Question

In the text and problems so far, we have found the e vectors for Question: Using the results of Problem 1, express the vector in Problem 4in spherical coordinates.

Short answer

The acceleration in cylindrical coordinates is $\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(2\dot{r}\dot{\theta }+r\dot{\theta }\right)e_{\theta }+\ddot{z}{e}_z$.

Step-by-step solution

Given Information

The spherical coordinates are given below.

$$\begin{gathered} \mathrm{x}=\mathrm{rcos\varphi sin\theta } \\ \mathrm{y}=\mathrm{rsin\varphi sin\theta } \\ \mathrm{z}=\mathrm{rcos\theta } \end{gathered}$$

Definition of a cylindrical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

Express the vector.

Assume the following quantity.

$$\begin{gathered} C=(i,j,k) \\ S=(e_r,e_{\theta },e_{\varphi }) \\ {e}_r=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +k\cos \theta \\ {e}_{\theta }=i\cos \varphi \cos \theta +j\sin \varphi \cos \theta +k\sin \theta \\ {e}_{\theta }=-i\sin \varphi +j\cos \varphi \end{gathered}$$

T is the transformation operator defined by $T{c}_j=s_j$.

$$\begin{gathered} T=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right) \\ V\left(E\right)=\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right) \\ V\left(F\right)=V\text{'} \\ V\left(F\right)=\left(\begin{array}{c}x{\text{'}}_1\\ x{\text{'}}_2\\ x{\text{'}}_3\end{array}\right) \end{gathered}$$

Let T be transition operator.

$$\begin{gathered} T{e}_j=f_j \\ =t_{ij}{e}_i \end{gathered}$$

The value V is given below.

$$\begin{gathered} V=x_i{e}_i \\ =x{\text{'}}_j{f}_j \\ =x{\text{'}}_{j}T{e}_j \\ =x{\text{'}}_j{t}_{ij}{e}_i \end{gathered}$$

The formula states that $V=TV\text{'}$.

$V\left(F\right)=T{\left(E\right)}^{-1}V\left(E\right)$

Assume the values mentioned below.

$$\begin{gathered} D=(i,j,k) \\ C=\left(e_r,e_{\theta },e_z\right) \end{gathered}$$

The value of matrix T(D) is given below.

$T\left(D\right)=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right)$

D and C are orthonormal bases. hence T(D) is orthogonal.

$$\begin{gathered} T{\left(D\right)}^{-1}=T{\left(D\right)}^T \\ T{\left(D\right)}^{-1}=\left(\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right) \end{gathered}$$

Proved earlier, $V\left(C\right)=T{\left(D\right)}^{-1}V\left(D\right)$

$$\begin{gathered} V\left(C\right)=\left(\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}r\sin \theta \\ -r\cos \theta \\ 1\end{array}\right) \\ V\left(C\right)=\left(\begin{array}{c}0\\ -r\\ 1\end{array}\right) \\ V=-e_{\theta }r+e_z \end{gathered}$$

Hence, the value of vector is $V=-e_{\theta }r+e_z$.