The co-ordinates \((x, y, z)\) of a particle with respect to a uniformly rotating frame may be related to those with respect to a fixed inertial frame, \(\left(x^{*}, y^{*}, z^{*}\right)\), by the transformation $$ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & \sin \omega t & 0 \\ -\sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x^{*} \\ y^{*} \\ z^{*} \end{array}\right] $$ (Here, we use matrix notation: this stands for three separate equations, \(x=\cos \omega t \cdot x^{*}+\sin \omega t \cdot y^{*}\) etc.) Write down the inverse relation giving \(\left(x^{*}, y^{*}, z^{*}\right)\) in terms of \((x, y, z)\). By differentiating with respect to \(t\), rederive the relation (5.15) between \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) and \(\ddot{\boldsymbol{r}}\). [Hint: Note that \(\ddot{\boldsymbol{r}}=(\ddot{x}, \ddot{y}, \ddot{z})\), while \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) is the vector obtained by applying the above transformation \(\left.\operatorname{to}\left(\ddot{x}^{*}, \ddot{y}^{*}, \ddot{z}^{*}\right) .\right]\)

Step by step solution

01

Find the Inverse Relation

To find the inverse relation, we need to find the inverse of the transformation matrix. The given transformation matrix is an orthogonal matrix, so its inverse is equal to its transpose. Therefore, the inverse relation can be obtained by transposing the given matrix: $$ \left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

02

Differentiate the Inverse Relation

Now, we differentiate the inverse relation with respect to time \(t\): $$ \frac{d}{dt}\left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\frac{d}{dt}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d}{dt}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

03

Differentiate the Matrix Elements

To find \(\frac{\text{d}^2 \textbf{r}}{\text{d}t^2}\) we need to differentiate both the elements of the inverse matrix and the coordinate vector \((x, y, z)\): $$ \frac{d^2}{dt^2}\left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\frac{d^2}{dt^2}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + 2\frac{d}{dt}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d}{dt}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d^2}{dt^2}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

04

Derive the Relation between \(\frac{\text{d}^2 \textbf{r}}{\text{d}t^2}\) and \(\ddot{\textbf{r}}\)

After differentiating the matrix elements, we need to apply the transformation to \((\ddot{x^*}, \ddot{y^*}, \ddot{z^*})\) and rederive the relation: $$ \frac{d^2 \textbf{r}}{dt^2} = \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} \ddot{x^*} \\ \ddot{y^*} \\ \ddot{z^*} \end{array}\right] $$ Now, we apply the transformation to \((\ddot{x^*}, \ddot{y^*}, \ddot{z^*})\) and substitute \(\ddot{\textbf{r}} = (\ddot{x}, \ddot{y}, \ddot{z})\): $$ \frac{d^2 \textbf{r}}{dt^2} = \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right] \ddot{\textbf{r}} $$ This is the rederived relation between \(\frac{d^2 \textbf{r}}{dt^2}\) and \(\ddot{\textbf{r}}\), as we were asked to find.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!