mathrm dr}{\mathrm dt} =\frac{\mathrm d}{\mathrm dt}\left(\frac{\delta r}{\delta t}+\omega \times r\right)$$

Result in the Book is as follows.

$$\frac{\delta ^2r}{\delta ^2t}+\frac{\delta \omega }{\delta t}\times r+2\omega \times \frac{\delta r}{\delta t}+\omega \times \left(\omega \times r\right)$$

I really can not find a any clue for it....

Could you help me? Please let me know...

Qmechanic
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user245835
  • 1

2 Answers2

0

Given a fixed frame $A$ with fixed basis $\hat e_i$ and a rotating frame $\tilde A$ with rotating basis $\hat {\tilde e}_i$, we have, by definition of the components of $\vec r$ in each basis: $$ \vec r \equiv \sum_i x_i \hat e_i \equiv \sum_i \tilde x_i \hat {\tilde e}_i $$

By definition: $$ \frac{\delta \vec r}{\delta t}\equiv \sum_i\frac{d\tilde x_i}{dt}\hat {\tilde e}_i $$

We also have: $$ \frac{d\hat{\tilde e}_i}{dt} = \vec \omega \times \hat {\tilde e}_i $$

Combining the above three equations leads to the relationship we already know: $$ \frac{d\vec r}{dt}=\frac{\delta \vec r}{\delta t} + \vec \omega \times \vec r $$

The above comes from differentiating $$ \vec r = \sum_i \tilde x_i \hat {\tilde e}_i\qquad (1) $$ wrt time.

We can also differentiate Eq (1) with respect to time twice: $$ \frac{d^2 \vec r}{dt^2} = \frac{d}{dt}\sum_i \left(\frac{d\tilde x_i}{dt}\hat{\tilde e}_i + \tilde x_i\vec\omega\times\hat {\tilde e}_i \right) $$ $$ =\sum_i\left(\frac{d^2 \tilde x_i}{dt^2}\hat{\tilde e}_i +\frac{d\tilde x_i}{dt}\vec \omega\times\hat{\tilde e}_i +\frac{d \tilde x_i}{dt}\vec\omega\times\hat {\tilde e}_i +\tilde x_i\vec\omega\times\vec \omega\times\hat{\tilde e}_i \right) $$

By definition: $$ \frac{\delta^2r}{\delta t^2} =\sum_i\frac{d^2 \tilde x_i}{dt^2}\hat{\tilde e}_i\;, $$ which shows that $$ \frac{d^2 \vec r}{dt^2} =\frac{\delta^2r}{\delta t^2} +\vec \omega\times\frac{\delta \vec r}{\delta t} +\vec\omega\times\frac{\delta \vec r}{\delta t} +\vec\omega\times\vec \omega\times\vec r $$

hft
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As an alternative answer to hft's, consider: \begin{align} \frac{d^2\mathbf{r}}{dt^2} &= \frac{d}{dt}\big(\frac{\delta\mathbf{r}}{\delta t} + \boldsymbol{\omega}\times\mathbf{r}\big)\\[5pt] &= \frac{d}{dt}\big(\frac{\delta\mathbf{r}}{\delta t} \big)+ \frac{d\boldsymbol{\omega}}{dt}\times\mathbf{r}+ \boldsymbol{\omega}\times \frac{d\mathbf{r}}{dt}\\[5pt] &= \frac{\delta^2\mathbf{r}}{\delta t^2} + \boldsymbol{\omega}\times\frac{\delta\mathbf{r}}{\delta t} + \frac{\delta\boldsymbol{\omega}}{\delta t}\times \mathbf{r} + \boldsymbol{\omega}\times \frac{\delta\mathbf{r}}{\delta t} +\boldsymbol{\omega}\times \boldsymbol{\omega}\times \mathbf{r}\\[5pt] &= \frac{\delta^2\mathbf{r}}{\delta t^2} + \frac{\delta\boldsymbol{\omega}}{\delta t}\times \mathbf{r} + 2 \boldsymbol{\omega}\times \frac{\delta\mathbf{r}}{\delta t} + \boldsymbol{\omega}\times \boldsymbol{\omega}\times \mathbf{r}. \end{align}

Sahand Tabatabaei
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