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Explanation for integration of Dr/Dt

  1. Mar 7, 2017 #1
    1. The problem statement, all variables and given/known data

    Why integration of $$\frac{D^2\mathbf r}{Dt^2}=−2\mathbf w \times \frac{D\mathbf r}{Dt}−g\mathbf R$$ gives us
    $$\frac{D\mathbf r}{Dt}= \mathbf v_0 −2\mathbf w×(\mathbf r−\mathbf r_0)−gt\mathbf R$$


    2. Relevant equations

    Consider a time-varying vector written in the body coordinate system, [itex]\xi(t) = R(t)\mathbf s(t).[/itex]
    $$\frac{d\xi}{dt} = R\frac{d\mathbf s}{dt}+\mathbf w \times \xi = \frac{D\xi}{Dt} +\mathbf w \times \xi.$$

    3. The attempt at a solution
    It looks for me like they incorporated [itex]-\mathbf v_0[/itex] for LHS and [itex]-\mathbf r_0[/itex] for [itex]\frac{D\mathbf r}{Dt}.[/itex]

    Ah! Probably they took definite integral [itex]\int_0^t[/itex]!
     
    Last edited: Mar 7, 2017
  2. jcsd
  3. Mar 12, 2017 #2

    Ssnow

    User Avatar
    Gold Member

    Hi, if you take ##\int_{0}^{t} \cdot ds## (as you said at the end) of both sides of ## \frac{D^{2} \mathbf{r}}{Ds^2}=-2\mathbf{w}\times \frac{D\mathbf{r}}{Ds}-g\mathbf{R}## you obtain in the first side ## \frac{D \mathbf{r}}{Ds}(t)-\frac{D \mathbf{r}}{Ds}(0)## that is problably ##\frac{D \mathbf{r}}{Ds}(t)-\mathbf{v}_{0}##, the second side is a simple integration and calling ##\mathbf{r}=\mathbf{r}(t)## and ##\mathbf{r}_{0}=\mathbf{r}(0)## you have the result ...

    Ssnow
     
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