Question

Bipolar.

Short answer

The required values are mentioned below.

$$\begin{gathered} {\tilde{S}}_u=\left(\ddot{u}-\frac{\sin hu}{\cos hu+\cos v}{\dot{u}}^2+\frac{2\sin v}{\cos hu+\cos v}\dot{u}\dot{v}+\frac{\sin hu}{\cos hu+\cos v}{\dot{v}}^2\right)\frac{a}{\cos hu+\cos v} \\ {\tilde{S}}_v=\left(\ddot{u}-\frac{\sin v}{\cos hu+\cos v}{\dot{u}}^2+\frac{2\sin hu}{\cos hu+\cos v}\dot{u}\dot{v}+\frac{\sin v}{\cos hu+\cos v}{\dot{v}}^2\right)\frac{a}{\cos hu+\cos v} \end{gathered}$$

Step-by-step solution

Given Information

A Bipolar.

Definition of 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.

Find all the values.

Velocity in bipolar is $\dot{s}=\dot{u}\frac{a}{\cos hu+\cos v}{\hat{e}}_u+\dot{\dot{v}\frac{a}{\cos hu+\cos v}{\hat{e}}_u}$.

Find the acceleration.

$$\begin{gathered} \ddot{s}=(\ddot{u}+{\mathrm{\Gamma }}_{\mathrm{ij}}^{\mathrm{u}}\dot{{\mathrm{q}}_{\mathrm{i}}{\dot{\mathrm{q}}}_{\mathrm{j}}){\mathrm{h}}_{\mathrm{u}}{\hat{\mathrm{e}}}_{\mathrm{u}}+}(\ddot{\mathrm{u}}+{\mathrm{\Gamma }}_{\mathrm{ij}}^{\mathrm{v}}\dot{{\mathrm{q}}_{\mathrm{i}}{\dot{\mathrm{q}}}_{\mathrm{j}}){\mathrm{h}}_{\mathrm{v}}{\hat{\mathrm{e}}}_{\mathrm{v}}} \\ \frac{{\partial }^2\mathrm{s}}{\partial {\mathrm{q}}_{\mathrm{i}}\partial {\mathrm{q}}_{\mathrm{j}}}={\mathrm{\Gamma }}_{\mathrm{ij}}^{\mathrm{k}}\frac{\partial \mathrm{s}}{\partial {\mathrm{q}}_{\mathrm{k}}} \end{gathered}$$

Solve further.

$$\begin{gathered} \frac{{\partial }^{2}s}{\partial u^2}=-\frac{a\sin hu}{\cos hu+\cos v}\frac{{\sin }^{2}v+1+\cos hu+\cos v}{{\left(\cos hu+\cos v\right)}^2}{\hat{e}}_x+\frac{a\sin v}{\cos hu+\cos v}\frac{{\sin }^{2}u-(1+\cos u+\cos v)}{{\left(\cos hu+\cos v\right)}^2}{\hat{e}}_y \\ \frac{{\partial }^{2}s}{\partial u^2}=-\frac{a\sin hu}{\cos u+\cos v}\frac{\partial s}{\partial u}-\frac{\sin v}{\cos hu+\cos v}\frac{\partial s}{\partial u} \\ \frac{{\partial }^{2}s}{\partial u\partial v}=-\frac{a\sin v}{\cos hu+\cos v}\frac{{\sin }^{2}u+(1+\cos hu+\cos v)}{{\left(\cos hu+\cos v\right)}^2}{\hat{e}}_x+\frac{a\sin hu}{\cos hu+\cos v}\frac{{\sin }^{2}v-1+\cos u+\cos v}{{\left(\cos hu+\cos v\right)}^2}{\hat{e}}_y \\ \frac{{\partial }^{2}s}{\partial u\partial v}=\frac{\sin v}{\cos hu+\cos v}\frac{\partial s}{\partial u}-\frac{\sin hu}{\cos hu+\cos v}\frac{\partial s}{\partial u} \end{gathered}$$

Solve further.

$$\begin{gathered} \frac{{\partial }^{2}s}{\partial v^2}=\frac{a\sin hu}{\cos hu+\cos v}\frac{{\sin }^{2}v+1+\cos hu+\cos v}{{\left(\cos hu+\cos v\right)}^2}{\hat{e}}_x+\frac{a\sin v}{\cos hu+\cos v}\frac{1+\cos hu\cos v-\sin h^{2}u}{{\left(\cos hu+\cos v\right)}^2}\hat{e} \\ \frac{{\partial }^{2}s}{\partial v^2}=\frac{a\sin hu}{\cos hu+\cos v}\frac{\partial s}{\partial u}+\frac{\sin v}{\cos hu+\cos v}\frac{\partial s}{\partial u} \end{gathered}$$

Find the components of acceleration.

$$\begin{gathered} {\tilde{S}}_u=\left(\ddot{u}-\frac{\sin hu}{\cos hu+\cos v}{\dot{u}}^2+\frac{2\sin v}{\cos hu+\cos v}\dot{u}\dot{v}+\frac{\sin hu}{\cos hu+\cos v}{\dot{v}}^2\right)\frac{a}{\cos hu+\cos v} \\ {\tilde{S}}_v=\left(\ddot{u}-\frac{\sin v}{\cos hu+\cos v}{\dot{u}}^2+\frac{2\sin hu}{\cos hu+\cos v}\dot{u}\dot{v}+\frac{\sin v}{\cos hu+\cos v}{\dot{v}}^2\right)\frac{a}{\cos hu+\cos v} \end{gathered}$$